Ratio & Proportion

6 methods · 24 questions

★ Signature method unlocked: the technique this chapter rewards most often, with 3 worked questions.

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Treat every percentage change as a decimal multiplier and multiply them together. Never add or subtract percentage rates directly, as each percentage is taken of a new base.

Trigger: A quantity undergoes multiple percentage changes in sequence, or a rate is applied repeatedly over time.

Instances:
(i) Increase of $r %$ : multiply by $1 + \frac{r}{100}$ ;
(ii) Decrease of $r %$ : multiply by $1 - \frac{r}{100}$ ;
(iii) Compound changes: multiply the factors, e.g. $+ 25%$ then $- 40%$ is $1.25 \times 0.60$ .

Linked questions (3)

NSAA 2019 Section 1, Part A (Mathematics questions) · Q4

The resistance to the motion of a car is directly proportional to the square of the speed of the car.

The car increases its speed by 20%.

What is the percentage increase in the resistance to the motion of the car?

A20%
B24%
C44%
D120%
E224%
F240%
G400%

Worked solution

Answer:C

Direct proportionality to the square of the speed means

R = k v^{2}

for some constant

k

.

Increasing the speed by 20% replaces

v

1.2 v

, so the new resistance is

k (1.2 v)^{2} = 1.44 k v^{2} = 1.44 R .

The resistance is therefore multiplied by

1.44

. The increase is

0.44 R

, that is 44% of the original resistance.

The answer is C.

NSAA 2018 Section 1, Part A (Mathematics questions) · Q11

The original price of an item is

p

The price is increased by 125%

The increased price is then decreased by 40% to

q

The relationship between

p

and

q

can be expressed as

m p = q

What is the value of

m

A $\frac{7}{20}$
B $\frac{17}{20}$
C $\frac{27}{20}$
D $\frac{33}{20}$
E $\frac{37}{20}$

Worked solution

Answer:C

An increase of 125% multiplies the price by

1 + 1.25 = 2.25

, so the increased price is

2.25 p

.

A decrease of 40% then multiplies by

1 - 0.4 = 0.6

, so

q = 0.6 \times 2.25 p = 1.35 p .

Comparing with

m p = q

gives

m = 1.35 = \frac{135}{100} = \frac{27}{20} .

Note that the two changes do not cancel: raising by 125% and lowering by 40% leaves the price larger than it started, which agrees with

m > 1

.

The answer is C.

NSAA 2016 Section 1, Part A (Mathematics questions) · Q5

A medical scanner is bought for £15 000.

The value of the scanner depreciates by 20% every year.

By how much has the scanner reduced in value after 2 years?

A£600
B£3000
C£5400
D£6000
E£9000
F£9600
G£12 000

Worked solution

Answer:C

Losing 20% of the value each year means multiplying by

0.8

each year, so after two years the scanner is worth

15000 \times 0. 8^{2} = 15000 \times 0.64 = 9600.

The question asks for the reduction in value, not the value that remains, so the answer is

15000 - 9600 = 5400.

Year by year: the first year loses

3000

and leaves

12000

; the second year loses one fifth of

12000

, that is

2400

; and

3000 + 2400 = 5400

.

Note that repeated depreciation is not the same as losing 40% in one step, which would have given

6000

.

The answer is C.