ESAT Paper 3 Mathematics Sample Questions

Correct Answer: E. 20.5

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:

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

2. Rationalise the denominator

Multiply top and bottom by the conjugate $3+\sqrt{2}$.

Denominator: $(3-\sqrt{2})(3+\sqrt{2}) = 3^{2} - 2 = 7$.

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

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

3. Scale to $6\ \text{min}$, dividing every term

Multiplying by $6$ gives $\dfrac{84 + 42\sqrt{2}}{7}$. The $7$ underneath divides the whole numerator, the surd term as well as the whole number, and it goes into each of them exactly:

$\dfrac{84}{7} = 12$ and $\dfrac{42}{7} = 6$,

so the output over $6\ \text{min}$ is $12 + 6\sqrt{2}\ \text{L}$.

Sanity check

$\sqrt{2} \approx 1.41$, so the quoted volume is about $5.41\ \text{L}$ and the quoted time about $1.59\ \text{min}$, a rate near $3.41\ \text{L}\,\text{min}^{-1}$. Over $6\ \text{min}$ that is near $20.5\ \text{L}$, and the exact value $12 + 6\sqrt{2}$ is $20.5$ to 3 significant figures.

The key is $20.5$.

Why the Other Options Are Wrong (❌):

• A. 1.76 · Rate inverted
  Divided the time by the volume instead of the volume by the time, taking $\dfrac{3-\sqrt{2}}{4+\sqrt{2}}$ as the rate and multiplying that by $6$ to get about $1.76$. 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 $12 + 6\sqrt{2}$ litres.
• B. 51.5 · Multiplied volume by time
  Multiplied the stated volume by the stated time rather than dividing, then scaled the product by $6$, giving about $51.5$. A volume multiplied by a time is measured in litre minutes and is not a volume at all. Dividing $(4+\sqrt{2})$ by $(3-\sqrt{2})$ first, then scaling, gives $12 + 6\sqrt{2}$ litres.
• C. 32.5 · Rate step skipped
  Read $(4+\sqrt{2})\ \text{L}$ as the volume delivered in one minute and scaled it straight to $6$ minutes, giving about $32.5$. That volume takes $(3-\sqrt{2})\ \text{min}$, so the division by the time was never carried out and the rate used is wrong by a factor of $(3-\sqrt{2})$.
• D. 71.4 · Partial cancelling
  Rationalised and scaled correctly as far as $\dfrac{84 + 42\sqrt{2}}{7}$, then cancelled the $7$ into the first term only and read the result as $12 + 42\sqrt{2}$, about $71.4$. A denominator divides every term of the numerator it sits under, so $42$ is reduced to $6$ at the same time as $84$ is reduced to $12$, leaving $12 + 6\sqrt{2}$.
• F. 4.64 · Conjugate applied to the denominator only
  Multiplied the denominator by the conjugate $3+\sqrt{2}$ and left the numerator as it was, so the rate read $\dfrac{4+\sqrt{2}}{7}$ and the scaled volume $\dfrac{6(4+\sqrt{2})}{7}$. Multiplying one half of a fraction on its own changes its value: the conjugate goes over both halves, which turns the numerator into $(4+\sqrt{2})(3+\sqrt{2}) = 14 + 7\sqrt{2}$ and leaves $12 + 6\sqrt{2}$ after scaling.
• G. 12.0 · Cross terms lost in the expansion
  Expanded $(4+\sqrt{2})(3+\sqrt{2})$ as $4\times3$ 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: $84 \div 7 = 12$. The cross terms are $4\sqrt{2}$ and $3\sqrt{2}$, and they are where the $7\sqrt{2}$ comes from, so the volume is $12 + 6\sqrt{2}$ and not a whole number of litres.
• H. 15.9 · Surd term dropped from the conjugate product
  Took $(3-\sqrt{2})(3+\sqrt{2})$ to be $3^{2}$, on the reading that every term containing $\sqrt{2}$ cancels in a conjugate product, so the whole numerator was divided by $3^{2}$ instead. What cancels in that product is the pair of middle terms, $+3\sqrt{2}$ and $-3\sqrt{2}$; the last term is $-(\sqrt{2})^{2} = -2$ and it stays, so the denominator is $3^{2} - 2 = 7$ and the volume is $12 + 6\sqrt{2}$.

Common Mistake (⚠️):
Dividing only the rational term of the numerator by $7$. Once the scaled quotient reads $\dfrac{84 + 42\sqrt{2}}{7}$, the $7$ underneath divides every term, so $42$ is reduced to $6$ at the same time as $84$ is reduced to $12$. Cancelling into the first term alone leaves $12 + 42\sqrt{2}$, about $71.4$, in place of $12 + 6\sqrt{2}$.

Takeaway (📌):
Rationalise with the conjugate, then divide the whole numerator. A denominator of the form $a - \sqrt{b}$ becomes $a^{2} - b$ in one subtraction, here $3^{2} - 2 = 7$, and every term written above it is then divided by that same number.

Correct Answer: C. 324

Step-by-Step Breakdown:

1. Complete the first corner panel 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, $\tfrac{2}{9}$, so the horizontal edge is $18 \times \tfrac{9}{2} = 81\,\mathrm{cm}$.

2. Find the first area

Those two edges meet at the right angle, so they serve as base and height directly: $\tfrac{1}{2} \times 18 \times 81 = 729\,\mathrm{cm^2}$.

3. Read off the length scale factor

Corresponding lengths of similar shapes are all in one ratio, and the two upright edges correspond, so $k = \dfrac{12}{18} = \tfrac{2}{3}$.

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 $k^2$: $729 \times \tfrac{4}{9} = 324\,\mathrm{cm^2}$.

The direct route agrees: the second corner panel has a horizontal edge of $54\,\mathrm{cm}$, and $\tfrac{1}{2} \times 12 \times 54 = 324\,\mathrm{cm^2}$.

The key is $324$.

Why the Other Options Are Wrong (❌):

• A. 486 · Length scale factor used on an area
  $729 \times \tfrac{2}{3} = 486$, the scale factor applied once, as though an area behaved like a length. Every corresponding length of the second corner panel is $\tfrac{2}{3}$ times the first, so it is $\tfrac{2}{3}$ times as tall AND $\tfrac{2}{3}$ times as wide, and the factor has to act twice.
• B. 216 · Volume scale factor used for an area
  $729 \times \tfrac{8}{27} = 216$. Cubing the scale factor is right for a capacity, not for a flat face, and the corner panel is cut from glass.
• D. 729 · Scaling step omitted
  $\tfrac{1}{2} \times 18 \times 81 = 729$ is the first corner panel's own area, returned without ever crossing to the second one.
• E. 16 · Tangent ratio inverted
  Reading the tangent as horizontal over upright makes the first corner panel's horizontal edge $18 \times \tfrac{2}{9} = 4\,\mathrm{cm}$, its area $\tfrac{1}{2} \times 18 \times 4 = 36\,\mathrm{cm^2}$, and the scaled area $36 \times \tfrac{4}{9} = 16$.

Common Mistake (⚠️):
Applying $k$ where $k^2$ belongs: $729 \times \tfrac{2}{3} = 486$ instead of $729 \times \tfrac{4}{9} = 324$. Both edges at the right angle are scaled by $\tfrac{2}{3}$, so their product carries that factor twice over.

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.

Correct Answer: B. 1

Step-by-Step Breakdown:

1. Gradient of the water main

$m = \dfrac{15 - (3)}{-2 - (4)} = \dfrac{12}{-6} = -2$

The water main falls $2$ 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 $-2$ as its gradient, which is $\tfrac{1}{2}$. Check: $(-2)\times\left(\tfrac{1}{2}\right) = -1$.

3. Fit the sewer connection through the crossing point

Write $y = \tfrac{1}{2}x + c$ and put in $(4,3)$:

$3 = \tfrac{1}{2}(4) + c = (2) + c$, so $c = 3 - (2) = 1$.

4. Read off the crossing and check

The sewer connection is $y = \tfrac{1}{2}x + c$ with $c = 1$, so it meets the $y$ axis at $(0,1)$ and the required coordinate is $1$.

Check: from $(0,1)$ to $(4,3)$ the change in $y$ is $3 - (1) = 2$ over a change in $x$ of $4$, and $\dfrac{2}{4} = \tfrac{1}{2}$ as intended, so the point $(4,3)$ does lie on the sewer connection. The sewer connection climbs gently while the water main falls steeply, which is what gradients of $\tfrac{1}{2}$ and $-2$ look like on a plan.

The key is $1$.

Why the Other Options Are Wrong (❌):

• A. 11 · Parallel used instead of perpendicular
  The sewer connection was given the water main's own gradient of $-2$: $y = -2x + c$ through $(4,3)$ gives $c = 3 - (-8) = 11$. That line has the water main's gradient and passes through $(4,3)$, which lies on the water main, so it is the water main itself, not a line crossing it at right angles.
• C. 5 · Sign not changed
  The gradient was inverted without being negated, using $-\tfrac{1}{2}$: $3 = -\tfrac{1}{2}(4) + c$ gives $c = 3 - (-2) = 5$. Since $(-2)\times\left(-\tfrac{1}{2}\right) = 1$, not $-1$, that line is not perpendicular to the water main.
• D. 16 · Wrong point substituted
  The perpendicular gradient $\tfrac{1}{2}$ was found correctly, then the sewer connection was fitted through the other point $(-2,15)$ instead of the crossing point: $15 = \tfrac{1}{2}(-2) + c$ gives $c = 15 - (-1) = 16$. The sewer connection meets the water main at $(4,3)$, so $(-2,15)$ does not lie on it at all.
• E. -2 · Wrong axis used
  The sewer connection $y = \tfrac{1}{2}x + c$ with $c = 1$ was found correctly, then $y$ was set to $0$ rather than $x$: $\tfrac{1}{2}x + (1) = 0$ gives $x = -2$, which is where the sewer connection meets the $x$ axis, not the $y$ axis.

Common Mistake (⚠️):
Giving the sewer connection the gradient $-2$ of the water main it crosses. A line with the same gradient is parallel, and two parallel lines never meet at right angles; the sewer connection needs the negative reciprocal $\tfrac{1}{2}$, and the test is that $(-2)\times\left(\tfrac{1}{2}\right) = -1$.

Takeaway (📌):
Flip the gradient and change its sign, then substitute the shared point. In $y = mx + c$ the constant $c$ already answers any question about the $y$ axis, so no further solving is needed.

Correct Answer: B. 10

Step-by-Step Breakdown:

1. Complete the square

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

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 $0\ ^\circ\text{C}$ from $t=2$ to $t=12$, a stretch of

Sanity check: at the midpoint $t=7$ the model gives $49-98+24=-25$, below freezing, while at $t=0$ it gives $24$ and at $t=14$ it gives $196-196+24=24$, both above. The cold spell is symmetric about $t=7$, $5$ hours on each side, which is what $t=7\pm5$ says.

The key is $10$.

Why the Other Options Are Wrong (❌):

• A. 5 · Half the interval given
  The working stopped at $\sqrt{25}=5$. That is the distance from the turning point at $t=7$ out to each root, so the full stretch below freezing is twice it: $2\times5=10$ hours.
• C. 7 · Turning point time reported
  The time of the minimum was given, from $-\dfrac{b}{2a}=\dfrac{14}{2}=7$. That is when the temperature is at its lowest, $-25\ ^\circ\text{C}$, and says nothing on its own about how long it stays below $0\ ^\circ\text{C}$.
• D. 12 · Later root quoted as the duration
  Solved $(t-7)^{2}=25$ correctly for $t=2$ and $t=12$, then read the later crossing as the length of the cold spell, measuring from $t=0$. The temperature is still above freezing until $t=2$, since $T=24$ at $t=0$, so the time spent below $0\ ^\circ\text{C}$ is $12-2=10$ hours.
• E. 2 · Root quoted instead of interval
  The roots were found correctly from $t=7\pm5$, giving $t=2$ and $t=12$, and the earlier one was reported. That is the hour at which the temperature first reaches $0\ ^\circ\text{C}$, not the length of time spent under it.

Common Mistake (⚠️):
Solving the quadratic and then quoting a root. The values $t=2$ and $t=12$ are the two moments at which the temperature passes through $0\ ^\circ\text{C}$; the question asks how long the temperature spends below that line, which is the gap between them, $12-2=10$ 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.

Correct Answer: F. 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

2. Fix the constant from the recorded design

3. Substitute the new value of $l$

4. Check it against the shape of the curve

$l$ fell from $35$ to $14$, and the curve falls, so $w$ must come out above $60$, and $150$ does. The product test also passes: $14\times150=2100$, the same as $35\times60$. 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
  $w$ was scaled by the same factor as $l$: $60\times\dfrac{14}{35}=24$. That is direct proportion, a straight line through the origin, and it contradicts a curve that falls as $l$ grows. The product test fails too: $14\times24=336$ rather than $2100$.
• B. 375 · Inverse square used
  An inverse square rule was used instead of the reciprocal: $60\times\left(\dfrac{5}{2}\right)^2=60\times\dfrac{25}{4}=375$. That curve does approach both axes, but it is not a hyperbola, and it fails the constant product test, since $14\times375=5250$ rather than $2100$.
• C. 100 · Divided by the change in the variable
  The constant was found correctly, $k=35\times60=2100$, but then divided by the change in $l$, which is $21$, instead of by $l$ itself: $\dfrac{2100}{21}=100$. The curve gives $w$ at $l=14$, so the division is $\dfrac{2100}{14}=150$.
• D. 81 · Constant sum instead of product
  The sum was held constant instead of the product: $35+60=95$, then $95-14=81$. A constant sum is a straight line of gradient $-1$, which would cut the $l$ axis at $95$ rather than run alongside it.
• E. 29400 · Constant relation rearranged by multiplying
  The relation $l\,w=k$ has to be divided through by $l$ to leave $w$ on its own. Writing $w=k\,l$ 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 $14^{2}$ above the value the curve gives. The constant is $k=35\times60=2100$, so the reading at $l=14$ is $\dfrac{2100}{14}=150$ and not $2100\times14$.

Common Mistake (⚠️):
Changing $w$ by the same factor as $l$. On a reciprocal curve the two factors are reciprocals of each other: $l$ is multiplied by $\dfrac{2}{5}$, so $w$ is multiplied by $\dfrac{5}{2}$, and the product $l\,w$ stays at $2100$.

Takeaway (📌):
A first-quadrant hyperbola with both axes as asymptotes is $y=k/x$. One point fixes $k$ as the product of its coordinates, and every other point is that product divided by the coordinate you know.