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R9

Define percentage as 'number of parts per hundred'; interpret percentages and percentage changes as fractions/decimals, multiplicatively; percentages > 100%; percentage change and simple interest

Percentages

Worked answers, methods and verified real exam appearances for R9 on Edexcel GCSE Maths 1MA1.

Explanation

  • A percentage is a number of parts per hundred, so p%=p100p\%=\dfrac{p}{100}. Percentage change is multiplicative: an increase of r%r\% uses multiplier 1+r1001+\dfrac{r}{100} and a decrease uses 1r1001-\dfrac{r}{100}.
  • Percentages above 100%100\% correspond to multipliers greater than 11.
  • To reverse a change, divide by the multiplier; do not apply the opposite percentage to the changed value.
  • Simple interest is calculated each year from the original principal, so the yearly interest is constant.
  • Examiners expect the original amount to be the denominator when calculating percentage change.

Worked example

After a 12%12\% decrease, a machine is worth £704. Find its value before the decrease.

  1. 1.A 12%12\% decrease leaves 100%12%=88%100\%-12\%=88\%.
  2. 2.Write 0.88×original=7040.88\times\text{original}=704.
  3. 3.Original =704÷0.88=800=704\div0.88=800.

Answer: £800.

Common mistakes

  • Don't add 12%12\% to the reduced value when reversing a 12%12\% decrease.
  • Don't use the new value rather than the original value as the denominator for percentage change.
  • Don't calculate simple interest from a growing balance as though it were compound interest.

Exam tip

For reverse percentage, write the multiplier equation and divide by the multiplier.

Worked practice

Q1
Tier 1 · Easy

1

Work out 35%35\% of 240240.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkMark scheme
1
  • 8484
235%=0.3535\%=0.35, so 0.35×240=840.35\times240=84.
Q2
Tier 2 · Standard

2

After a 12%12\% decrease, a machine is valued at £704. Work out its value before the decrease.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
QuestionAnswerMarkMark scheme
2
  • £800
3A 12%12\% decrease leaves 88%=0.8888\%=0.88 of the original value. Divide by the multiplier: £704/0.88=£800704/0.88=£800.
Q3
Tier 3 · Hard

3

A saver deposits £2500 in an account paying 3.6%3.6\% simple interest each year. After 55 years, a fee equal to 2%2\% of the final balance is charged. Calculate the amount left after the fee.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
QuestionAnswerMarkMark scheme
3
  • £2891
5The yearly simple interest is 0.036×£2500=£900.036\times£2500=£90. Over 55 years this is £450, giving £2950. The fee is 0.02×£2950=£590.02\times£2950=£59, so £2950£59=£28912950-£59=£2891 remains.
Q4
Tier 1 · Easy

4

Work out 135%135\% of 8080.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
QuestionAnswerMarkMark scheme
4
  • 108108
2135%=1.35135\%=1.35, so 1.35×80=1081.35\times80=108.
Q5
Tier 2 · Standard

5

The price of an item increases from £72 to £81. Work out the percentage increase.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
QuestionAnswerMarkMark scheme
5
  • 12.5%12.5\%
3The increase is £81£72=£981-£72=£9. As a fraction of the original price this is 972=0.125\dfrac{9}{72}=0.125, so the percentage increase is 12.5%12.5\%.
Q6
Tier 3 · Hard

6

An investment earns £378 in simple interest over 33 years at a rate of 4.5%4.5\% per year. Work out the original amount invested.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
QuestionAnswerMarkMark scheme
6
  • £2800
3Over 33 years the simple interest is 3×4.5%=13.5%3\times4.5\%=13.5\% of the original amount. Therefore the original amount is £378÷0.135=£2800378\div0.135=£2800.
Q7
Tier 2 · Standard

7

A workshop makes 840840 parts in one month. This is 140%140\% of the number made in the previous month. Work out the number of parts made in the previous month.

(2)

(Total for Question 7 is 2 marks)

Mark scheme

Mark scheme for question 7
QuestionAnswerMarkMark scheme
7
  • 600600 parts
2140%=1.4140\%=1.4, so 1.4×previous number=8401.4\times\text{previous number}=840. The previous number was 840÷1.4=600840\div1.4=600.
Q8
Tier 3 · Hard

8

Nadia invests £1800 at 4%4\% simple interest per year. She invests another amount at 2.5%2.5\% simple interest per year. After 33 years, the total interest from the two investments is £396. Work out the second amount invested.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
QuestionAnswerMarkMark scheme
8
  • £2400
4The first investment earns 1800×0.04×3=£2161800\times0.04\times3=£216. The second investment therefore earns £396£216=£180396-£216=£180. Over 33 years its simple-interest rate is 3×2.5%=7.5%3\times2.5\%=7.5\%, so the second amount is £180÷0.075=£2400180\div0.075=£2400.
Q9
Tier 3 · Hard

9

A quantity is increased by 20%20\% and then decreased by p%p\%. Its final value is 108%108\% of its original value. Work out pp.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
QuestionAnswerMarkMark scheme
9
  • p=10p=10
4The multipliers satisfy 1.20(1p100)=1.081.20\left(1-\dfrac{p}{100}\right)=1.08. Hence 1p100=1.08÷1.20=0.91-\dfrac{p}{100}=1.08\div1.20=0.9, so p=10p=10.
Q10
Tier 3 · Hard

10

A tank is initially 80%80\% full. Then 15%15\% of the water in it is used. After 102102 litres are added, the tank is 92%92\% full. Work out the capacity of the tank.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
QuestionAnswerMarkMark scheme
10
  • 425425 litres
4Let the capacity be CC litres. After 15%15\% is used, 0.8C×0.85=0.68C0.8C\times0.85=0.68C litres remain. Therefore 0.68C+102=0.92C0.68C+102=0.92C, so 0.24C=1020.24C=102 and C=425C=425 litres.

Verified exam appearances

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Other points in R Ratio, proportion and rates of change

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