ESAT Practice Set 1A Mathematics Worked Solutions

Correct Answer: B. 18

Fastest Approach (🚀):
Skip the constant. $m$ is multiplied by $\dfrac{4}{3}$, so on a reciprocal curve $\theta$ is multiplied by $\dfrac{3}{4}$, and $\dfrac{3}{4}$ of $24$ is $18$.

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 trial

3. Substitute the new value of $m$

4. Check it against the shape of the curve

$m$ rose from $36$ to $48$, and the curve falls, so $\theta$ must come out below $24$, and $18$ does. The product test also passes: $48\times18=864$, the same as $36\times24$. Finally the value is positive and finite, as it must be for a curve that never reaches either axis.

The key is $18$.

Why the Other Options Are Wrong (❌):

• A. 32 · Direct proportion assumed
  $\theta$ was scaled by the same factor as $m$: $24\times\dfrac{48}{36}=32$. That is direct proportion, a straight line through the origin, and it contradicts a curve that falls as $m$ grows. The product test fails too: $48\times32=1536$ rather than $864$.
• C. 13.5 · Inverse square used
  An inverse square rule was used instead of the reciprocal: $24\times\left(\dfrac{3}{4}\right)^2=24\times\dfrac{9}{16}=13.5$. That curve does approach both axes, but it is not a hyperbola, and it fails the constant product test, since $48\times13.5=648$ rather than $864$.
• D. 72 · Divided by the change in the variable
  The constant was found correctly, $k=36\times24=864$, but then divided by the change in $m$, which is $12$, instead of by $m$ itself: $\dfrac{864}{12}=72$. The curve gives $\theta$ at $m=48$, so the division is $\dfrac{864}{48}=18$.
• E. 12 · Constant sum instead of product
  The sum was held constant instead of the product: $36+24=60$, then $60-48=12$. A constant sum is a straight line of gradient $-1$, which would cut the $m$ axis at $60$ rather than run alongside it.

Common Mistake (⚠️):
Changing $\theta$ by the same factor as $m$. On a reciprocal curve the two factors are reciprocals of each other: $m$ is multiplied by $\dfrac{4}{3}$, so $\theta$ is multiplied by $\dfrac{3}{4}$, and the product $m\,\theta$ stays at $864$.

Takeaway (📌):
The Cheat Code: 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.

Correct Answer: C. 39.0

Fastest Approach (🚀):
Reverse one bearing and take both differences before touching the scale. Two angles of $70^{\circ}$, at $W$ and at $L$, settle the whole shape at once, and only then is there one multiplication by $3000\ \mathrm{m}$ to do.

Step-by-Step Breakdown:

1. Read the scale

A scale of $1:300\,000$ means that $1\ \mathrm{cm}$ on the chart stands for $300\,000\ \mathrm{cm}$ on the ground, and $300\,000\ \mathrm{cm} = 3000\ \mathrm{m}$.

2. Find the angle at $P$

Both bearings given at $P$ are measured clockwise from the same north line, so the angle between the chart lines $PW$ and $PL$ is their difference:

$075^{\circ} - 035^{\circ} = 40^{\circ}$.

3. Find the angle at $W$

The bearing $035^{\circ}$ points from $P$ towards $W$, so it must be reversed before it can be used at $W$: the bearing of $P$ from $W$ is $035^{\circ} + 180^{\circ} = 215^{\circ}$. The lighthouse lies on $145^{\circ}$ from $W$, so the angle inside the triangle at $W$ is $215^{\circ} - 145^{\circ} = 70^{\circ}$. The third angle is then $180^{\circ} - 40^{\circ} - 70^{\circ} = 70^{\circ}$, the same as the angle at $W$. Equal angles at $W$ and $L$ mean the sides opposite them are equal, and those sides are $PL$ and $PW$, so $PL$ measures the same $13\ \mathrm{cm}$ on the chart as $PW$.

4. Convert the chart length to a ground distance

$13 \times 3000\ \mathrm{m} = 39000\ \mathrm{m} = 39.0\ \mathrm{km}$.

Check the conversion by undoing it: $39000\ \mathrm{m}$ divided by $3000\ \mathrm{m}$ per centimetre returns the $13\ \mathrm{cm}$ that was given, so the scale has been applied the right way round.

The key is $39.0$.

Why the Other Options Are Wrong (❌):

• A. 78.0 · Route through the middle mark instead of the straight line
  Follows the route $P$ to $W$ to $L$ instead of the straight line $PL$, counting both legs as $13\ \mathrm{cm}$ of chart: $13 + 13 = 26\ \mathrm{cm}$, and $26 \times 3000\ \mathrm{m} = 78000\ \mathrm{m} = 78.0\ \mathrm{km}$. The question asks for the single side $PL$, which the two equal angles at $W$ and $L$ fix at $13\ \mathrm{cm}$.
• B. 13.0 · Chart length quoted as the real distance
  Quotes the chart length itself, $13$, and attaches kilometres to it, so the scale of $1:300\,000$ is never applied. On the chart $PL$ is $13\ \mathrm{cm}$; on the ground each of those centimetres stands for $3000\ \mathrm{m}$.
• D. 3.9 · Centimetres to metres by the wrong power of ten
  Multiplies correctly, $13 \times 300\,000 = 3900000\ \mathrm{cm}$ on the ground, then divides by $1000$ instead of $100$ to reach metres, giving $3900\ \mathrm{m}$ and so $3.9\ \mathrm{km}$. There are $100\ \mathrm{cm}$ in a metre, so the ground distance is $39000\ \mathrm{m}$.
• E. 3900.0 · Scale read as centimetres to metres
  Reads the scale as $1\ \mathrm{cm}$ to $300\,000\ \mathrm{m}$, treating the second figure as metres when a ratio scale carries no units at all: $13 \times 300\,000 = 3900000\ \mathrm{m}$, which is $3900.0\ \mathrm{km}$. The second figure is in the same unit as the first, so $1\ \mathrm{cm}$ stands for $300\,000\ \mathrm{cm}$, which is $3000\ \mathrm{m}$.

Common Mistake (⚠️):
Using the bearing $035^{\circ}$ at $W$ without reversing it, so the angle at $W$ is taken as the gap between $145^{\circ}$ and $035^{\circ}$ instead of between $145^{\circ}$ and $215^{\circ}$. A bearing is measured at the point it is taken from, and $035^{\circ}$ was taken at $P$; at $W$ the line back to $P$ runs on $215^{\circ}$. Without the reversal the angle at $W$ comes out as something other than $70^{\circ}$, the two base angles no longer match, and nothing fixes $PL$ at $13\ \mathrm{cm}$.

Takeaway (📌):
The Cheat Code: two bearings from one point give the angle there by subtraction; a bearing from the far end of a line needs $180^{\circ}$ added or removed before it can be compared with anything. Find the angles, spot the equal pair, and only then multiply by the scale.

Correct Answer: A. 2720

Fastest Approach (🚀):
Average the parallel sides, $\frac{11 + 7}{2} = 9$, and multiply by the height $6$ to reach $54$ in one line. Take the $20$ off before scaling up, so the only sizeable multiplication left is $34 \times 80$.

Step-by-Step Breakdown:

1. Area of the trapezium cross-section

The parallel sides are $11\ \mathrm{cm}$ and $7\ \mathrm{cm}$ and the perpendicular height between them is $6\ \mathrm{cm}$, so

$A = \frac{1}{2}(11 + 7)(6) = \frac{1}{2}(18)(6) = 54\ \mathrm{cm}^{2}$.

2. Area of the water passage's cross-section

The water passage is rectangular, so its cross-section is $5 \times 4 = 20\ \mathrm{cm}^{2}$.

3. Area of cast iron in one cross-section

The water passage runs the whole length of the section, so the same rectangle is missing from every cross-section and the cast iron occupies

$54 - 20 = 34\ \mathrm{cm}^{2}$.

4. Multiply by the length of the prism

A right prism has volume equal to its cross-sectional area times its length, so

$V = 34 \times 80 = 2720\ \mathrm{cm}^{3}$.

Check by dividing back: $2720 \div 80 = 34\ \mathrm{cm}^{2}$, the cross-sectional area found in step 3.

The key is $2720$.

Why the Other Options Are Wrong (❌):

• B. 4320 · Incomplete Method
  Treats the cast-iron channel section as solid and forgets that the water passage is empty space: $54 \times 80 = 4320$. The $20\ \mathrm{cm}^{2}$ of every cross-section that the water passage occupies holds no cast iron.
• C. 7040 · Formula Error
  Drops the factor $\frac{1}{2}$ from the trapezium formula: $(11 + 7)(6) = 108$ instead of $54$, then $(108 - 20) \times 80 = 7040$. A trapezium is the average of its parallel sides times its height, not their sum.
• D. 1600 · Answered the wrong quantity
  Finds the space the water passage takes up rather than the cast iron around it: $20 \times 80 = 1600$. That is the volume removed, and the question asks for what is left.
• E. 2880 · Area confused with perimeter
  Uses the water passage's perimeter in place of its area: $2(5 + 4) = 18$ instead of $5 \times 4 = 20$, so the cross-section becomes $54 - 18 = 36$ and $36 \times 80 = 2880$. A perimeter is a length and cannot be taken off an area.

Common Mistake (⚠️):
Leaving the factor $\frac{1}{2}$ out of the trapezium formula, so the cross-section is taken as $(11 + 7) \times 6 = 108\ \mathrm{cm}^{2}$ rather than $54\ \mathrm{cm}^{2}$, which inflates every figure computed after it and lands on $7040$.

Takeaway (📌):
The Cheat Code: for a prism with a hole running through it, finish the cross-section completely, hole included, and multiply by the length once at the very end. One multiplication beats subtracting two volumes.

Correct Answer: E. 1250

Fastest Approach (🚀):
Cancel the two coating figures and both parts come out as perfect squares, $16$ and $25$, so the lengths read off by inspection as $4 : 5$ and the capacities as $64 : 125$. The capacity you are given divides exactly by $64$, since $640 \div 64 = 10$, which turns the last line into the single product $10 \times 125$.

Step-by-Step Breakdown:

1. Turn the two coating figures into a ratio of surface areas

The coat is the same thickness on both storage silos, so the volume of weatherproof paint used is proportional to the area it covers. The two outer surfaces are therefore in the ratio $128 : 200$, which is $16 : 25$ in its lowest terms.

2. Square root the area ratio to get the length scale factor

For mathematically similar solids corresponding areas are in the ratio $k^{2}$, where $k$ is the length scale factor. Here $k^{2} = \frac{25}{16}$, and both parts are perfect squares, so $k = \frac{5}{4}$ and corresponding lengths are in the ratio $4 : 5$.

3. Cube the length scale factor to get the capacity ratio

Corresponding volumes are in the ratio $k^{3}$, so the capacities are in the ratio $4^{3} : 5^{3} = 64 : 125$.

4. Apply that ratio to the capacity you are given

$640 \div 64 = 10$, an exact division, so the larger capacity is $10 \times 125 = 1250$.

Check: $64 \times 1250 = 80000$ and $125 \times 640 = 80000$, so $640 : 1250$ is exactly $64 : 125$.

The key is $1250\,\mathrm{m}^{3}$.

Why the Other Options Are Wrong (❌):

• A. 800 · length factor used for a capacity
  The length scale factor $k = \frac{5}{4}$ is found correctly and then applied only once: $640 \times \frac{5}{4} = 800$. That factor turns a length into the corresponding length, and a capacity needs it three times over.
• B. 1000 · area factor used for a capacity
  Treats the capacity as proportional to the weatherproof paint used: $640 \times \frac{200}{128} = 640 \times \frac{25}{16} = 1000$. A coat of fixed thickness measures area, so that fraction is $k^{2}$, and a capacity scales as $k^{3}$.
• C. 3200 · length ratio part used as a multiplier
  Reads the length ratio $4 : 5$ as an instruction to multiply by $5$: $640 \times 5 = 3200$. A ratio scales by the fraction $\frac{5}{4}$, not by its second part on its own, and that fraction still has to be cubed.
• D. 16000 · area ratio part used as a multiplier
  Cancels the two coating figures to $16 : 25$ and then multiplies the capacity by the $25$: $640 \times 25 = 16000$. Two things go wrong at once: a ratio scales by the fraction $\frac{25}{16}$ rather than by one of its parts, and that fraction is the area factor rather than the volume factor.

Common Mistake (⚠️):
Treating the capacity as proportional to the weatherproof paint and so multiplying by $\frac{200}{128} = \frac{25}{16}$, which gives $1000$. A coat at a fixed thickness measures surface area, and area and volume scale by different powers of the same length factor, so the coating ratio has to be square rooted before it can be cubed.

Takeaway (📌):
The Cheat Code: Anything that coats a similar solid, paint, plating or fabric, scales as $k^{2}$; anything that fills it, capacity, mass or the cost of a full load, scales as $k^{3}$. Work out $k$ first from whichever ratio you are handed, then raise it to the power the question is actually asking for.

Correct Answer: C. 150

Fastest Approach (🚀):
Use the area once and then abandon it. $60\pi = \tfrac{1}{2}r(10\pi)$ gives $r = 12$ immediately, and from there the curved edge is $\tfrac{10\pi}{24\pi} = \tfrac{5}{12}$ of the circumference, so the angle is $\tfrac{5}{12}$ of $360^\circ$.

Step-by-Step Breakdown:

1. Link the area to the curved edge

Write $r$ for the radius, which here is the length of each straight edge, and $\ell$ for the arc, which is the curved edge. Since the sector is the fraction $\tfrac{\theta}{360}$ of the disc,

so area, radius and arc are tied together without the angle appearing.

2. Solve for the radius

Each straight edge is $12\ \text{m}$ long.

3. Compare the arc with the whole circumference

A full circle of radius $12\ \text{m}$ has circumference $2\pi(12) = 24\pi\ \text{m}$, so the curved edge covers

of the way round.

4. Turn the fraction into an angle, then check

Check against the area, which has not yet been used twice: $\tfrac{5}{12}$ of the disc is $\tfrac{5}{12} \times \pi(12)^2 = \tfrac{5}{12} \times 144\pi = 60\pi\ \text{m}^2$, exactly the area stated.

The key is $150$.

Why the Other Options Are Wrong (❌):

• A. 75 · Disc area written as $2\pi r^2$
  Found $r = 12\ \text{m}$ correctly from $A = \tfrac{1}{2} r \ell$, then wrote the area of the whole disc as $2\pi r^2 = 2\pi(144) = 288\pi$ instead of $\pi r^2 = 144\pi$. The sector then looks like $\tfrac{60\pi}{288\pi} = \tfrac{5}{24}$ of the circle, giving $\tfrac{5}{24} \times 360^\circ = 75^\circ$. The $2\pi r$ belongs to the circumference; the area carries no factor of $2$.
• B. 12 · Stopped at the radius
  Stopped at the intermediate value: $A = \tfrac{1}{2} r \ell$ gives $r = \tfrac{2 \times 60\pi}{10\pi} = 12$, and $12$ was carried straight into the answer. That $12$ is the length of each straight edge in metres, not an angle in degrees; the angle still has to be taken from the arc, $\tfrac{10\pi}{2\pi(12)} \times 360^\circ = 150^\circ$.
• D. 60 · Compared a length with an area
  Divided the arc by the area directly, $\tfrac{10\pi}{60\pi} = \tfrac{1}{6}$, and read that as the fraction of a full turn: $\tfrac{1}{6} \times 360^\circ = 60^\circ$. A length divided by an area is not a fraction of anything.
• E. 210 · Gave the angle outside the sector
  Reached $150^\circ$ and then gave the rest of the full turn, $360^\circ - 150^\circ = 210^\circ$. That is the angle on the far side of the two straight edges, the part of the circle the sector does not occupy, and the question asks for the angle measured through the sector itself.

Common Mistake (⚠️):
Dividing the two given quantities into one another, $\tfrac{10\pi}{60\pi}$, as though the ratio of an arc to an area were the fraction of a turn. An arc is a length and a sector is an area, so the two are only comparable once the radius has been extracted from them, and the fraction of a turn must be measured against a quantity of the same kind: the arc against the full circumference $2\pi r$, or the area against the full disc $\pi r^2$.

Takeaway (📌):
The Cheat Code: $A = \tfrac{1}{2} r \ell$ converts an arc and an area into the radius in one line, with no $\pi$ to cancel and no angle to carry. With $r$ in hand, the angle is just the arc written as a fraction of $2\pi r$, multiplied by $360^\circ$.