Engineering and Science Admissions Test

ESAT Maths 1 Ratio & proportion questions

9 original ESAT Maths 1 ratio & proportion questions in the real exam style, each with a step-by-step worked solution and an explanation of why the tempting wrong answers are wrong. Try each one before opening the solution.

Methods on this page: Proportion: find the constant, or scale by the ratioRatios: make the shared part match, then compareGrowth and decay: multiply by the same factor each timePercentage change as multipliers

Question 1ESAT Maths 1Warm-up

y

is directly proportional to the square root of

x

. When

x = 16

y = 4

. What is

y

when

x = 9

A $2$
B $\dfrac{9}{4}$
C $3$
D $4$
E $\dfrac{16}{3}$

Show the answer and worked solution

Answer: C · Method: Proportion: find the constant, or scale by the ratio

$y = kx^{1/2}$ , so when $x$ is multiplied by $\tfrac{9}{16}$ , $y$ is multiplied by $\left(\tfrac{9}{16}\right)^{1/2} = \dfrac{3}{4}$ .
So $y = 4 \times \dfrac{3}{4} = 3$ .

Why the tempting wrong answers are wrong

B: Use the power in the relationship, not a straight line.
E: Check direct vs inverse: does $y$ go up or down as $x$ goes up?

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Question 2ESAT Maths 1Warm-up

y

is directly proportional to the square of

x

. When

x = 3

y = 8

. What is

y

when

x = 4

A $\dfrac{9}{2}$
B $\dfrac{32}{3}$
C $\dfrac{110}{9}$
D $\dfrac{119}{9}$
E $\dfrac{128}{9}$
F $\dfrac{137}{9}$
G $\dfrac{146}{9}$
H $\dfrac{155}{9}$

Show the answer and worked solution

Answer: E · Method: Proportion: find the constant, or scale by the ratio

$y = kx^{2}$ , so when $x$ is multiplied by $\tfrac{4}{3}$ , $y$ is multiplied by $\left(\tfrac{4}{3}\right)^{2} = \dfrac{16}{9}$ .
So $y = 8 \times \dfrac{16}{9} = \dfrac{128}{9}$ .

Why the tempting wrong answers are wrong

A: Check direct vs inverse: does $y$ go up or down as $x$ goes up?
B: Use the power in the relationship, not a straight line.

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Question 3ESAT Maths 1Warm-up

y

is inversely proportional to

x

. When

x = 4

y = 24

. What is

y

when

x = 8

A $11$
B $12$
C $13$
D $14$
E $48$

Show the answer and worked solution

Answer: B · Method: Proportion: find the constant, or scale by the ratio

$y = kx^{-1}$ , so when $x$ is multiplied by $\tfrac{8}{4}$ , $y$ is multiplied by $\left(\tfrac{8}{4}\right)^{-1} = \dfrac{1}{2}$ .
So $y = 24 \times \dfrac{1}{2} = 12$ .

Why the tempting wrong answers are wrong

E: Check direct vs inverse: does $y$ go up or down as $x$ goes up?

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Question 4ESAT Maths 1Warm-up

a : b = 6 : 1

and

b : c = 7 : 4

. What is

a : c

in its simplest form?

A $23 : 2$
B $2 : 21$
C $3 : 2$
D $21 : 2$
E $24 : 7$

Show the answer and worked solution

Answer: D · Method: Ratios: make the shared part match, then compare

Make the $b$ parts match: the LCM of $1$ and $7$ is $7$ .
$a : b = 42 : 7$ and $b : c = 7 : 4$ , so $a : c = 42 : 4 = 21 : 2$ .

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Question 5ESAT Maths 1Warm-up

A car worth £2000 loses 50% of its value each year. What is it worth after 2 years?

A $\pounds 0$
B $\pounds 490$
C $\pounds 498$
D $\pounds 499$
E $\pounds 500$
F $\pounds 501$
G $\pounds 502$
H $\pounds 1000$

Show the answer and worked solution

Answer: E · Method: Growth and decay: multiply by the same factor each time

Each year the value is multiplied by $0.5$ .
After 2 years: $2000 \times 0.5^{2} = 500$ .

Why the tempting wrong answers are wrong

A: That's simple interest: compound changes multiply each year.
H: That's after one year.

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Question 6ESAT Maths 1Warm-up

£5000 is invested at 10% compound interest per year. What is the investment worth after 2 years?

A $\pounds 5500$
B $\pounds 6000$
C $\pounds 6049$
D $\pounds 6050$
E $\pounds 6051$
F $\pounds 6052$
G $\pounds 6060$

Show the answer and worked solution

Answer: D · Method: Growth and decay: multiply by the same factor each time

Each year the value is multiplied by $1.1$ .
After 2 years: $5000 \times 1.1^{2} = 6050$ .

Why the tempting wrong answers are wrong

A: That's after one year.
B: That's simple interest: compound changes multiply each year.

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Question 7ESAT Maths 1Warm-up

y

is inversely proportional to the square of

x

. When

x = 5

y = 16

. What is

y

when

x = 8

A $\dfrac{17}{4}$
B $\dfrac{21}{4}$
C $\dfrac{25}{4}$
D $\dfrac{29}{4}$
E $\dfrac{33}{4}$
F $\dfrac{37}{4}$
G $10$
H $\dfrac{1024}{25}$

Show the answer and worked solution

Answer: C · Method: Proportion: find the constant, or scale by the ratio

$y = kx^{-2}$ , so when $x$ is multiplied by $\tfrac{8}{5}$ , $y$ is multiplied by $\left(\tfrac{8}{5}\right)^{-2} = \dfrac{25}{64}$ .
So $y = 16 \times \dfrac{25}{64} = \dfrac{25}{4}$ .

Why the tempting wrong answers are wrong

G: Use the power in the relationship, not a straight line.
H: Check direct vs inverse: does $y$ go up or down as $x$ goes up?

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Question 8ESAT Maths 1Standard

Last year a manager's salary was 80% of the director's salary. This year the manager's salary rose by 10% and the director's rose by 25%. What percentage of the director's new salary is the manager's new salary?

A $60\tfrac{2}{5}\%$
B $65\%$
C $65\tfrac{2}{5}\%$
D $68\%$
E $70\tfrac{2}{5}\%$
F $75\tfrac{2}{5}\%$
G $80\tfrac{2}{5}\%$
H $90\tfrac{10}{11}\%$

Show the answer and worked solution

Answer: E · Method: Percentage change as multipliers

Take the director's old salary as 100, so the manager's was 80.
New salaries: manager $80 \times 1.1 = 88$ , director $1.25 \times 100 = 125$ .
So the manager earns $70\tfrac{2}{5}\%$ of the director's new salary.

Why the tempting wrong answers are wrong

B: Percentage rises multiply. You can't add and subtract them from the 80%.
D: The two rises apply to different salaries: use the ratio of the multipliers.

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Question 9ESAT Maths 1Standard

£120 is shared between Ali and Ben in the ratio

3 : 7

. Ali then gives Ben £5. What fraction of the £120 does Ali now have?

A $\dfrac{7}{40}$
B $\dfrac{13}{60}$
C $\dfrac{31}{120}$
D $\dfrac{36}{125}$
E $\dfrac{3}{10}$
F $\dfrac{41}{120}$
G $\dfrac{31}{89}$
H $\dfrac{23}{60}$

Show the answer and worked solution

Answer: C · Method: Ratios: make the shared part match, then compare

Ali gets $\tfrac{3}{10} \times 120 = 36$ and Ben gets $84$ .
After giving £5, Ali has £31, which is $\dfrac{31}{120}$ of the total.

Why the tempting wrong answers are wrong

E: That's Ali's share before giving any away.
G: That's Ali's money compared with Ben's, not with the total.

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ESAT Maths 1 Ratio & proportion questions

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