ESAT Paper 3 Mathematics Sample Questions

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, $\frac{2}{9}$, so the horizontal edge is $18\times \frac{9}{2}=81\mathrm{c}\mathrm{m}$.

2. Find the first area

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

3. Read off the length scale factor

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

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 $k^2$: $729\times \frac{4}{9}=324\mathrm{c}{\mathrm{m}}^2$.

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

The key is $324$.

Why the Other Options Are Wrong (❌):

• A. 486 · Length scale factor used on an area
  $729\times \frac{2}{3}=486$, the scale factor applied once, as though an area behaved like a length. Every corresponding length of the second corner panel is $\frac{2}{3}$ times the first, so it is $\frac{2}{3}$ times as tall AND $\frac{2}{3}$ times as wide, and the factor has to act twice.
• B. 216 · Volume scale factor used for an area
  $729\times \frac{8}{27}=216$. 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
  $\frac{1}{2}\times 18\times 81=729$ 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 $18\times \frac{2}{9}=4\mathrm{c}\mathrm{m}$, its area $\frac{1}{2}\times 18\times 4=36\mathrm{c}{\mathrm{m}}^2$, and the scaled area $36\times \frac{4}{9}=16$.

Common Mistake (⚠️):
Applying $k$ where $k^2$ belongs: $729\times \frac{2}{3}=486$ instead of $729\times \frac{4}{9}=324$. Both edges at the right angle are scaled by $\frac{2}{3}$, 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

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

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 $\frac{1}{2}$. Check: $(-2)\times \left(\frac{1}{2}\right)=-1$.

3. Fit the sewer connection through the crossing point

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

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

4. Read off the crossing and check

The sewer connection is $y=\frac{1}{2}x+c$ 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 ${\displaystyle \frac{2}{4}}=\frac{1}{2}$ 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 $\frac{1}{2}$ 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 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 $-\frac{1}{2}$: $3=-\frac{1}{2}(4)+c$ gives $c=3-(-2)=5$. Since $(-2)\times (-\frac{1}{2})=1$, not $-1$, that line is not perpendicular to the water main.
• D. 16 · Wrong point substituted
  The perpendicular gradient $\frac{1}{2}$ was found correctly, then the sewer connection was fitted through the other point $(-2,15)$ instead of the crossing point: $15=\frac{1}{2}(-2)+c$ 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 $y=\frac{1}{2}x+c$ with $c=1$ was found correctly, then $y$ was set to $0$ rather than $x$: $\frac{1}{2}x+(1)=0$ 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 $\frac{1}{2}$, and the test is that $(-2)\times \left(\frac{1}{2}\right)=-1$.

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

$$T=t^2-14t+24$$$$=(t-7{)}^2-49+24$$$$=(t-7{)}^2-25$$

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

$$(t-7{)}^2=25\Rightarrow t-7=\pm 5\Rightarrow t$$$$=2\text{ or }t=12$$

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

$$12-2=10\text{hours}$$

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 $t=7\pm 5$ 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: $2\times 5=10$ hours.
• C. 7 · Turning point time reported
  The time of the minimum was given, from $-{\displaystyle \frac{b}{2a}}={\displaystyle \frac{14}{2}}=7$. 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 $t=7\pm 5$, 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

$$w=\frac{k}{l},\text{that is}lw=k$$

2. Fix the constant from the recorded design

$$k=35\times 60=2100$$

3. Substitute the new value of $l$

$$w=\frac{2100}{14}=150$$

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: $14\times 150=2100$, the same as $35\times 60$. 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
  $w$ was scaled by the same factor as $l$: $60\times {\displaystyle \frac{14}{35}}=24$. 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: $14\times 24=336$ rather than $2100$.
• B. 375 · Inverse square used
  An inverse square rule was used instead of the reciprocal: $60\times {\left({\displaystyle \frac{5}{2}}\right)}^2=60\times {\displaystyle \frac{25}{4}}=375$. That curve does approach both axes, but it is not a hyperbola, and it fails the constant product test, since $14\times 375=5250$ rather than $2100$.
• C. 100 · Divided by the change in the variable
  The constant was found correctly, $k=35\times 60=2100$, but then divided by the change in $l$, which is $21$, instead of by $l$ itself: ${\displaystyle \frac{2100}{21}}=100$. The curve gives $w$ at $l=14$, so the division is ${\displaystyle \frac{2100}{14}}=150$.
• 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 $lw=k$ has to be divided through by $l$ to leave $w$ on its own. Writing $w=kl$ 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 $k=35\times 60=2100$, so the reading at $l=14$ is ${\displaystyle \frac{2100}{14}}=150$ and not $2100\times 14$.

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 ${\displaystyle \frac{2}{5}}$, so $w$ is multiplied by ${\displaystyle \frac{5}{2}}$, and the product $lw$ stays at $2100$.

Takeaway (📌):
A first-quadrant curve hugging both axes 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.