ESAT Practice Set 1B Mathematics Worked Solutions

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 $25\ \text{m s}^{-1}$: $\tfrac{1}{2} \times 140 \times 25 = 1750\ \text{m}$ 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 $40\ \text{s}$:

2. Distance at constant speed

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

3. Distance while braking

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

4. Add the three areas

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 $\tfrac{1}{2} \times (120 + 20) \times 25 = 1750\ \text{m}$.

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 $120\ \text{s}$: $\tfrac{1}{2} \times 120 \times 25 = 1500$. 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: $25 \times 40 + \tfrac{1}{2} \times 25 \times 20 + 25 \times 60 = 1000 + 250 + 1500 = 2750$. 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: $25 \times 40 + \tfrac{1}{2} \times 25 \times 20 + \tfrac{1}{2} \times 25 \times 60 = 1000 + 250 + 750 = 2000$. 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 (📌):
The Cheat Code: 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: 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, $\tfrac{4}{3}$, so the horizontal edge is $36 \times \tfrac{3}{4} = 27\,\mathrm{cm}$.

2. Find the first area

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

3. Read off the length scale factor

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

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 $k^2$: $486 \times 4 = 1944\,\mathrm{cm^2}$.

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

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
  $\tfrac{1}{2} \times 36 \times 27 = 486$ 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 $36 \times \tfrac{4}{3} = 48\,\mathrm{cm}$, its area $\tfrac{1}{2} \times 36 \times 48 = 864\,\mathrm{cm^2}$, 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 (📌):
The Cheat Code: 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 $\tfrac{1}{2}$, and moving from $(2,7)$ to the $y$ axis changes $x$ by $-2$, so $y$ changes by $\tfrac{1}{2}\times(-2) = -1$ 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

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

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

3. Fit the cable duct through the crossing point

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

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

4. Read off the crossing and check

The cable duct is $y = \tfrac{1}{2}x + c$ 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 $\dfrac{1}{2} = \tfrac{1}{2}$ 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 $\tfrac{1}{2}$ 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 parallel to the gas main, so it never crosses it at right angles.
• B. 8 · Sign not changed
  The gradient was inverted without being negated, using $-\tfrac{1}{2}$: $7 = -\tfrac{1}{2}(2) + c$ gives $c = 7 - (-1) = 8$. Since $(-2)\times\left(-\tfrac{1}{2}\right) = 1$, not $-1$, that line is not perpendicular to the gas main.
• D. -4 · Wrong point substituted
  The perpendicular gradient $\tfrac{1}{2}$ was found correctly, then the cable duct was fitted through the other point $(6,-1)$ instead of the crossing point: $-1 = \tfrac{1}{2}(6) + c$ 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 $y = \tfrac{1}{2}x + c$ with $c = 6$ was found correctly, then $y$ was set to $0$ rather than $x$: $\tfrac{1}{2}x + (6) = 0$ 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 $\tfrac{1}{2}$ multiply to $-1$: pairing $-2$ with $2$ gives $-4$ and pairing $-2$ with $-\tfrac{1}{2}$ gives $1$, so neither pair is perpendicular.

Takeaway (📌):
The Cheat Code: 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. 228

Fastest Approach (🚀):
Cancel $\pi$ on sight and compare areas rather than rearranging: the full ring is $\pi\left(400 - 100\right) = 300\pi$, the sweep covers $190\pi$, so $\theta$ is $\dfrac{190}{300}$ 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

$\text{area} = \dfrac{\theta}{360}\pi R^2 - \dfrac{\theta}{360}\pi r^2 = \dfrac{\theta}{360}\pi\left(20^2 - 10^2\right)$

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

Step 3: Solve for the angle

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

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

Step 4: Check

$\dfrac{228}{360}$ of the complete ring is $\dfrac{228}{360} \times 300\pi = 190\pi\ \text{cm}^2$, 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 $20$: $\dfrac{\theta}{360}\pi\left(400\right) = 190\pi$ gives $\theta = \dfrac{190 \times 360}{400} = 171^\circ$. 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: $\dfrac{\theta}{360}\pi\left(225\right) = 190\pi$ gives $\theta = \dfrac{190 \times 360}{225} = 304^\circ$. Areas depend on $r^2$, so averaging the radii does not reproduce the ring: $\left(\dfrac{20 + 10}{2}\right)^2 = 225$ 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: $\dfrac{\theta}{180}\pi\left(300\right) = 190\pi$ gives $\theta = \dfrac{190 \times 180}{300} = 114^\circ$. 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 $\pi\left(400 - 100\right) = 300\pi$, and $300\pi - 190\pi = 110\pi$ is the part NOT cleaned, so $\dfrac{110}{300} \times 360 = 132^\circ$ 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 $\left(20 - 10\right)^2 = 100$ 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 (📌):
The Cheat Code: A swept ring is one sector minus another through the SAME angle: area $= \dfrac{\theta}{360}\pi\left(R^2 - r^2\right)$. Difference the squares, never the radii.

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

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 $t=20$, a stretch of

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 $t=11\pm9$ 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: $2\times9=18$ hours.
• C. 11 · Turning point time reported
  The time of the minimum was given, from $-\dfrac{b}{2a}=\dfrac{22}{2}=11$. 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 $t=11\pm9$, 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 (📌):
The Cheat Code: 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.