Engineering and Science Admissions Test

ESAT Maths 1 Number questions

12 original ESAT Maths 1 number 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: Convert units: square or cube the conversion factorHCF and LCM from prime factorsPercentage change as multipliersUse prime factorisationBounds: half a unit either side; divide by the lower bound for an upper boundRecurring decimals: multiply by a power of 10 and subtractCount systematically: multiply the choices at each step

Question 1ESAT Maths 1Warm-up

A cube has edges of length

7\ \text{km}

. What is its volume in

\text{m}^3

A $3.43 \times 10^{-7}$
B $3.43 \times 10^{5}$
C $3.43 \times 10^{8}$
D $3.43 \times 10^{10}$
E $3.43 \times 10^{11}$
F $3.43 \times 10^{12}$
G $3.43 \times 10^{13}$
H $3.43 \times 10^{14}$

Show the answer and worked solution

Answer: E · Method: Convert units: square or cube the conversion factor

$1\ \text{km} = 10^{3}\ \text{m}$ , so $1\ \text{km}^{3} = 10^{9}\ \text{m}^{3}$ .
Volume $= 7^3 = 343\ \text{km}^3$ $= 343 \times 10^{9} = 3.43 \times 10^{11}\ \text{m}^{3}$ .

Why the tempting wrong answers are wrong

A: Check the direction: is the answer in m bigger or smaller than the number in the original unit?
B: The conversion factor must be cubed as well.

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Question 2ESAT Maths 1Warm-up

Two lights flash together at the same moment. One flashes every 10 seconds and the other every 4 seconds. How many seconds later do they next flash together?

A $2$
B $14$
C $19$
D $20$
E $21$
F $40$

Show the answer and worked solution

Answer: D · Method: HCF and LCM from prime factors

$10 = 2 \times 5$ and $4 = 2 \times 2$ share the factor $2$ .
The next time is the lowest common multiple: $\text{LCM}(10, 4) = 20$ seconds.

Why the tempting wrong answers are wrong

A: That's the highest common factor.
F: Multiplying works only when the numbers share no factor. You want the lowest common multiple.

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Question 3ESAT Maths 1Warm-up

The price of a jacket is increased by 20%, and the new price is then reduced by 20%. It now costs

\pounds 96

. What was the original price?

A $\pounds 92.16$
B $\pounds 96$
C $\pounds 99$
D $\pounds 100$
E $\pounds 101$
F $\pounds 102$
G $\pounds 110$

Show the answer and worked solution

Answer: D · Method: Percentage change as multipliers

The price was multiplied by $1.2$ and then by $0.8$ .
So the original price is $96 \div (1.2 \times 0.8) = 100$ .

Why the tempting wrong answers are wrong

A: To undo a 20% increase, divide by $1.2$ rather than taking 20% off.
B: Percentage changes multiply: you can't just combine 20% and 20% into one change of 0%.

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Question 4ESAT Maths 1Warm-up

What is the smallest positive integer

n

for which

60n

is a perfect square?

A $13$
B $14$
C $15$
D $16$
E $17$
F $30$
G $450$

Show the answer and worked solution

Answer: C · Method: Use prime factorisation

$60 = 2^{2} \times 3^{1} \times 5^{1}$ .
A perfect square needs every prime's power to be a multiple of 2, so multiply by $15$ .

Why the tempting wrong answers are wrong

G: That makes a perfect cube.

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Question 5ESAT Maths 1Warm-up

A rectangle has length

100\ \text{cm}

, correct to the nearest centimetre, and width

2\ \text{cm}

, correct to the nearest centimetre. What is the lower bound for its area, in

\text{cm}^2

A $148.25$
B $149.25$
C $150.25$
D $151.25$
E $199$
F $200$
G $251.25$

Show the answer and worked solution

Answer: B · Method: Bounds: half a unit either side; divide by the lower bound for an upper bound

The measurements lie in $99.5 \le l < 100.5$ and $1.5 \le w < 2.5$ .
The lower bound of the area uses the lower bounds of both: $99.5 \times 1.5 = 149.25$ .

Why the tempting wrong answers are wrong

E: Both measurements were rounded, so both need their bounds.
F: That uses the rounded values. Bounds come from the half-unit either side.
G: For the lower bound of a product, use the lower bound of both measurements.

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Question 6ESAT Maths 1Warm-up

Which fraction is equal to

0.\dot{3}

A $\dfrac{1}{33}$
B $\dfrac{3}{10}$
C $\dfrac{1}{3}$
D $\dfrac{4}{9}$
E $\dfrac{4}{3}$

Show the answer and worked solution

Answer: C · Method: Recurring decimals: multiply by a power of 10 and subtract

Let $x = 0.\dot{3}$ . Then $10x - x = 9x$ is a whole number, so $x = \dfrac{1}{3}$ .

Why the tempting wrong answers are wrong

B: A recurring decimal isn't the same as the terminating decimal with those digits.

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Question 7ESAT Maths 1Warm-up

A code is made of 2 letters chosen from A, B, C, followed by 3 digits (0–9). No letter or digit may be used more than once. How many different codes are possible?

A $4319$
B $4320$
C $4321$
D $6000$
E $8640$
F $9000$

Show the answer and worked solution

Answer: B · Method: Count systematically: multiply the choices at each step

Letters: $3 \times 2$ . Digits: $10 \times 9 \times 8$ . Total: $4320$ .

Why the tempting wrong answers are wrong

F: Without repeats, each later choice has one option fewer.

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Question 8ESAT Maths 1Warm-up

A cube has edges of length

3\ \text{cm}

. What is its volume in

\text{m}^3

A $2.7 \times 10^{-7}$
B $2.7 \times 10^{-5}$
C $2.7 \times 10^{-4}$
D $2.7 \times 10^{-3}$
E $2.7 \times 10^{-1}$
F $2.7 \times 10^{7}$

Show the answer and worked solution

Answer: B · Method: Convert units: square or cube the conversion factor

$1\ \text{cm} = 10^{-2}\ \text{m}$ , so $1\ \text{cm}^{3} = 10^{-6}\ \text{m}^{3}$ .
Volume $= 3^3 = 27\ \text{cm}^3$ $= 27 \times 10^{-6} = 2.7 \times 10^{-5}\ \text{m}^{3}$ .

Why the tempting wrong answers are wrong

E: The conversion factor must be cubed as well.
F: Check the direction: is the answer in m bigger or smaller than the number in the original unit?

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Question 9ESAT Maths 1Standard

A = 2^{4} \times 3 \times 5^{2}

and

B = 2^{3} \times 3^{2} \times 7^{2}

. What is the highest common factor of

A

and

B

A $23$
B $2^{3} \times 3$
C $5^{2}$
D $26$
E $2^{3} \times 3^{2}$
F $2^{4} \times 3^{2} \times 5^{2} \times 7^{2}$
G $2^{5} \times 3^{2} \times 5^{2} \times 7^{2}$
H $2^{7} \times 3^{3} \times 5^{2} \times 7^{2}$

Show the answer and worked solution

Answer: B · Method: HCF and LCM from prime factors

For the HCF take the lower power of each prime that appears.
That gives $2^{3} \times 3$ .

Why the tempting wrong answers are wrong

F: That's the LCM: for the HCF take the lower power of each prime.
H: That's the product of the two numbers.

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Question 10ESAT Maths 1Standard

A runner covers

100\ \text{m}

, correct to the nearest 10 m, in

4\ \text{s}

, correct to the nearest second. What is the upper bound for the runner's average speed, in

\text{m s}^{-1}

A $\dfrac{70}{3}$
B $25$
C $26.25$
D $30$
E $31$

Show the answer and worked solution

Answer: D · Method: Bounds: half a unit either side; divide by the lower bound for an upper bound

The measurements lie in $95 \le d < 105$ and $3.5 \le t < 4.5$ .
The upper bound of a quotient divides the upper bound of the top by the lower bound of the bottom: $\dfrac{105}{3.5} = 30$ .

Why the tempting wrong answers are wrong

A: For the upper bound of a division, use the upper bound of the top and the lower bound of the bottom.
B: That uses the rounded values. Bounds come from the half-unit either side.
C: Both measurements were rounded, so both need their bounds.

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Question 11ESAT Maths 1Standard

What is

2 \times 0.6\dot{1}

as a fraction in its simplest form?

A $\dfrac{10}{9}$
B $\dfrac{11}{9}$
C $\dfrac{37}{30}$
D $\dfrac{20}{9}$
E $\dfrac{61}{5}$

Show the answer and worked solution

Answer: B · Method: Recurring decimals: multiply by a power of 10 and subtract

Let $x = 0.6\dot{1}$ . Then $100x - 10x$ removes the recurring part, giving $x = \dfrac{11}{18}$ .
So $2 \times 0.6\dot{1} = \dfrac{11}{9}$ .

Why the tempting wrong answers are wrong

E: A recurring decimal isn't the same as the terminating decimal with those digits.

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Question 12ESAT Maths 1Standard

How many three-digit whole numbers even and have three different digits?

A $320$
B $328$
C $337$
D $338$
E $346$

Show the answer and worked solution

Answer: B · Method: Count systematically: multiply the choices at each step

Count the choices digit by digit, taking the most restricted position first and splitting into cases where a choice affects later ones.
There are $328$ such numbers.

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ESAT Maths 1 Number questions

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