ESAT Mathematics 1 Practice Questions by Topic

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 $640÷64=10$, 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

$640÷64=10$, 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 $1250m^3$.

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 (📌):
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: B. 1944

Fastest Approach (🚀):
The two edges at the right angle are the base and the height, so the first area is half their product and the sloping edge is never needed. The whole question then reduces to $486\times 4$, and the option reading $486$ is the one that never crossed between the two brackets.

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, $\frac{4}{3}$, so the horizontal edge is $36\times \frac{3}{4}=27\mathrm{c}\mathrm{m}$.

2. Find the first area

Those two edges meet at the right angle, so they serve as base and height directly: $\frac{1}{2}\times 36\times 27=486\mathrm{c}{\mathrm{m}}^2$.

3. Read off the length scale factor

Corresponding lengths of similar shapes are all in one ratio, and the two upright edges correspond, so $k={\displaystyle \frac{72}{36}}=2$.

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 $k^2$: $486\times 4=1944\mathrm{c}{\mathrm{m}}^2$.

The direct route agrees: the second bracket has a horizontal edge of $54\mathrm{c}\mathrm{m}$, and $\frac{1}{2}\times 72\times 54=1944\mathrm{c}{\mathrm{m}}^2$.

The key is $1944$.

Why the Other Options Are Wrong (❌):

• A. 972 · Length scale factor used on an area
  $486\times 2=972$, the scale factor applied once, as though an area behaved like a length. Every corresponding length of the second bracket is $2$ times the first, so it is $2$ times as tall AND $2$ times as wide, and the factor has to act twice.
• C. 3888 · Volume scale factor used for an area
  $486\times 8=3888$. Cubing the scale factor is right for a capacity, not for a flat face, and the bracket is cut from steel sheet.
• D. 486 · Scaling step omitted
  $\frac{1}{2}\times 36\times 27=486$ is the first bracket's own area, returned without ever crossing to the second one.
• E. 3456 · Tangent ratio inverted
  Reading the tangent as horizontal over upright makes the first bracket's horizontal edge $36\times \frac{4}{3}=48\mathrm{c}\mathrm{m}$, its area $\frac{1}{2}\times 36\times 48=864\mathrm{c}{\mathrm{m}}^2$, and the scaled area $864\times 4=3456$.

Common Mistake (⚠️):
Applying $k$ where $k^2$ belongs: $486\times 2=972$ instead of $486\times 4=1944$. Both edges at the right angle are scaled by $2$, 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: C. 6

Fastest Approach (🚀):
The cable duct has gradient $\frac{1}{2}$, and moving from $(2,7)$ to the $y$ axis changes $x$ by $-2$, so $y$ changes by $\frac{1}{2}\times (-2)=-1$ and the crossing is at $7+(-1)=6$. No equation has to be written down at all.

Step-by-Step Breakdown:

1. Gradient of the gas main

$m={\displaystyle \frac{-1-(7)}{6-(2)}}={\displaystyle \frac{-8}{4}}=-2$

The gas main falls $2$ metres for every metre across.

2. Gradient of the cable duct

Perpendicular gradients multiply to $-1$, so the cable duct 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 cable duct through the crossing point

Write $y=\frac{1}{2}x+c$ and put in $(2,7)$:

$7=\frac{1}{2}(2)+c=(1)+c$, so $c=7-(1)=6$.

4. Read off the crossing and check

The cable duct is $y=\frac{1}{2}x+c$ with $c=6$, so it meets the $y$ axis at $(0,6)$ and the required coordinate is $6$.

Check: from $(0,6)$ to $(2,7)$ the change in $y$ is $7-(6)=1$ over a change in $x$ of $2$, and ${\displaystyle \frac{1}{2}}=\frac{1}{2}$ as intended, so the point $(2,7)$ does lie on the cable duct. The cable duct climbs gently while the gas main falls steeply, which is what gradients of $\frac{1}{2}$ and $-2$ look like on a plan.

The key is $6$.

Why the Other Options Are Wrong (❌):

• A. 11 · Parallel used instead of perpendicular
  The cable duct was given the gas main's own gradient of $-2$: $y=-2x+c$ through $(2,7)$ gives $c=7-(-4)=11$. That line is the gas main itself, $y=-2x+11$, which also passes through $(6,-1)$, so it cannot cross the gas main at right angles.
• B. 8 · Sign not changed
  The gradient was inverted without being negated, using $-\frac{1}{2}$: $7=-\frac{1}{2}(2)+c$ gives $c=7-(-1)=8$. Since $(-2)\times (-\frac{1}{2})=1$, not $-1$, that line is not perpendicular to the gas main.
• D. -4 · Wrong point substituted
  The perpendicular gradient $\frac{1}{2}$ was found correctly, then the cable duct was fitted through the other point $(6,-1)$ instead of the crossing point: $-1=\frac{1}{2}(6)+c$ gives $c=-1-(3)=-4$. The cable duct meets the gas main at $(2,7)$, so $(6,-1)$ does not lie on it at all.
• E. -12 · Wrong axis used
  The cable duct $y=\frac{1}{2}x+c$ with $c=6$ was found correctly, then $y$ was set to $0$ rather than $x$: $\frac{1}{2}x+(6)=0$ gives $x=-12$, which is where the cable duct meets the $x$ axis, not the $y$ axis.

Common Mistake (⚠️):
Taking the negative of the gradient, or its reciprocal, but not both. Only $-2$ and $\frac{1}{2}$ multiply to $-1$: pairing $-2$ with $2$ gives $-4$ and pairing $-2$ with $-\frac{1}{2}$ 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.

Correct Answer: A. 18

Fastest Approach (🚀):
Skip the roots entirely. Half of $22$ is $11$, and ${11}^2-40=81$, so the completed square is $(t-11{)}^2-81$ and the width below zero is $2\sqrt{81}=18$ hours in a single line.

Step-by-Step Breakdown:

1. Complete the square

$$T=t^2-22t+40$$$$=(t-11{)}^2-121+40$$$$=(t-11{)}^2-81$$

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

$$(t-11{)}^2=81\Rightarrow t-11=\pm 9\Rightarrow t$$$$=2\text{ or }t=20$$

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 $t=20$, a stretch of

$$20-2=18\text{hours}$$

Sanity check: at the midpoint $t=11$ the model gives $121-242+40=-81$, below freezing, while at $t=0$ it gives $40$ and at $t=22$ it gives $484-484+40=40$, both above. The cold spell is symmetric about $t=11$, $9$ hours on each side, which is what $t=11\pm 9$ 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 $t=11$ out to each root, so the full stretch below freezing is twice it: $2\times 9=18$ hours.
• C. 11 · Turning point time reported
  The time of the minimum was given, from $-{\displaystyle \frac{b}{2a}}={\displaystyle \frac{22}{2}}=11$. That is when the temperature is at its lowest, $-81{}^{\circ }\text{C}$, and says nothing on its own about how long it stays below $0{}^{\circ }\text{C}$.
• D. 20 · Later root quoted as the duration
  Solved $(t-11{)}^2=81$ correctly for $t=2$ and $t=20$, 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 $T=40$ at $t=0$, so the time spent below $0{}^{\circ }\text{C}$ is $20-2=18$ hours.
• E. 2 · Root quoted instead of interval
  The roots were found correctly from $t=11\pm 9$, giving $t=2$ and $t=20$, 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 $t=20$ 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, $20-2=18$ hours.

Takeaway (📌):
For a quadratic in the form $(t-p{)}^2-q$ whose squared term has coefficient $1$, the roots are $p\pm \sqrt{q}$ when $q>0$, 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: B. 18

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

$$\theta =\frac{k}{m},\text{that is}m\theta =k$$

2. Fix the constant from the recorded run

$$k=36\times 24=864$$

3. Substitute the new value of $m$

$$\theta =\frac{864}{48}=18$$

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: $48\times 18=864$, the same as $36\times 24$. 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 $m$: $24\times {\displaystyle \frac{48}{36}}=32$. 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: $48\times 32=1536$ rather than $864$.
• C. 13.5 · Inverse square used
  An inverse square rule was used instead of the reciprocal: $24\times {\left({\displaystyle \frac{3}{4}}\right)}^2=24\times {\displaystyle \frac{9}{16}}=13.5$. That curve does approach both axes, but it is not a hyperbola, and it fails the constant product test, since $48\times 13.5=648$ rather than $864$.
• D. 72 · Divided by the change in the variable
  The constant was found correctly, $k=36\times 24=864$, but then divided by the change in $m$, which is $12$, instead of by $m$ itself: ${\displaystyle \frac{864}{12}}=72$. The curve gives $\theta$ at $m=48$, so the division is ${\displaystyle \frac{864}{48}}=18$.
• 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 ${\displaystyle \frac{4}{3}}$, so $\theta$ is multiplied by ${\displaystyle \frac{3}{4}}$, and the product $m\theta$ stays at $864$.

Takeaway (📌):
A curve of the shape $y=k/x$ keeps the product $xy$ constant; hugging both axes alone does not prove that shape, since $y=k/x^2$ hugs them too. 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: B. 1750

Fastest Approach (🚀):
Treat the whole journey as one trapezium instead of three pieces. Parallel sides $120\text{s}$ and $20\text{s}$, height $25{\text{m s}}^{-1}$: $\frac{1}{2}\times 140\times 25=1750\text{m}$ in one line.

Step-by-Step Breakdown:

1. Distance while accelerating

On a speed-time graph the distance is the area beneath the line. The first stage is a triangle rising from $0$ to $25{\text{m s}}^{-1}$ over $40\text{s}$:

$$\frac{1}{2}\times 40\times 25=500\text{m}.$$

2. Distance at constant speed

The middle stage is a rectangle, height $25{\text{m s}}^{-1}$ and width $20\text{s}$:

$$25\times 20=500\text{m}.$$

3. Distance while braking

The last stage is a triangle falling from $25{\text{m s}}^{-1}$ to rest over $60\text{s}$:

$$\frac{1}{2}\times 60\times 25=750\text{m}.$$

4. Add the three areas

$$500+500+750=1750\text{m}.$$

Check: the whole shape is one trapezium of height $25{\text{m s}}^{-1}$ with parallel sides $40+20+60=120\text{s}$ and $20\text{s}$, giving $\frac{1}{2}\times (120+20)\times 25=1750\text{m}$.

The key is $1750$.

Why the Other Options Are Wrong (❌):

• A. 3000 · Triangle halving omitted
  Both sloping stages were treated as rectangles, so neither was halved: $$25\times 40+500+25\times 60$$$$=1000+500+1500=3000$$. This counts every second of accelerating and braking as though the train were already at $25{\text{m s}}^{-1}$, which overstates both of those stages.
• C. 1500 · Graph shape misread as one triangle
  The whole graph was taken as a single triangle over the full $120\text{s}$: $\frac{1}{2}\times 120\times 25=1500$. That halves the constant-speed stage as well, but for those $20\text{s}$ the train was at $25{\text{m s}}^{-1}$ the whole time, so its area is the full rectangle $500$, not half of it.
• D. 2750 · Halving applied to the wrong stage
  The half was put on the constant-speed stage instead of on the two sloping ones: $$25\times 40+\frac{1}{2}\times 25\times 20+25\times 60$$$$=1000+250+1500=2750$$. It is the triangles that carry the half, because only while accelerating or braking is the average speed half of $25{\text{m s}}^{-1}$.
• E. 2000 · Durations paired with the wrong shapes
  The $40\text{s}$ of acceleration was treated as the rectangle and the $20\text{s}$ at constant speed as the first triangle: $$25\times 40+\frac{1}{2}\times 25\times 20$$$$+\frac{1}{2}\times 25\times 60$$$$=1000+250+750=2000$$. The rectangle belongs to the stage at constant speed, which lasts $20\text{s}$, and the first triangle to the $40\text{s}$ of acceleration.

Common Mistake (⚠️):
Remembering that a half belongs somewhere and attaching it to the wrong stage. The half belongs to the two triangles, the $40\text{s}$ of acceleration and the $60\text{s}$ of braking, and never to the $20\text{s}$ at constant speed, whose area is the full rectangle $500\text{m}$.

Takeaway (📌):
On a speed-time graph the distance is the area, so cut the shape into triangles and rectangles and add. A stage rising from rest, or falling to rest, contributes half of the rectangle that boxes it.

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 $3000m$ to do.

Step-by-Step Breakdown:

1. Read the scale

A scale of $1:300000$ means that $1\mathrm{c}\mathrm{m}$ on the chart stands for $300000\mathrm{c}\mathrm{m}$ on the ground, and $300000\mathrm{c}\mathrm{m}=3000m$.

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 $13\mathrm{c}\mathrm{m}$ on the chart as $PW$.

4. Convert the chart length to a ground distance

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

Check the conversion by undoing it: $39000m$ divided by $3000m$ per centimetre returns the $13\mathrm{c}\mathrm{m}$ 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 $13\mathrm{c}\mathrm{m}$ of chart: $13+13=26\mathrm{c}\mathrm{m}$, and $26\times 3000m=78000m=78.0\mathrm{k}\mathrm{m}$. The question asks for the single side $PL$, which the two equal angles at $W$ and $L$ fix at $13\mathrm{c}\mathrm{m}$.
• 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 $1:300000$ is never applied. On the chart $PL$ is $13\mathrm{c}\mathrm{m}$; on the ground each of those centimetres stands for $3000m$.
• D. 3.9 · Centimetres to metres by the wrong power of ten
  Multiplies correctly, $13\times 300000=3900000\mathrm{c}\mathrm{m}$ on the ground, then divides by $1000$ instead of $100$ to reach metres, giving $3900m$ and so $3.9\mathrm{k}\mathrm{m}$. There are $100\mathrm{c}\mathrm{m}$ in a metre, so the ground distance is $39000m$.
• E. 3900.0 · Scale read as centimetres to metres
  Reads the scale as $1\mathrm{c}\mathrm{m}$ to $300000m$, treating the second figure as metres when a ratio scale carries no units at all: $13\times 300000=3900000m$, which is $3900.0\mathrm{k}\mathrm{m}$. The second figure is in the same unit as the first, so $1\mathrm{c}\mathrm{m}$ stands for $300000\mathrm{c}\mathrm{m}$, which is $3000m$.

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 $13\mathrm{c}\mathrm{m}$.

Takeaway (📌):
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 $11\mathrm{c}\mathrm{m}$ and $7\mathrm{c}\mathrm{m}$ and the perpendicular height between them is $6\mathrm{c}\mathrm{m}$, so

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

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

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

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

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

4. Multiply by the length of the prism

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

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

Check by dividing back: $2720÷80=34{\mathrm{c}\mathrm{m}}^2$, 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 $20{\mathrm{c}\mathrm{m}}^2$ 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 $(108-20)\times 80=7040$. The area of 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 $(11+7)\times 6=108{\mathrm{c}\mathrm{m}}^2$ rather than $54{\mathrm{c}\mathrm{m}}^2$, which inflates every figure computed after it and lands on $7040$.

Takeaway (📌):
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: C. 150

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

$$A=\frac{\theta }{360}\pi r^2$$$$=\frac{1}{2}r(\frac{\theta }{360}\times 2\pi r)=\frac{1}{2}r\ell$$

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

2. Solve for the radius

$$60\pi =\frac{1}{2}\times r\times 10\pi \Rightarrow r$$$$=\frac{2\times 60\pi }{10\pi }=\frac{120}{10}=12\text{m}$$

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

$$\frac{10\pi }{24\pi }=\frac{5}{12}$$

of the way round.

4. Turn the fraction into an angle, then check

$$\theta =\frac{5}{12}\times {360}^{\circ }={150}^{\circ }$$

Check against the area, which so far has been used only once: $\frac{5}{12}$ of the disc is $\frac{5}{12}\times \pi (12{)}^2=\frac{5}{12}\times 144\pi =60\pi {\text{m}}^2$, 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 $A=\frac{1}{2}r\ell$, 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 $\frac{60\pi }{288\pi }=\frac{5}{24}$ of the circle, giving $\frac{5}{24}\times {360}^{\circ }={75}^{\circ }$. 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: $A=\frac{1}{2}r\ell$ gives $r=\frac{2\times 60\pi }{10\pi }=12$, 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, $\frac{10\pi }{2\pi (12)}\times {360}^{\circ }={150}^{\circ }$.
• D. 60 · Compared a length with an area
  Divided the arc by the area directly, $\frac{10\pi }{60\pi }=\frac{1}{6}$, and read that as the fraction of a full turn: $\frac{1}{6}\times {360}^{\circ }={60}^{\circ }$. 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, $\frac{10\pi }{60\pi }$, 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 (📌):
$A=\frac{1}{2}r\ell$ converts an arc and an area into the radius in one line, with no angle to carry. With $r$ in hand, the angle is just the arc written as a fraction of $2\pi r$, multiplied by ${360}^{\circ }$.

Correct Answer: A. 228

Fastest Approach (🚀):
Cancel $\pi$ on sight and compare areas rather than rearranging: the full ring is $\pi (400-100)=300\pi$, the sweep covers $190\pi$, so $\theta$ is ${\displaystyle \frac{190}{300}}$ of ${360}^{\circ }$, which is ${228}^{\circ }$.

Step-by-Step Breakdown:

Step 1: Identify the region

The part of the arm that touches the surface reaches out to $R=20\text{cm}$ but starts at $r=10\text{cm}$, so the cleaned region is a sector of radius $20\text{cm}$ with the sector of radius $10\text{cm}$ through the same angle $\theta$ removed.

Step 2: Write its area

$$\text{area}={\displaystyle \frac{\theta }{360}}\pi R^2-{\displaystyle \frac{\theta }{360}}\pi r^2$$$$={\displaystyle \frac{\theta }{360}}\pi ({20}^2-{10}^2)$$

$={\displaystyle \frac{\theta }{360}}\pi (400-100)={\displaystyle \frac{\theta }{360}}\left(300\pi \right)$.

Step 3: Solve for the angle

${\displaystyle \frac{\theta }{360}}\left(300\pi \right)=190\pi$

Cancel $\pi$: ${\displaystyle \frac{\theta }{360}}={\displaystyle \frac{190}{300}}$, so $\theta ={\displaystyle \frac{360\times 190}{300}}={228}^{\circ }$.

Step 4: Check

${\displaystyle \frac{228}{360}}$ of the complete ring is ${\displaystyle \frac{228}{360}}\times 300\pi =190\pi {\text{cm}}^2$, which is the area given.

The key is $228$.

Why the Other Options Are Wrong (❌):

• B. 171 · Region not subtracted
  Forgot to remove the inner sector and used the whole disc of radius $20$: ${\displaystyle \frac{\theta }{360}}\pi \left(400\right)=190\pi$ gives $\theta ={\displaystyle \frac{190\times 360}{400}}={171}^{\circ }$. The part of the arm within $10\text{cm}$ of the pivot touches nothing and cleans no area, so $100$ has to come off $400$ before the angle is found.
• C. 304 · Conceptual Error
  Used the mid radius $15\text{cm}$ as a single sector: ${\displaystyle \frac{\theta }{360}}\pi \left(225\right)=190\pi$ gives $\theta ={\displaystyle \frac{190\times 360}{225}}={304}^{\circ }$. Areas depend on $r^2$, so averaging the radii does not reproduce the ring: ${\left({\displaystyle \frac{20+10}{2}}\right)}^2=225$ is not $400-100=300$.
• D. 114 · Formula misuse
  Divided by $180$ instead of $360$, as though a sweep were a fraction of a half turn: ${\displaystyle \frac{\theta }{180}}\pi \left(300\right)=190\pi$ gives $\theta ={\displaystyle \frac{190\times 180}{300}}={114}^{\circ }$. A complete ring corresponds to a full turn of ${360}^{\circ }$, so this is exactly half the true angle.
• E. 132 · Complement taken
  Subtracted the given area from the complete ring and worked with the remainder: the whole ring is $\pi (400-100)=300\pi$, and $300\pi -190\pi =110\pi$ is the part NOT cleaned, so ${\displaystyle \frac{110}{300}}\times 360={132}^{\circ }$ is the angle of the unswept part of the ring, not the angle of the sweep.

Common Mistake (⚠️):
Subtracting the radii and then squaring, using ${(20-10)}^2=100$ in place of ${20}^2-{10}^2=300$. Those are not the same number, and only the difference of the squares comes from subtracting one sector area from the other.

Takeaway (📌):
A swept ring is one sector minus another through the SAME angle: area $={\displaystyle \frac{\theta }{360}}\pi (R^2-r^2)$. Difference the squares, never the radii.