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ESAT Paper 2 Mathematics Sample Questions

Five questions from ESAT Paper 2, written to the depth of a full paper, with a worked solution for every one. This is a sample: a full module runs to 27 questions, written for one school at a time. Full timed papers, 27 questions a module, are in ESAT practice, free to start. Part of the ESAT preparation guide.

Question 1

A dosing pump moves coolant at a constant rate. A measured $(3 + \sqrt{2}) L$ takes $(4 - \sqrt{2}) min$ to come through it. Correct to 3 significant figures, what volume in litres comes through in $4 min$ ?

A. 2.34
B. 45.7
C. 17.7
D. 6.83
E. 43.6
F. 1.26
G. 4.00
H. 5.97

Difficulty: ★★★☆☆ Target Time: $90 s$ Tests: N/A

Key Idea (💡): Two ideas meet here. A steady rate is a volume divided by a time, so a volume and a time both written as surd expressions give a rate that is a quotient of surds. A quotient like that is cleared by the conjugate, because $a - \sqrt{b}$ and $a + \sqrt{b}$ multiply to the rational number $a^{2} - b$ . Scaling the cleared rate by a time is then multiplication, followed by one division that has to reach every term of the numerator rather than the first term alone.

ESAT specification (UAT-UK): M2.11

Reveal the answer & worked solution: commit to an option first

Correct Answer: D. 6.83

Step-by-Step Breakdown:

1. Write the rate as a quotient

The delivery is steady, so the rate is the volume divided by the time:

$rate = \frac{3 + \sqrt{2}}{4 - \sqrt{2}} L {min}^{- 1}$ .

2. Rationalise the denominator

Multiply top and bottom by the conjugate $4 + \sqrt{2}$ .

Denominator: $(4 - \sqrt{2}) (4 + \sqrt{2}) = 4^{2} - 2 = 14$ .

Numerator: $(3 + \sqrt{2}) (4 + \sqrt{2})$ $= 12 + 2 + (3 + 4) \sqrt{2} = 14 + 7 \sqrt{2}$ .

So the rate is $\frac{14 + 7 \sqrt{2}}{14} L {min}^{- 1}$ .

3. Scale to $4 min$ , dividing every term

Multiplying by $4$ gives $\frac{56 + 28 \sqrt{2}}{14}$ . The $14$ underneath divides the whole numerator, the surd term as well as the whole number, and it goes into each of them exactly:

$\frac{56}{14} = 4$ and $\frac{28}{14} = 2$ ,

so the output over $4 min$ is $4 + 2 \sqrt{2} L$ .

Sanity check

$\sqrt{2} \approx 1.41$ , so the quoted volume is about $4.41 L$ and the quoted time about $2.59 min$ , a rate near $1.71 L {min}^{- 1}$ . Over $4 min$ that is near $6.83 L$ , and the exact value $4 + 2 \sqrt{2}$ is $6.83$ to 3 significant figures.

The key is $6.83$ .

Why the Other Options Are Wrong (❌):

A. 2.34 · Rate inverted
Divided the time by the volume instead of the volume by the time, taking $\frac{4 - \sqrt{2}}{3 + \sqrt{2}}$ as the rate and multiplying that by $4$ to get about $2.34$ . That quotient counts minutes per litre, so multiplying it by a number of minutes cannot give a volume. Taking the quotient the other way up and scaling it gives $4 + 2 \sqrt{2}$ litres.
B. 45.7 · Multiplied volume by time
Multiplied the stated volume by the stated time rather than dividing, then scaled the product by $4$ , giving about $45.7$ . A volume multiplied by a time is measured in litre minutes and is not a volume at all. Dividing $(3 + \sqrt{2})$ by $(4 - \sqrt{2})$ first, then scaling, gives $4 + 2 \sqrt{2}$ litres.
C. 17.7 · Rate step skipped
Read $(3 + \sqrt{2}) L$ as the volume delivered in one minute and scaled it straight to $4$ minutes, giving about $17.7$ . That volume takes $(4 - \sqrt{2}) min$ , so the division by the time was never carried out and the rate used is wrong by a factor of $(4 - \sqrt{2})$ .
E. 43.6 · Partial cancelling
Rationalised and scaled correctly as far as $\frac{56 + 28 \sqrt{2}}{14}$ , then cancelled the $14$ into the first term only and read the result as $4 + 28 \sqrt{2}$ , about $43.6$ . A denominator divides every term of the numerator it sits under, so $28$ is reduced to $2$ at the same time as $56$ is reduced to $4$ , leaving $4 + 2 \sqrt{2}$ .
F. 1.26 · Conjugate applied to the denominator only
Multiplied the denominator by the conjugate $4 + \sqrt{2}$ and left the numerator as it was, so the rate read $\frac{3 + \sqrt{2}}{14}$ and the scaled volume $\frac{4 (3 + \sqrt{2})}{14}$ . Multiplying one half of a fraction on its own changes its value: the conjugate goes over both halves, which turns the numerator into $(3 + \sqrt{2}) (4 + \sqrt{2}) = 14 + 7 \sqrt{2}$ and leaves $4 + 2 \sqrt{2}$ after scaling.
G. 4.00 · Cross terms lost in the expansion
Expanded $(3 + \sqrt{2}) (4 + \sqrt{2})$ as $3 \times 4$ plus $\sqrt{2} \times \sqrt{2}$ and dropped the two cross terms, so the numerator read $14$ with no surd left in it and the answer came out rational: $56 \div 14 = 4$ . The cross terms are $3 \sqrt{2}$ and $4 \sqrt{2}$ , and they are where the $7 \sqrt{2}$ comes from, so the volume is $4 + 2 \sqrt{2}$ and not a whole number of litres.
H. 5.97 · Surd term dropped from the conjugate product
Took $(4 - \sqrt{2}) (4 + \sqrt{2})$ to be $4^{2}$ , on the reading that the conjugate makes every surd disappear, so the whole numerator was divided by $4^{2}$ instead. What cancels in that product is the pair of middle terms, $+ 4 \sqrt{2}$ and $- 4 \sqrt{2}$ ; the last term is $- (\sqrt{2})^{2} = - 2$ and it stays, so the denominator is $4^{2} - 2 = 14$ and the volume is $4 + 2 \sqrt{2}$ .

Common Mistake (⚠️):
Forgetting that a denominator divides the surd term as well as the whole number. The scaled numerator here is $56 + 28 \sqrt{2}$ over $14$ , and both parts are reduced, $56$ to $4$ and $28$ to $2$ . Reducing $56$ only gives $4 + 28 \sqrt{2}$ , about $43.6$ , rather than $4 + 2 \sqrt{2}$ .

Takeaway (📌):
A rate written as one surd expression over another is not usable until the bottom is rational. Multiply by the conjugate, read off the integer denominator $14$ , scale by the time, and divide every term of what is left. The exact output comes out as $4 + 2 \sqrt{2}$ litres, and only the final comparison with the options needs a decimal at all.

Question 2

A right-angled triangular bracket has an upright edge of $40 c m$ , and the angle between its sloping and horizontal edges has tangent $\frac{5}{2}$ . A second bracket, mathematically similar and cut from the same steel sheet, has an upright edge of $10 c m$ . Find the area of the second one in $c m^{2}$ .

A. 20
B. 80
C. 5
D. 320
E. 125

Difficulty: ★★★☆☆ Target Time: $75 s$ Tests: N/A

Key Idea (💡): A tangent is a quotient of two lengths, so it is unchanged by similarity: both shapes here have the same angles and the same ratio of upright to horizontal edge. What does change is size, through a single length scale factor $k$ taken from any pair of corresponding edges. Lengths carry one factor of $k$ , areas carry $k^{2}$ because an area is built from two lengths, and volumes carry $k^{3}$ . Finish one shape with trigonometry, then cross to the other exactly once.

ESAT specification (UAT-UK): M3.10

Reveal the answer & worked solution: commit to an option first

Correct Answer: A. 20

Step-by-Step Breakdown:

1. Complete the first bracket with the tangent

The angle lies between the sloping edge and the horizontal edge, so the upright edge is opposite it and the horizontal edge is adjacent to it. That makes the tangent the upright divided by the horizontal, $\frac{5}{2}$ , so the horizontal edge is $40 \times \frac{2}{5} = 16 c m$ .

2. Find the first area

Those two edges meet at the right angle, so they serve as base and height directly: $\frac{1}{2} \times 40 \times 16 = 320 c m^{2}$ .

3. Read off the length scale factor

Corresponding lengths of similar shapes are all in one ratio, and the two upright edges correspond, so $k = \frac{10}{40} = \frac{1}{4}$ .

4. Scale the area by $k^{2}$

An area is a product of two lengths and each of them carries a factor of $k$ , so the area carries $k^{2}$ : $320 \times \frac{1}{16} = 20 c m^{2}$ .

The direct route agrees: the second bracket has a horizontal edge of $4 c m$ , and $\frac{1}{2} \times 10 \times 4 = 20 c m^{2}$ .

The key is $20$ .

Why the Other Options Are Wrong (❌):

B. 80 · Length scale factor used on an area
$320 \times \frac{1}{4} = 80$ , the scale factor applied once, as though an area behaved like a length. Every corresponding length of the second bracket is $\frac{1}{4}$ times the first, so it is $\frac{1}{4}$ times as tall AND $\frac{1}{4}$ times as wide, and the factor has to act twice.
C. 5 · Volume scale factor used for an area
$320 \times \frac{1}{64} = 5$ . Cubing the scale factor is right for a capacity, not for a flat face, and the bracket is cut from steel sheet.
D. 320 · Scaling step omitted
$\frac{1}{2} \times 40 \times 16 = 320$ is the first bracket's own area, returned without ever crossing to the second one.
E. 125 · Tangent ratio inverted
Reading the tangent as horizontal over upright makes the first bracket's horizontal edge $40 \times \frac{5}{2} = 100 c m$ , its area $\frac{1}{2} \times 40 \times 100 = 2000 c m^{2}$ , and the scaled area $2000 \times \frac{1}{16} = 125$ .

Common Mistake (⚠️):
Multiplying the first area by the length scale factor instead of by its square, so $320$ becomes $320 \times \frac{1}{4} = 80$ rather than $20$ . The scale factor turns one length into one length, and an area is built from two lengths, so the factor has to act twice.

Takeaway (📌):
Finish one shape completely, then cross over once. Trigonometric ratios are identical in similar shapes, so use them where the numbers already are, and only then apply $k$ , $k^{2}$ or $k^{3}$ according to whether you are carrying a length, an area or a volume.

Question 3

A surveyor's plan uses coordinates in metres. A straight water main passes through $(5, - 4)$ and $(10, 21)$ , and a straight sewer connection meets it at right angles at $(5, - 4)$ . At what $y$ coordinate does the sewer connection cross the $y$ axis?

A. -3
B. -29
C. -5
D. 23
E. -15

Difficulty: ★★★☆☆ Target Time: $60 s$ Tests: N/A

Key Idea (💡): A gradient comes from two points as rise over run, both differences taken in the same order. The gradient of a perpendicular line is the negative reciprocal, so it is flipped and its sign is changed, and the check is that the two gradients multiply to $- 1$ . Once the gradient of the sewer connection is known, substituting the crossing point into $y = m x + c$ fixes $c$ , and $c$ is exactly the $y$ intercept the question asks for, so nothing further has to be solved.

ESAT specification (UAT-UK): M4.10

Reveal the answer & worked solution: commit to an option first

Correct Answer: A. -3

Step-by-Step Breakdown:

1. Gradient of the water main

$m = \frac{21 - (- 4)}{10 - (5)} = \frac{25}{5} = 5$

The water main climbs $5$ metres for every metre across.

2. Gradient of the sewer connection

Perpendicular gradients multiply to $- 1$ , so the sewer connection has the negative reciprocal of $5$ as its gradient, which is $- \frac{1}{5}$ . Check: $(5) \times (- \frac{1}{5}) = - 1$ .

3. Fit the sewer connection through the crossing point

Write $y = - \frac{1}{5} x + c$ and put in $(5, - 4)$ :

$- 4 = - \frac{1}{5} (5) + c = (- 1) + c$ , so $c = - 4 - (- 1) = - 3$ .

4. Read off the crossing and check

The sewer connection is $y = - \frac{1}{5} x + c$ with $c = - 3$ , so it meets the $y$ axis at $(0, - 3)$ and the required coordinate is $- 3$ .

Check: from $(0, - 3)$ to $(5, - 4)$ the change in $y$ is $- 4 - (- 3) = - 1$ over a change in $x$ of $5$ , and $\frac{- 1}{5} = - \frac{1}{5}$ as intended, so the point $(5, - 4)$ does lie on the sewer connection. The sewer connection falls gently while the water main climbs steeply, which is what gradients of $- \frac{1}{5}$ and $5$ look like on a plan.

The key is $- 3$ .

Why the Other Options Are Wrong (❌):

B. -29 · Parallel used instead of perpendicular
The sewer connection was given the water main's own gradient of $5$ : $y = 5 x + c$ through $(5, - 4)$ gives $c = - 4 - (25) = - 29$ . That line is parallel to the water main, so it never crosses it at right angles.
C. -5 · Sign not changed
The gradient was inverted without being negated, using $\frac{1}{5}$ : $- 4 = \frac{1}{5} (5) + c$ gives $c = - 4 - (1) = - 5$ . Since $(5) \times (\frac{1}{5}) = 1$ , not $- 1$ , that line is not perpendicular to the water main.
D. 23 · Wrong point substituted
The perpendicular gradient $- \frac{1}{5}$ was found correctly, then the sewer connection was fitted through the other point $(10, 21)$ instead of the crossing point: $21 = - \frac{1}{5} (10) + c$ gives $c = 21 - (- 2) = 23$ . The sewer connection meets the water main at $(5, - 4)$ , so $(10, 21)$ does not lie on it at all.
E. -15 · Wrong axis used
The sewer connection $y = - \frac{1}{5} x + c$ with $c = - 3$ was found correctly, then $y$ was set to $0$ rather than $x$ : $- \frac{1}{5} x + (- 3) = 0$ gives $x = - 15$ , which is where the sewer connection meets the $x$ axis, not the $y$ axis.

Common Mistake (⚠️):
Taking the negative of the gradient, or its reciprocal, but not both. Only $5$ and $- \frac{1}{5}$ multiply to $- 1$ : pairing $5$ with $- 5$ gives $- 25$ and pairing $5$ with $\frac{1}{5}$ gives $1$ , so neither pair is perpendicular.

Takeaway (📌):
Perpendicular means negative reciprocal, both operations, and the product of the two gradients must come to $- 1$ . Substitute the point the two lines share, never the other one, and read $c$ straight off as the $y$ intercept.

Question 4

After the door is shut, the temperature in a cold store is given in degrees Celsius by $T = t^{2} - 20 t + 19$ , with $t$ measured in hours. For how long, in hours, is the temperature below $0^{\circ} C$ ?

A. 18
B. 9
C. 10
D. 19
E. 1

Difficulty: ★★★☆☆ Target Time: $55 s$ Tests: N/A

Key Idea (💡): Completing the square rewrites a quadratic whose squared term has coefficient $1$ as $(t - p)^{2} - q$ , and that single form carries two pieces of information: the turning point sits at $t = p$ with value $- q$ , and the roots sit at $t = p \pm \sqrt{q}$ . Because the curve is symmetric about its turning point, the two roots lie the same distance either side of $t = p$ . An upward parabola is negative only between its roots, so the length of time the expression stays below zero is exactly the gap between them.

ESAT specification (UAT-UK): M4.11

Reveal the answer & worked solution: commit to an option first

Correct Answer: A. 18

Step-by-Step Breakdown:

1. Complete the square

$T = t^{2} - 20 t + 19$ $= (t - 10)^{2} - 100 + 19$ $= (t - 10)^{2} - 81$

2. Find the times at which $T = 0$

$(t - 10)^{2} = 81 \Rightarrow t - 10 = \pm 9 \Rightarrow t$ $= 1 or t = 19$

3. Decide where the model is negative, then measure the interval

The coefficient of $t^{2}$ is positive, so the parabola opens upwards and $T$ is negative between the two roots and positive outside them. The temperature is therefore below $0^{\circ} C$ from $t = 1$ to $t = 19$ , a stretch of

$19 - 1 = 18 hours$

Sanity check: at the midpoint $t = 10$ the model gives $100 - 200 + 19 = - 81$ , below freezing, while at $t = 0$ it gives $19$ and at $t = 20$ it gives $400 - 400 + 19 = 19$ , both above. The cold spell is symmetric about $t = 10$ , $9$ hours on each side, which is what $t = 10 \pm 9$ says.

The key is $18$ .

Why the Other Options Are Wrong (❌):

B. 9 · Half the interval given
The working stopped at $\sqrt{81} = 9$ . That is the distance from the turning point at $t = 10$ out to each root, so the full stretch below freezing is twice it: $2 \times 9 = 18$ hours.
C. 10 · Turning point time reported
The time of the minimum was given, from $- \frac{b}{2 a} = \frac{20}{2} = 10$ . That is when the temperature is at its lowest, $- 81^{\circ} C$ , and says nothing on its own about how long it stays below $0^{\circ} C$ .
D. 19 · Later root quoted as the duration
Solved $(t - 10)^{2} = 81$ correctly for $t = 1$ and $t = 19$ , then read the later crossing as the length of the cold spell, measuring from $t = 0$ . The temperature is still above freezing until $t = 1$ , since $T = 19$ at $t = 0$ , so the time spent below $0^{\circ} C$ is $19 - 1 = 18$ hours.
E. 1 · Root quoted instead of interval
The roots were found correctly from $t = 10 \pm 9$ , giving $t = 1$ and $t = 19$ , and the earlier one was reported. That is the hour at which the temperature first reaches $0^{\circ} C$ , not the length of time spent under it.

Common Mistake (⚠️):
Stopping at $\sqrt{81} = 9$ . That is the distance from the turning point at $t = 10$ to either root, so it is only half of the cold spell; the full stretch below freezing runs from $t = 1$ to $t = 19$ , which is $2 \times 9 = 18$ hours.

Takeaway (📌):
For a quadratic in the form $(t - p)^{2} - q$ whose squared term has coefficient $1$ , the roots are $p \pm \sqrt{q}$ , so the width of the negative stretch is $2 \sqrt{q}$ . One completion of the square delivers the turning point and the interval together.

Question 5

A data logger presents the pressure $p$ of a fixed mass of trapped gas, in kPa, plotted against its volume $V$ , in ${c m}^{3}$ , at constant temperature. The curve is a rectangular hyperbola: it falls as $V$ increases and draws ever closer to each axis without touching either. One reading recorded $V = 40$ together with $p = 60$ . According to the curve, what is $p$ , in kPa, at $V = 16$ ?

A. 24
B. 375
C. 100
D. 84
E. 150
F. 38400

Difficulty: ★★★☆☆ Target Time: $50 s$ Tests: N/A

Key Idea (💡): A curve that lies in the first quadrant and has both axes as asymptotes is the graph of the reciprocal function, $p = \frac{k}{V}$ . Its defining property is that the product of the two coordinates is the same at every point, so one recorded pair fixes $k$ and every other point then follows by a single division. Because $k$ is not zero, neither coordinate can ever reach zero, which is exactly why the curve closes on the axes without touching them.

ESAT specification (UAT-UK): M4.12

Reveal the answer & worked solution: commit to an option first

Correct Answer: E. 150

Step-by-Step Breakdown:

1. Name the curve from its description

A hyperbola in the first quadrant with both axes as asymptotes is the reciprocal graph, so every point of the curve satisfies

$p = \frac{k}{V}, that is V p = k$

2. Fix the constant from the recorded reading

$k = 40 \times 60 = 2400$

3. Substitute the new value of $V$

$p = \frac{2400}{16} = 150$

4. Check it against the shape of the curve

$V$ fell from $40$ to $16$ , and the curve falls, so $p$ must come out above $60$ , and $150$ does. The product test also passes: $16 \times 150 = 2400$ , the same as $40 \times 60$ . Finally the value is positive and finite, as it must be for a curve that never reaches either axis.

The key is $150$ .

Why the Other Options Are Wrong (❌):

A. 24 · Direct proportion assumed
$p$ was scaled by the same factor as $V$ : $60 \times \frac{16}{40} = 24$ . That is direct proportion, a straight line through the origin, and it contradicts a curve that falls as $V$ grows. The product test fails too: $16 \times 24 = 384$ rather than $2400$ .
B. 375 · Inverse square used
An inverse square rule was used instead of the reciprocal: $60 \times {(\frac{5}{2})}^{2} = 60 \times \frac{25}{4} = 375$ . That curve does approach both axes, but it is not a hyperbola, and it fails the constant product test, since $16 \times 375 = 6000$ rather than $2400$ .
C. 100 · Divided by the change in the variable
The constant was found correctly, $k = 40 \times 60 = 2400$ , but then divided by the change in $V$ , which is $24$ , instead of by $V$ itself: $\frac{2400}{24} = 100$ . The curve gives $p$ at $V = 16$ , so the division is $\frac{2400}{16} = 150$ .
D. 84 · Constant sum instead of product
The sum was held constant instead of the product: $40 + 60 = 100$ , then $100 - 16 = 84$ . A constant sum is a straight line of gradient $- 1$ , which would cut the $V$ axis at $100$ rather than run alongside it.
F. 38400 · Constant relation rearranged by multiplying
The relation $V p = k$ has to be divided through by $V$ to leave $p$ on its own. Writing $p = k V$ instead multiplies the constant by the new reading, which describes a straight line climbing away from the axes rather than a curve closing on them, and it lands a factor of $16^{2}$ above the value the curve gives. The constant is $k = 40 \times 60 = 2400$ , so the reading at $V = 16$ is $\frac{2400}{16} = 150$ and not $2400 \times 16$ .

Common Mistake (⚠️):
Treating a falling curve as a falling straight line. A straight line through the recorded reading would cross the $V$ axis at some finite value, where $p$ would be zero, and the stated asymptotes rule that out: the reciprocal model is forced.

Takeaway (📌):
Asymptotes on both axes means the product of the coordinates is constant. Multiply the pair you are given, divide by the new value, and you are done in two operations.

Where to go next

Next: ESAT Paper 3 Maths, five more questions at the same standard.
Five questions at test pace in Maths: practice set 1A and set 1B, each worked in full.
Twenty sample questions in Maths across the four papers, five on each module page.
Every paper and practice set across all five ESAT subjects is indexed on the ESAT preparation guide.
Teaching a cohort rather than sitting the test? ESAT mock papers for schools carry all 27 questions in every module, with worked solutions and mark schemes, and each pack is written for one school and issued to no other.

For individual preparation. These questions are free for a candidate working on their own. They are not licensed for a school, college or tutoring provider to print and run with a class, to republish or resell, or to fold into a course sold by another provider. A school sitting a timed mock with a class needs a paper written for its own candidates and used nowhere else: a commissioned pack, which comes with mark schemes and a syllabus reference.

ESAT Paper 2 Mathematics Sample Questions

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