8300: course teaching notes

Version: Version 1.0, 12 September 2014; first examination 2017

These are original course-owned teaching notes. Objective-level exceptions are listed in the review, and lessons are tier labelled.

3.1 · Exact arithmetic and estimation · Foundation

Can we pack without leftovers?

• A supplier packs 72 pencils and 90 pens into identical gift bags. How can we avoid leftovers?
• This lesson studies prime factor 质因数: A prime number that divides the integer exactly.

Choose the mathematical structure

• Prime factors reveal shared structure. Use the smallest common prime powers for the HCF and the largest for the LCM. Estimate before calculating; use brackets to preserve the order of operations.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• The HCF divides both numbers; the LCM is a multiple of both. They answer different questions. A decimal estimate is not an exact fraction.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• For a non-calculator paper, keep fractions exact and show cancellation. For a calculator paper, enter the full expression and compare with your estimate.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Exact arithmetic and estimation · Higher

Can we pack without leftovers?

• A supplier packs 72 pencils and 90 pens into identical gift bags. How can we avoid leftovers?
• This lesson studies prime factor 质因数: A prime number that divides the integer exactly.

Choose the mathematical structure

• Prime factors reveal shared structure. Use the smallest common prime powers for the HCF and the largest for the LCM. Estimate before calculating; use brackets to preserve the order of operations.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• The HCF divides both numbers; the LCM is a multiple of both. They answer different questions. A decimal estimate is not an exact fraction.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• For a non-calculator paper, keep fractions exact and show cancellation. For a calculator paper, enter the full expression and compare with your estimate.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Integer indices and standard form · Foundation

How small is a microscopic length?

• A microscope records a length of 0.000072 metres. A compact representation must keep its size correct.
• This lesson studies index 指数: The power to which a base is raised.

Choose the mathematical structure

• Use integer powers, square and cube roots and standard form with 1≤a<10. In multiplying powers with the same base, add indices; in division, subtract them.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Index laws do not turn a sum into a single power: 2^3+2^4=24, not 2^7. Do not round a surd when an exact answer is requested.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• This Foundation/Core lesson excludes fractional powers and surd rationalisation. Estimate a result before using a calculator and retain the required precision.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Indices, surds and standard form · Higher

How small is a microscopic length?

• A microscope records a length of 0.000072 metres. A compact representation must keep its size correct.
• This lesson studies index 指数: The power to which a base is raised.

Choose the mathematical structure

• For the same positive base, multiplication adds indices and division subtracts them. A negative index means reciprocal; a fractional index represents a root. Standard form has 1≤a<10.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Index laws do not turn a sum into a single power: 2^3+2^4=24, not 2^7. Do not round a surd when an exact answer is requested.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Check powers of ten against the original quantity. Use surds for exact geometry, and round only the final length when the question asks for a decimal.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Accuracy, bounds and compound measures · Higher

Is the printed measurement exact?

• A rectangular panel is labelled 8.0 cm by 5.0 cm, each to the nearest 0.1 cm. Its true area is not fixed at 40 cm².
• This lesson studies lower bound 下界: The smallest possible value consistent with a stated rounding rule.

Choose the mathematical structure

• A value rounded to the nearest unit u lies from stated value-u/2 up to, but usually not including, stated value+u/2. For positive quantities, combine extremes according to the operation.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• An upper bound is not automatically achieved. Dividing upper distance by upper time does not give the largest speed. Keep enough digits in intermediate calculations.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Distinguish measurement uncertainty from arithmetic rounding. A sensible reported precision cannot be finer than the measurements justify.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Signed numbers, place value and operation order · Foundation

Which way does the balance move?

• An account starts 12 yuan overdrawn. A payment of 35 yuan arrives, then a 9-yuan charge is taken. Which direction does each change move the balance?
• This lesson studies reciprocal · ⁨recíproco⁩ 倒数: A number that multiplies a nonzero original number to give 1.

Choose the mathematical structure

• Larger numbers lie farther right on the number line. Adding a negative moves left; subtracting a negative moves right. Multiply or divide signs first, then magnitudes. Evaluate brackets, powers and roots before multiplication/division, then addition/subtraction, working left to right within equal priority.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Subtraction is not commutative: 3-8 and 8-3 differ. A negative sign outside a square is not inside its base: -3²=-9, but (-3)²=9. Zero has no reciprocal.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Use a signed starting balance, a positive credit and a negative debit. A shop buying for 48 yuan and selling for 60 makes 12 yuan profit; reversing the prices gives a 12-yuan loss. Estimate the sign before calculating.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Signed numbers, place value and operation order · Higher

Which way does the balance move?

• An account starts 12 yuan overdrawn. A payment of 35 yuan arrives, then a 9-yuan charge is taken. Which direction does each change move the balance?
• This lesson studies reciprocal · ⁨recíproco⁩ 倒数: A number that multiplies a nonzero original number to give 1.

Choose the mathematical structure

• Larger numbers lie farther right on the number line. Adding a negative moves left; subtracting a negative moves right. Multiply or divide signs first, then magnitudes. Evaluate brackets, powers and roots before multiplication/division, then addition/subtraction, working left to right within equal priority.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Subtraction is not commutative: 3-8 and 8-3 differ. A negative sign outside a square is not inside its base: -3²=-9, but (-3)²=9. Zero has no reciprocal.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Use a signed starting balance, a positive credit and a negative debit. A shop buying for 48 yuan and selling for 60 makes 12 yuan profit; reversing the prices gives a 12-yuan loss. Estimate the sign before calculating.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Exact fractions and mixed-number operations · Foundation

Can we keep the recipe exact?

• A recipe uses 1½ cups for one batch. How much remains after using ¾ of a cup, and why should the answer stay exact?
• This lesson studies improper fraction 假分数: A fraction whose numerator is at least as large as its positive denominator.

Choose the mathematical structure

• Convert mixed numbers to improper fractions. Add or subtract using a common denominator; multiply numerators and denominators; divide by a nonzero fraction by multiplying its reciprocal. Cancel common factors, not added terms. These rules also apply to negative fractions.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Adding denominators does not preserve the unit size: 1/2+1/3 is not 2/5. Cancel only factors of a whole numerator and denominator. A division by zero is undefined.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• For a non-calculator question show the common denominator or reciprocal step. Convert 19/12 to 1 7/12 if a mixed number is requested; round only when the question explicitly needs a decimal.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Exact fractions and mixed-number operations · Higher

Can we keep the recipe exact?

• A recipe uses 1½ cups for one batch. How much remains after using ¾ of a cup, and why should the answer stay exact?
• This lesson studies improper fraction 假分数: A fraction whose numerator is at least as large as its positive denominator.

Choose the mathematical structure

• Convert mixed numbers to improper fractions. Add or subtract using a common denominator; multiply numerators and denominators; divide by a nonzero fraction by multiplying its reciprocal. Cancel common factors, not added terms. These rules also apply to negative fractions.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Adding denominators does not preserve the unit size: 1/2+1/3 is not 2/5. Cancel only factors of a whole numerator and denominator. A division by zero is undefined.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• For a non-calculator question show the common denominator or reciprocal step. Convert 19/12 to 1 7/12 if a mixed number is requested; round only when the question explicitly needs a decimal.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Systematic lists and possibility grids · Foundation

Have we listed every choice?

• A cafe offers tea or juice with apple, banana or melon. How can a list prove that no lunch choice was missed?
• This lesson studies systematic list 系统列表: A list made in a fixed order so every allowed outcome appears exactly once.

Choose the mathematical structure

• Fix one first choice and list every allowed second choice before moving to the next first choice. Use a table to check that no outcome is missing or duplicated. If combinations are forbidden, remove those entries from the list.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• State whether order matters and whether repetition is allowed. Choosing A then B can be different from B then A for a code, but not for an unordered two-person team. A restriction can make a simple product invalid.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Foundation should justify answers with a complete ordered list or grid. Higher may compress a verified structure using the product rule, but must still check restrictions.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Systematic lists and the product rule · Higher

Have we listed every choice?

• A cafe offers tea or juice with apple, banana or melon. How can a list prove that no lunch choice was missed?
• This lesson studies systematic list 系统列表: A list made in a fixed order so every allowed outcome appears exactly once.

Choose the mathematical structure

• Fix one first choice and list every permitted second choice before moving to the next first choice. A table gives the same structure. When every one of m first choices allows n second choices, the Higher product rule gives mn outcomes. If restrictions change the options, count the allowed rows separately.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• State whether order matters and whether repetition is allowed. Choosing A then B can be different from B then A for a code, but not for an unordered two-person team. A restriction can make a simple product invalid.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Foundation should justify answers with a complete ordered list or grid. Higher may compress a verified structure using the product rule, but must still check restrictions.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Exact surds and rationalising denominators · Higher

Can a root stay exact?

• A square has area 12 cm². Can its side be written exactly without a long calculator decimal?
• This lesson studies surd 根式: An irrational root kept in exact form rather than replaced by a rounded decimal.

Choose the mathematical structure

• Extract square factors: √(a²b)=a√b for a≥0,b≥0. Add only matching root parts. Multiply roots with nonnegative radicands. To rationalise a denominator, multiply numerator and denominator by the same suitable root or conjugate.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• √(a+b) generally differs from √a+√b. Match the radicand before collecting terms. Rationalising changes the form, not the value; multiplying only the denominator changes the value.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA N8 Higher requires exact surds and rationalisation. Foundation retains exact fractions and multiples of π; do not assign this surd lesson to Foundation.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Terminating decimals and fractions · Foundation

Does the display tell the whole number?

• A calculator shows 0.333333. Is the display the exact value of one third, or only the digits that fit on the screen?
• This lesson studies terminating decimal 有限小数: A decimal that ends after a finite number of digits.

Choose the mathematical structure

• Use the place value of the last digit to write a power-of-ten denominator, then simplify. Compare numbers using matching decimal places or a common denominator.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Keep the place value: 0.06 is six hundredths, not six tenths. A fraction may exceed 1. Simplify the entire numerator and denominator by the same common factor.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA N10 Foundation covers terminating decimals and fractions, including ordering. Recurring-decimal conversion is reserved for the Higher lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Terminating and recurring decimal conversions · Higher

Does the display tell the whole number?

• A calculator shows 0.333333. Is the display the exact value of one third, or only the digits that fit on the screen?
• This lesson studies recurring decimal 循环小数: A decimal with a digit or block of digits that repeats forever.

Choose the mathematical structure

• A finite decimal uses a power-of-ten denominator before simplifying. For a recurring decimal, multiply by powers of ten so the repeated tails align, then subtract. Use matching decimal places to compare numbers. A displayed rounded decimal need not be the exact fraction.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Align the recurring tails before subtracting. 0.333333 is finite and differs from 0.333333... . A non-recurring prefix needs a second power of ten; blindly dividing every digit block by 99 fails.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA N10 Higher includes recurring conversion. Both tiers convert and order terminating decimals and fractions. Check a conversion by long division; 3 divided by 8 gives 0.375.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Metric and compound-unit conversions · Foundation

Why must the factor be squared?

• A floor tile measures 0.4 m by 0.4 m. Why is its area not 0.16×100 square centimetres?
• This lesson studies conversion factor 换算因子: The multiplier that expresses the same quantity in another unit.

Choose the mathematical structure

• Convert each dimension. Since 1 m=100 cm, 1 m²=10,000 cm² and 1 m³=1,000,000 cm³. Also 1 litre=1000 cm³, 1 kg=1000 g and 1 hour=3600 seconds. For a compound unit convert numerator and denominator, keeping the physical quantity unchanged.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• The area multiplier is the square of the length multiplier; the volume multiplier is its cube. An hour is 60 minutes, not 100. Converting only the numerator of a speed gives an inconsistent unit.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA N13 and R1 include metric length, area, volume, capacity and compound measures. Use given conversion factors for unfamiliar imperial units; do not guess them.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Metric and compound-unit conversions · Higher

Why must the factor be squared?

• A floor tile measures 0.4 m by 0.4 m. Why is its area not 0.16×100 square centimetres?
• This lesson studies conversion factor 换算因子: The multiplier that expresses the same quantity in another unit.

Choose the mathematical structure

• Convert each dimension. Since 1 m=100 cm, 1 m²=10,000 cm² and 1 m³=1,000,000 cm³. Also 1 litre=1000 cm³, 1 kg=1000 g and 1 hour=3600 seconds. For a compound unit convert numerator and denominator, keeping the physical quantity unchanged.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• The area multiplier is the square of the length multiplier; the volume multiplier is its cube. An hour is 60 minutes, not 100. Converting only the numerator of a speed gives an inconsistent unit.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA N13 and R1 include metric length, area, volume, capacity and compound measures. Use given conversion factors for unfamiliar imperial units; do not guess them.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Estimation, rounding and simple error intervals · Foundation

Would the estimate catch a typo?

• A calculator reports 49.8×19.7÷10.2. Could 962 be a reasonable answer, or does a quick estimate expose an input error?
• This lesson studies significant figure 有效数字: A digit counted from the first nonzero digit when reporting numerical precision.

Choose the mathematical structure

• For a rough check use easy nearby numbers: 50×20÷10=100. Decimal places count digits after the point; significant figures start at the first nonzero digit. Round only the final result. Nearest-unit rounding has an interval extending half a unit either way; positive truncation keeps values from the stated value up to the next unit.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Zeros before the first nonzero digit do not count as significant figures. Rounding and truncation give different intervals. Do not turn an approximate check into an exact answer, or round every intermediate result.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA N14–N16 require accuracy interpretation at both tiers. Foundation uses simple intervals and limits of accuracy. Higher combines upper and lower bounds for calculated quantities in the separate bounds lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Estimation, rounding and simple error intervals · Higher

Would the estimate catch a typo?

• A calculator reports 49.8×19.7÷10.2. Could 962 be a reasonable answer, or does a quick estimate expose an input error?
• This lesson studies significant figure 有效数字: A digit counted from the first nonzero digit when reporting numerical precision.

Choose the mathematical structure

• For a rough check use easy nearby numbers: 50×20÷10=100. Decimal places count digits after the point; significant figures start at the first nonzero digit. Round only the final result. Nearest-unit rounding has an interval extending half a unit either way; positive truncation keeps values from the stated value up to the next unit.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Zeros before the first nonzero digit do not count as significant figures. Rounding and truncation give different intervals. Do not turn an approximate check into an exact answer, or round every intermediate result.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA N14–N16 require accuracy interpretation at both tiers. Foundation uses simple intervals and limits of accuracy. Higher combines upper and lower bounds for calculated quantities in the separate bounds lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Ratio shares and fraction operators · Foundation

Which fraction describes the whole?

• A club shares 84 yuan between two teams in the ratio 2:5. Does the first team receive two fifths of the total?
• This lesson studies ratio · ⁨razón⁩ 比: A comparison that states the relative numbers of equal-sized parts.

Choose the mathematical structure

• Add the ratio parts to find the whole: 2:5 has 7 equal parts. Divide the total by 7, then multiply by the required part count. A fraction acts as a multiplier; a percentage p acts as p/100. Keep part-to-part and part-to-whole comparisons separate.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A ratio 2:5 does not give the first share as 2/5 of the total. The total has 7 parts. A multiplier over 1 increases a positive quantity; it does not automatically mean a probability.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA N11/N12 and R3–R8 link ratio parts to fractions and operators. Check the shares add to the stated total and simplify a ratio by dividing every part by the same positive factor.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.1 · Ratio shares and fraction operators · Higher

Which fraction describes the whole?

• A club shares 84 yuan between two teams in the ratio 2:5. Does the first team receive two fifths of the total?
• This lesson studies ratio · ⁨razón⁩ 比: A comparison that states the relative numbers of equal-sized parts.

Choose the mathematical structure

• Add the ratio parts to find the whole: 2:5 has 7 equal parts. Divide the total by 7, then multiply by the required part count. A fraction acts as a multiplier; a percentage p acts as p/100. Keep part-to-part and part-to-whole comparisons separate.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A ratio 2:5 does not give the first share as 2/5 of the total. The total has 7 parts. A multiplier over 1 increases a positive quantity; it does not automatically mean a probability.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA N11/N12 and R3–R8 link ratio parts to fractions and operators. Check the shares add to the stated total and simplify a ratio by dividing every part by the same positive factor.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.1. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Equations, identities and rearrangement · Foundation

When do two plans cost the same?

• Two mobile plans cost 20+3x and 44+x yuan for x GB. When do they cost the same?
• This lesson studies identity 恒等式: An equality that holds for every allowed value of its variable.

Choose the mathematical structure

• An equation asks which inputs satisfy an equality; an identity holds for all allowed inputs. Preserve equality by applying the same operation to both sides. State restrictions before dividing by a variable.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Cancelling a term is not the same as cancelling a factor. In (x²+2x)/x, factor the numerator and retain x≠0. Check a rearrangement by substitution.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Define the unknown and set up a linear equation. Check the answer by substitution. Restrict this Foundation/Core lesson to simple expressions and equations.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Equations, identities and rearrangement · Higher

When do two plans cost the same?

• Two mobile plans cost 20+3x and 44+x yuan for x GB. When do they cost the same?
• This lesson studies identity 恒等式: An equality that holds for every allowed value of its variable.

Choose the mathematical structure

• An equation asks which inputs satisfy an equality; an identity holds for all allowed inputs. Preserve equality by applying the same operation to both sides. State restrictions before dividing by a variable.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Cancelling a term is not the same as cancelling a factor. In (x²+2x)/x, factor the numerator and retain x≠0. Check a rearrangement by substitution.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Set up the equation from units and the meaning of the unknown. A negative or fractional solution may be algebraically correct but impossible for a count.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Factorising and solving simple quadratics · Foundation

Which widths make enough space?

• A rectangular enclosure has area x(10-x). What widths give at least 21 square metres?
• This lesson studies root 零点: An input for which the expression has value zero.

Choose the mathematical structure

• Expand brackets and factorise simple quadratics. Solve by setting each factor equal to zero, and use a graph to interpret the roots.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Multiplying an inequality by a negative number reverses its direction. A sketch must show which side of each root satisfies the inequality. Geometry may restrict x further.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• This Foundation/Core lesson uses factorisation and graphical roots; the discriminant, quadratic formula and quadratic inequalities are reserved for the advanced tier.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Quadratics and inequalities · Higher

Which widths make enough space?

• A rectangular enclosure has area x(10-x). What widths give at least 21 square metres?
• This lesson studies discriminant 判别式: The quantity b²-4ac that determines the real roots of ax²+bx+c=0.

Choose the mathematical structure

• Factor where possible; otherwise complete the square or use the quadratic formula. A quadratic inequality needs the sign on intervals, not only the roots. The discriminant identifies repeated or missing real roots.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Multiplying an inequality by a negative number reverses its direction. A sketch must show which side of each root satisfies the inequality. Geometry may restrict x further.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A11/A18 Higher interprets roots, intercepts and turning points, completes the square and uses the quadratic formula, including equations needing rearrangement. Factorisation and substitution check each result.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Two linear simultaneous equations · Foundation

Can one equation determine two counts?

• Two ticket types raise a total amount. A single equation cannot identify both unknown counts.
• This lesson studies elimination 消元法: Combining equations to remove one variable while retaining the same solutions.

Choose the mathematical structure

• Multiply equations to make a variable cancel, or substitute an expression from one equation into the other. Solve the remaining linear equation and recover the second variable. The intersection is a point satisfying both original equations.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Check the ordered pair in both original equations. Multiplying an equation means multiplying every term, including the constant. Parallel distinct lines have no common solution.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A19 Foundation solves two linear equations by elimination or substitution and interprets the graph intersection. Linear/quadratic systems belong to Higher.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Linear and linear/quadratic simultaneous equations · Higher

Can one equation determine two counts?

• Two ticket types raise a total amount. A single equation cannot identify both unknown counts.
• This lesson studies elimination 消元法: Combining equations to remove one variable while retaining the same solutions.

Choose the mathematical structure

• For two linear equations, use elimination or substitution and check both equations. For a line and a quadratic, substitute the linear relation first; then solve the resulting quadratic. For inequalities, shade the region satisfying every condition.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• One equation checked is not enough. A line can meet a quadratic twice, so retain both solutions unless the context removes one. Inequality boundaries may be included or excluded according to the sign.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A19 Higher includes two linear equations and linear/quadratic systems. Elimination suits linear pairs; substitution reduces a line/curve pair to a quadratic. Retain every solution and check both original equations.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Straight lines and gradients · Foundation

How does a path's slope become an equation?

• A path rises 6 metres over a horizontal distance of 3 metres. Its gradient connects a diagram to an equation.
• This lesson studies gradient 斜率: The change in y divided by the corresponding change in x.

Choose the mathematical structure

• Gradient is change in y divided by change in x. A straight line has y=mx+c, where c is its y-intercept. Parallel lines have equal gradients.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A vertical line has no finite gradient; do not force it into y=mx+c. Read the signs of a circle's centre carefully. The radius to a tangent is perpendicular to the tangent.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Plot a straight line using two checked points and label its intercept. Perpendicular-gradient formulae and circle equations are not part of this Foundation/Core lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Coordinate geometry and tangents · Higher

How does a path's slope become an equation?

• A path rises 6 metres over a horizontal distance of 3 metres. Its gradient connects a diagram to an equation.
• This lesson studies gradient 斜率: The change in y divided by the corresponding change in x.

Choose the mathematical structure

• A line through (x₁,y₁) with gradient m has y-y₁=m(x-x₁). Parallel lines have equal gradients. Finite perpendicular gradients multiply to -1. A circle has (x-a)²+(y-b)²=r².
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A vertical line has no finite gradient; do not force it into y=mx+c. Read the signs of a circle's centre carefully. The radius to a tangent is perpendicular to the tangent.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Before solving a line-circle intersection, predict whether there are zero, one or two intersections. Substitution produces a quadratic whose discriminant checks the prediction.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Arithmetic sequences and nth terms · Foundation

Does the change add or multiply?

• A saving plan adds ¥30 more each week; a population model grows by 5% each year. Equal differences and equal ratios need different models.
• This lesson studies common difference 公差: The constant added between consecutive terms of an arithmetic sequence.

Choose the mathematical structure

• Find a constant difference for an arithmetic sequence. Its nth term is a+(n-1)d. A term-to-term rule describes how to reach the next term; a position-to-term rule gives a term directly.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• The first term has index 1, so the exponent is n-1. A sequence is a list; a series is a sum. A geometric sequence can alternate in sign and still converge.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Generate several terms and check a proposed nth-term rule. Infinite geometric series and advanced sum formulae are excluded from this Foundation/Core lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Arithmetic and finite geometric sequences · Higher

Does the change add or multiply?

• A saving plan adds ¥30 more each week; a population model grows by 5% each year. Equal differences and equal ratios need different models.
• This lesson studies common ratio 公比: The constant multiplier between consecutive terms of a geometric sequence.

Choose the mathematical structure

• An arithmetic sequence adds a constant difference and has nth term a+(n-1)d. A geometric sequence multiplies by a constant ratio and has nth term ar^(n-1). Check the starting position and keep a surd ratio exact when one is supplied.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A geometric ratio is not a common difference. The exponent is n-1 when a is the first term at position 1. A finite pattern does not itself justify an infinite-sum formula.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A23–A25 Higher uses arithmetic rules, geometric progression patterns and quadratic nth terms. Arithmetic-series formulae and infinite-series sums are excluded from this GCSE lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Algebraic notation, substitution and vocabulary · Foundation

What does each symbol do?

• A delivery charge is 4 yuan plus 3 yuan for each kilometre. What do the separate terms in 4+3d describe?
• This lesson studies coefficient 系数: The numerical factor multiplying a variable term.

Choose the mathematical structure

• A term is a part joined by addition or subtraction. In 4+3d, 4 is a constant and 3 is the coefficient of d. The expression has no equality sign; 4+3d=19 is an equation. A formula connects named quantities. In ab, multiplication is understood; a²b means a×a×b, not a×b×b.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Put a negative substituted number in brackets before squaring. A coefficient is not an exponent. An expression can be evaluated but cannot be solved unless a condition or equation is supplied.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A1–A3 require precise notation and vocabulary, including formulae from other subjects. Identify the inputs and units before substitution; keep fraction coefficients exact.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Algebraic notation, substitution and vocabulary · Higher

What does each symbol do?

• A delivery charge is 4 yuan plus 3 yuan for each kilometre. What do the separate terms in 4+3d describe?
• This lesson studies coefficient 系数: The numerical factor multiplying a variable term.

Choose the mathematical structure

• A term is a part joined by addition or subtraction. In 4+3d, 4 is a constant and 3 is the coefficient of d. The expression has no equality sign; 4+3d=19 is an equation. A formula connects named quantities. In ab, multiplication is understood; a²b means a×a×b, not a×b×b.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Put a negative substituted number in brackets before squaring. A coefficient is not an exponent. An expression can be evaluated but cannot be solved unless a condition or equation is supplied.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A1–A3 require precise notation and vocabulary, including formulae from other subjects. Identify the inputs and units before substitution; keep fraction coefficients exact.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Collecting, expanding and factorising expressions · Foundation

Can two area descriptions agree?

• An L-shaped area has a symbolic width. Can two different expressions describe the same total area?
• This lesson studies factorise 因式分解: Rewrite an expression as a product of factors.

Choose the mathematical structure

• Collect like terms, distribute over brackets and reverse expansion by factorising. Take out a common factor first. Expand products of two binomials and factorise simple monic quadratics and differences of squares.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• x² and x are unlike terms. A factor multiplies a whole expression; it is not a separate added term. Check a proposed factorisation by expanding it.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A4 Foundation includes collecting, single brackets, common factors, two-binomial expansion and monic quadratic factorisation. Higher adds non-monic factorisation, more binomials, surds and algebraic fractions.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Collecting, expanding and factorising expressions · Higher

Can two area descriptions agree?

• An L-shaped area has a symbolic width. Can two different expressions describe the same total area?
• This lesson studies factorise 因式分解: Rewrite an expression as a product of factors.

Choose the mathematical structure

• Collect only like terms. Distribute multiplication over every term in a bracket, including signs. Factorising reverses expansion. Use common factors first; then pairs for a quadratic. Algebraic fractions may cancel common factors, with forbidden denominator values stated.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• x² and x are unlike terms. Cancelling across a sum is invalid; factor the whole expression first. A simplified fraction must retain values excluded by its original denominator.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A4 Foundation includes collecting, single brackets, common factors, two-binomial expansion and monic quadratic factorisation. Higher adds non-monic factorisation, more binomials, surds and algebraic fractions.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Rearranging formulae and checking the subject · Foundation

Can we work backwards from the area?

• A triangle has known area and height, but its base is missing. How can the area formula be reversed?
• This lesson studies subject 公式主项: The quantity written alone on one side of a formula.

Choose the mathematical structure

• Undo operations on both sides in reverse order. In A=bh/2, multiply both sides by 2 then divide by nonzero h to obtain b=2A/h. When the desired variable occurs twice, collect it before dividing. State restrictions introduced by division.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Doing an operation to only one side breaks the equality. The subject is a variable, not a numerical answer. Check the rearranged result by putting the found value into the original formula.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A5 includes standard and given formulae from other subjects, in words and symbols. Choose a valid unit conversion before substituting; a rearrangement does not itself change units.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Rearranging formulae and checking the subject · Higher

Can we work backwards from the area?

• A triangle has known area and height, but its base is missing. How can the area formula be reversed?
• This lesson studies subject 公式主项: The quantity written alone on one side of a formula.

Choose the mathematical structure

• Undo operations on both sides in reverse order. In A=bh/2, multiply both sides by 2 then divide by nonzero h to obtain b=2A/h. When the desired variable occurs twice, collect it before dividing. State restrictions introduced by division.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Doing an operation to only one side breaks the equality. The subject is a variable, not a numerical answer. Check the rearranged result by putting the found value into the original formula.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A5 includes standard and given formulae from other subjects, in words and symbols. Choose a valid unit conversion before substituting; a rearrangement does not itself change units.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Identities, equivalence and algebraic arguments · Foundation

Does one successful test prove a claim?

• Two students test x=1 and get equal results. Does that prove their expressions agree for every x?
• This lesson studies equivalent expression 等价表达式: An expression with the same value as another for every allowed input.

Choose the mathematical structure

• An equation may hold only for some values, while an identity holds for every allowed input. Establish equivalent expressions by valid expansion or factorisation. A few matching inputs are checks rather than a general argument.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Start from an expression or the assumptions, not from the conclusion as though it were already true. An example can disprove an all-values claim, but one confirming example cannot prove it.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A3/A6 Foundation distinguishes expression, equation and identity and argues equivalence using algebra. General parity and divisibility proofs belong to the Higher variant.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Identities, equivalence and algebraic arguments · Higher

Does one successful test prove a claim?

• Two students test x=1 and get equal results. Does that prove their expressions agree for every x?
• This lesson studies equivalent expression 等价表达式: An expression with the same value as another for every allowed input.

Choose the mathematical structure

• An equation may be true only at certain values. An identity is true at every allowed value. Establish equivalence by valid expansion or factorisation; testing a few inputs is only a check. Higher proofs use a general integer or algebraic variable and a conclusion tied to its definition.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Start from an expression or the assumptions, not from the conclusion as though it were already true. An example can disprove an all-values claim, but one confirming example cannot prove it.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A6 Foundation distinguishes equation/identity and argues equivalence. Higher extends this to algebraic proofs. State integer restrictions when using parity or consecutive integers.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Function machines and reversing operations · Foundation

Which input produced the output?

• A machine doubles a number then adds three. Which starting number would produce eleven?
• This lesson studies input · ⁨entrada⁩ 输入值: The value supplied to a rule before its operations are carried out.

Choose the mathematical structure

• Follow the operations in their stated order. To recover a starting input, undo the final operation first. A table pairs each input with its output. The same starting input must have only one output for the rule to be a function.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Reversing the rule does not mean repeating it. The last forward step is the first reverse step. Different operation orders can produce different outputs.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A7 Foundation interprets simple functions as input-output rules. Higher also uses formal inverse and composite function notation in the separate function lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Function machines and reversing operations · Higher

Which input produced the output?

• A machine doubles a number then adds three. Which starting number would produce eleven?
• This lesson studies input · ⁨entrada⁩ 输入值: The value supplied to a rule before its operations are carried out.

Choose the mathematical structure

• Follow the operations in their stated order. To recover a starting input, undo the final operation first. A table pairs each input with its output. The same starting input must have only one output for the rule to be a function.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Reversing the rule does not mean repeating it. The last forward step is the first reverse step. Different operation orders can produce different outputs.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A7 Foundation interprets simple functions as input-output rules. Higher also uses formal inverse and composite function notation in the separate function lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Domains, inverses and composition · Higher

Which inputs are allowed?

• A square-root model returns a real output only for some inputs. Its formula alone does not specify a complete function.
• This lesson studies domain 定义域: The set of allowed inputs to a function.

Choose the mathematical structure

• State the domain and range. For an inverse, first ensure the function is one-to-one on its domain. Composition fg means apply g first, then f; the intermediate output must be an allowed input to f.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Squaring can introduce extraneous solutions. Restricting a parabola's domain is essential before claiming an inverse. A horizontal translation inside f has the opposite sign to the graph's movement.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Check f(f inverse(x))=x on the inverse domain. Use a sketch to test whether a horizontal line meets the original graph more than once.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Coordinate quadrants and basic graph families · Foundation

What happens near the forbidden input?

• A table of a cube and a reciprocal gives very different values near zero. Can their graphs have the same shape?
• This lesson studies asymptote 渐近线: A line approached by a graph without a finite crossing in the stated example.

Choose the mathematical structure

• Coordinates are ordered (x,y). Signs identify the four quadrants. Use a value table, intercepts and symmetry to sketch a line, quadratic, cubic or reciprocal. For y=1/x, zero is excluded and the axes are asymptotes. Approximate intersections give graphical solutions.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Join a reciprocal branch smoothly on its own side of zero; never draw a segment through the excluded input. A graph table needs enough points to reveal a turning point or shape.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A8/A11/A12 Foundation includes all quadrants, lines, quadratics, simple cubics and reciprocals. Higher exponential and degree-based trigonometric families are treated separately.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Coordinate quadrants and basic graph families · Higher

What happens near the forbidden input?

• A table of a cube and a reciprocal gives very different values near zero. Can their graphs have the same shape?
• This lesson studies asymptote 渐近线: A line approached by a graph without a finite crossing in the stated example.

Choose the mathematical structure

• Coordinates are ordered (x,y). Signs identify the four quadrants. Use a value table, intercepts and symmetry to sketch a line, quadratic, cubic or reciprocal. For y=1/x, zero is excluded and the axes are asymptotes. Approximate intersections give graphical solutions.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Join a reciprocal branch smoothly on its own side of zero; never draw a segment through the excluded input. A graph table needs enough points to reveal a turning point or shape.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A8/A11/A12 Foundation includes all quadrants, lines, quadratics, simple cubics and reciprocals. Higher exponential and degree-based trigonometric families are treated separately.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Exponential and degree-based trigonometric graphs · Higher

Which graph repeats and which keeps growing?

• A tide repeats, while a doubling population keeps growing. Which graph feature distinguishes these patterns?
• This lesson studies period · ⁨período⁩ 周期: The input interval after which a repeating graph has the same values again.

Choose the mathematical structure

• For y=k^x with k>0,k≠1, the y-intercept is 1; k>1 gives growth and 0<k<1 gives decay. In degrees, sine and cosine repeat every 360° with range [-1,1]; tangent repeats every 180° and is undefined at 90°+180°n. Label axes in degrees.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Degrees and radians are different units. A steep tangent branch is not a finite point at its asymptote. Exponential growth is not repeated addition of a constant amount.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A12 Higher requires exponential functions with positive bases and sine/cosine/tangent for angles of any size in degrees. No differentiation or radian sector formula is introduced here.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Translations and reflections of function graphs · Higher

Does changing the input move the graph left or right?

• A graph starts with its minimum at (0,0). How does changing the input to x-3 move the minimum?
• This lesson studies graph translation 图像平移: Move every point of a graph by the same horizontal and vertical displacement.

Choose the mathematical structure

• y=f(x)+a shifts the output up by a. y=f(x-a) shifts the graph right by a. y=-f(x) reflects in the x-axis; y=f(-x) reflects in the y-axis. Transform the points as well as the formula, checking a known feature.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A positive number added inside the input, f(x+3), moves the graph left, not right. A reflection in the y-axis changes x; a reflection in the x-axis changes y.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A13 Higher covers translations and reflections of a given function. Use features and transformed points to justify the sketch; do not substitute a geometric enlargement for this graph transformation.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Distance-time graphs and contextual intersections · Foundation

Where is the waiting time on the graph?

• A delivery rider travels, waits, then travels again. Which parts of a distance-time graph reveal the waiting time?
• This lesson studies stationary 静止的: Not changing position over the time interval described.

Choose the mathematical structure

• Read the axis quantities and units before interpreting shape. On a distance-time graph, slope is speed for a segment with increasing distance; a horizontal segment means no distance change. An intersection of two charge graphs gives equal costs. A curved section requires a local rather than one fixed slope.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A horizontal distance graph does not mean fast motion. A graph height gives distance, while slope gives its rate. Average speed includes every elapsed interval, including waiting.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A14 uses real contexts and graphical solutions. Foundation can interpret a plotted non-standard function; Higher also interprets exponential models and nonlinear graph estimates in the next lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Distance-time graphs and contextual intersections · Higher

Where is the waiting time on the graph?

• A delivery rider travels, waits, then travels again. Which parts of a distance-time graph reveal the waiting time?
• This lesson studies stationary 静止的: Not changing position over the time interval described.

Choose the mathematical structure

• Read the axis quantities and units before interpreting shape. On a distance-time graph, slope is speed for a segment with increasing distance; a horizontal segment means no distance change. An intersection of two charge graphs gives equal costs. A curved section requires a local rather than one fixed slope.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A horizontal distance graph does not mean fast motion. A graph height gives distance, while slope gives its rate. Average speed includes every elapsed interval, including waiting.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A14 uses real contexts and graphical solutions. Foundation can interpret a plotted non-standard function; Higher also interprets exponential models and nonlinear graph estimates in the next lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Graphical gradients and area estimates · Higher

Which graphical quantity answers the question?

• A vehicle speeds up along a curve. Is the slope between its start and finish enough to describe its speed halfway?
• This lesson studies tangent gradient 切线斜率: The slope of a tangent, used to estimate a curve’s rate at one input.

Choose the mathematical structure

• A chord gives an average rate between two inputs; a tangent estimates the local rate. Read two well-separated points on the drawn tangent to reduce measurement error. On a velocity-time graph, area represents displacement. Split it into triangles/trapezia or estimate curved area using narrow strips.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Tangent estimates and curve chords use different pairs of points. Area under a distance-time graph is not distance travelled. A strip estimate is approximate unless the graph is linear on each strip.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A15 and R15 Higher require graph-based rates and area interpretation. This lesson uses graphical reasoning and geometric areas, not symbolic differentiation/integration.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Origin-centred circles and tangent equations · Higher

Which line just touches the circular path?

• A circular path has a sensor at (3,4). Which straight line just touches the path at that point?
• This lesson studies radius gradient 半径斜率: The slope of a radius joining the centre to a point on the circle.

Choose the mathematical structure

• For an origin-centred circle, x²+y²=r². A point is on the circle if its squared coordinates sum to r². A tangent is perpendicular to the radius there. Use a negative reciprocal gradient when both gradients are finite; handle horizontal/vertical cases directly.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• The radius and tangent share a point but different directions. Do not use the negative reciprocal of zero; a horizontal radius has a vertical tangent. Check the tangent passes through its contact point.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A16 Higher concerns circles centred at the origin. Offset-centre circle formulae and circle calculus are not needed for this lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Numerical iteration for equation roots · Higher

Will repeated calculation settle down?

• An equation has no convenient factorisation. Can repeated calculation locate a solution without guessing random values?
• This lesson studies iteration 迭代: Repeat a rule, using the previous output as the next input.

Choose the mathematical structure

• Rearrange an equation as x=g(x), choose x₀ and use x_(n+1)=g(x_n). Record sufficient working precision. A stable-looking sequence must still be checked in the original equation; not every rearrangement converges. A sign change across continuous inputs can check a rounded root.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Use the previous iterate, not the starting value every time. Keep more digits than the final answer. A sign change needs continuity; a jump across a vertical asymptote does not guarantee a root.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A20 Higher uses numerical iteration and suffix notation. Use the calculator to follow the stated rule, then report the requested rounding with a check. Newton derivatives are outside this GCSE lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Linear inequalities and number-line solutions · Foundation

Which purchases fit within the budget?

• A student can spend at most 17 yuan on a fixed 5-yuan charge plus 3 yuan per item. Is one equality enough to describe every allowed purchase?
• This lesson studies solution set 解集: All inputs that satisfy the stated condition.

Choose the mathematical structure

• Solve a linear inequality using the same balance operations as an equation. Multiplying or dividing by a negative reverses its direction. Use a closed endpoint for ≤ or ≥ and an open endpoint for < or >. Intersect restrictions to find their common permitted inputs.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A reversed sign is needed only when multiplying or dividing by a negative, not when adding a negative. Include the physical domain: negative or fractional item counts may be meaningless.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A22 Foundation requires one-variable linear inequalities and number lines. Higher quadratic and two-variable regions are in a separate lesson; avoid replacing the inequality with a single boundary value.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Linear inequalities and number-line solutions · Higher

Which purchases fit within the budget?

• A student can spend at most 17 yuan on a fixed 5-yuan charge plus 3 yuan per item. Is one equality enough to describe every allowed purchase?
• This lesson studies solution set 解集: All inputs that satisfy the stated condition.

Choose the mathematical structure

• Solve a linear inequality using the same balance operations as an equation. Multiplying or dividing by a negative reverses its direction. Use a closed endpoint for ≤ or ≥ and an open endpoint for < or >. Intersect restrictions to find their common permitted inputs.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A reversed sign is needed only when multiplying or dividing by a negative, not when adding a negative. Include the physical domain: negative or fractional item counts may be meaningless.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A22 Foundation requires one-variable linear inequalities and number lines. Higher quadratic and two-variable regions are in a separate lesson; avoid replacing the inequality with a single boundary value.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Quadratic inequalities and two-variable regions · Higher

Where do all the restrictions overlap?

• A design must lie above one line and below another. How can a graph show every feasible pair at once?
• This lesson studies boundary line 边界线: The equality line separating values that satisfy an inequality from those that do not.

Choose the mathematical structure

• For a quadratic inequality, locate roots and test the sign in each interval. For a two-variable inequality, draw its equality boundary; use a dashed line if equality is excluded and a solid line if included. Test a point on each side and intersect the allowed regions. Write the domain or set notation clearly.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• For an upward-opening quadratic, positive values lie outside the roots, not between them. A boundary intersection is not necessarily included. Use a test point not on the boundary itself.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A22 Higher includes quadratic inequalities in one variable and linear inequalities in two variables. The graphical overlap is the solution region, not one selected point.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Figurate, geometric and Fibonacci-type sequences · Foundation

Does the dot pattern add or multiply?

• A growing dot pattern adds a longer row each time. How can a diagram distinguish it from repeated multiplication?
• This lesson studies recursive rule 递推规则: A rule that generates new terms from preceding terms.

Choose the mathematical structure

• Record positions and terms separately. Triangular numbers add 1,2,3,...; square and cube numbers use n² and n³. A geometric sequence multiplies by a common positive ratio. A Fibonacci-type sequence starts with stated terms and then adds its two predecessors. A supplied recursive rule must include enough initial values.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A nonconstant first difference does not mean a pattern is random. A geometric ratio is not a common difference. Check the starting index when writing a position rule.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A23/A24 Foundation includes these patterns and positive rational geometric ratios. Higher also permits surd ratios and derives quadratic nth terms in the next lesson. Infinite-series sums are not part of this GCSE lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Figurate, geometric and Fibonacci-type sequences · Higher

Does the dot pattern add or multiply?

• A growing dot pattern adds a longer row each time. How can a diagram distinguish it from repeated multiplication?
• This lesson studies recursive rule 递推规则: A rule that generates new terms from preceding terms.

Choose the mathematical structure

• Record positions and terms separately. Triangular numbers add 1,2,3,...; square and cube numbers use n² and n³. A geometric sequence multiplies by a common positive ratio. A Fibonacci-type sequence starts with stated terms and then adds its two predecessors. A supplied recursive rule must include enough initial values.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A nonconstant first difference does not mean a pattern is random. A geometric ratio is not a common difference. Check the starting index when writing a position rule.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A23/A24 Foundation includes these patterns and positive rational geometric ratios. Higher also permits surd ratios and derives quadratic nth terms in the next lesson. Infinite-series sums are not part of this GCSE lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Quadratic nth terms and surd-ratio progressions · Higher

What does the second difference reveal?

• The terms 3,8,15,24 grow by 5,7,9. What rule predicts the next term without building every earlier term?
• This lesson studies second difference 二阶差分: The difference between consecutive first differences of a sequence.

Choose the mathematical structure

• A constant second difference signals a quadratic rule an²+bn+c. Its value is 2a. Subtract an² from the terms; fit the remaining linear rule and test at several positions. A geometric sequence with a surd ratio still multiplies by the same exact factor; keep root values exact.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• The second difference equals 2a, not a. A quadratic rule must be checked against the first terms; there can be a nonzero constant c. A finite pattern alone does not prove a unique rule without the stated sequence family.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A25 Higher derives quadratic nth terms; A24 permits surd-ratio sequences. Foundation recognises and generates quadratic patterns without this general derivation.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.2 · Algebraic fractions and excluded values · Higher

A cancelled fraction can look like a straight line and still have one missing point. Cancelling a factor does not put an undefined input back into the original formula.

• A cancelled fraction can look like a straight line and still have one missing point. Cancelling a factor does not put an undefined input back into the original formula.
• This lesson studies excluded value 排除值: An input for which an original denominator or divisor is zero.

Choose the mathematical structure

• Factor every numerator and denominator before cancelling a common factor. A term joined by addition is not a cancellable factor. Record exclusions from the original expression first. For addition or subtraction use a common denominator, retaining brackets around the entire numerator. Multiply factored numerators and denominators. To divide, multiply by the reciprocal and exclude inputs making the divisor zero as well as inputs making any original fraction undefined.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Never cancel the x in (x+2)/x. A simplified denominator does not show all original restrictions. In a division task, check when the divisor is zero. Cross-multiplication can produce an excluded candidate; a formal root is not automatically a solution.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA A4 Higher includes algebraic fractions and valid cancellation. The equation example checks whether a candidate is inside the original domain. Explain each factor cancellation, retain original restrictions and check every equation candidate. These operations extend the existing manipulation example into a complete worked arithmetic sequence.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.3 · Percentages, ratio and proportional reasoning · Foundation

Can we recover the original price?

• A coat is reduced by 20% to ¥240. The discount applies to the original price, not to the sale price.
• This lesson studies multiplier 乘数: A factor that performs a percentage change in one multiplication.

Choose the mathematical structure

• A p% increase has multiplier 1+p/100; a decrease has multiplier 1-p/100. Reverse a percentage by dividing by the multiplier. In a ratio, first find the total number of parts.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A percentage uses a stated base. Subtracting the percentages loses that base. For compound change, multiply the multipliers; do not add the percentages.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Use percentage multipliers and divide a total into ratio parts. This Foundation/Core lesson uses linear proportional contexts, not the advanced regression methods.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.3. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.3 · Percentages, ratio and proportional reasoning · Higher

Can we recover the original price?

• A coat is reduced by 20% to ¥240. The discount applies to the original price, not to the sale price.
• This lesson studies multiplier 乘数: A factor that performs a percentage change in one multiplication.

Choose the mathematical structure

• A p% increase has multiplier 1+p/100; a decrease has multiplier 1-p/100. Reverse a percentage by dividing by the multiplier. In a ratio, first find the total number of parts.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A percentage uses a stated base. Subtracting the percentages loses that base. For compound change, multiply the multipliers; do not add the percentages.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• For direct proportion use y=kx; for inverse proportion use y=k/x. Calculate k from a known pair before using a new value. State what you held constant.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.3. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.3 · Ratios, equivalent proportions and mixtures · Foundation

A recipe uses concentrate and water in the ratio 2:5. A 700 ml batch and a doubled batch should taste the same, although their volumes differ.

• A recipe uses concentrate and water in the ratio 2:5. A 700 ml batch and a doubled batch should taste the same, although their volumes differ.
• This lesson studies part-to-whole ratio 部分与整体的比: A comparison of one share with the combined quantity.

Choose the mathematical structure

• Put quantities in the same units before simplifying a ratio. Divide every part by the same nonzero factor. For a:b the whole has a+b parts; the first share is a/(a+b) of the whole but a/b of the other share. Equivalent ratios have the same multiplicative relationship: a/b=c/d gives ad=bc when denominators are nonzero.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Adding the same amount to both shares generally changes the ratio. Do not divide by 2+7 when the stated ratio is already part:whole 2:7; the whole is seven parts. A concentration fraction uses total volume as denominator.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA R3–R8 connects ratio notation, fractions, equivalent proportions and linear functions. Check that both shares sum to the stated total, and name which quantity is compared with which. Use a ratio table to scale a mixture without assuming a fixed additive difference.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.3. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.3 · Ratios, equivalent proportions and mixtures · Higher

A recipe uses concentrate and water in the ratio 2:5. A 700 ml batch and a doubled batch should taste the same, although their volumes differ.

• A recipe uses concentrate and water in the ratio 2:5. A 700 ml batch and a doubled batch should taste the same, although their volumes differ.
• This lesson studies part-to-whole ratio 部分与整体的比: A comparison of one share with the combined quantity.

Choose the mathematical structure

• Put quantities in the same units before simplifying a ratio. Divide every part by the same nonzero factor. For a:b the whole has a+b parts; the first share is a/(a+b) of the whole but a/b of the other share. Equivalent ratios have the same multiplicative relationship: a/b=c/d gives ad=bc when denominators are nonzero.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Adding the same amount to both shares generally changes the ratio. Do not divide by 2+7 when the stated ratio is already part:whole 2:7; the whole is seven parts. A concentration fraction uses total volume as denominator.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA R3–R8 connects ratio notation, fractions, equivalent proportions and linear functions. Check that both shares sum to the stated total, and name which quantity is compared with which. Use a ratio table to scale a mixture without assuming a fixed additive difference.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.3. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.3 · Scale drawings and maps · Foundation

A map at 1:25,000 shows a 4 cm route. The written scale compares centimetres with centimetres, not centimetres with kilometres.

• A map at 1:25,000 shows a 4 cm route. The written scale compares centimetres with centimetres, not centimetres with kilometres.
• This lesson studies scale 比例尺: The ratio of a drawing length to its corresponding real length.

Choose the mathematical structure

• In a scale 1:n, one drawing unit represents n of the same real unit. Multiply a drawing length by n to obtain the real length; divide a real length by n to draw it. Then convert the unit. Measure only when the diagram explicitly supplies an accurate scale.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A ratio compares matching units. Never measure a diagram marked not to scale. A photocopied numerical scale can become invalid even though its scale bar still works.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA R2 includes maps, scale factors and geometric problems. Label drawing and real dimensions separately; reverse the calculation to verify the scale. Use an exact ratio until the context asks for rounding.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.3. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.3 · Scale drawings and maps · Higher

A map at 1:25,000 shows a 4 cm route. The written scale compares centimetres with centimetres, not centimetres with kilometres.

• A map at 1:25,000 shows a 4 cm route. The written scale compares centimetres with centimetres, not centimetres with kilometres.
• This lesson studies scale 比例尺: The ratio of a drawing length to its corresponding real length.

Choose the mathematical structure

• In a scale 1:n, one drawing unit represents n of the same real unit. Multiply a drawing length by n to obtain the real length; divide a real length by n to draw it. Then convert the unit. Measure only when the diagram explicitly supplies an accurate scale.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A ratio compares matching units. Never measure a diagram marked not to scale. A photocopied numerical scale can become invalid even though its scale bar still works.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA R2 includes maps, scale factors and geometric problems. Label drawing and real dimensions separately; reverse the calculation to verify the scale. Use an exact ratio until the context asks for rounding.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.3. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.3 · Direct and inverse proportional relationships · Foundation

Buying twice as much fabric at a fixed price doubles the bill. Sharing one fixed job among twice as many equally productive workers halves its duration.

• Buying twice as much fabric at a fixed price doubles the bill. Sharing one fixed job among twice as many equally productive workers halves its duration.
• This lesson studies constant of proportionality 比例常数: The fixed multiplier in a stated proportional relationship.

Choose the mathematical structure

• Direct proportion y=kx preserves y/x for nonzero x and gives a straight line through the origin. Inverse proportion y=k/x preserves xy and gives a reciprocal curve, with x=0 excluded. State the fixed assumptions: a start-up charge breaks direct proportion, while changing productivity can break inverse proportion. Higher constructs the equation from a known pair; Foundation interprets and uses a given equation.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• An increasing graph need not be direct proportion. An inverse relationship is proportional to 1/x, not to -x. Keep a product constant for inverse proportion and a quotient constant for direct proportion.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA R10/R13/R14 covers numerical, algebraic and graphical direct/inverse proportion. Interpret the units of k and the gradient. In Foundation use the supplied equations; Higher also constructs and interprets them from the context.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.3. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.3 · Direct and inverse proportional relationships · Higher

Buying twice as much fabric at a fixed price doubles the bill. Sharing one fixed job among twice as many equally productive workers halves its duration.

• Buying twice as much fabric at a fixed price doubles the bill. Sharing one fixed job among twice as many equally productive workers halves its duration.
• This lesson studies constant of proportionality 比例常数: The fixed multiplier in a stated proportional relationship.

Choose the mathematical structure

• Direct proportion y=kx preserves y/x for nonzero x and gives a straight line through the origin. Inverse proportion y=k/x preserves xy and gives a reciprocal curve, with x=0 excluded. State the fixed assumptions: a start-up charge breaks direct proportion, while changing productivity can break inverse proportion. Higher constructs the equation from a known pair; Foundation interprets and uses a given equation.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• An increasing graph need not be direct proportion. An inverse relationship is proportional to 1/x, not to -x. Keep a product constant for inverse proportion and a quotient constant for direct proportion.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA R10/R13/R14 covers numerical, algebraic and graphical direct/inverse proportion. Interpret the units of k and the gradient. In Foundation use the supplied equations; Higher also constructs and interprets them from the context.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.3. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.3 · Rates, unit prices, density and pressure · Foundation

Two packages have different sizes and prices. Comparing the sticker prices alone cannot tell which gives more material for each yuan.

• Two packages have different sizes and prices. Comparing the sticker prices alone cannot tell which gives more material for each yuan.
• This lesson studies density · ⁨densidad⁩ 密度: Mass per unit volume, with both units stated.

Choose the mathematical structure

• A compound rate divides one quantity by another: speed=d/t, pay=earnings/time, unit price=cost/amount, density=mass/volume and pressure=force/area. Rearrange these equations before substituting. Convert each dimension separately; converting cm² to m² uses a squared length factor. Comparisons must use the same units and conditions.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Volume conversions cube the length factor and area conversions square it. A density of 2.7 g/cm³ is not 2.7 kg/m³. A cheaper package may cost more per kilogram; include only comparable products.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA R1/R11 includes speed, pay, pricing, density and pressure in numerical and algebraic contexts. Show the rearrangement with named quantities and carry units through the answer; use an inverse check such as density×volume=mass.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.3. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.3 · Rates, unit prices, density and pressure · Higher

Two packages have different sizes and prices. Comparing the sticker prices alone cannot tell which gives more material for each yuan.

• Two packages have different sizes and prices. Comparing the sticker prices alone cannot tell which gives more material for each yuan.
• This lesson studies density · ⁨densidad⁩ 密度: Mass per unit volume, with both units stated.

Choose the mathematical structure

• A compound rate divides one quantity by another: speed=d/t, pay=earnings/time, unit price=cost/amount, density=mass/volume and pressure=force/area. Rearrange these equations before substituting. Convert each dimension separately; converting cm² to m² uses a squared length factor. Comparisons must use the same units and conditions.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Volume conversions cube the length factor and area conversions square it. A density of 2.7 g/cm³ is not 2.7 kg/m³. A cheaper package may cost more per kilogram; include only comparable products.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA R1/R11 includes speed, pay, pricing, density and pressure in numerical and algebraic contexts. Show the rearrangement with named quantities and carry units through the answer; use an inverse check such as density×volume=mass.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.3. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.3 · Length, area and volume scale factors · Foundation

A model doubles every length. Each rectangular face doubles in two directions, while its volume doubles in three directions.

• A model doubles every length. Each rectangular face doubles in two directions, while its volume doubles in three directions.
• This lesson studies area scale factor 面积比例因子: The squared length factor between similar shapes.

Choose the mathematical structure

• For similar shapes with corresponding length factor k, area factor is k² and volume factor is k³. Name the direction of the comparison: model to real or real to model. Recover k from an area ratio by a square root and from a volume ratio by a cube root. A change in only one dimension does not create similar solids. Higher links corresponding sides to equal trigonometric ratios.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Areas do not scale by k, and volumes do not scale by k². Similarity needs every corresponding length to share the same factor. Reversing a ratio requires the reciprocal factor.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA R12 Foundation includes ratios and scale factors for lengths, areas and volumes. Higher adds links to similarity including trigonometric ratios. Check dimensions and compare a simple box or rectangle before applying the general factor.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.3. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.3 · Length, area and volume scale factors · Higher

A model doubles every length. Each rectangular face doubles in two directions, while its volume doubles in three directions.

• A model doubles every length. Each rectangular face doubles in two directions, while its volume doubles in three directions.
• This lesson studies area scale factor 面积比例因子: The squared length factor between similar shapes.

Choose the mathematical structure

• For similar shapes with corresponding length factor k, area factor is k² and volume factor is k³. Name the direction of the comparison: model to real or real to model. Recover k from an area ratio by a square root and from a volume ratio by a cube root. A change in only one dimension does not create similar solids. Higher links corresponding sides to equal trigonometric ratios.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Areas do not scale by k, and volumes do not scale by k². Similarity needs every corresponding length to share the same factor. Reversing a ratio requires the reciprocal factor.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA R12 Foundation includes ratios and scale factors for lengths, areas and volumes. Higher adds links to similarity including trigonometric ratios. Check dimensions and compare a simple box or rectangle before applying the general factor.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.3. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.3 · Simple interest, compound growth and decay · Foundation

A 1000-yuan deposit earns 10% a year. Simple interest adds the same 100 yuan each year; compound interest earns interest on earlier interest too.

• A 1000-yuan deposit earns 10% a year. Simple interest adds the same 100 yuan each year; compound interest earns interest on earlier interest too.
• This lesson studies compound interest 复利: Interest calculated on the updated balance each period.

Choose the mathematical structure

• A percentage r corresponds to decimal r/100; increases use multiplier 1+r/100 and decreases 1-r/100. Simple interest on principal P for n periods at rate r is Prn/100, giving balance P(1+rn/100). Compound balance is P(1+r/100)^n. Decay uses the corresponding decreasing multiplier. Use matching rate and period units, and divide by the multiplier to recover an original value.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Use the original value as the denominator for percentage change. Repeated 20% decreases do not subtract 40% of the initial amount. A rate per year cannot be treated as a rate per month without a specified conversion.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA R9/R16 includes percentage comparisons, original values, simple interest and repeated growth/decay. State whether the task asks for interest alone or total balance, and interpret whole-period threshold answers.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.3. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.3 · Simple interest, compound growth and decay · Higher

A 1000-yuan deposit earns 10% a year. Simple interest adds the same 100 yuan each year; compound interest earns interest on earlier interest too.

• A 1000-yuan deposit earns 10% a year. Simple interest adds the same 100 yuan each year; compound interest earns interest on earlier interest too.
• This lesson studies compound interest 复利: Interest calculated on the updated balance each period.

Choose the mathematical structure

• A percentage r corresponds to decimal r/100; increases use multiplier 1+r/100 and decreases 1-r/100. Simple interest on principal P for n periods at rate r is Prn/100, giving balance P(1+rn/100). Compound balance is P(1+r/100)^n. Decay uses the corresponding decreasing multiplier. Use matching rate and period units, and divide by the multiplier to recover an original value.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Use the original value as the denominator for percentage change. Repeated 20% decreases do not subtract 40% of the initial amount. A rate per year cannot be treated as a rate per month without a specified conversion.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA R9/R16 includes percentage comparisons, original values, simple interest and repeated growth/decay. State whether the task asks for interest alone or total balance, and interpret whole-period threshold answers.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.3. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.3 · Iterative processes with repeated deposits · Higher

A saver receives 10% interest and then adds 50 yuan each year. The deposit comes after interest, so the order changes the balance.

• A saver receives 10% interest and then adds 50 yuan each year. The deposit comes after interest, so the order changes the balance.
• This lesson studies recurrence relation 递推关系: A rule obtaining a later value from the preceding value.

Choose the mathematical structure

• Write an update rule and an initial value. Apply the operations in their stated order, using the previous output as the next input. A recurrence B next=1.1B+50 differs from 1.1(B+50). Tables can locate a first whole-period threshold; verify both the preceding and crossing values. Iteration can model growth, decay or other repeated processes, not just solve equations.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Preserve operation order and use the updated value each time. Do not round early or confuse the initial value with the first updated value. A continuous fractional-period estimate does not answer a whole-period question by itself.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA R16 Higher extends growth/decay to general iterative processes. State the model assumptions, initial value, recurrence and threshold interpretation. Use substitution to check a proposed fixed point without calculus.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.3. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Angles, lengths and area · Foundation

Does volume scale like length?

• A model has lengths one third of the real object. How much smaller are its area and volume?
• This lesson studies scale factor 相似比: The multiplier that relates corresponding lengths in similar shapes.

Choose the mathematical structure

• Use angle facts with a stated reason. Similar shapes have equal corresponding angles and proportional corresponding lengths. Areas of rectangles and triangles come from their dimensions; compound shapes can be split into simpler parts.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Equal angles alone establish similarity, not equal size. Use corresponding lengths in the same order. Convert linear units before calculating area or volume, or square/cube the conversion factor correctly.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Use a labelled sketch and appropriate units. This Foundation/Core lesson does not test area/volume scale factors or advanced circle-theorem proofs.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Angle reasoning, similarity and mensuration · Higher

Does volume scale like length?

• A model has lengths one third of the real object. How much smaller are its area and volume?
• This lesson studies scale factor 相似比: The multiplier that relates corresponding lengths in similar shapes.

Choose the mathematical structure

• For similar shapes with length scale factor k, areas scale by k² and volumes by k³. State angle reasons explicitly. A circle's tangent is perpendicular to the radius at the contact point.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Equal angles alone establish similarity, not equal size. Use corresponding lengths in the same order. Convert linear units before calculating area or volume, or square/cube the conversion factor correctly.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• A geometric proof should name the relevant theorem, identify the equal angle or ratio, and draw the conclusion. A scale drawing is evidence only when the task permits measurement.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Right-angled triangles in two dimensions · Foundation

Which side does the ladder need?

• A ladder reaches a height of 4 m while its foot is 3 m from a wall. Which sides are known, and which angle do we need?
• This lesson studies hypotenuse 斜边: The side opposite the right angle in a right-angled triangle.

Choose the mathematical structure

• Use Pythagoras in a right triangle and use sine, cosine or tangent with the sides labelled relative to the chosen angle.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Label sides relative to the chosen angle. Pythagoras needs a right angle. A calculator angle mode error can produce a plausible but wrong result. Keep unrounded values for later steps.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• This Foundation/Core lesson uses right-angled triangles only. Sine and cosine rules for non-right triangles belong to the advanced-tier lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Right triangles and non-right triangles · Higher

Which side does the ladder need?

• A ladder reaches a height of 4 m while its foot is 3 m from a wall. Which sides are known, and which angle do we need?
• This lesson studies hypotenuse 斜边: The side opposite the right angle in a right-angled triangle.

Choose the mathematical structure

• In a right triangle a²+b²=c²; sinθ=opposite/hypotenuse, cosθ=adjacent/hypotenuse and tanθ=opposite/adjacent. For other triangles, use the sine or cosine rule, or area=ab sin C/2.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Label sides relative to the chosen angle. Pythagoras needs a right angle. A calculator angle mode error can produce a plausible but wrong result. Keep unrounded values for later steps.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Use a plan or elevation for a three-dimensional problem before applying a triangle rule. Explain why the chosen triangle contains the required length or angle.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Vectors and transformation geometry · Foundation

Why is displacement shorter than the walk?

• Walking 4 m east and 3 m north gives a displacement of 5 m, even though the travelled distance is 7 m.
• This lesson studies resultant 合向量: The vector sum representing the combined displacement or force.

Choose the mathematical structure

• Add corresponding vector components and subtract position vectors to find a displacement. A translation moves every point by the same vector; a scalar multiple changes length and possibly direction.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• The order of subtraction matters: BA=-AB. Proving parallelism needs a scalar-multiple relation; a sketch alone is insufficient. Negative enlargement reverses position about its centre.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Draw arrows with direction and identify the starting and ending points. Scalar products, spatial line equations and advanced angle calculations are excluded from this Foundation/Core lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Vectors and transformation geometry · Higher

Why is displacement shorter than the walk?

• Walking 4 m east and 3 m north gives a displacement of 5 m, even though the travelled distance is 7 m.
• This lesson studies resultant 合向量: The vector sum representing the combined displacement or force.

Choose the mathematical structure

• Add corresponding vector components and subtract position vectors to find a displacement. A translation moves every point by the same vector; a scalar multiple changes length and possibly direction.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• The order of subtraction matters: BA=-AB. Proving parallelism needs a scalar-multiple relation; a sketch alone is insufficient. Negative enlargement reverses position about its centre.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Draw arrows with direction and identify the starting and ending points. Scalar products, spatial line equations and advanced angle calculations are excluded from this Foundation/Core lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Reflections, rotations and enlargements · Foundation

Where is the enlargement centre?

• A logo is enlarged around a point away from the origin. Multiplying its coordinates alone puts it in the wrong place.
• This lesson studies centre of enlargement 位似中心: The point from which each point's displacement is multiplied by a scale factor.

Choose the mathematical structure

• A translation adds a vector. A reflection reverses signed perpendicular distance from a mirror line. A rotation needs a centre, angle and direction. For enlargement from C, use new P=C+k(P-C).
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Specify a reflection line, rotation centre/angle/direction or enlargement centre/factor. Fractional positive enlargement factors reduce a shape without reversing its position about the centre.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G7 Foundation includes single reflections, rotations, translations and enlargements with positive integer/fractional factors. Negative factors and combinations of isometries belong to separate Higher teaching.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Reflections, rotations and enlargements · Higher

Where is the enlargement centre?

• A logo is enlarged around a point away from the origin. Multiplying its coordinates alone puts it in the wrong place.
• This lesson studies centre of enlargement 位似中心: The point from which each point's displacement is multiplied by a scale factor.

Choose the mathematical structure

• A translation adds a vector. A reflection reverses signed perpendicular distance from a mirror line. A rotation needs a centre, angle and direction. For enlargement from C, use new P=C+k(P-C).
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A rotation needs its centre and direction, not just an angle. A negative enlargement factor places the image on the opposite side of the centre. A translation does not change orientation or size.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Describe a transformation completely before constructing the image. Check corresponding distances and angles. For combined transformations, apply them in the stated order; they usually do not commute.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Constructions, loci and geometric conditions · Foundation

Where can both conditions hold?

• A router must be equally far from two rooms and within reach of a power point. Each condition creates a different set of possible positions.
• This lesson studies locus 轨迹: The set of all points satisfying a stated geometric condition.

Choose the mathematical structure

• Points equally distant from A and B lie on the perpendicular bisector of AB. Points at fixed distance r from C lie on a circle. Points equally distant from two intersecting lines lie on their angle bisectors.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• The perpendicular bisector concerns distance to two points; the angle bisector concerns distance to two lines. A sketch is not a ruler-and-compass construction: preserve arcs as evidence of the method.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G2 uses ruler-and-compass constructions, including a 60° angle, perpendiculars, bisectors and intersections of loci. Preserve construction arcs and justify the equidistance condition.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Constructions, loci and geometric conditions · Higher

Where can both conditions hold?

• A router must be equally far from two rooms and within reach of a power point. Each condition creates a different set of possible positions.
• This lesson studies locus 轨迹: The set of all points satisfying a stated geometric condition.

Choose the mathematical structure

• Points equally distant from A and B lie on the perpendicular bisector of AB. Points at fixed distance r from C lie on a circle. Points equally distant from two intersecting lines lie on their angle bisectors.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• The perpendicular bisector concerns distance to two points; the angle bisector concerns distance to two lines. A sketch is not a ruler-and-compass construction: preserve arcs as evidence of the method.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G2 uses ruler-and-compass constructions, including a 60° angle, perpendiculars, bisectors and intersections of loci. Preserve construction arcs and justify the equidistance condition.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Circle theorems and reasoned proofs · Higher

Which angles see the same chord?

• Two observers see the same chord from the circle. Their angles are linked by a theorem rather than by the apparent size of the drawing.
• This lesson studies cyclic quadrilateral 圆内接四边形: A quadrilateral whose four vertices lie on one circle.

Choose the mathematical structure

• The angle at the centre is twice the angle at the circumference on the same arc. Angles in the same segment are equal. Opposite angles of a cyclic quadrilateral sum to 180°. A radius is perpendicular to a tangent at contact.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Identify the same chord and the correct arc before using a theorem. Two visible right angles do not prove a quadrilateral cyclic without a valid converse argument. A diagram need not be to scale.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Write one reason alongside each angle calculation. For the alternate-segment theorem, name the tangent and chord, then identify the angle in the opposite segment. Use auxiliary radii only when they help the proof.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Shape vocabulary, properties and symmetry · Foundation

A square tile is also a rectangle and a rhombus. A shape may satisfy more than one definition, so the name alone does not always give every property.

• A square tile is also a rectangle and a rhombus. A shape may satisfy more than one definition, so the name alone does not always give every property.
• This lesson studies line of symmetry 对称轴: A mirror line that maps a whole shape onto itself.

Choose the mathematical structure

• A point marks a position; a line extends in both directions and a segment has two endpoints. A plane is a flat two-dimensional surface. A vertex is a corner; an edge is a boundary segment where solid faces meet. A polygon is a closed plane shape made of straight sides; regular means all sides and all interior angles are equal. Points label vertices; AB names a side and angle ABC has vertex B. Parallel lines have the same direction; perpendicular lines meet at 90°. Reflection symmetry uses a mirror line; rotational symmetry counts matches during one full turn, including 360°.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Equal-looking lengths need stated equal-length marks or a deduction. Regular does not mean equal sides alone. An axis of symmetry is a full line, not just an internal diagonal.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G1/G4 requires conventional names, notation and derived shape properties. State the property that justifies a classification and allow overlapping classes, such as square and rectangle.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Shape vocabulary, properties and symmetry · Higher

A square tile is also a rectangle and a rhombus. A shape may satisfy more than one definition, so the name alone does not always give every property.

• A square tile is also a rectangle and a rhombus. A shape may satisfy more than one definition, so the name alone does not always give every property.
• This lesson studies line of symmetry 对称轴: A mirror line that maps a whole shape onto itself.

Choose the mathematical structure

• A point marks a position; a line extends in both directions and a segment has two endpoints. A plane is a flat two-dimensional surface. A vertex is a corner; an edge is a boundary segment where solid faces meet. A polygon is a closed plane shape made of straight sides; regular means all sides and all interior angles are equal. Points label vertices; AB names a side and angle ABC has vertex B. Parallel lines have the same direction; perpendicular lines meet at 90°. Reflection symmetry uses a mirror line; rotational symmetry counts matches during one full turn, including 360°.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Equal-looking lengths need stated equal-length marks or a deduction. Regular does not mean equal sides alone. An axis of symmetry is a full line, not just an internal diagonal.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G1/G4 requires conventional names, notation and derived shape properties. State the property that justifies a classification and allow overlapping classes, such as square and rectangle.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Angle facts and polygon reasoning · Foundation

A surveyor measures one angle where a road crosses two parallel boundaries. Parallelism allows another angle to be deduced without measuring it.

• A surveyor measures one angle where a road crosses two parallel boundaries. Parallelism allows another angle to be deduced without measuring it.
• This lesson studies corresponding angles 同位角: Angles in matching positions at a transversal crossing two lines.

Choose the mathematical structure

• Angles around one point sum to 360°; angles on a straight line sum to 180°; vertically opposite angles are equal. When the crossed lines are parallel, corresponding and alternate angles are equal and co-interior angles sum to 180°. A triangle has angle sum 180°. Split an n-sided polygon into n-2 triangles to derive its interior sum (n-2)×180°.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Corresponding/alternate equalities require parallel lines. Name the theorem rather than using informal letter-shape labels. An interior angle is not the same as an exterior turning angle.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G3/G6 expects reasons and derivations. Mark the given parallelism, identify the angle positions and build a chain of justified equalities rather than reading angles from a sketch.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Angle facts and polygon reasoning · Higher

A surveyor measures one angle where a road crosses two parallel boundaries. Parallelism allows another angle to be deduced without measuring it.

• A surveyor measures one angle where a road crosses two parallel boundaries. Parallelism allows another angle to be deduced without measuring it.
• This lesson studies corresponding angles 同位角: Angles in matching positions at a transversal crossing two lines.

Choose the mathematical structure

• Angles around one point sum to 360°; angles on a straight line sum to 180°; vertically opposite angles are equal. When the crossed lines are parallel, corresponding and alternate angles are equal and co-interior angles sum to 180°. A triangle has angle sum 180°. Split an n-sided polygon into n-2 triangles to derive its interior sum (n-2)×180°.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Corresponding/alternate equalities require parallel lines. Name the theorem rather than using informal letter-shape labels. An interior angle is not the same as an exterior turning angle.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G3/G6 expects reasons and derivations. Mark the given parallelism, identify the angle positions and build a chain of justified equalities rather than reading angles from a sketch.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Triangle congruence and geometric proof · Foundation

Two triangular braces need to fit the same frame. Matching three angles fixes their shape but does not guarantee that their sizes agree.

• Two triangular braces need to fit the same frame. Matching three angles fixes their shape but does not guarantee that their sizes agree.
• This lesson studies congruence 全等: Equal shape and size, with matching lengths and angles.

Choose the mathematical structure

• Use SSS (three sides), SAS (two sides and their included angle), ASA (two angles and the corresponding side), or RHS (right angle, hypotenuse and one other side). Match vertices in the same order. AAA establishes similarity, not congruence. SSA generally permits more than one triangle. A proof needs a given fact, a valid criterion and a matching-part conclusion.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• The SAS angle must lie between the named sides. RHS uses the hypotenuse, not two arbitrary sides. A proof diagram supports the argument; it cannot establish equality just by appearance.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G5/G6 uses basic congruence criteria and simple geometric proofs, including isosceles base angles and Pythagoras. State every matching pair and explain which criterion applies.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Triangle congruence and geometric proof · Higher

Two triangular braces need to fit the same frame. Matching three angles fixes their shape but does not guarantee that their sizes agree.

• Two triangular braces need to fit the same frame. Matching three angles fixes their shape but does not guarantee that their sizes agree.
• This lesson studies congruence 全等: Equal shape and size, with matching lengths and angles.

Choose the mathematical structure

• Use SSS (three sides), SAS (two sides and their included angle), ASA (two angles and the corresponding side), or RHS (right angle, hypotenuse and one other side). Match vertices in the same order. AAA establishes similarity, not congruence. SSA generally permits more than one triangle. A proof needs a given fact, a valid criterion and a matching-part conclusion.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• The SAS angle must lie between the named sides. RHS uses the hypotenuse, not two arbitrary sides. A proof diagram supports the argument; it cannot establish equality just by appearance.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G5/G6 uses basic congruence criteria and simple geometric proofs, including isosceles base angles and Pythagoras. State every matching pair and explain which criterion applies.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Circle parts and geometric definitions · Foundation

A round window can be split by a straight chord or by two radii. These cuts create different regions, even though both use a curved boundary.

• A round window can be split by a straight chord or by two radii. These cuts create different regions, even though both use a curved boundary.
• This lesson studies segment 弓形: A region bounded by a chord and its corresponding arc.

Choose the mathematical structure

• A circle consists of points a fixed radius from its centre. A diameter is a chord through the centre and has length 2r. A chord joins two circumference points; an arc is part of the circumference. A sector lies between two radii and an arc. A segment lies between a chord and an arc. A tangent touches at one point and is perpendicular to the radius there.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A chord need not pass through the centre; only a diameter must. Sector and segment have different straight boundaries. Do not confuse circumference length with the shaded area inside a circle.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G9 includes all named circle parts at Foundation. Use the given radius and position labels to identify the part; Higher circle-theorem proofs are developed separately.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Circle parts and geometric definitions · Higher

A round window can be split by a straight chord or by two radii. These cuts create different regions, even though both use a curved boundary.

• A round window can be split by a straight chord or by two radii. These cuts create different regions, even though both use a curved boundary.
• This lesson studies segment 弓形: A region bounded by a chord and its corresponding arc.

Choose the mathematical structure

• A circle consists of points a fixed radius from its centre. A diameter is a chord through the centre and has length 2r. A chord joins two circumference points; an arc is part of the circumference. A sector lies between two radii and an arc. A segment lies between a chord and an arc. A tangent touches at one point and is perpendicular to the radius there.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A chord need not pass through the centre; only a diameter must. Sector and segment have different straight boundaries. Do not confuse circumference length with the shaded area inside a circle.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G9 includes all named circle parts at Foundation. Use the given radius and position labels to identify the part; Higher circle-theorem proofs are developed separately.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Geometric reasoning on coordinate axes · Foundation

A rectangular garden is drawn using corner coordinates. The axes allow lengths and perpendicular edges to be checked even if the sketch looks distorted.

• A rectangular garden is drawn using corner coordinates. The axes allow lengths and perpendicular edges to be checked even if the sketch looks distorted.
• This lesson studies midpoint 中点: The point halfway between the endpoints of a segment.

Choose the mathematical structure

• Find horizontal or vertical length by subtracting coordinates; use Pythagoras for a diagonal. Midpoint coordinates are averages of the endpoints. Equal coordinate changes identify translations. To prove a quadrilateral property, justify side directions and lengths, rather than naming a shape from its appearance.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A coordinate difference can be negative even though a length is nonnegative. The diagonal is not the sum of the two edge lengths. One pair of equal sides alone does not establish a rectangle.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G11 solves geometric problems on axes. Keep point labels, coordinate arithmetic and geometric reasons linked in the written argument; use the diagram to choose a method, then verify it numerically.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Geometric reasoning on coordinate axes · Higher

A rectangular garden is drawn using corner coordinates. The axes allow lengths and perpendicular edges to be checked even if the sketch looks distorted.

• A rectangular garden is drawn using corner coordinates. The axes allow lengths and perpendicular edges to be checked even if the sketch looks distorted.
• This lesson studies midpoint 中点: The point halfway between the endpoints of a segment.

Choose the mathematical structure

• Find horizontal or vertical length by subtracting coordinates; use Pythagoras for a diagonal. Midpoint coordinates are averages of the endpoints. Equal coordinate changes identify translations. To prove a quadrilateral property, justify side directions and lengths, rather than naming a shape from its appearance.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A coordinate difference can be negative even though a length is nonnegative. The diagonal is not the sum of the two edge lengths. One pair of equal sides alone does not establish a rectangle.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G11 solves geometric problems on axes. Keep point labels, coordinate arithmetic and geometric reasons linked in the written argument; use the diagram to choose a method, then verify it numerically.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Solid properties, plans and elevations · Foundation

A building plan shows the footprint but hides its height. Front and side elevations add information that a single view cannot show.

• A building plan shows the footprint but hides its height. Front and side elevations add information that a single view cannot show.
• This lesson studies elevation 立面图: An orthographic view from a stated horizontal direction.

Choose the mathematical structure

• A face is a flat boundary polygon; curved surfaces must be named separately. Edges join faces and vertices are corners. A prism has two congruent parallel end faces and a constant cross-section. A pyramid joins one polygon base to an apex. Plan is the view from above; front and side elevations are direct views without perspective. Give the viewing direction and align corresponding widths, depths and heights.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Do not draw perspective diagonals in an orthographic elevation. A plan alone cannot give height. Different hidden arrangements can share the same plan and elevations; do not claim a unique reconstruction without enough information.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G12/G13 covers the named solids and construction/interpretation of views. Label dimensions and directions, preserve alignment between views and explain any hidden-space ambiguity.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Solid properties, plans and elevations · Higher

A building plan shows the footprint but hides its height. Front and side elevations add information that a single view cannot show.

• A building plan shows the footprint but hides its height. Front and side elevations add information that a single view cannot show.
• This lesson studies elevation 立面图: An orthographic view from a stated horizontal direction.

Choose the mathematical structure

• A face is a flat boundary polygon; curved surfaces must be named separately. Edges join faces and vertices are corners. A prism has two congruent parallel end faces and a constant cross-section. A pyramid joins one polygon base to an apex. Plan is the view from above; front and side elevations are direct views without perspective. Give the viewing direction and align corresponding widths, depths and heights.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Do not draw perspective diagonals in an orthographic elevation. A plan alone cannot give height. Different hidden arrangements can share the same plan and elevations; do not claim a unique reconstruction without enough information.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G12/G13 covers the named solids and construction/interpretation of views. Label dimensions and directions, preserve alignment between views and explain any hidden-space ambiguity.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Measuring angles and three-figure bearings · Foundation

A walking route has bearing 070° from A to B. Turning around changes the reference point and reverses the direction; it does not keep the same bearing.

• A walking route has bearing 070° from A to B. Turning around changes the reference point and reverses the direction; it does not keep the same bearing.
• This lesson studies bearing 方位角: A direction measured clockwise from north at the starting point.

Choose the mathematical structure

• Draw a north line at the starting point, then measure clockwise to the route. Write three digits: east 090°, south 180°, west 270°, north 000°. NE, SE, SW and NW are 045°,135°,225°,315°. For the reverse bearing add 180° and reduce modulo 360°. Use a ruler for a stated-scale length and the correct protractor scale for a stated angle.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A bearing is measured at the departure point, not the destination. Read clockwise from north rather than the acute angle to the nearest compass axis. The back bearing differs by 180°, even when the diagram is oblique.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G15 includes measured lengths/angles, maps, the eight compass directions and three-figure bearings. State which point supplies north and distinguish a numerical bearing from a measured scale distance.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Measuring angles and three-figure bearings · Higher

A walking route has bearing 070° from A to B. Turning around changes the reference point and reverses the direction; it does not keep the same bearing.

• A walking route has bearing 070° from A to B. Turning around changes the reference point and reverses the direction; it does not keep the same bearing.
• This lesson studies bearing 方位角: A direction measured clockwise from north at the starting point.

Choose the mathematical structure

• Draw a north line at the starting point, then measure clockwise to the route. Write three digits: east 090°, south 180°, west 270°, north 000°. NE, SE, SW and NW are 045°,135°,225°,315°. For the reverse bearing add 180° and reduce modulo 360°. Use a ruler for a stated-scale length and the correct protractor scale for a stated angle.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A bearing is measured at the departure point, not the destination. Read clockwise from north rather than the acute angle to the nearest compass axis. The back bearing differs by 180°, even when the diagram is oblique.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G15 includes measured lengths/angles, maps, the eight compass directions and three-figure bearings. State which point supplies north and distinguish a numerical bearing from a measured scale distance.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Combined isometries and invariants · Higher

Reflecting a logo and then sliding it can differ from sliding it and then reflecting it. The order of instructions is part of the transformation.

• Reflecting a logo and then sliding it can differ from sliding it and then reflecting it. The order of instructions is part of the transformation.
• This lesson studies invariant 不变量: A property preserved by a specified transformation.

Choose the mathematical structure

• Apply each transformation to the current image in the stated order. Rotations, reflections and translations preserve lengths and angles, so their combinations also preserve them. Reflections reverse orientation; rotations and translations preserve it. Two reflections in parallel lines give a translation; in intersecting lines they give a rotation through twice the directed angle between the mirrors.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Preserved length does not mean every point stays fixed. Do not commute transformations unless a checked argument permits it. A single reflection reverses orientation, while two reflections restore it.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G8 Higher requires changes and invariance under combinations of rigid transformations. Describe the resulting map completely and test it on more than one point before making a whole-shape claim.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Circle angles and proof chains · Higher

A circle theorem connects angles standing on one chord. The same-looking angle on the other side of the chord can instead be supplementary.

• A circle theorem connects angles standing on one chord. The same-looking angle on the other side of the chord can instead be supplementary.
• This lesson studies angle at the centre 圆心角: An angle formed by two radii meeting at the centre.

Choose the mathematical structure

• The central angle on an arc is twice a circumference angle standing on that same arc. A diameter therefore gives a 90° circumference angle. Angles in the same segment are equal. Opposite angles of a cyclic quadrilateral sum to 180°. State the chord, arc and segment before applying a theorem.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Distinguish the reflex central angle from the smaller one and identify which arc does not contain the circumference vertex. Opposite segments can give supplementary rather than equal angles. A theorem must be tied to named points.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G10 Higher requires application and proof. Draw auxiliary radii, use isosceles base angles and point sums, then extend the proof to the intended configuration rather than inferring equality from the drawing.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Tangents, chords and the alternate segment · Higher

Two paths from an outside point just touch a circular pond. Their equal lengths follow from right-triangle congruence, not from a symmetric-looking sketch.

• Two paths from an outside point just touch a circular pond. Their equal lengths follow from right-triangle congruence, not from a symmetric-looking sketch.
• This lesson studies alternate segment theorem 弦切角定理: The tangent–chord angle equals the angle subtended by that chord in the opposite segment.

Choose the mathematical structure

• A tangent is perpendicular to its contact radius. Tangents from one external point are equal. The perpendicular from the centre to a chord bisects it. The tangent–chord angle equals the angle on that chord in the alternate segment. Use equal radii, right triangles and the central-angle theorem to prove these relationships.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A tangent–chord angle and a radius–chord angle are different. Equal tangent lengths refer to one external point. The chord is bisected by a perpendicular from the centre, not by any line that happens to cross it.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G10 Higher includes these theorem proofs and related results. Give the congruence criterion or angle chain explicitly and identify the relevant chord, contact point and alternate segment.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Areas, perimeters and composite plane shapes · Foundation

A sloping garden fence can have a long side but a small perpendicular height. The side length and the height serve different jobs in an area formula.

• A sloping garden fence can have a long side but a small perpendicular height. The side length and the height serve different jobs in an area formula.
• This lesson studies perpendicular height 垂直高度: The distance between a base and its parallel opposite level, measured at right angles.

Choose the mathematical structure

• Triangle area is bh/2, parallelogram area bh and trapezium area (a+b)h/2 for parallel sides a,b. Heights are perpendicular to the selected base. Perimeter adds only the outside boundary. Split a composite shape into non-overlapping parts, or subtract a missing region from a containing shape.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Use perpendicular height rather than a sloping side. A shared internal division line is not part of the perimeter. An area answer has squared units; a perimeter answer has length units.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G16/G17 includes triangle/parallelogram/trapezium area and composite perimeter/area. Mark the parallel bases, perpendicular height and counted outside edges before calculating.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Areas, perimeters and composite plane shapes · Higher

A sloping garden fence can have a long side but a small perpendicular height. The side length and the height serve different jobs in an area formula.

• A sloping garden fence can have a long side but a small perpendicular height. The side length and the height serve different jobs in an area formula.
• This lesson studies perpendicular height 垂直高度: The distance between a base and its parallel opposite level, measured at right angles.

Choose the mathematical structure

• Triangle area is bh/2, parallelogram area bh and trapezium area (a+b)h/2 for parallel sides a,b. Heights are perpendicular to the selected base. Perimeter adds only the outside boundary. Split a composite shape into non-overlapping parts, or subtract a missing region from a containing shape.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Use perpendicular height rather than a sloping side. A shared internal division line is not part of the perimeter. An area answer has squared units; a perimeter answer has length units.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G16/G17 includes triangle/parallelogram/trapezium area and composite perimeter/area. Mark the parallel bases, perpendicular height and counted outside edges before calculating.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Prism volume and cylinder measurement · Foundation

A tank has the same triangular end shape all along its length. Its volume depends on that end area and the distance the end shape extends.

• A tank has the same triangular end shape all along its length. Its volume depends on that end area and the distance the end shape extends.
• This lesson studies cross-section 横截面: A slice perpendicular to a prism’s length that remains constant along it.

Choose the mathematical structure

• For a right prism, volume is constant cross-sectional area times perpendicular length. Cuboid volume is lwh. A cylinder is a circular prism: V=πr²h. Total closed-cylinder surface area includes two circular ends and the curved surface 2πrh. Open containers omit the specified faces.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Use cross-sectional area, not perimeter, in the volume formula. Distinguish cylinder radius from diameter and identify whether end faces are present. A length times an area gives cubic units.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G16/G17 includes cuboids, right prisms and cylinders. Draw the constant end shape, calculate it first and label the extrusion length. Preserve exact multiples of pi when asked.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Prism volume and cylinder measurement · Higher

A tank has the same triangular end shape all along its length. Its volume depends on that end area and the distance the end shape extends.

• A tank has the same triangular end shape all along its length. Its volume depends on that end area and the distance the end shape extends.
• This lesson studies cross-section 横截面: A slice perpendicular to a prism’s length that remains constant along it.

Choose the mathematical structure

• For a right prism, volume is constant cross-sectional area times perpendicular length. Cuboid volume is lwh. A cylinder is a circular prism: V=πr²h. Total closed-cylinder surface area includes two circular ends and the curved surface 2πrh. Open containers omit the specified faces.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Use cross-sectional area, not perimeter, in the volume formula. Distinguish cylinder radius from diameter and identify whether end faces are present. A length times an area gives cubic units.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G16/G17 includes cuboids, right prisms and cylinders. Draw the constant end shape, calculate it first and label the extrusion length. Preserve exact multiples of pi when asked.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Circle lengths, areas and composite boundaries · Foundation

A semicircular window needs glass and a frame. Glass uses area; the frame includes a curved arc and the straight diameter.

• A semicircular window needs glass and a frame. Glass uses area; the frame includes a curved arc and the straight diameter.
• This lesson studies circumference 圆周长: The total length of a circle’s boundary.

Choose the mathematical structure

• Circle circumference is 2πr=πd and area is πr². A semicircle has half the circle area, but its complete perimeter includes the diameter as well as half the circumference. For composite shapes count every exposed boundary once, and separate straight edges from arcs.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Radius must be halved from a given diameter before squaring. Half a circle’s circumference is only its arc, not the complete semicircle perimeter. An internal join is not exposed boundary.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G17 includes exact pi answers and composite circle perimeters/areas. Give both the exact expression and the required final rounded value, keeping the meaning and units clear.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Circle lengths, areas and composite boundaries · Higher

A semicircular window needs glass and a frame. Glass uses area; the frame includes a curved arc and the straight diameter.

• A semicircular window needs glass and a frame. Glass uses area; the frame includes a curved arc and the straight diameter.
• This lesson studies circumference 圆周长: The total length of a circle’s boundary.

Choose the mathematical structure

• Circle circumference is 2πr=πd and area is πr². A semicircle has half the circle area, but its complete perimeter includes the diameter as well as half the circumference. For composite shapes count every exposed boundary once, and separate straight edges from arcs.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Radius must be halved from a given diameter before squaring. Half a circle’s circumference is only its arc, not the complete semicircle perimeter. An internal join is not exposed boundary.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G17 includes exact pi answers and composite circle perimeters/areas. Give both the exact expression and the required final rounded value, keeping the meaning and units clear.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Spheres, cones, pyramids and frustums · Foundation

A cone-shaped cup is cut flat at its top. The missing small cone must be removed from the original cone before the capacity is known.

• A cone-shaped cup is cut flat at its top. The missing small cone must be removed from the original cone before the capacity is known.
• This lesson studies frustum 截锥体: The remaining solid after a cone or pyramid is cut parallel to its base.

Choose the mathematical structure

• Pyramid and cone volumes are one third of base area times perpendicular height. A sphere has volume 4πr³/3 and area 4πr². A cone’s curved area is πrl using slant height l; its volume uses perpendicular height h. For a frustum subtract the removed similar solid. Composite surface area counts only exposed faces; joined faces are hidden.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Perpendicular height and slant height are not interchangeable. A frustum is not a full cone of the leftover height. Shared composite faces do not contribute exposed area.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G17 additional Foundation includes spheres, pyramids, cones, composite solids and frustums. Use similarity to find missing removed dimensions, then subtract volumes or exposed areas with matching units.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Spheres, cones, pyramids and frustums · Higher

A cone-shaped cup is cut flat at its top. The missing small cone must be removed from the original cone before the capacity is known.

• A cone-shaped cup is cut flat at its top. The missing small cone must be removed from the original cone before the capacity is known.
• This lesson studies frustum 截锥体: The remaining solid after a cone or pyramid is cut parallel to its base.

Choose the mathematical structure

• Pyramid and cone volumes are one third of base area times perpendicular height. A sphere has volume 4πr³/3 and area 4πr². A cone’s curved area is πrl using slant height l; its volume uses perpendicular height h. For a frustum subtract the removed similar solid. Composite surface area counts only exposed faces; joined faces are hidden.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Perpendicular height and slant height are not interchangeable. A frustum is not a full cone of the leftover height. Shared composite faces do not contribute exposed area.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G17 additional Foundation includes spheres, pyramids, cones, composite solids and frustums. Use similarity to find missing removed dimensions, then subtract volumes or exposed areas with matching units.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Arcs, sectors and reverse angle calculations · Foundation

A circular fan opens through 120°. It covers one third of a full turn, so its arc and area are each one third of the corresponding whole circle.

• A circular fan opens through 120°. It covers one third of a full turn, so its arc and area are each one third of the corresponding whole circle.
• This lesson studies sector 扇形: A circle region bounded by two radii and their intervening arc.

Choose the mathematical structure

• For an angle θ in degrees, fraction of a turn is θ/360. Arc length is this fraction of 2πr and sector area is this fraction of πr². Sector perimeter adds the two radii. Rearrange the same fraction to recover an angle from an arc or an area. Keep degree and length units separate.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• An arc length is not a sector perimeter. Use the angle as a fraction of 360°, not 180°. Radius is squared only for area, not arc length.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G18 includes arc lengths, sector angles and areas in degree-based geometry. Show the full-circle quantity and fraction before multiplying, then check that the result fits the angle.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Arcs, sectors and reverse angle calculations · Higher

A circular fan opens through 120°. It covers one third of a full turn, so its arc and area are each one third of the corresponding whole circle.

• A circular fan opens through 120°. It covers one third of a full turn, so its arc and area are each one third of the corresponding whole circle.
• This lesson studies sector 扇形: A circle region bounded by two radii and their intervening arc.

Choose the mathematical structure

• For an angle θ in degrees, fraction of a turn is θ/360. Arc length is this fraction of 2πr and sector area is this fraction of πr². Sector perimeter adds the two radii. Rearrange the same fraction to recover an angle from an arc or an area. Keep degree and length units separate.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• An arc length is not a sector perimeter. Use the angle as a fraction of 360°, not 180°. Radius is squared only for area, not arc length.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G18 includes arc lengths, sector angles and areas in degree-based geometry. Show the full-circle quantity and fraction before multiplying, then check that the result fits the angle.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Exact trigonometric values from special triangles · Foundation

A calculator shows sin30° as 0.5, but a special triangle explains why the value is exactly one half and how the other values connect.

• A calculator shows sin30° as 0.5, but a special triangle explains why the value is exactly one half and how the other values connect.
• This lesson studies exact value 精确值: A value retained as a fraction or surd rather than a rounded decimal.

Choose the mathematical structure

• Bisect an equilateral triangle of side 2 to obtain a 30–60–90 triangle with sides 1,√3,2. A right isosceles triangle has sides 1,1,√2. Apply opposite/hypotenuse, adjacent/hypotenuse and opposite/adjacent to derive sine, cosine and tangent. Use degree angles and retain surds exactly; tan90° is undefined.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Sine and cosine interchange when the chosen acute angle changes to its complement. Do not turn √3 into a rounded decimal when an exact answer is required. Tangent at 90° is not zero.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G21 is additional Foundation and includes the listed exact values. Derive them from labelled special triangles, then use them in G20 right-triangle calculations.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Exact trigonometric values from special triangles · Higher

A calculator shows sin30° as 0.5, but a special triangle explains why the value is exactly one half and how the other values connect.

• A calculator shows sin30° as 0.5, but a special triangle explains why the value is exactly one half and how the other values connect.
• This lesson studies exact value 精确值: A value retained as a fraction or surd rather than a rounded decimal.

Choose the mathematical structure

• Bisect an equilateral triangle of side 2 to obtain a 30–60–90 triangle with sides 1,√3,2. A right isosceles triangle has sides 1,1,√2. Apply opposite/hypotenuse, adjacent/hypotenuse and opposite/adjacent to derive sine, cosine and tangent. Use degree angles and retain surds exactly; tan90° is undefined.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Sine and cosine interchange when the chosen acute angle changes to its complement. Do not turn √3 into a rounded decimal when an exact answer is required. Tangent at 90° is not zero.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G21 is additional Foundation and includes the listed exact values. Derive them from labelled special triangles, then use them in G20 right-triangle calculations.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Sine rule, cosine rule and triangle area · Higher

A triangular plot has no right angle. Two known sides and their included angle can determine the remaining side without inventing a perpendicular side length.

• A triangular plot has no right angle. Two known sides and their included angle can determine the remaining side without inventing a perpendicular side length.
• This lesson studies included angle 夹角: The angle between the two named sides.

Choose the mathematical structure

• Sine rule pairs opposite sides/angles: a/sinA=b/sinB=c/sinC. Cosine rule a²=b²+c²-2bc cosA uses the angle opposite a. Area is ab sinC/2 when C lies between a and b. Choose the rule from the known information. An inverse sine may give an acute angle and an obtuse supplement; check the angle sum and supplied sides before accepting either.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Do not pair a side with its adjacent angle in the sine rule. The area angle must be included. A calculator’s first inverse-sine answer need not be the only possible triangle.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G22/G23 Higher includes unknown sides/angles and areas of general triangles. Write the chosen rule before substitution and state any second possible configuration.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Right triangles in three-dimensional shapes · Higher

A cuboid’s space diagonal is longer than its base diagonal. Two different right triangles are needed to connect its width, depth and height.

• A cuboid’s space diagonal is longer than its base diagonal. Two different right triangles are needed to connect its width, depth and height.
• This lesson studies projection 投影: The view or component of a spatial length on a selected plane.

Choose the mathematical structure

• Identify a plane containing the wanted length or angle. Calculate a base diagonal first, then combine it with the perpendicular height. For an angle between a line and a plane, use the angle between the line and its orthogonal projection onto that plane. Mark right angles; general-triangle rules are used only when the selected triangle is not right-angled.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Do not combine unrelated lengths as if they met at a right angle. The angle to a plane uses the base projection, not an arbitrary base edge. A pyramid’s face slant is different from its sloping edge.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G20 Higher extends right-triangle and, where possible, general-triangle reasoning into 3D. Draw the relevant section triangle separately and name which spatial points it represents.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.4 · Vector geometry and midpoint arguments · Higher

A triangle’s two side midpoints create a new segment. Vectors can prove that it is parallel to the third side and exactly half its length.

• A triangle’s two side midpoints create a new segment. Vectors can prove that it is parallel to the third side and exactly half its length.
• This lesson studies collinear 共线: Lying on one straight line.

Choose the mathematical structure

• Add/subtract components and multiply a vector by a scalar. Position vectors locate points from one origin; displacement AB=b-a joins two points. A scalar multiple gives parallel directions; to establish collinearity, also connect the displacements to a common point. Midpoints average position vectors. Give a chain of vector equalities with clear start and end points.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Keep AB=b-a, not a-b. Parallelism alone does not locate a line. A numerical example can check the algebra but cannot replace the general midpoint proof.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA G25 Higher uses vectors for geometric arguments/proofs, while G24 and the basic G25 operations also belong to Foundation. This proof lesson avoids scalar products and spatial line equations.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.4. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.5 · Probability trees and outcomes · Foundation

What changes after the first draw?

• A bag has 3 red and 2 blue counters. Taking two without replacement changes the chance of the second colour.
• This lesson studies conditional probability 条件概率: The probability of an event after restricting the sample space to a stated condition.

Choose the mathematical structure

• A probability lies between 0 and 1. Exhaustive, mutually exclusive outcomes have probabilities summing to 1. Multiply successive branch probabilities and add separate routes to an outcome.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Mutually exclusive means no overlap; independent means that knowing one event does not change the other's probability. Two disjoint events with positive probability are not independent.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Use a frequency table or a simple tree before calculating. Formal conditional probability formulae are outside this Foundation/Core support lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.5. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.5 · Probability, trees and conditional reasoning · Higher

What changes after the first draw?

• A bag has 3 red and 2 blue counters. Taking two without replacement changes the chance of the second colour.
• This lesson studies conditional probability 条件概率: The probability of an event after restricting the sample space to a stated condition.

Choose the mathematical structure

• Multiply along a tree branch and add disjoint branches. With replacement, the composition stays fixed. Conditional probability is P(A given B)=P(A∩B)/P(B), for P(B)>0.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Mutually exclusive means no overlap; independent means that knowing one event does not change the other's probability. Two disjoint events with positive probability are not independent.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• A two-way table makes the restricted denominator visible. Before using a product P(A)P(B), justify independence from the context or the supplied information.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.5. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.5 · Experimental probability, fairness and expected outcomes · Foundation

A coin gives six heads in its first ten tosses. This does not mean it must give four tails next, or prove that the coin is unfair.

• A coin gives six heads in its first ten tosses. This does not mean it must give four tails next, or prove that the coin is unfair.
• This lesson studies relative frequency 相对频率: The observed count of an outcome divided by the number of trials.

Choose the mathematical structure

• Record each trial consistently and total the counts. Relative frequency estimates probability as outcome frequency/trials. For a stated probability p, expected count in n future trials is np; it is a long-run average, not a guarantee. Unbiased trials with larger samples usually give more stable estimates, but cannot force exact agreement with theory. Probabilities lie from 0 to 1 and a mutually exclusive exhaustive list sums to 1.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A larger biased sample can still be misleading. Expected does not mean certain. Mutually exclusive events cannot happen together; if categories overlap, do not simply add their probabilities as separate outcomes.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA P1–P5 uses frequency tables/trees, randomness/fairness, expected counts, the probability scale and empirical/theoretical comparison. State whether a value is observed, estimated or theoretical.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.5. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.5 · Experimental probability, fairness and expected outcomes · Higher

A coin gives six heads in its first ten tosses. This does not mean it must give four tails next, or prove that the coin is unfair.

• A coin gives six heads in its first ten tosses. This does not mean it must give four tails next, or prove that the coin is unfair.
• This lesson studies relative frequency 相对频率: The observed count of an outcome divided by the number of trials.

Choose the mathematical structure

• Record each trial consistently and total the counts. Relative frequency estimates probability as outcome frequency/trials. For a stated probability p, expected count in n future trials is np; it is a long-run average, not a guarantee. Unbiased trials with larger samples usually give more stable estimates, but cannot force exact agreement with theory. Probabilities lie from 0 to 1 and a mutually exclusive exhaustive list sums to 1.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• A larger biased sample can still be misleading. Expected does not mean certain. Mutually exclusive events cannot happen together; if categories overlap, do not simply add their probabilities as separate outcomes.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA P1–P5 uses frequency tables/trees, randomness/fairness, expected counts, the probability scale and empirical/theoretical comparison. State whether a value is observed, estimated or theoretical.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.5. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.5 · Systematic possibilities and equally likely outcomes · Foundation

Two fair dice can total seven in six different ways, but total two in only one way. The possible totals are not equally likely.

• Two fair dice can total seven in six different ways, but total two in only one way. The possible totals are not equally likely.
• This lesson studies sample space 样本空间: The complete set of outcomes in an experiment.

Choose the mathematical structure

• List outcomes systematically using a grid, table or tree. For equally likely outcomes, probability is favourable outcomes/total outcomes. Keep ordered outcomes distinct when the experiments have labelled first and second stages. A grid shows completeness and prevents duplicate counting; unequal probabilities need weights rather than a simple count.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Counting totals rather than equally likely dice pairs gives wrong weights. State whether order matters. A possibility table lists outcomes; its entries are not automatically equiprobable.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA P6/P7 uses tables, grids and trees to enumerate theoretical possibilities. Explain why the selected elementary outcomes have equal probability before dividing counts.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.5. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.5 · Systematic possibilities and equally likely outcomes · Higher

Two fair dice can total seven in six different ways, but total two in only one way. The possible totals are not equally likely.

• Two fair dice can total seven in six different ways, but total two in only one way. The possible totals are not equally likely.
• This lesson studies sample space 样本空间: The complete set of outcomes in an experiment.

Choose the mathematical structure

• List outcomes systematically using a grid, table or tree. For equally likely outcomes, probability is favourable outcomes/total outcomes. Keep ordered outcomes distinct when the experiments have labelled first and second stages. A grid shows completeness and prevents duplicate counting; unequal probabilities need weights rather than a simple count.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Counting totals rather than equally likely dice pairs gives wrong weights. State whether order matters. A possibility table lists outcomes; its entries are not automatically equiprobable.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA P6/P7 uses tables, grids and trees to enumerate theoretical possibilities. Explain why the selected elementary outcomes have equal probability before dividing counts.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.5. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.5 · Venn diagrams, unions and two-way counts · Foundation

A class survey asks about cycling and swimming. Some students do both, so adding the two group totals counts those students twice.

• A class survey asks about cycling and swimming. Some students do both, so adding the two group totals counts those students twice.
• This lesson studies intersection 交集: The outcomes belonging to both named sets.

Choose the mathematical structure

• Place the overlap in a Venn diagram first, then fill the only-regions and neither-region. Union means at least one named set; intersection means both; complement means outside a named set in the stated universal group. Two-way tables classify every observation by one category from each of two variables. Check row, column and grand totals.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Do not add the overlap twice. Neither is outside both circles, not just outside their overlap. A universal group must be stated before taking a complement.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA P4/P6 uses exhaustive events, systematic sets and Venn/table representations. Translate the words both, either/at least one, only and neither into the correct counted regions.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.5. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.5 · Venn diagrams, unions and two-way counts · Higher

A class survey asks about cycling and swimming. Some students do both, so adding the two group totals counts those students twice.

• A class survey asks about cycling and swimming. Some students do both, so adding the two group totals counts those students twice.
• This lesson studies intersection 交集: The outcomes belonging to both named sets.

Choose the mathematical structure

• Place the overlap in a Venn diagram first, then fill the only-regions and neither-region. Union means at least one named set; intersection means both; complement means outside a named set in the stated universal group. Two-way tables classify every observation by one category from each of two variables. Check row, column and grand totals.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Do not add the overlap twice. Neither is outside both circles, not just outside their overlap. A universal group must be stated before taking a complement.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA P4/P6 uses exhaustive events, systematic sets and Venn/table representations. Translate the words both, either/at least one, only and neither into the correct counted regions.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.5. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.5 · Conditional probability with tables and expected frequencies · Higher

The probability a randomly selected student swims changes when we learn that the student cycles. The given condition changes which students are possible.

• The probability a randomly selected student swims changes when we learn that the student cycles. The given condition changes which students are possible.
• This lesson studies conditional probability 条件概率: A probability within the group selected by given information.

Choose the mathematical structure

• Conditioning restricts the denominator to the given group. In a table, divide the intersection count by the condition’s row/column total. P(A given B)=P(A and B)/P(B) when P(B)>0. Reversing the condition usually changes the denominator. Expected-frequency trees help interpret percentages without treating the two conditions as interchangeable.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Given positive and positive given condition are different questions. Use the conditioned group total, not the grand total. A rare starting category can make false-positive counts significant even when the detection rate is high.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA P9 Higher requires conditional calculations and interpretation using two-way tables, trees and Venn diagrams. Name the restricted group and retain expected counts until the final ratio.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.5. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.6 · Centre, spread and data displays · Foundation

Can one average tell the whole story?

• Two groups have the same median but different spread. One summary cannot describe both location and consistency.
• This lesson studies range 极差: The maximum observed value minus the minimum observed value.

Choose the mathematical structure

• The mean is total divided by count. The median is the central value after sorting. The range is maximum minus minimum. Use frequency tables, bar charts and suitable comparisons.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• The tallest histogram bar need not contain the most observations. A grouped mean is an estimate. Correlation does not prove causation, and extrapolation extends beyond the observed range.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• A bar chart uses separate bars for categories. Unequal-class-width histograms and formal density calculations are outside this Foundation/Core lesson.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.6. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.6 · Data summaries, histograms and interpretation · Higher

Can one average tell the whole story?

• Two groups have the same median but different spread. One summary cannot describe both location and consistency.
• This lesson studies frequency density · ⁨densidad de frecuencia⁩ 频率密度: Frequency divided by class width, used as histogram height.

Choose the mathematical structure

• Compare an appropriate average and spread in context. A histogram uses area for frequency, so height=frequency/class width. Grouped estimates assume representative values within intervals.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• The tallest histogram bar need not contain the most observations. A grouped mean is an estimate. Correlation does not prove causation, and extrapolation extends beyond the observed range.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Choose a display that fits the data type. Give both a numerical comparison and what it means for the population; do not infer more precision than the sample supports.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.6. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.6 · Cumulative frequency and box plots · Higher

Which route is more consistent?

• Two bus routes have similar typical times but different reliability. Quartiles show the middle half of the journeys.
• This lesson studies interquartile range 四分位距: The difference between the upper and lower quartiles.

Choose the mathematical structure

• A cumulative frequency counts observations below successive class boundaries. Read quartiles at one quarter, one half and three quarters of the total frequency. A box plot represents minimum, lower quartile, median, upper quartile and maximum.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Plot against class boundaries rather than midpoints. Grouped quartiles are estimates. The range is sensitive to extremes; the IQR describes only the middle half.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• Explain a comparison in the context of the measured quantity. An outlier rule may use Q1-1.5IQR and Q3+1.5IQR; use the rule specified in the task rather than assuming every graph follows it.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Higher · 3.6. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.6 · Samples, populations and justified comparisons · Foundation

A school asks the first twenty students leaving a sports club about exercise. A large share of active answers may describe the club rather than the whole school.

• A school asks the first twenty students leaving a sports club about exercise. A large share of active answers may describe the club rather than the whole school.
• This lesson studies population 总体: The complete group about which a statistical claim is made.

Choose the mathematical structure

• Define the target population and variables before sampling. A sample is a subset; a census includes the whole population. Random selection reduces systematic selection bias but does not eliminate sampling variability or non-response. Primary data is collected for the present investigation; secondary data was collected by another source or purpose. Discrete data is counted; continuous data is measured.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Sample size alone cannot fix biased selection. A precise calculated estimate need not be accurate for the target population. Recording continuous measurements as integers does not change the underlying variable type.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA S1/S4/S5 uses samples to describe populations and recognises limitations/data types. State what the sample supports and what might make extrapolating to the whole population unreliable.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course

• 8300 · Foundation · 3.6. Match the target tier and specification before assigning extensions.
• Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.

3.6 · Samples, populations and justified comparisons · Higher

A school asks the first twenty students leaving a sports club about exercise. A large share of active answers may describe the club rather than the whole school.

• A school asks the first twenty students leaving a sports club about exercise. A large share of active answers may describe the club rather than the whole school.
• This lesson studies population 总体: The complete group about which a statistical claim is made.

Choose the mathematical structure

• Define the target population and variables before sampling. A sample is a subset; a census includes the whole population. Random selection reduces systematic selection bias but does not eliminate sampling variability or non-response. Primary data is collected for the present investigation; secondary data was collected by another source or purpose. Discrete data is counted; continuous data is measured.
• State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.

Work through a checked case

• Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.

Test a tempting shortcut

• Sample size alone cannot fix biased selection. A precise calculated estimate need not be accurate for the target population. Recording continuous measurements as integers does not change the underlying variable type.
• When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.

Interpret a new situation

• AQA S1/S4/S5 uses samples to describe populations and recognises limitations/data types. State what the sample supports and what might make extrapolating to the whole population unreliable.
• A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.

Use this in your course