Pearson Edexcel International GCSE Mathematics A 4MA1: Percentages, ratio and proportion – Study Guide

This study guide teaches sections 1.6 (Percentages) and 1.7 (Ratio and proportion) of topic 1, Numbers and the number system, in the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), first assessed in June 2018 with papers available in January and June. Section 1.6 has Foundation content that both tiers need plus two statements marked Higher tier only. Section 1.7 is the same for both tiers.

Each tier sits two 2-hour papers of 100 marks, and a calculator may be used on every paper. Neither formulae sheet prints a percentage or compound interest formula, so learn the methods below.

The other pages for this unit are the revision notes and the practice questions. For the whole of topic 1, see the Numbers and the number system study guide, the topic 1 revision notes and the topic 1 practice questions. The course hub is Edexcel IGCSE Mathematics, the printable checklist lists every statement, and the free 10-minute diagnostics show where to start.

What this unit covers

Spec	What you must be able to do	Tier
1.6A	Understand that “percentage” means “number of parts per 100”	Both
1.6B	Express a given number as a percentage of another number	Both
1.6C	Express a percentage as a fraction and as a decimal	Both
1.6D	Understand the multiplicative nature of percentages as operators	Both
1.6E	Solve simple percentage problems, including percentage increase and decrease	Both
1.6F	Use reverse percentages	Both
1.6G	Use compound interest and depreciation	Both
1.6 (H) A	Use repeated percentage change	Higher tier only
1.6 (H) B	Solve compound interest problems	Higher tier only
1.7A	Use ratio notation, including reduction to simplest form and links to fraction notation	Both
1.7B	Divide a quantity in a given ratio or ratios	Both
1.7C	Use the process of proportionality to evaluate unknown quantities	Both
1.7D	Calculate an unknown quantity from quantities that vary in direct proportion	Both
1.7E	Solve word problems about ratio and proportion, including maps and scale diagrams	Both

Inverse proportion and forms such as y ∝ x² belong to section 2.5 (Proportion), which is Higher tier only and sits in topic 2. They are not part of this unit.

Percentages (1.6A–C)

A percentage is a number of parts per 100. So 35% means 35 out of every 100.

Percentage to fraction and decimal. Write the percentage over 100, then simplify. Divide by 100 for the decimal.

35%   = 35/100  = 7/20   = 0.35
12.5% = 12.5/100 = 1/8   = 0.125
4.5%  = 4.5/100 = 9/200  = 0.045
135%  = 135/100 = 27/20  = 1.35

A percentage over 100 is allowed: 135% of a quantity is more than the whole quantity.

One number as a percentage of another. Write the first as a fraction of the second, then multiply by 100. Put both quantities in the same units first.

Worked example. A student scores 63 out of 75. Then express 750 g as a percentage of 2.5 kg.

63/75 x 100 = 84%
2.5 kg = 2500 g
750/2500 x 100 = 30%

Percentages as multipliers (1.6D–E)

Finding a percentage of an amount is a multiplication. The specification’s own wording is “the multiplicative nature of percentages as operators”.

35% of 260 = 0.35 x 260 = 91

To increase or decrease, add to or take from 100% first, then convert to a multiplier.

Change	Percentage of original	Multiplier
Increase by 15%	115%	1.15
Decrease by 8%	92%	0.92
Increase by 2.5%	102.5%	1.025
Decrease by 30%	70%	0.70

Increase 64 by 15%:  64 x 1.15 = 73.6
Decrease 950 by 8%:  950 x 0.92 = 874

Percentage change. Find the actual change, then divide by the original value.

percentage change = (change / original) x 100

Worked example. A rail fare falls from £250 to £215. Calculate the percentage decrease.

change = 250 - 215 = 35
35/250 x 100 = 14% decrease

Reverse percentages (1.6F)

Here you know the value after a percentage change and need the original. The final value is the original times the multiplier, so divide by the multiplier.

original = final value / multiplier

Worked example. A phone bill is £403.20 after a 12% increase. Find the bill before the increase.

multiplier = 1.12
403.20 / 1.12 = £360
check: 360 x 1.12 = 403.20

Worked example. In a sale with 25% off, a lamp costs £63. Find the original price.

multiplier = 0.75
63 / 0.75 = £84

Do not find 25% of £63 and add it on. That gives £78.75, because the 25% was taken from the original price, not from £63.

Compound interest and depreciation (1.6G)

With compound interest, each year’s interest is added to the amount, and the next year’s interest is worked out on the new total. Depreciation is the same idea with a decrease each year. For n years at r% per year:

value after n years = starting amount x (multiplier)^n
growth multiplier       = 1 + r/100
depreciation multiplier = 1 - r/100

Worked example. £5000 is invested at 4% per year compound interest for 3 years. Find the value and the total interest.

5000 x 1.04^3 = 5624.32
interest = 5624.32 - 5000 = £624.32

Simple interest at the same rate would give 3 x (4% of 5000) = £600. Compound gives more because the interest earns interest.

Worked example. A machine costs £18 000 and depreciates by 15% per year. Find its value after 2 years.

18000 x 0.85^2 = £13 005

Read the question for what it asks: the value, or the interest (or the loss), which is the change. Keep full calculator values until the last step.

Repeated percentage change (1.6, Higher tier only)

When several percentage changes follow each other, multiply the multipliers. Each change acts on the result of the one before, so you never add the percentages.

Worked example. A share price rises by 25% and then falls by 12%. Find the overall percentage change.

1.25 x 0.88 = 1.1
1.1 means 110% of the original: an overall increase of 10%

Adding the percentages (25 − 12 = 13%) is wrong, because the 12% fall is taken from a larger amount.

Solving compound interest problems (1.6, Higher tier only)

At Higher tier you work with the same formula in any direction: find the original amount, the rate or the number of years.

Worked example: find the original. A car is worth £14 580 after 3 years of 10% depreciation per year. Find its value when new.

original x 0.9^3 = 14580
original = 14580 / 0.729 = £20 000

Worked example: find the rate. £2500 grows to £2704 in 2 years at a fixed rate of compound interest. Find the rate.

2500 x m^2 = 2704
m^2 = 1.0816
m = √1.0816 = 1.04
rate = 4% per year

Worked example: find the number of years. £3000 is invested at 7% per year compound interest. After how many whole years will it first be worth more than £4000? Try values of n and show them.

n = 4: 3000 x 1.07^4 = 3932.39   (not yet)
n = 5: 3000 x 1.07^5 = 4207.66   (more than 4000)
answer: 5 years

Write down the values either side of the target. A bare “5” may lose the method marks.

Ratio notation (1.7A)

A ratio compares quantities in the same units. Simplify by dividing every part by the highest common factor.

Worked example. Simplify 45 cm : 2 m.

2 m = 200 cm
45 : 200   divide by 5   ->   9 : 40

The form 1 : n. Divide both parts by the first part. Here n need not be a whole number.

8 : 28  ->  1 : 3.5

Links to fractions. If A : B = 3 : 5, there are 8 parts in total. A is 3/8 of the total, B is 5/8 of the total, and A is 3/5 of B.

Combining ratios. If a : b = 2 : 3 and b : c = 4 : 5, make b the same in both. The LCM of 3 and 4 is 12.

a : b = 8 : 12     b : c = 12 : 15
a : b : c = 8 : 12 : 15

Dividing in a ratio (1.7B)

Find the value of one part, then multiply. What the question gives you decides how you find one part.

Given the total. Share 1260 in the ratio 2 : 3 : 4.

total parts = 9
one part = 1260 / 9 = 140
shares: 280, 420, 560      (check: sum = 1260)

Given the difference. Two numbers are in the ratio 7 : 3 and differ by 48.

difference = 7 - 3 = 4 parts = 48
one part = 12
numbers: 84 and 36

Given one share. The ratio is 5 : 2 and the larger share is 85.

5 parts = 85, so one part = 17
smaller share = 2 x 17 = 34

Proportion (1.7C–D)

The unitary method. Find the value for one unit, then scale up.

Worked example. 6 pens cost £8.70. Find the cost of 11 pens.

1 pen = 8.70 / 6 = £1.45
11 pens = 11 x 1.45 = £15.95

Best buy. Compare the same unit for both options. A 400 g pack costs £2.60 and a 750 g pack costs £4.65.

400 g: 2.60 / 4   = £0.65 per 100 g
750 g: 4.65 / 7.5 = £0.62 per 100 g
the 750 g pack is better value

Direct proportion. “s varies directly as t” means s = kt for a constant k. If t doubles, s doubles, and s/t is always the same.

Worked example. s varies directly as t, and s = 14 when t = 4. Complete the table.

t	4	10	?
s	14	?	24.5

k = 14 / 4 = 3.5,  so s = 3.5t
t = 10:   s = 3.5 x 10 = 35
s = 24.5: t = 24.5 / 3.5 = 7

Maps and scale diagrams (1.7E)

A map scale 1 : 25 000 means 1 cm on the map is 25 000 cm in real life. Convert units at the end: 100 cm = 1 m and 100 000 cm = 1 km.

Worked example. On a 1 : 25 000 map, two villages are 6.4 cm apart. Find the real distance in km.

6.4 x 25000 = 160 000 cm = 1.6 km

Worked example. A road is 3.5 km long. How long is it on a 1 : 50 000 map?

3.5 km = 350 000 cm
350000 / 50000 = 7 cm

Scale diagrams. A floor plan is drawn at 1 : 200. A room measures 4.6 cm by 3.2 cm on the plan. Its real size is 4.6 x 200 = 920 cm = 9.2 m by 3.2 x 200 = 640 cm = 6.4 m.

Common errors

• Dividing by the new value instead of the original in percentage change.
• Taking the percentage off the final value in a reverse percentage question instead of dividing by the multiplier.
• Adding percentage changes instead of multiplying multipliers (Higher tier).
• Using simple interest when the question says compound, or giving the final value when it asks for the interest.
• Dividing by the number of shares (people) instead of the total number of parts.
• Mixing units in a ratio, such as cm with m or g with kg.
• Multiplying by the map scale when you should divide, or the other way round. Ask: should the real distance be bigger than the map distance?
• Rounding the multiplier or an intermediate value too early.

Next steps

Condense this into the revision notes, then test yourself with the practice questions, which have mark-by-mark answers.

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification, Issue 2, November 2017, Pearson Education Limited (first assessment June 2018). Topic 1, Numbers and the number system, sections 1.6 Percentages and 1.7 Ratio and proportion (Foundation Tier content and Higher Tier content).