ESAT Paper 4 Maths Worked Solutions

Correct Answer: A. $\sqrt{\frac{1}{(10 - \sqrt{b})^2} + \frac{b^2}{2}}$

Fastest Approach (🚀):
Diagonal of base = $\sqrt{b^2+b^2} = b\sqrt{2}$. Half diagonal = $b\frac{\sqrt{2}}{2} = \frac{b}{\sqrt{2}}$. Pythagoras: $l^2 = h^2 + (\text{half-diagonal})^2 = (\frac{1}{10-\sqrt{b}})^2 + (\frac{b}{\sqrt{2}})^2$. Taking the square root gives option A immediately.

Step-by-Step Breakdown:

• The problem describes a square-based pyramid with a base dimension of $b$ and a perpendicular height of $\frac{1}{10-\sqrt{b}}$.
• Visualizing the Geometry: Imagine dropping a plumb line straight down from the apex to the center of the square base. This line forms the perpendicular height. A line from the center of the base out to any of the four corners forms the base of a right-angled triangle, and the slant edge is the hypotenuse connecting the corner back to the apex.
• To find the slant edge length (often denoted as $l$), we must first determine the horizontal distance from the center of the square base to one of its corners.
• The diagonal of a square with side $b$ is $b\sqrt{2}$, so the distance from the center to a corner is half of that: $\frac{b\sqrt{2}}{2}$.
• Applying the Pythagorean theorem to our 3D triangle gives $l^2 = h^2 + \left(\frac{b\sqrt{2}}{2}\right)^2$.
• Substituting the given height yields $l^2 = \left(\frac{1}{10-\sqrt{b}}\right)^2 + \frac{2b^2}{4}$, which simplifies to $l^2 = \frac{1}{(10-\sqrt{b})^2} + \frac{b^2}{2}$.
• Taking the square root gives $l = \sqrt{\frac{1}{(10-\sqrt{b})^2} + \frac{b^2}{2}}$.
• Correct Answer: Option A.

Common Mistake (⚠️):
Confusing the half-diagonal with the distance from the center to the midpoint of an edge (which would be $b/2$), or squaring the expressions incorrectly.

Correct Answer: D. 84

Fastest Approach (🚀):
Time to work: $T_1 = 5/6\text{ h}$. Distance $d = x \times (5/6) = 5x/6$.
Return trip: Speed = $60\text{ km/h}$. Distance is $d$. Time $T_2 = d / 60$.
Total Time = $5/6 + d/60$.
Average Speed = $2d / (5/6 + d/60) = 70$.
Divide by $d$: $2 / (5/(6d) + 1/60) = 70$.
$5/(6d) + 1/60 = 2/70 = 1/35$.
$5/(6d) = 1/35 - 1/60 = (12-7)/420 = 5/420 = 1/84$.
So $6d = 5 \times 84 = 420 \implies d = 70\text{ km}$.
Since $d = 5x/6$, we have $70 = 5x/6 \implies x = 70 \times 6 / 5 = 84\text{ km/h}$.

Step-by-Step Breakdown:

• Part 1: The Outbound Trip

James travels to work at a speed of $x$ km/h for a duration of 50 minutes. Converting 50 minutes to hours gives $\frac{5}{6}$ hours. Thus, the one-way distance $D$ is $x \times \frac{5}{6} = \frac{5x}{6}$ km.

• Part 2: The Return Trip

On the return trip, he travels the same distance $D$ at 60 km/h. The time taken for the return trip in hours is $t_{\text{return}} = \frac{D}{60} = \frac{5x/6}{60} = \frac{x}{72}$ hours.

• His average speed for the entire round trip is given as 70 km/h.
• Average speed is the total distance ($2D$) divided by the total time ($t_{\text{outbound}} + t_{\text{return}}$): $\frac{2(5x/6)}{5/6 + x/72} = 70$.
• Simplifying the fraction: $\frac{5x/3}{(60+x)/72} = 70 \implies \frac{120x}{60+x} = 70$.
• Multiplying both sides by the denominator gives $120x = 4200 + 70x$.
• Solving for $x$ yields $50x = 4200 \implies x = 84$.
• Correct Answer: Option D.

Common Mistake (⚠️):
Assuming average speed is the arithmetic mean of the two speeds ($(x + 60)/2 = 70 \implies x = 80$), which is only true if the times are equal, not the distances.

Correct Answer: D. $(2x)!$

Fastest Approach (🚀):
Take $\ln$ and use $\ln(N!) \approx N \ln N$:
A: $\ln(x^x) = x \ln x$
B: $\ln((x^2)!) \approx x^2 \ln(x^2) = 2x^2 \ln x$ (Largest! Dominates by a factor of $x$)
C: $\ln((x!)^2) = 2 \ln(x!) \approx 2x \ln x - 2x$
D: $\ln((2x)!) \approx 2x \ln(2x) = 2x \ln x + 2x \ln 2$
Ranking from largest: B $\gg$ D $>$ C $>$ A.
The second largest is D.

Step-by-Step Breakdown:

• We are asked to determine which of the expressions $x^x$, $(x^2)!$, $(x!)^2$, and $(2x)!$ is the second largest for large values of $x$.
• To formalize this, we can consider the limit of the ratios of these terms as $x \to \infty$. A function $f(x)$ grows faster than $g(x)$ if $\lim_{x \to \infty} \frac{f(x)}{g(x)} = \infty$.
• By using Stirling's approximation ($n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n$), factorial functions grow much faster than standard exponential functions.
• The expression $(x^2)!$ has the largest argument inside the factorial (scaling as $x^{2x^2}$), making it by far the largest term.
• Next, we compare $(2x)!$ and $(x!)^2$. Using the ratio test conceptually, $(2x)!$ expands to $(2x)(2x-1)\dots(x+1) \times x!$, which contains $x$ terms all strictly greater than $x$. Thus, $(2x)!$ is vastly larger than $(x!) \times (x!)$.
• More formally, $(2x)! \approx \left(\frac{2x}{e}\right)^{2x} = 4^x \left(\frac{x}{e}\right)^{2x}$, whereas $(x!)^2 \approx \left(\frac{x}{e}\right)^{2x}$.
• The expression $x^x$ is the smallest.
• The descending order of growth is $(x^2)! > (2x)! > (x!)^2 > x^x$, making $(2x)!$ the second largest.
• Correct Answer: Option D.

Common Mistake (⚠️):
Guessing that $(x^2)!$ and $(2x)!$ are similar, or confusing $(x!)^2$ with $(x^2)!$. Missing the 'second largest' specification in the prompt and picking B.

Correct Answer: D. $-\frac{15}{2} \leq x \leq \frac{3}{2}$

Fastest Approach (🚀):
Since it's $\leq$, the solution must use $\leq$ (closed interval), instantly eliminating A, B, and C. Option D is the only possible answer format.
To verify: $4x^2 + 24x - 45 \leq 0$. Factors to $(2x - 3)(2x + 15) \leq 0$. Roots are $3/2$ and $-15/2$. Region is $[-15/2, 3/2]$.

Step-by-Step Breakdown:

• The provided inequality is $x^2 + 6x \le \frac{45}{4}$.
• Multiplying the entire inequality by 4 to remove the fraction gives $4x^2 + 24x - 45 \le 0$.
• We can factorize this quadratic equation into $(2x + 15)(2x - 3) \le 0$.
• Setting the factors to zero gives the roots $x = -\frac{15}{2}$ and $x = \frac{3}{2}$.
• Visualizing the curve: If you quickly sketch the quadratic curve $y = 4x^2 + 24x - 45$, it forms a U-shape (a positive parabola) intersecting the x-axis at $-7.5$ and $1.5$.
• We are looking for where the curve is $\le 0$, which corresponds to the region below or on the x-axis.
• This region lies exactly between the two roots: $-7.5 \le x \le 1.5$.
• Correct Answer: Option D.

Common Mistake (⚠️):
Solving the roots correctly but choosing the wrong inequality signs (e.g. strict inequalities $<$ when the prompt uses $\leq$).

Correct Answer: C. 5

Fastest Approach (🚀):
We want $(3/4)^n < 1/4 = 0.25$.
$n=1: 3/4 = 0.75$
$n=2: 9/16 \approx 0.56$
$n=3: 27/64 \approx 0.42$
$n=4: 81/256 \approx 0.32$ (Note $81 \times 4 = 324 > 256$, so $> 1/4$)
$n=5: 243/1024 \approx 0.237$ (Note $243 \times 4 = 972 < 1024$, so $< 1/4$)
Day 5 is the first day it drops below $25\%$.

Step-by-Step Breakdown:
A pack of frozen peas costs 2.50 on the starting day (day 0).
Each subsequent day, the price is reduced by 25%, meaning the new price is multiplied by 0.75.

• The peas are disposed of once the price drops below 25% of the starting value. We must solve for $n$ in the inequality $0.75^n < 0.25$.
• Logarithmic Method: Instead of calculating manually day-by-day, we can take the logarithm of both sides.
• $\log(0.75^n) < \log(0.25) \implies n \log(0.75) < \log(0.25)$.
• Since $\log(0.75)$ is negative, dividing both sides by it flips the inequality sign: $n > \frac{\log(0.25)}{\log(0.75)}$.
• Note that $0.75 = 3/4$ and $0.25 = 1/4$. Thus, $n > \frac{\log(1) - \log(4)}{\log(3) - \log(4)} = \frac{-0.602}{0.477 - 0.602} = \frac{-0.602}{-0.125} = 4.816$.
• Since $n$ must be an integer, the first day that satisfies this is day 5.
• Correct Answer: Option C.

Common Mistake (⚠️):
Wasting time calculating exact monetary values (e.g., $£2.50 \times 0.75 = £1.875$) and making arithmetic errors, rather than working with simple fractions.

Correct Answer: A. $a$ increases by 44%

Fastest Approach (🚀):
New $x, y, z$ are $0.6 \times$ original. Squared, they contribute a factor of $0.6^2 = 0.36$. New $k$ is $2 \times$ original. Squared, it contributes a factor of $2^2 = 4$. Overall multiplier = $0.36 \times 4 = 1.44$. A multiplier of $1.44$ represents a $44\%$ increase.

Step-by-Step Breakdown:

• We are given the equation $a = (x^2 + y^2 + z^2)k^2$.
• The value of $k$ is doubled, meaning $k$ becomes $2k$.
• The variables $x, y$, and $z$ are decreased by 40%. Decreasing by 40% means retaining 60%, yielding a multiplier of 0.6.
• Substituting these changes gives the new value: $a_{\text{new}} = ((0.6x)^2 + (0.6y)^2 + (0.6z)^2)(2k)^2$.
• Factoring out the constants gives $a_{\text{new}} = 0.36(x^2 + y^2 + z^2) \times 4k^2$.
• Simplifying this yields $a_{\text{new}} = 1.44 \times (x^2 + y^2 + z^2)k^2 = 1.44a$.
• A multiplier of $1.44$ represents a 44% increase over the original value.
• Correct Answer: Option A.

Common Mistake (⚠️):
Misinterpreting 'decreased by 40%' as 'multiplied by 0.4', which would give $0.16 \times 4 = 0.64$ (a $36\%$ decrease). Or, forgetting to square the multipliers.

Correct Answer: B. 328/625

Fastest Approach (🚀):
$p = 2/5$, $q = 3/5$.
$P(0) = (3/5)^4 = 81/625$.
$P(1) = 4 \times (2/5)^1 \times (3/5)^3 = 4 \times (2/5) \times (27/125) = 216/625$.
$P(\geq 2) = 1 - (81 + 216)/625 = 1 - 297/625 = 328/625$. Option B.

Step-by-Step Breakdown:

• The probability of rain on any given day is 40%, or $p = 0.4$. The probability of no rain is $q = 0.6$. The number of trials is $n = 4$.
• We want the probability of at least two rainy days: $P(X \ge 2) = 1 - P(X=0) - P(X=1)$.
• Formal Binomial Formula: The probability of exactly $k$ successes in $n$ trials is $P(X=k) = \binom{n}{k} p^k q^{n-k}$.
• For 0 rainy days: $P(X=0) = \binom{4}{0} (0.4)^0 (0.6)^4 = 1 \times 1 \times 0.1296 = 0.1296$.
• For exactly 1 rainy day: $P(X=1) = \binom{4}{1} (0.4)^1 (0.6)^3 = 4 \times 0.4 \times 0.216 = 0.3456$.
• Subtracting these from 1 gives $P(X \ge 2) = 1 - 0.1296 - 0.3456 = 0.5248$.
• Converting the decimal $0.5248$ to a fraction yields $\frac{5248}{10000}$, which simplifies to $\frac{328}{625}$.
• Correct Answer: Option B.

Common Mistake (⚠️):
Forgetting the multiplier $\binom{4}{1} = 4$ for $P(1)$, calculating it simply as $(2/5)(3/5)^3 = 54/625$, which leads to $1 - (81+54)/625 = 490/625$.

Correct Answer: A. 1/5

Fastest Approach (🚀):
Ways to get $\geq 9$:
9: (3,6), (4,5), (5,4), (6,3) $\rightarrow 4$ ways.
10: (4,6), (5,5), (6,4) $\rightarrow 3$ ways.
11: (5,6), (6,5) $\rightarrow 2$ ways.
12: (6,6) $\rightarrow 1$ way.
Total = $4+3+2+1 = 10$ possible outcomes.
Ways to get 11 = 2.
Probability = $2/10 = 1/5$. Option A.

Step-by-Step Breakdown:

• Two fair six-sided dice are rolled. The condition is that the total score is 9 or greater.
• Sample Space Grid: Imagine a $6 \times 6$ grid for the dice outcomes. Only the bottom-right triangular corner of this grid contains sums $\ge 9$.
• The valid pairs in this sub-grid are:
• Sum of 9: (3,6), (4,5), (5,4), (6,3)
• Sum of 10: (4,6), (5,5), (6,4)
• Sum of 11: (5,6), (6,5)
• Sum of 12: (6,6)

Counting them up, there are exactly 10 valid outcomes in our restricted sample space.
We must find the conditional probability that the score is exactly 11.

• Looking at our list, the only outcomes that sum to 11 are (5,6) and (6,5) — giving 2 target outcomes.
• Therefore, the probability is $\frac{2}{10}$, which simplifies to $\frac{1}{5}$.
• Correct Answer: Option A.

Common Mistake (⚠️):
Calculating the regular probability of scoring 11 out of all 36 possible outcomes ($2/36 = 1/18$) by ignoring the 'Given that' condition.

Correct Answer: D. 9/58

Fastest Approach (🚀):
Assume 1000 people.
$2\%$ have condition $\rightarrow 20$ sick, $980$ healthy.
Sick get tested: $10\%$ false means $90\%$ correct $\rightarrow 20 \times 0.90 = 18$ test Positive.
Healthy get tested: $10\%$ false means $10\%$ wrong $\rightarrow 980 \times 0.10 = 98$ test Positive.
Total Positives = $18 + 98 = 116$.
Probability actually sick = $18 / 116 = 9 / 58$. Option D.

Step-by-Step Breakdown:

• A medical condition affects 2 in every 100 people ($P(D) = 0.02$). The diagnostic test has a 10% false rate ($P(+|\text{No } D) = 0.10$ and $P(-|D) = 0.10$, meaning $P(+|D) = 0.90$).
• Probability Tree Approach: Let's trace a population of 1000 people to make it intuitive.
• Branch 1 (Condition): 20 people have the disease, 980 people do not.
• Branch 2 (Test for the 20 with disease): 90% test positive = 18 true positives.
• Branch 3 (Test for the 980 without disease): 10% test positive = 98 false positives.
• We are given that a person tests positive. The total pool of positive results is $18 \text{ (true)} + 98 \text{ (false)} = 116$ people.
• The probability they actually have the condition is the true positives over the total positives: $\frac{18}{116}$.
• Simplifying the fraction $\frac{18}{116}$ yields $\frac{9}{58}$.
• Correct Answer: Option D.

Common Mistake (⚠️):
Calculating $P(\text{Positive} \mid \text{Condition})$ which is $90\%$, or multiplying $0.90 \times 0.02 = 0.018$ (which is just the probability of getting a true positive out of the whole population).

Correct Answer: B. $4R^2(2 + \sqrt{3})$

Fastest Approach (🚀):
Width = $R + 2R + R = 4R$.
Height = top radius ($R$) + height of equilateral triangle between centers + bottom radius ($R$).
Equilateral triangle side = $2R$. Height = $\sqrt{(2R)^2 - R^2} = \sqrt{3R^2} = R\sqrt{3}$.
Total height = $R + R\sqrt{3} + R = R(2 + \sqrt{3})$.
Area = Width $\times$ Height = $4R \times R(2 + \sqrt{3}) = 4R^2(2 + \sqrt{3})$. Option B.

Step-by-Step Breakdown:

• The problem involves three equal circles of radius $r$ packed tangent to each other inside a rectangle.
• Equilateral Triangle Concept: When three identical circles are mutually tangent, their centers form a perfect equilateral triangle. The length of each side of this triangle is exactly $2r$ (the sum of the radii of two touching circles).
• The height of an equilateral triangle of side length $2r$ is $\sqrt{(2r)^2 - r^2} = r\sqrt{3}$.
• The total height of the rectangle containing them must cover the radius of the top circle ($r$), the vertical height of the triangle ($r\sqrt{3}$), and the radius of the bottom circles ($r$).
• Total height = $r + r\sqrt{3} + r = r(2 + \sqrt{3})$.
• The total width of the rectangle is simply the diameter of two side-by-side circles: $4r$.
• The area of the rectangle is $\text{Width} \times \text{Height} = 4r \times r(2 + \sqrt{3}) = 4r^2(2 + \sqrt{3})$.
• Correct Answer: Option B.

Common Mistake (⚠️):
Assuming the height of the rectangle is $4R$ (thinking the bottom circle sits squarely beneath the top ones as a perfect square of $4$ circles, giving an area of $16R^2$).

Correct Answer: D. Reflection about the y-axis

Fastest Approach (🚀):
Track vertices A and B.
Initially: A is at top-left, B is at top-right. The line AD passes through A and the origin.
Reflect across AD: A stays at A. B (top-right) swaps with F (left).
Rotate $60^{\circ}$ clockwise: A (top-left) moves to top-right (original position of B). The new B (which was F on the left) moves to top-left (original position of A).
So, A and B have simply swapped places horizontally. C and F, and D and E also swap horizontally. This is exactly a reflection across the y-axis.
To restore it, apply the inverse: another reflection across the y-axis. Option D.

Step-by-Step Breakdown:

• Setting up. Label the hexagon $ABCDEF$ centred on the origin, with $AD$ a long diagonal through the centre.
• Step 1 — Reflect in the line $AD$. A reflection in a line through the centre fixes the two vertices on that line ($A$ and $D$) and swaps the vertices on either side: $B \leftrightarrow F$ and $C \leftrightarrow E$.
• Step 2 — Rotate $60^{\circ}$ clockwise about the origin. A regular hexagon has rotational symmetry of order 6, so a $60^{\circ}$ turn advances every vertex by exactly one position around the figure.
• Composing the two. Track a single vertex to see the net effect. $A$ is fixed by the reflection, then the rotation carries it to the position originally held by $B$. Meanwhile $B$ is sent to $F$ by the reflection, and the rotation then carries it to the position originally held by $A$.
• So $A$ and $B$ have exchanged positions, and the same argument applied to the remaining vertices shows $C \leftrightarrow F$ and $D \leftrightarrow E$. Every vertex has been swapped with its mirror image in the vertical axis.
• Identifying the net transformation. A reflection composed with a rotation is itself a reflection, and the vertex correspondence above is exactly the map $(x, y) \mapsto (-x, y)$ — a reflection in the $y$-axis.
• Undoing it. A reflection is a self-inverse transformation: applying it twice returns every point to where it started. Therefore the original configuration is restored by a second reflection about the $y$-axis.
• Correct Answer: Option D.

Common Mistake (⚠️):
Applying the transformations in the wrong order, or not realizing that the inverse of a single reflection is just the reflection itself.

Correct Answer: E. Only 2

Fastest Approach (🚀):
1: CS $\implies$ Maths.
2: Eng $\implies$ Phys.
3: Not Phys $\implies$ Not Maths.
Contrapositive of 3: Maths $\implies$ Phys.
Chain them: CS $\implies$ Maths $\implies$ Phys.
So, CS $\implies$ Phys. This is statement 2.
Eng $\implies$ Phys is isolated; it doesn't chain backwards to Maths or CS. Thus, none of the other statements (which involve Eng $\implies$ Maths, CS $\implies$ Eng, etc.) are necessary.

Step-by-Step Breakdown:

• Translate each statement into an implication.
• Statement 1: $\text{CS} \implies \text{Maths}$
• Statement 2: $\text{Eng} \implies \text{Physics}$
• Statement 3: $\lnot\text{Physics} \implies \lnot\text{Maths}$
• Take the contrapositive of statement 3. The contrapositive of $\lnot Q \implies \lnot P$ is $P \implies Q$, and it is logically equivalent to the original. So statement 3 is equivalent to $\text{Maths} \implies \text{Physics}$.
• Chain what can be chained. Statement 1 ends where the rewritten statement 3 begins, so they link:

$$\text{CS} \implies \text{Maths} \implies \text{Physics}$$
giving $\text{CS} \implies \text{Physics}$. That is exactly proposed consequence 2, so consequence 2 is necessary.

• Check the remaining proposals.
• 1 ($\text{Eng} \implies \text{Maths}$): statement 2 gives $\text{Eng} \implies \text{Physics}$, but implications cannot be run backwards — knowing someone likes Physics tells us nothing about Maths. Not necessary.
• 3 ($\text{Maths} \implies \lnot\text{Eng}$): nothing in the premises forbids a person from liking both. Not necessary.
• 4 ($\text{Eng} \implies \lnot\text{CS}$): again nothing excludes someone liking both subjects. Not necessary.
• 5 ($\text{CS} \implies \text{Eng}$): there is no premise with Engineering as its conclusion, so this cannot be derived. Not necessary.

Only proposal 2 follows of necessity.
Correct Answer: Option E.

Common Mistake (⚠️):
Assuming the converse is true (e.g., since Maths $\implies$ Phys, thinking Phys $\implies$ Maths), which would incorrectly validate statement 1.

Correct Answer: B. $a^2(\frac{\pi}{3} - \frac{\sqrt{3}}{4})$

Fastest Approach (🚀):
Sector angle is $2\pi/3$.
Area of sector = $\frac{1}{2}a^2(\frac{2\pi}{3}) = \frac{\pi a^2}{3}$.
Area of triangle = $\frac{1}{2}a^2 \sin(\frac{2\pi}{3}) = \frac{1}{2}a^2(\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}a^2}{4}$.
Area of segment = Sector - Triangle = $\frac{\pi a^2}{3} - \frac{\sqrt{3}a^2}{4} = a^2(\frac{\pi}{3} - \frac{\sqrt{3}}{4})$. Option B.

Step-by-Step Breakdown:

• Identify the shaded region. The shaded area is a circular segment: the region cut off by a chord, bounded by that chord and the arc it subtends. The circle has radius $a$ and the sector angle marked in the diagram is $\theta = \frac{2\pi}{3}$ (that is, $120^{\circ}$).
• The segment formula. A segment is what remains when the triangle formed by the two radii and the chord is removed from the sector:

$$A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}}$$

• Area of the sector. For a sector of radius $r$ and angle $\theta$ measured in radians,

$$A_{\text{sector}} = \tfrac{1}{2} r^2 \theta = \tfrac{1}{2} a^2 \left(\tfrac{2\pi}{3}\right) = \frac{\pi a^2}{3}$$

• Area of the triangle. The triangle has two sides equal to the radius $a$ with the angle $\theta$ between them, so use $A = \tfrac{1}{2}ab\sin C$:

$$A_{\text{triangle}} = \tfrac{1}{2} a^2 \sin\left(\tfrac{2\pi}{3}\right) = \tfrac{1}{2} a^2 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}\,a^2}{4}$$
Note $\sin\left(\tfrac{2\pi}{3}\right) = \sin\left(\tfrac{\pi}{3}\right) = \tfrac{\sqrt{3}}{2}$, since $\tfrac{2\pi}{3}$ lies in the second quadrant where sine is positive.

• Subtract.

$$A_{\text{segment}} = \frac{\pi a^2}{3} - \frac{\sqrt{3}\,a^2}{4} = a^2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)$$

• Correct Answer: Option B.

Common Mistake (⚠️):
Using $\sin(2\pi/3) = 1/2$ instead of $\sqrt{3}/2$, or forgetting the $1/2$ in the triangle area formula.

Correct Answer: D. $a(\frac{2\pi}{3} + \frac{\sqrt{3}}{4})$ horizontally, $-\frac{3a}{4}$ vertically

Fastest Approach (🚀):
Initial dot position relative to center: $(0, a/2)$.
Translation of center after $1/3$ rev: $\Delta x_c = \frac{1}{3} \times 2\pi a = \frac{2\pi a}{3}$.
Rotation of dot: $1/3$ rev clockwise is $-120^{\circ}$. New rel position: $(x,y) = (\frac{a}{2}\sin(120^{\circ}), \frac{a}{2}\cos(120^{\circ})) = (\frac{a\sqrt{3}}{4}, -\frac{a}{4})$.
Final absolute position: $x = \frac{2\pi a}{3} + \frac{a\sqrt{3}}{4}$, $y = a - \frac{a}{4} = \frac{3a}{4}$.
Initial absolute position: $x=0$, $y = a + \frac{a}{2} = \frac{3a}{2}$.
Displacement: $\Delta x = \frac{2\pi a}{3} + \frac{a\sqrt{3}}{4}$, $\Delta y = \frac{3a}{4} - \frac{3a}{2} = -\frac{3a}{4}$. Option D.

Step-by-Step Breakdown:

• Set up coordinates. Let the circle have radius $a$ and roll clockwise along a horizontal surface. The dot is marked halfway between the centre and the top of the circle, so relative to the centre it starts at $\left(0, \tfrac{a}{2}\right)$. The centre starts at height $a$, so the dot's initial absolute position is $\left(0, \tfrac{3a}{2}\right)$.
• Step 1 — Translation of the centre. Rolling without slipping means the horizontal distance covered equals the arc length unrolled. For $\tfrac{1}{3}$ of a revolution:

$$\Delta x_{\text{centre}} = \tfrac{1}{3}(2\pi a) = \frac{2\pi a}{3}$$
The centre stays at constant height $a$ throughout.

• Step 2 — Rotation of the dot about the centre. One third of a revolution clockwise is a turn of $120^{\circ}$. Applying a clockwise rotation of $120^{\circ}$ to the offset vector $\left(0, \tfrac{a}{2}\right)$:

$$\left(\tfrac{a}{2}\sin 120^{\circ},\; \tfrac{a}{2}\cos 120^{\circ}\right) = \left(\frac{a\sqrt{3}}{4},\; -\frac{a}{4}\right)$$

• Step 3 — Combine translation and rotation. The dot's final absolute position is the new centre position plus the new offset:

$$x_{\text{final}} = \frac{2\pi a}{3} + \frac{a\sqrt{3}}{4}, \qquad y_{\text{final}} = a - \frac{a}{4} = \frac{3a}{4}$$

• Step 4 — Take the displacement. The question asks for the position relative to the starting position, so subtract the initial coordinates $\left(0, \tfrac{3a}{2}\right)$:

$$\Delta x = a\left(\frac{2\pi}{3} + \frac{\sqrt{3}}{4}\right), \qquad \Delta y = \frac{3a}{4} - \frac{3a}{2} = -\frac{3a}{4}$$

• Correct Answer: Option D.

Common Mistake (⚠️):
Selecting option A or B by miscalculating the vertical displacement (forgetting to subtract the initial height of $3a/2$ and just giving the new relative y-coordinate).

Correct Answer: D. 60/11

Fastest Approach (🚀):
Minute hand speed = $360^{\circ}/60\text{ min} = 6^{\circ}/\text{min}$.
Hour hand speed = $360^{\circ}/(12\times60)\text{ min} = 0.5^{\circ}/\text{min}$.
Relative speed = $6 - 0.5 = 5.5 = 11/2 ^{\circ}/\text{min}$.
Initial gap at 1:00 = $1/12$ of $360^{\circ} = 30^{\circ}$.
Time to catch up = Distance / Relative Speed = $30 / (11/2) = 60 / 11$ minutes. Option D.

Step-by-Step Breakdown:

• Angular speeds of the two hands. The minute hand sweeps a full $360^{\circ}$ in 60 minutes, so it moves at $6^{\circ}$ per minute. The hour hand sweeps $360^{\circ}$ in 12 hours, that is in 720 minutes, so it moves at $0.5^{\circ}$ per minute.
• Relative speed. Because both hands travel in the same direction, treat this as a catch-up problem and use the difference:

$$6 - 0.5 = 5.5 = \frac{11}{2}\ \text{degrees per minute}$$

• The initial gap. At exactly 1 pm the minute hand points at 12 and the hour hand points at 1. One hour mark is $\tfrac{360^{\circ}}{12} = 30^{\circ}$, so the minute hand starts $30^{\circ}$ behind.
• Time to close the gap. The hands meet when the minute hand has made up the whole $30^{\circ}$:

$$t = \frac{\text{gap}}{\text{relative speed}} = \frac{30}{11/2} = \frac{60}{11}\ \text{minutes}$$

• Sense check. $\tfrac{60}{11} \approx 5.45$ minutes, slightly more than the 5 minutes it takes the minute hand to reach the 1 — which is right, because the hour hand has crept forward in the meantime.
• Correct Answer: Option D.

Common Mistake (⚠️):
Thinking the minute hand just has to move to the '1' (which takes 5 minutes), ignoring that the hour hand has moved forward during those 5 minutes.

Correct Answer: C. 1/2

Fastest Approach (🚀):
Probability of picking any coin is $1/3$.
$P(H) = P(H|F_1)\cdot 1/3 + P(H|F_2)\cdot 1/3 + P(H|TH)\cdot 1/3 = (1/2)(1/3) + (1/2)(1/3) + (1)(1/3) = 1/6 + 1/6 + 1/3 = 4/6 = 2/3$.
The part coming from the two-headed coin is $P(H \cap TH) = 1 \times 1/3 = 1/3$.
So $P(TH | H) = (1/3) / (2/3) = 1/2$. Option C.

Step-by-Step Breakdown:

• Define the events. Let $TH$ be the event that the two-headed coin was chosen and $H$ the event that the toss showed heads. The box holds 3 coins, so each is equally likely to be picked: $P(\text{any given coin}) = \tfrac{1}{3}$.
• Conditional probabilities of heads. A fair coin shows heads with probability $\tfrac{1}{2}$; the two-headed coin shows heads with probability $1$.
• Total probability of heads. Sum over the three equally likely coins:

$$P(H) = \tfrac{1}{2}\cdot\tfrac{1}{3} + \tfrac{1}{2}\cdot\tfrac{1}{3} + 1\cdot\tfrac{1}{3} = \tfrac{1}{6} + \tfrac{1}{6} + \tfrac{1}{3} = \tfrac{2}{3}$$

• The joint probability. The chance of both picking the two-headed coin and seeing heads is

$$P(TH \cap H) = 1 \times \tfrac{1}{3} = \tfrac{1}{3}$$

• Apply Bayes' theorem.

$$P(TH \mid H) = \frac{P(TH \cap H)}{P(H)} = \frac{1/3}{2/3} = \frac{1}{2}$$

• Why it is not $\tfrac{1}{3}$. Seeing heads is evidence: it is twice as likely under the two-headed coin as under a fair one, so it lifts the probability from the prior $\tfrac{1}{3}$ up to $\tfrac{1}{2}$.
• Correct Answer: Option C.

Common Mistake (⚠️):
Answering 1/3, which is the prior probability of picking the coin before the coin toss outcome (Heads) is known.

Correct Answer: C. 15/16

Fastest Approach (🚀):
Total possible outcomes for 5 coins = $2^5 = 32$.
Outcomes with NO adjacent identical flips are strictly alternating: HTHTH and THTHT. There are exactly 2 such outcomes.
Probability of alternating = $2 / 32 = 1 / 16$.
Probability of at least two in a row = $1 - 1/16 = 15/16$. Option C.

Step-by-Step Breakdown:
We want the probability that, somewhere among 5 coin tosses, two consecutive tosses show the same result (e.g. HH or TT appearing at least once in the sequence).
Complementary approach. It is far easier to count the sequences where this never happens — i.e. every adjacent pair of tosses differs — and subtract from 1.

• Total outcomes. Each of the 5 coins independently lands heads or tails, giving $2^5 = 32$ equally likely sequences.

Counting alternating sequences. A sequence with no two adjacent flips the same must strictly alternate between H and T. For 5 tosses there are exactly two such sequences: HTHTH and THTHT.
Complement probability. $P(\text{no two the same in a row}) = \frac{2}{32} = \frac{1}{16}$.

• Final answer. $P(\text{at least two the same in a row}) = 1 - \frac{1}{16} = \frac{15}{16}$.
• Correct Answer: Option C.

Common Mistake (⚠️):
Trying to count all the ways to get 2, 3, 4, or 5 in a row and overcounting overlapping sequences.

Correct Answer: D. $\frac{p+q-\sqrt{p^2+q^2}}{2}$

Fastest Approach (🚀):
Legs are $p$ and $q$. Hypotenuse is $\sqrt{p^2+q^2}$.
Using the standard formula for the inradius of a right triangle: $r = \frac{\text{sum of legs} - \text{hypotenuse}}{2}$.
Since the center is $(c, c)$ and it touches both axes, the radius $r = c$.
So $c = \frac{p + q - \sqrt{p^2+q^2}}{2}$. Option D.

Step-by-Step Breakdown:

• Recognise the configuration. $P = (p, 0)$ lies on the $x$-axis and $Q = (0, q)$ lies on the $y$-axis, so $OPQ$ is a right-angled triangle with the right angle at the origin. The circle touches both axes and the segment $PQ$, which makes it the inscribed circle of that triangle.
• Centre and radius. A circle tangent to both axes in the first quadrant has its centre equidistant from each, at $(c, c)$, with radius equal to that same distance. Hence $r = c$.
• Side lengths. The two legs are $OP = p$ and $OQ = q$, and by Pythagoras the hypotenuse is

$$PQ = \sqrt{p^2 + q^2}$$

• Inradius of a right-angled triangle. For a right triangle with legs $a$, $b$ and hypotenuse $h$, the inradius is

$$r = \frac{a + b - h}{2}$$
This follows from equating two expressions for the area, $A = rs$ with $s$ the semi-perimeter, and $A = \tfrac{1}{2}ab$.

• Substitute.

$$c = r = \frac{p + q - \sqrt{p^2 + q^2}}{2}$$

• Sense check with a 3–4–5 triangle. Taking $p = 3$, $q = 4$ gives $c = \frac{3 + 4 - 5}{2} = 1$, the well-known inradius of that triangle.
• Correct Answer: Option D.

Common Mistake (⚠️):
Choosing option B (with a plus sign) which represents the radius of one of the excircles, not the incircle.

Correct Answer: E. 1/10

Fastest Approach (🚀):
Since $1000$ is a multiple of $10$, every final digit ($0-9$) appears with equal probability. There are $10$ possible digits, so the probability of any specific one is $1/10$. Option E.

Step-by-Step Breakdown:

• Describe the sample space. The integers from 1 to 1000 inclusive give $1000$ equally likely outcomes.
• Count the favourable outcomes. The numbers ending in 1 form the arithmetic sequence

$$1,\ 11,\ 21,\ \dots,\ 991$$
Using the inclusive counting rule for an arithmetic progression with common difference 10:
$$n = \frac{991 - 1}{10} + 1 = 99 + 1 = 100$$

• Compute the probability.

$$P = \frac{100}{1000} = \frac{1}{10}$$

• The quick argument. Because $1000$ is an exact multiple of $10$, the range splits into 100 complete blocks of ten consecutive integers. Each block contains exactly one number ending in 1, so the final digits are uniformly distributed and any chosen digit has probability $\tfrac{1}{10}$.
• Correct Answer: Option E.

Common Mistake (⚠️):
Overthinking and trying to count specific digit occurrences in all positions, confusing it with questions like 'how many times does the digit 1 appear in numbers 1-1000'.

Correct Answer: G. x < 28

Fastest Approach (🚀):
$-8 < 6 - x/2$
Subtract 6: $-14 < -x/2$
Multiply by -2 (and flip sign): $28 > x$
Rewrite: $x < 28$.
Option G.

Step-by-Step Breakdown:

• The inequality.

$$-8 < 6 - \frac{x}{2}$$

• Step 1 — Isolate the term in $x$. Subtract 6 from both sides:

$$-14 < -\frac{x}{2}$$

• Step 2 — Clear the fraction. Multiply both sides by $-2$. Multiplying an inequality by a negative number reverses the direction of the inequality sign:

$$28 > x$$

• Step 3 — Rewrite in standard form. Reading the statement from the other side gives

$$x < 28$$

• Verify with a test value. Take $x = 0$, which satisfies $x < 28$. The right-hand side becomes $6 - 0 = 6$, and indeed $-8 < 6$. Now take $x = 30$, which does not satisfy it: the right-hand side is $6 - 15 = -9$, and $-8 < -9$ is false. The boundary is confirmed.
• Correct Answer: Option G.

Common Mistake (⚠️):
Forgetting to flip the inequality sign when multiplying by $-2$, leading to $x > 28$.

Correct Answer: E. 15 cm

Fastest Approach (🚀):
$120 = \frac{1}{2}(x - 1 + x + 5)x = \frac{1}{2}(2x + 4)x = (x+2)x = x^2 + 2x$.
$x^2 + 2x - 120 = 0$.
Factors of $-120$ that sum to $2$ are $12$ and $-10$. So $(x+12)(x-10) = 0$.
Since lengths must be positive, $x = 10$.
RS = $x + 5 = 15$. Option E.

Step-by-Step Breakdown:

• Area formula for a trapezium. With parallel sides $a$ and $b$ and perpendicular height $h$,

$$A = \tfrac{1}{2}(a + b)h$$

• Substitute the given expressions. Here $a = PQ = x - 1$, $b = RS = x + 5$, $h = QR = x$, and $A = 120$:

$$120 = \tfrac{1}{2}\big((x - 1) + (x + 5)\big)x$$

• Simplify. The bracket collapses neatly:

$$120 = \tfrac{1}{2}(2x + 4)x = (x + 2)x = x^2 + 2x$$

• Form and solve the quadratic.

$$x^2 + 2x - 120 = 0$$
Seek two numbers multiplying to $-120$ and summing to $+2$: these are $+12$ and $-10$, so
$$(x + 12)(x - 10) = 0 \implies x = -12 \ \text{ or } \ x = 10$$

• Apply the physical constraint. A side length cannot be negative, so $x = -12$ is rejected and $x = 10$.
• Answer the question actually asked. The question wants $RS$, not $x$:

$$RS = x + 5 = 10 + 5 = 15\ \text{cm}$$

• Correct Answer: Option E.

Common Mistake (⚠️):
Stopping after finding $x = 10$ and choosing option B (10 cm), forgetting that the question asks for the length of RS, not $x$.

Correct Answer: C. $125 / (2\pi)$

Fastest Approach (🚀):
Height $h = 10$. Circumference $C = 5$.
$2\pi r = 5 \implies r = 5 / (2\pi)$.
Volume $V = \pi r^2 h = \pi (5 / 2\pi)^2 \times 10 = \pi (25 / 4\pi^2) \times 10 = 250 / 4\pi = 125 / 2\pi$. Option C.

Step-by-Step Breakdown:

• Picture the fold. The sheet measures $10\,\text{m}$ by $5\,\text{m}$ and is rolled up by joining the edges $PS$ and $QR$. We are told the resulting cylinder has height $10\,\text{m}$, so the $10\,\text{m}$ side becomes the vertical height and the $5\,\text{m}$ side wraps around to form the circular cross-section.
• The wrapped edge becomes the circumference. Therefore

$$2\pi r = 5 \implies r = \frac{5}{2\pi}$$

• Volume of a cylinder.

$$V = \pi r^2 h$$

• Substitute.

$$V = \pi \left(\frac{5}{2\pi}\right)^2 (10) = \pi \cdot \frac{25}{4\pi^2} \cdot 10 = \frac{250}{4\pi}$$

• Simplify.

$$V = \frac{125}{2\pi}\ \text{m}^3$$

• Common trap. Do not treat $5\,\text{m}$ as the diameter or the radius — it is the circumference, which is why the $2\pi$ ends up in the denominator.
• Correct Answer: Option C.

Common Mistake (⚠️):
Using $10\text{m}$ as the circumference and $5\text{m}$ as the height, which would give $V = \pi (10/2\pi)^2 \times 5 = 125/\pi$, leading to option E.

Correct Answer: E. $N = \frac{720}{180 - X}$

Fastest Approach (🚀):
Exterior angle $\angle RQT = 360/n$.
Triangle $QRT$ is isosceles with $RQ = RT$. Therefore, the base angles are equal: $\angle RTQ = \angle RQT = 360/n$.
Sum of angles in triangle: $x + \angle RQT + \angle RTQ = 180$.
$x + 360/n + 360/n = 180$.
$720/n = 180 - x$.
$n = 720 / (180 - x)$. Option E.

Step-by-Step Breakdown:

• The exterior angle of a regular polygon. The exterior angles of any polygon sum to $360^{\circ}$, so for a regular polygon with $n$ sides each exterior angle is $\tfrac{360}{n}$.
• Locate that angle in the diagram. $PQT$ is a straight line, so $\angle RQT$ is the exterior angle of the polygon at vertex $Q$:

$$\angle RQT = \frac{360}{n}$$

• Use the isosceles triangle. We are told $RQ = RT$, so triangle $QRT$ is isosceles with the two equal sides meeting at $R$. The base angles opposite those equal sides are therefore equal:

$$\angle RTQ = \angle RQT = \frac{360}{n}$$

• Sum the angles of triangle $QRT$. With $x = \angle QRT$ the apex angle,

$$x + \frac{360}{n} + \frac{360}{n} = 180$$

• Rearrange for $n$.

$$\frac{720}{n} = 180 - x \implies n = \frac{720}{180 - x}$$

• Correct Answer: Option E.

Common Mistake (⚠️):
Assuming $RQ = QT$ instead of $RQ = RT$, which would mean $\angle QRT = \angle QTR = x$, leading to $2x + 360/n = 180 \implies n = 360/(180-2x)$.

Correct Answer: F. 16

Fastest Approach (🚀):
$x \propto 1/\sqrt{y} \implies x\sqrt{y} = k$.
$k = 8 \times \sqrt{9} = 8 \times 3 = 24$.
When $x=6$: $6\sqrt{y} = 24 \implies \sqrt{y} = 4 \implies y = 16$. Option F.

Step-by-Step Breakdown:

• Write the proportionality as an equation. "$x$ is inversely proportional to the square root of $y$" means

$$x = \frac{k}{\sqrt{y}} \quad \text{equivalently} \quad x\sqrt{y} = k$$
for some constant $k$.

• Find the constant from the given pair. Substituting $x = 8$ when $y = 9$:

$$k = 8 \times \sqrt{9} = 8 \times 3 = 24$$

• Use the constant for the new value. With $x = 6$:

$$6\sqrt{y} = 24 \implies \sqrt{y} = 4$$

• Solve for $y$. Squaring both sides:

$$y = 16$$

• Sense check. $x$ fell from 8 to 6, a factor of $\tfrac{3}{4}$. Since $x \propto y^{-1/2}$, $y$ must change by a factor of $\left(\tfrac{3}{4}\right)^{-2} = \tfrac{16}{9}$, and indeed $9 \times \tfrac{16}{9} = 16$.
• Correct Answer: Option F.

Common Mistake (⚠️):
Squaring incorrectly (e.g. $\sqrt{y}=4 \implies y=2$) or setting up direct proportionality ($x = k\sqrt{y}$).

Correct Answer: B. 2√3 km

Fastest Approach (🚀):
A is East of B. So line BA points West ($270^{\circ}$) and AB points East ($090^{\circ}$).
Bearing of C from B is $060^{\circ}$. The angle inside the triangle at B is $90^{\circ}(\text{East}) - 60^{\circ}(\text{Bearing}) = 30^{\circ}$.
Bearing of C from A is $330^{\circ}$. The angle inside the triangle at A is $330^{\circ}(\text{Bearing}) - 270^{\circ}(\text{West}) = 60^{\circ}$.
Since angles at A and B are $60^{\circ}$ and $30^{\circ}$, the angle at C is $90^{\circ}$.
We have a right triangle with hypotenuse $AB = 4$.
We want $BC$. $BC$ is opposite to angle A ($60^{\circ}$).
$BC = AB \sin(60^{\circ}) = 4 \times (\sqrt{3}/2) = 2\sqrt{3}$. Option B.

Step-by-Step Breakdown:

• Set up the directions. Bearings are measured clockwise from North. $A$ is due East of $B$, so from $B$ the direction to $A$ is the bearing $090^{\circ}$, and from $A$ the direction back to $B$ is $270^{\circ}$.
• Interior angle at $B$. The bearing of $C$ from $B$ is $060^{\circ}$, while $A$ lies on bearing $090^{\circ}$ from $B$. The angle $\angle ABC$ between those two directions is

$$90^{\circ} - 60^{\circ} = 30^{\circ}$$

• Interior angle at $A$. The bearing of $C$ from $A$ is $330^{\circ}$, while $B$ lies on bearing $270^{\circ}$ from $A$. The angle $\angle BAC$ is

$$330^{\circ} - 270^{\circ} = 60^{\circ}$$

• Interior angle at $C$. Angles in a triangle sum to $180^{\circ}$:

$$\angle ACB = 180^{\circ} - 30^{\circ} - 60^{\circ} = 90^{\circ}$$
So $ABC$ is right-angled at $C$, and $AB = 4\,\text{km}$ is the hypotenuse.

• Find $BC$. In the right-angled triangle, $BC$ is adjacent to the $30^{\circ}$ angle at $B$:

$$BC = AB\cos(30^{\circ}) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3}\ \text{km}$$

• Correct Answer: Option B.

Common Mistake (⚠️):
Using the Sine Rule blindly without noticing it's a right-angled triangle, leading to messy algebra, or misinterpreting bearings (e.g. thinking bearing is from East rather than North).

Correct Answer: B. $5 - 2\sqrt{2}\sqrt{3}$

Fastest Approach (🚀):
$(\sqrt{3} - \sqrt{2})^2 = (\sqrt{3})^2 - 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2 = 3 - 2\sqrt{3}\sqrt{2} + 2 = 5 - 2\sqrt{2}\sqrt{3}$. Option B.

Step-by-Step Breakdown:

• We need to simplify the expression $(\sqrt{3} - \sqrt{2})^2$.
• Expand using the binomial identity $(a-b)^2 = a^2 - 2ab + b^2$, with $a=\sqrt{3}$ and $b=\sqrt{2}$:

$$(\sqrt{3}-\sqrt{2})^2 = (\sqrt{3})^2 - 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2$$

• Simplify each term. $(\sqrt{3})^2=3$ and $(\sqrt{2})^2=2$, with a cross term of $2\sqrt{2}\sqrt{3}$:

$$= 3 - 2\sqrt{2}\sqrt{3} + 2$$

• Combine the constants.

$$= 5 - 2\sqrt{2}\sqrt{3}$$

• Common trap. It is tempting to square each surd separately and ignore the cross term, giving $3-2=1$ (Option D) — but $(a-b)^2 \ne a^2-b^2$; the middle term $-2ab$ must not be dropped.
• Correct Answer: Option B.

Common Mistake (⚠️):
Squaring individual terms and ignoring the cross term: $(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$ (Option D).

Correct Answer: C. $p = 1$

Fastest Approach (🚀):
$P(2) = 8 + 4p + 2q + p^2 = 0 \implies 2q = -p^2 - 4p - 8$.
$P(1) = 1 + p + q + p^2 = -3.5 \implies 2 + 2p + 2q + 2p^2 = -7$.
Substitute $2q$: $2 + 2p + (-p^2 - 4p - 8) + 2p^2 = -7$.
Simplify: $p^2 - 2p - 6 = -7 \implies p^2 - 2p + 1 = 0$.
Factor: $(p - 1)^2 = 0 \implies p = 1$. Option C.

Step-by-Step Breakdown:

• Set up two equations from the two substitutions. Let $f(x) = x^3 + px^2 + qx + p^2$.
• Substituting $x = 2$ and $f(2) = 0$:

$$8 + 4p + 2q + p^2 = 0 \qquad (1)$$

• Substituting $x = 1$ and $f(1) = -3.5$:

$$1 + p + q + p^2 = -3.5 \qquad (2)$$

• Eliminate $q$. Rearranging (2) gives

$$q = -4.5 - p - p^2$$

• Substitute this into (1):

$$8 + 4p + 2(-4.5 - p - p^2) + p^2 = 0$$
$$8 + 4p - 9 - 2p - 2p^2 + p^2 = 0$$
$$-p^2 + 2p - 1 = 0$$

• Solve. Multiplying through by $-1$ gives a perfect square:

$$p^2 - 2p + 1 = 0 \implies (p - 1)^2 = 0$$

• The repeated root gives a single value, $p = 1$.
• Correct Answer: Option C.

Common Mistake (⚠️):
Making arithmetic errors with the $-3.5$ or dropping a negative sign during expansion, leading to a quadratic with two distinct roots like $(p-1)(p+3)=0$ (Option B).