ESAT Paper 3 Maths Worked Solutions

Correct Answer: A. $a = -5, b = 14$

Fastest Approach (🚀):
Sum of roots: $2 + 5 = 7 = -(a - 2) \implies a = -5$.
Product of roots: $2 \times 5 = 10 = b - 4 \implies b = 14$.
Matches Option A.

Step-by-Step Breakdown:

1. Setting Up the Root Equation

The points of intersection between the parabola $y = x^2 + ax + b$ and the straight line $y = 2x + 4$ satisfy:
$x^2 + ax + b = 2x + 4$
$x^2 + (a - 2)x + (b - 4) = 0$

2. Applying Vieta's Formulas

We are given that the intersection occurs at $x_1 = 2$ and $x_2 = 5$. These are the roots of the quadratic equation above.

Using the sum of roots identity ($x_1 + x_2 = -\frac{B}{A}$):
$2 + 5 = -(a - 2) \implies 7 = 2 - a \implies a = -5$

Using the product of roots identity ($x_1 x_2 = \frac{C}{A}$):
$2 \times 5 = b - 4 \implies 10 = b - 4 \implies b = 14$

3. Conclusion & Option Matching

The values are $a = -5$ and $b = 14$.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $a = -4, b = 10$ — Fails Vieta
  Neither Vieta relation holds: the roots sum to $7$, so $-(a-2)=7$, and they multiply to $10$, so $b-4=10$.
• C. $a = -5, b = 10$ — Constant Not Shifted
  Right $a$, wrong $b$. The product of the roots equals $b-4$, so $b-4=10$ gives $b=14$.
• D. $a = -4, b = 14$ — Sum Slip
  Right $b$, wrong $a$. The sum of the roots equals $-(a-2)=7$, so $a=-5$.
• E. $a = -3, b = 13$ — Fails the Check
  Substituting $x=2$ gives $4-6+13=11$, against the line's value of $8$.

Common Mistake (⚠️):
Finding $y$-coordinates first by plugging $x=2$ and $x=5$ into $y=2x+4$ and setting up simultaneous linear equations, which takes double the time compared to using Vieta's formulas directly.

Takeaway (📌):
Vieta's Shortcut: For any intersection of $f(x)=g(x)$ where roots are known, immediately write the difference equation $f(x)-g(x)=0$ and equate coefficients to root sums/products.

Correct Answer: D. $x^2$

Fastest Approach (🚀):
For $x = 0.5$: $x^2 = 0.25$, $x = 0.5$, $\sqrt{x} \approx 0.71$, $1/x = 2$, $1/(1-x) = 2$. Smallest is $x^2$.
Matches Option D.

Step-by-Step Breakdown:

1. Comparing Fractional behaviour

For any real number $x$ strictly between $0$ and $1$ ($x \in (0,1)$):

1. Squaring a fraction ($x^2$): Since $0 < x < 1$, multiplying by $x$ decreases the magnitude: $x^2 < x$.
2. Taking square root ($\sqrt{x}$): Square root of a fraction increases magnitude: $\sqrt{x} > x$.
3. Reciprocating ($\frac{1}{x}$): Reciprocal of a fraction $< 1$ yields a value $> 1$.
4. Reciprocating complement ($\frac{1}{1-x}$): Since $0 < 1-x < 1$, $\frac{1}{1-x} > 1$.

2. Ordering the Expressions

$x^2 < x < \sqrt{x} < 1 < \frac{1}{x}$, and $\frac{1}{1-x} > 1$ also.

Thus, $x^2$ has the smallest value for all $x \in (0,1)$.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\sqrt{x}$ — Root Direction
  On $0<x<1$, taking a root moves a number towards $1$, so $\sqrt x>x>x^{2}$.
• B. $\frac{1}{x}$ — Above One
  Greater than $1$ throughout - dividing by a proper fraction enlarges.
• C. $x$ — Second Smallest
  Smaller than $\sqrt x$, but still larger than $x^{2}$ on this interval.
• E. $\frac{1}{1-x}$ — Above One
  Also above $1$, since $0<1-x<1$.

Common Mistake (⚠️):
Assuming $\sqrt{x}$ is smaller than $x$, confusing whole number behaviour with fraction behaviour.

Takeaway (📌):
For $0 < x < 1$: Squaring REDUCES value ($x^2 < x$); Square root INCREASES value ($\sqrt{x} > x$).

Correct Answer: C. $\frac{1}{64}$

Fastest Approach (🚀):
$p^2 / 4 = 9/64 \implies p = 3/4 \implies P(T)_{\text{biased}} = 1/4 \implies P(4T) = (1/4) \times (1/16) = 1/64$.
Matches Option C.

Step-by-Step Breakdown:

1. Defining Probabilities

• Two fair coins: $P(H) = \frac{1}{2}, \quad P(T) = \frac{1}{2}$
• Two biased coins: Let $P(H) = p$ and $P(T) = 1 - p$

2. Solving for Bias Parameter $p$

The probability of throwing four heads is:
$P(4H) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times p \times p = \frac{p^2}{4}$

We are given $P(4H) = \frac{9}{64}$:
$\frac{p^2}{4} = \frac{9}{64} \implies p^2 = \frac{36}{64} = \frac{9}{16} \implies p = \frac{3}{4}$

3. Calculating Probability of Four Tails

For the biased coins, $P(T) = 1 - p = 1 - \frac{3}{4} = \frac{1}{4}$.
$P(4T) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) = \frac{1}{4} \times \frac{1}{16} = \frac{1}{64}$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $\frac{1}{32}$ — Factor of Two
  Twice the answer. The fair pair contributes $\left(\tfrac12\right)^{2}=\tfrac14$ and the biased pair $\left(\tfrac14\right)^{2}=\tfrac1{16}$.
• B. $\frac{1}{16}$ — Coins Omitted
  The biased pair alone. The two fair coins have to land tails as well.
• D. $\frac{3}{64}$ — Bias Reversed
  Uses $P(\text{tail})=\tfrac34$ on a biased coin. The bias favours heads - $p=\tfrac34$ - so its tail probability is $\tfrac14$.
• E. $\frac{1}{128}$ — Extra Factor
  Half the answer: one factor of $\tfrac12$ too many.

Common Mistake (⚠️):
Taking $p = 9/64$ directly without accounting for the two fair coins.

Takeaway (📌):
Break multi-coin probability problems into independent coin probabilities and multiply.

Correct Answer: C. $b = \pm \sqrt{\frac{2a^2 + 5}{4a^2 - 1}}$

Fastest Approach (🚀):
$4a^2 b^2 - 2a^2 = b^2 + 5 \implies b^2(4a^2-1) = 2a^2+5 \implies b = \pm \sqrt{\frac{2a^2+5}{4a^2-1}}$.
Matches Option C.

Step-by-Step Breakdown:

1. Clearing the Denominator

Multiply both sides by $(4b^2 - 2)$:
$a^2 (4b^2 - 2) = b^2 + 5$
$4a^2 b^2 - 2a^2 = b^2 + 5$

2. Collecting Terms in $b^2$

Move all terms containing $b^2$ to the left-hand side and remaining terms to the right:
$4a^2 b^2 - b^2 = 2a^2 + 5$
$b^2 (4a^2 - 1) = 2a^2 + 5$

3. Isolating $b$

$b^2 = \frac{2a^2 + 5}{4a^2 - 1}$
$b = \pm \sqrt{\frac{2a^2 + 5}{4a^2 - 1}}$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $b = \pm \sqrt{\frac{2a^2 + 5}{4a^2 + 1}}$ — Denominator Sign
  Sign slip in the denominator. Bringing $b^{2}$ across gives $b^{2}(4a^{2}-1)$, not $4a^{2}+1$.
• B. $b = \pm \sqrt{\frac{2a^2 - 5}{4a^2 - 1}}$ — Numerator Sign
  Sign slip in the numerator. Moving $-2a^{2}$ across gives $+2a^{2}+5$.
• D. $b = \pm \sqrt{\frac{2a^2 + 5}{4a^2 - 2}}$ — Term Not Collected
  Keeps the original $-2$ where collecting $b^{2}$ leaves $-1$.
• E. $b = \pm \sqrt{\frac{2a^2 - 5}{4a^2 + 1}}$ — Both Signs
  Both signs wrong at once.

Common Mistake (⚠️):
Sign errors when expanding $-2a^2$ or forgetting the $\pm$ sign on taking square roots.

Takeaway (📌):
Group the target squared variable on one side and factor it out: $b^2(\text{coeff}) = \text{remainder}$.

Correct Answer: D. $\frac{51}{125}$

Fastest Approach (🚀):
$P(B|H) = 42/50$, $P(B|H^c) = 6/50 \implies P(B) = \frac{2}{5} \times \frac{42}{50} + \frac{3}{5} \times \frac{6}{50} = \frac{84 + 18}{250} = \frac{102}{250} = \frac{51}{125}$.
Matches Option D.

Step-by-Step Breakdown:

1. Extracting Given Probabilities

Let $H$ be the event that an adult likes ham sandwiches, and $B$ be the event that an adult likes bacon sandwiches.

• $P(H) = \frac{2}{5} \implies P(H^c) = 1 - \frac{2}{5} = \frac{3}{5}$
• $P(B|H) = \frac{42}{50} = \frac{21}{25}$

2. Finding $P(B|H^c)$

We are given that $P(B|H)$ is $7$ times higher than $P(B|H^c)$:
$P(B|H) = 7 \times P(B|H^c) \implies P(B|H^c) = \frac{P(B|H)}{7} = \frac{42/50}{7} = \frac{6}{50} = \frac{3}{25}$

3. Applying Law of Total Probability

$P(B) = P(B|H) P(H) + P(B|H^c) P(H^c)$
$P(B) = \left(\frac{42}{50}\right) \left(\frac{2}{5}\right) + \left(\frac{6}{50}\right) \left(\frac{3}{5}\right) = \frac{84}{250} + \frac{18}{250} = \frac{102}{250} = \frac{51}{125}$

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\frac{36}{125}$ — Too Small
  Below the ham-liking branch on its own, which already contributes $\tfrac25\times\tfrac{21}{25}=\tfrac{42}{125}$.
• B. $\frac{39}{125}$ — Too Small
  Also below that $\tfrac{42}{125}$ floor.
• C. $\frac{42}{125}$ — Branch Omitted
  The ham-liking branch only. The other $\tfrac35$ of adults add $\tfrac35\times\tfrac{3}{25}=\tfrac{9}{125}$.
• E. $\frac{58}{125}$ — Too Large
  Above the total the two branches can reach: $\tfrac{42}{125}+\tfrac{9}{125}=\tfrac{51}{125}$.

Common Mistake (⚠️):
Adding conditional probabilities directly without weighting by $P(H)$ and $P(H^c)$.

Takeaway (📌):
Law of total probability $P(A) = \sum P(A|B_i) P(B_i)$ weights conditional probabilities by subgroup proportions.

Correct Answer: B. $7.5^\circ$

Fastest Approach (🚀):
Hour hand at 8:45 is at $8.75 \times 30^\circ = 262.5^\circ$. Minute hand is at $270^\circ$. Angle $= 270 - 262.5 = 7.5^\circ$.
Matches Option B.

Step-by-Step Breakdown:

1. Position of Minute Hand at 8:45

A clock face is $360^\circ$, divided into 60 minute marks ($6^\circ$ per minute).
At 45 minutes, the minute hand points to 9:
$\theta_{\text{minute}} = 45 \times 6^\circ = 270^\circ$

2. Position of Hour Hand at 8:45

Each hour mark is $30^\circ$ ($360^\circ / 12$). The hour hand moves at $0.5^\circ$ per minute ($30^\circ / 60$).
At 8:45 (8 hours + 45 minutes):
$\theta_{\text{hour}} = (8 \times 30^\circ) + (45 \times 0.5^\circ) = 240^\circ + 22.5^\circ = 262.5^\circ$

3. Calculating Angle Between Hands

$\text{Angle} = |\theta_{\text{minute}} - \theta_{\text{hour}}| = 270^\circ - 262.5^\circ = 7.5^\circ$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $5^\circ$ — Hand Rate
  Understates the hour hand's travel. It moves $0.5^\circ$ per minute, so in $45$ minutes it advances $22.5^\circ$ past the eight.
• C. $10^\circ$ — Not Consistent
  No consistent pair of hand positions gives $10^\circ$: the hands sit at $262.5^\circ$ and $270^\circ$.
• D. $12.5^\circ$ — Hand Rate
  Places the hour hand at $257.5^\circ$, only thirty-five minutes of travel past the eight.
• E. $15^\circ$ — Hand Rate
  Places it at $255^\circ$ - half the required advance.

Common Mistake (⚠️):
Assuming the hour hand stays stationary on the 8 o'clock mark, getting $30^\circ$.

Takeaway (📌):
The hour hand advances by $0.5^\circ$ for every minute that passes.

Correct Answer: A. $-5 < x < 1$

Fastest Approach (🚀):
$x^2 + 4x - 5 < 0 \implies (x+5)(x-1) < 0 \implies -5 < x < 1$.
Matches Option A.

Step-by-Step Breakdown:

1. Rearranging Inequality

Move all terms to the left-hand side:
$x^2 + 4x - 5 < 0$

2. Factoring the Quadratic

Find two numbers that multiply to $-5$ and add to $4$ ($+5$ and $-1$):
$(x + 5)(x - 1) < 0$

3. Determining the Solution Interval

The roots of the parabola are $x = -5$ and $x = 1$.
Since the coefficient of $x^2$ is positive ($+1$), the parabola opens upward.
The expression is strictly negative ($< 0$) between the two roots:
$-5 < x < 1$

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $-5 < x < -1$ — Fails the Check
  $x=0$ satisfies $0<5$, so the solution set has to contain zero.
• C. $-3 < x < 1$ — Wrong Root
  Right upper endpoint, wrong lower one: the roots of $x^{2}+4x-5$ are $-5$ and $1$.
• D. $-2 < x < 3$ — Fails the Check
  $x=2$ gives $4<-3$, which is false, so $3$ cannot be an endpoint.
• E. $-4 < x < 2$ — Fails the Check
  $x=1.5$ gives $2.25<-1$, false. Both endpoints are wrong.

Common Mistake (⚠️):
Writing $x < -5$ or $x > 1$, which is the solution for $x^2 + 4x - 5 > 0$.

Takeaway (📌):
For an upward-opening quadratic $A(x-r_1)(x-r_2) < 0$, solution is ALWAYS between the roots $r_1 < x < r_2$.

Correct Answer: C. $6 - \frac{3}{x}$

Fastest Approach (🚀):
$7 + \frac{-(x-3)(x+3)}{x(x-3)} = 7 - \frac{x+3}{x} = 7 - 1 - \frac{3}{x} = 6 - \frac{3}{x}$.
Matches Option C.

Step-by-Step Breakdown:

1. Factoring Numerator and Denominator

• Numerator: $9 - x^2 = (3 - x)(3 + x) = -(x - 3)(x + 3)$
• Denominator: $x^2 - 3x = x(x - 3)$

2. Canceling Common Binomial Factor

$\frac{9 - x^2}{x^2 - 3x} = \frac{-(x - 3)(x + 3)}{x(x - 3)} = \frac{-(x + 3)}{x} = -1 - \frac{3}{x}$

3. Combining with Constant 7

$7 + \left(-1 - \frac{3}{x}\right) = (7 - 1) - \frac{3}{x} = 6 - \frac{3}{x}$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $6 + \frac{3}{x}$ — Sign Error
  Right constant, wrong sign on the fraction: $-\tfrac{x+3}{x}=-1-\tfrac3x$.
• B. $7 - \frac{3}{x}$ — Constant Not Reduced
  Keeps the $7$ intact. The term $-\tfrac{x+3}{x}$ contributes a $-1$ as well as the $-\tfrac3x$.
• D. $7 + \frac{3}{x}$ — Both Signs
  Both faults at once: the $-1$ dropped and the fraction's sign reversed.
• E. $6 + \frac{4}{x}$ — Wrong Numerator
  The $4$ has no source - after cancelling $(x-3)$ the numerator is $x+3$.

Common Mistake (⚠️):
Sign error when factoring $(9-x^2)$, writing $(x-3)(x+3)$ instead of $-(x-3)(x+3)$.

Takeaway (📌):
Remember $(a-b) = -(b-a)$; factoring out $-1$ allows canceling opposing binomials.

Correct Answer: A. $a$ is inversely proportional to the square root of $b$

Fastest Approach (🚀):
$b \propto c^{-6} = (c^3)^{-2} \propto a^{-2} \implies a \propto b^{-1/2} \implies a \propto 1/\sqrt{b}$.
Matches Option A.

Step-by-Step Breakdown:

1. Setting Up Proportionality Equations

• $a \propto c^3 \implies a = k_1 c^3$
• $b \propto \frac{1}{c^6} \implies b = \frac{k_2}{c^6} = k_2 (c^3)^{-2}$

2. Eliminating Variable $c$

Substitute $c^3 = \frac{a}{k_1}$ into the equation for $b$:
$b = k_2 \left(\frac{a}{k_1}\right)^{-2} = k_2 k_1^2 a^{-2} = \frac{K}{a^2}$

3. Expressing $a$ in Terms of $b$

$a^2 = \frac{K}{b} \implies a = \frac{\sqrt{K}}{\sqrt{b}}$

Therefore, $a$ is inversely proportional to the square root of $b$.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $a$ is directly proportional to the square root of $b$ — Direction Reversed
  $b\propto a^{-2}$, so $a\propto b^{-1/2}$. The relationship is inverse, not direct.
• C. $a$ is inversely proportional to the square of $b$ — Power Inverted
  $a^{2}\propto\tfrac1b$. Take the square root of both sides, rather than squaring them.
• D. $a$ is directly proportional to the square of $b$ — Both Wrong
  Wrong direction and wrong power together.
• E. $a$ is inversely proportional to $b$ — Wrong Power
  Right direction, wrong power. $(c^{3})^{-2}=c^{-6}$ gives $b\propto a^{-2}$, hence $a\propto b^{-1/2}$.

Common Mistake (⚠️):
Miscalculating $(c^3)^{-2}$ as $c^{-5}$ instead of $c^{-6}$.

Takeaway (📌):
Use power rules $(c^m)^n = c^{m \cdot n}$ when combining proportional variable links.

Correct Answer: A. $\sqrt{\frac{1}{2}a^2 + l^2}$

Fastest Approach (🚀):
$d = \sqrt{(a/2)^2 + (a/2)^2 + l^2} = \sqrt{\frac{a^2}{2} + l^2}$.
Matches Option A.

Step-by-Step Breakdown:

1. Assigning 3D Coordinates

Place the square base of side length $a$ in the $xy$-plane with one corner at the origin $(0,0,0)$.
The vertices of the base lie at $(0,0,0), (a,0,0), (a,a,0), (0,a,0)$.
The top face lies in the plane $z = l$. Its centre point $C$ is directly above the centre of the base, so its coordinates are $\left(\frac{a}{2}, \frac{a}{2}, l\right)$.

2. Applying the 3D Distance Formula

The distance from $(0,0,0)$ to $\left(\frac{a}{2}, \frac{a}{2}, l\right)$ is:
$d = \sqrt{\left(\frac{a}{2} - 0\right)^2 + \left(\frac{a}{2} - 0\right)^2 + (l - 0)^2}$
$d = \sqrt{\frac{a^2}{4} + \frac{a^2}{4} + l^2} = \sqrt{\frac{a^2}{2} + l^2}$

3. Conclusion & Option Matching

The distance is $\sqrt{\frac{a^2}{2} + l^2}$.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $\sqrt{\frac{1}{3}a^2 + l^2}$ — Offset Miscounted
  A third would arise from a body diagonal split three ways. Here the two horizontal offsets are each $\tfrac a2$, giving $2\times\tfrac{a^{2}}{4}=\tfrac{a^{2}}{2}$.
• C. $\sqrt{\frac{1}{2}a^2 - l^2}$ — Sign Error
  Subtracts the height. Pythagoras in three dimensions adds all three squared components.
• D. $\sqrt{l^2 - \frac{1}{2}a^2}$ — Sign Error
  Also a subtraction, and it turns negative for a short cuboid - which no distance can.
• E. $\sqrt{\frac{1}{4}a^2 + l^2}$ — Offset Miscounted
  Counts one horizontal offset of $\tfrac a2$ where there are two, one along each base edge.

Common Mistake (⚠️):
Calculating distance to top corner rather than top face centre, getting $\sqrt{2a^2+l^2}$.

Takeaway (📌):
In 3D geometry problems, set up Cartesian axes at the origin $(0,0,0)$ to convert spatial problems into straightforward algebra.

Correct Answer: B. $\frac{\sqrt{3}}{2}l$

Fastest Approach (🚀):
$h = l \sin 60^\circ = \frac{\sqrt{3}}{2}l$.
Matches Option B.

Step-by-Step Breakdown:

1. Geometric Construction

Consider an equilateral triangle with side length $l$. The line from a vertex perpendicular to the opposite side is both an altitude and a median.
This splits the base $l$ into two halves of length $\frac{l}{2}$, forming a right-angled triangle.

2. Applying Pythagoras / Trigonometry

Method A (Pythagorean Theorem):
$h^2 + \left(\frac{l}{2}\right)^2 = l^2 \implies h^2 = l^2 - \frac{l^2}{4} = \frac{3}{4}l^2$
$h = \frac{\sqrt{3}}{2}l$

Method B (Trigonometry):
$h = l \sin(60^\circ) = l \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2}l$

3. Conclusion & Option Matching

The length of the perpendicular bisector is $\frac{\sqrt{3}}{2}l$.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\frac{1}{2}l$ — Wrong Leg
  Half the base, which is the other leg of the right triangle, not the altitude.
• C. $\frac{1}{\sqrt{2}}l$ — Wrong Triangle
  $\tfrac{1}{\sqrt2}\approx0.707$ belongs to a right isosceles triangle. Here $\sqrt{1-\tfrac14}=\tfrac{\sqrt3}{2}\approx0.866$.
• D. $\frac{2}{\sqrt{3}}l$ — Exceeds Side
  $\tfrac{2}{\sqrt3}\approx1.15$, longer than the side itself. An altitude is always shorter than the sides it falls between.
• E. $\frac{1}{\sqrt{3}}l$ — Centroid Distance
  $\tfrac{l}{\sqrt3}\approx0.577l$ is the distance from a vertex to the centre - two thirds of the altitude.

Common Mistake (⚠️):
Dividing by $\sqrt{3}$ instead of multiplying, giving $\frac{l}{\sqrt{3}}$ (which is the distance to the centroid, not the full altitude).

Takeaway (📌):
Memorize height of equilateral triangle $h = \frac{\sqrt{3}}{2}l$ and area $A = \frac{\sqrt{3}}{4}l^2$.

Correct Answer: D. 69% increase

Fastest Approach (🚀):
$P' = \frac{1.5 \times 0.9}{0.8} P = 1.6875 P \implies +68.75\% \approx 69\% \text{ increase}$.
Matches Option D.

Step-by-Step Breakdown:

1. Expressing Pressure in Terms of Variables

From the ideal gas equation $PV = nRT$, isolate $P$:
$P = \frac{nRT}{V}$

2. Applying Multipliers

• $n$ increases by $50\% \implies n' = 1.50 n$
• $T$ decreases by $10\% \implies T' = 0.90 T$
• $V$ decreases by $1/5 = 20\% \implies V' = 0.80 \text{V}$

Substitute into the pressure equation:
$P' = \frac{(1.50 n) R (0.90 T)}{0.80 V} = \left(\frac{1.50 \times 0.90}{0.80}\right) \frac{nRT}{V}$
$P' = \left(\frac{1.35}{0.80}\right) P = 1.6875 P$

3. Calculating Percentage Increase

$\text{Percentage change} = (1.6875 - 1) \times 100\% = +68.75\% \approx 69\% \text{ increase}$

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. 69% decrease — Direction Reversed
  Right magnitude, wrong direction: $\tfrac{1.5\times0.9}{0.8}=1.6875$, which is greater than one.
• B. 33% decrease — Percentages Added
  Adds and subtracts the percentages instead of multiplying the scale factors.
• C. 14% increase — Not Consistent
  $P=\tfrac{nRT}{V}$, so the multiplier is $\tfrac{1.5\times0.9}{0.8}=1.6875$. Nothing in the numbers produces $1.14$.
• E. 14% decrease — Not Consistent
  Wrong direction as well, with the same unexplained magnitude.

Common Mistake (⚠️):
Subtracting percentages linearly $(50 - 10 - 20 = 20\%)$ instead of multiplying scale factors.

Takeaway (📌):
Proportional changes in formulas $P \propto \frac{nT}{V}$ must be evaluated by multiplying scale factors: $k_P = \frac{k_n k_T}{k_V}$.

Correct Answer: E. $18 - 4\sqrt{3}$

Fastest Approach (🚀):
$3 \times 7 + 3\sqrt{3} - 7\sqrt{3} - (\sqrt{3})^2 = 21 - 4\sqrt{3} - 3 = 18 - 4\sqrt{3}$.
Matches Option E.

Step-by-Step Breakdown:

1. Area Formula & Expansion

The area of a rectangle is $\text{length} \times \text{width}$:
$\text{Area} = (3 - \sqrt{3})(7 + \sqrt{3})$

2. Expanding Using FOIL

$\text{First}: 3 \times 7 = 21$
$\text{Outer}: 3 \times \sqrt{3} = 3\sqrt{3}$
$\text{Inner}: -\sqrt{3} \times 7 = -7\sqrt{3}$
$\text{Last}: -\sqrt{3} \times \sqrt{3} = -3$

Combine like terms:
$\text{Area} = 21 + 3\sqrt{3} - 7\sqrt{3} - 3 = (21 - 3) + (3\sqrt{3} - 7\sqrt{3}) = 18 - 4\sqrt{3}$

3. Conclusion & Option Matching

The area of the rectangle is $18 - 4\sqrt{3}$.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $21 + 3\sqrt{3}$ — Terms Dropped
  Only the first two products, $3\times7$ and $3\sqrt3$. The $-7\sqrt3$ and the $-3$ are both missing.
• B. $18 + 4\sqrt{3}$ — Surd Sign
  Right rational part, wrong surd sign: $3\sqrt3-7\sqrt3=-4\sqrt3$.
• C. $15 - 3\sqrt{3}$ — Arithmetic Slip
  Neither part matches. Expanding gives $21-3=18$ and $3\sqrt3-7\sqrt3=-4\sqrt3$.
• D. $21 - 4\sqrt{3}$ — Term Dropped
  Right surd term, but $(-\sqrt3)(+\sqrt3)=-3$ has not been taken off the $21$.

Common Mistake (⚠️):
Writing $(-\sqrt{3})(\sqrt{3}) = +3$ instead of $-3$, leading to $24 - 4\sqrt{3}$.

Takeaway (📌):
Watch minus signs carefully when multiplying surds: $(-\sqrt{a})(+\sqrt{a}) = -a$.

Correct Answer: B. $a = \frac{4h}{h-2}$

Fastest Approach (🚀):
Divide by $a \neq 0$: $ah = 2a + 4h \implies a(h-2) = 4h \implies a = \frac{4h}{h-2}$.
Matches Option B.

Step-by-Step Breakdown:

1. Formulating Volume and Surface Area

For a square-based prism of base area $a^2$ (side length $a$) and height $h$:

• Volume: $V = a^2 h$
• Total Surface Area: $S = 2 \times (\text{base area}) + 4 \times (\text{side area}) = 2a^2 + 4ah$

2. Equating Surface Area and Volume

$a^2 h = 2a^2 + 4ah$

Since $a > 0$, divide both sides by $a$:
$ah = 2a + 4h$

3. Rearranging for $a$

Collect terms in $a$ on the left-hand side:
$ah - 2a = 4h$
$a(h - 2) = 4h$
$a = \frac{4h}{h - 2}$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $a = \frac{4h}{h+2}$ — Denominator Sign
  Sign in the denominator. Collecting terms gives $a(h-2)=4h$, because $2a$ moves to the left-hand side.
• C. $a = \frac{2h}{h-2}$ — Faces Miscounted
  Wrong numerator: the four side faces contribute $4ah$, so dividing through by $a$ leaves $4h$.
• D. $a = \frac{2h}{h-1}$ — Both Wrong
  Both parts wrong. Two square faces give $2a^{2}$ and four sides give $4ah$, so $a(h-2)=4h$.
• E. $a = \frac{2h}{h+2}$ — Both Wrong
  Wrong numerator and wrong denominator sign together.

Common Mistake (⚠️):
Forgetting to include both top and bottom base faces ($2a^2$) in total surface area, writing $S = a^2 + 4ah$.

Takeaway (📌):
Total surface area of a closed prism always includes two bases: $S = 2A_{\text{base}} + P_{\text{base}} h$.

Correct Answer: C. Reflection in the y-axis

Fastest Approach (🚀):
Point $(1,1) \xrightarrow{180^\circ} (-1,-1) \xrightarrow{\text{refl } x} (-1,1)$. To return $(-1,1) \to (1,1)$, reflect in the $y$-axis.
Matches Option C.

Step-by-Step Breakdown:

1. Tracking Coordinate Transformations

Consider a general point $(x,y)$ on the triangle $PQR$:

1. Rotation of $180^\circ$ clockwise around the origin:

$(x,y) \longrightarrow (-x, -y)$

1. Reflection in the $x$-axis:

$(-x, -y) \longrightarrow (-x, y)$

2. Determining the Inverse Transformation

The overall transformed position of the point is $(-x, y)$.
To return from $(-x, y)$ back to the original coordinates $(x, y)$, we must negate the $x$-coordinate while leaving $y$ unchanged:
$(-x, y) \longrightarrow (x, y)$

This mapping corresponds to a reflection in the $y$-axis.

3. Conclusion & Option Matching

The transformation required to return to the original position is a reflection in the $y$-axis.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Rotation clockwise about origin by 180 degrees — Wrong Transform
  Would send $(-1,1)$ to $(1,-1)$, not back to $(1,1)$.
• B. Rotation clockwise about origin by 90 degrees — Coincidence
  Happens to return that one vertex, but not the others: $(2,0)$ maps to $(-2,0)$ and then to $(0,2)$.
• D. Reflection in the x-axis — Wrong Axis
  Sends $(-1,1)$ to $(-1,-1)$. The two transformations compose to $(x,y)\to(-x,y)$, which only a $y$-axis reflection undoes.
• E. Reflection in the line y=x — Wrong Transform
  Sends $(-1,1)$ to $(1,-1)$.

Common Mistake (⚠️):
Applying transformations in the wrong order or confusing reflection in $x$-axis with reflection in $y$-axis.

Takeaway (📌):
Use a sample point like $(1,1)$ to visually track composite transformations step-by-step.

Correct Answer: A. 43

Fastest Approach (🚀):
$S_1 = 30 \times 26 = 780$, $S_2 = 34 \times 28 = 952 \implies \Delta S = 172 \implies \bar{x}_{\text{new}} = 172/4 = 43$.
Matches Option A.

Step-by-Step Breakdown:

1. Calculating Initial Total Age

$\text{Total age of 30 workers} = 30 \times 26 = 780$

2. Calculating New Total Age

Four new members join, so the group size becomes $30 + 4 = 34$.
The new mean age is $26 + 2 = 28$.
$\text{New total age} = 34 \times 28 = 952$

3. Finding Mean Age of New Workers

$\text{Sum of ages of 4 new workers} = 952 - 780 = 172$
$\text{Mean age of new workers} = \frac{172}{4} = 43$

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. 48 — Fails the Check
  Gives a total of $780+192=972$ over $34$ workers - a mean of $28.6$, not $28$.
• C. 63 — Fails the Check
  Gives $\tfrac{780+252}{34}=30.4$.
• D. 45 — Fails the Check
  Gives $\tfrac{780+180}{34}=28.2$. Close, but the rise is stated as exactly $2$.
• E. 38 — Fails the Check
  Gives $\tfrac{780+152}{34}=27.4$, a rise of only $1.4$.

Common Mistake (⚠️):
Adding $2$ to the old mean and assuming the new workers have average age $26+2=28$.

Takeaway (📌):
Always convert mean problems to sum problems: $\text{Sum} = n \times \bar{x}$.

Correct Answer: B. 115

Fastest Approach (🚀):
$a b = 6$. Since $y=23b$ is odd, $b=3, a=2 \implies x=46, y=69 \implies x+y = 115$.
Matches Option B.

Step-by-Step Breakdown:

1. GCD and LCM Properties

Let $d = \text{GCD}(x,y) = 23$.
We can write $x = 23a$ and $y = 23b$, where $a$ and $b$ are coprime positive integers with $a < b$.

The LCM is given by:
$\text{LCM}(x,y) = 23 \times a \times b = 138$
$a \times b = \frac{138}{23} = 6$

2. Finding Co-prime Pairs $(a,b)$

The factor pairs of $6$ with $a < b$ are:

1. $(a,b) = (1,6) \implies x = 23, y = 138$ (even $y$)
2. $(a,b) = (2,3) \implies x = 46, y = 69$ (odd $y$)

Since $y$ is an odd number, we must select $(a,b) = (2,3)$.

3. Calculating $x+y$

$x = 23 \times 2 = 46$
$y = 23 \times 3 = 69$
$x + y = 46 + 69 = 115$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. 161 — Parity Ignored
  Takes $(a,b)=(1,6)$, so $y=138$ - an even number, and the question requires $y$ odd.
• C. 92 — Not a Valid Pair
  No admissible pair gives this. With $ab=6$ and $y$ odd, the only choice is $(a,b)=(2,3)$, so $x+y=46+69$.
• D. 207 — Product Wrong
  $9\times23$, which needs $a+b=9$. The constraint is $ab=\tfrac{138}{23}=6$.
• E. 69 — One Value Only
  $y$ on its own. The question asks for $x+y$.

Common Mistake (⚠️):
Selecting $(a,b) = (1,6)$ without checking that $y$ must be odd, leading to $x+y=161$.

Takeaway (📌):
Express $x, y$ in terms of GCD $d$: $x=da, y=db$. Then $\text{LCM} = d \cdot a \cdot b$.

Correct Answer: C. $\frac{3^3-2^3}{3^3+3^2+3+1}$

Fastest Approach (🚀):
Option C $= \frac{27-8}{27+9+3+1} = \frac{19}{40} = 0.475 < 0.5 = \text{Option A}$. Smallest is Option C.

Matches Option C.

Step-by-Step Breakdown:

1. Evaluating Each Option

• Option A: $\sin(45^\circ)^2 = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} = 0.500$
• Option B: $\frac{\sqrt[3]{28}}{6}$. Since $3^3 = 27$, $\sqrt[3]{28} \approx 3.036$. So $\frac{3.036}{6} \approx 0.506$
• Option C: $\frac{3^3 - 2^3}{3^3 + 3^2 + 3 + 1} = \frac{27 - 8}{27 + 9 + 3 + 1} = \frac{19}{40} = 0.475$
• Option D: $2^{-\frac{\pi}{4}} \approx 2^{-0.785} > 2^{-1} = 0.500$ (around $0.580$)
• Option E: $\frac{\tan 60^\circ - \tan 30^\circ}{2} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{2} = \frac{\frac{2}{\sqrt{3}}}{2} = \frac{1}{\sqrt{3}} \approx 0.577$

2. Comparison

Comparing the values:
$0.475 < 0.500 < 0.506 < 0.577 < 0.580$

The smallest value is Option C.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $\sin(45^\circ)^2$ — Not Smallest
  $\left(\tfrac{1}{\sqrt2}\right)^{2}=0.5$, above C's $\tfrac{19}{40}=0.475$.
• B. $\frac{\sqrt[3]{28}}{6}$ — Not Smallest
  $\sqrt[3]{28}\approx3.04$, so this is about $0.506$.
• D. $2^{-\frac{\pi}{4}}$ — Exponent Misjudged
  $-\tfrac{\pi}{4}\approx-0.785$, and $2^{-0.785}\approx0.58$. A negative exponent smaller than $1$ in size keeps the value above $\tfrac12$.
• E. $\frac{\tan(60^\circ)-\tan(30^\circ)}{2}$ — Not Smallest
  $\tfrac{\sqrt3-\tfrac{1}{\sqrt3}}{2}=\tfrac{1}{\sqrt3}\approx0.577$.

Common Mistake (⚠️):
Assuming $2^{-\pi/4}$ is less than $0.5$ because $-\pi/4$ is negative.

Takeaway (📌):
Compare fractional expressions directly: $19/40 = 0.475 < 1/2 = 0.5$ provides an instant lower bound.

Correct Answer: F. 23/60

Fastest Approach (🚀):
$A_{\text{square}} = 1$. Outer shapes sum to $37/60 \implies \text{Area} = 23/60$.
Matches Option F.

Step-by-Step Breakdown:

1. Setting Up Coordinates

Let $Z = (0,0)$, $Y = (1,0)$, $X = (1,1)$, $W = (0,1)$ on a square of side length $1$.

• $WM:MX = 1:2 \implies M = \left(\frac{1}{3}, 1\right)$
• $XN:NY = 3:1 \implies N = \left(1, \frac{1}{4}\right)$
• $YP:PZ = 4:1 \implies P = \left(\frac{1}{5}, 0\right)$

2. Calculating Outer Areas

The triangle $MNP$ is inside the square. We subtract the 3 surrounding outer shapes:

1. $\triangle MXN$: base $MX = 2/3$, height $XN = 3/4$.

$\text{Area}_1 = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{1}{4} = \frac{15}{60}$

1. $\triangle NYP$: base $NY = 1/4$, height $YP = 4/5$.

$\text{Area}_2 = \frac{1}{2} \times \frac{1}{4} \times \frac{4}{5} = \frac{1}{10} = \frac{6}{60}$

1. Trapezium $WMPZ$: base $ZP = 1/5$, top $WM = 1/3$, height $WZ = 1$.

$\text{Area}_3 = \frac{1}{2} \left(\frac{1}{5} + \frac{1}{3}\right) \times 1 = \frac{1}{2} \left(\frac{8}{15}\right) = \frac{4}{15} = \frac{16}{60}$

3. Calculating Area of $\triangle MNP$

$\text{Area}(MNP) = 1 - \left(\frac{15}{60} + \frac{6}{60} + \frac{16}{60}\right) = 1 - \frac{37}{60} = \frac{23}{60}$

Matches Option F.

Why the Other Options Are Wrong (❌):

• A. 1/3 — Corners Miscounted
  $\tfrac{20}{60}$. The three corner triangles remove $\tfrac{37}{60}$ of the unit square between them, leaving $\tfrac{23}{60}$.
• B. 2/5 — Arithmetic Slip
  $\tfrac{24}{60}$ - one sixtieth over.
• C. 9/20 — Corners Miscounted
  $\tfrac{27}{60}$, which would need the corner triangles to total only $\tfrac{33}{60}$.
• D. 1/30 — Magnitude
  $\tfrac{2}{60}$ - far too small for a triangle with a vertex on each of three different sides of a unit square.
• E. 19/60 — Ratio Misread
  One sixtieth short, which is what reading $WM:MX=1:2$ as $WM=\tfrac12$ rather than $\tfrac13$ produces.

Common Mistake (⚠️):
Using incorrect ratio splits (e.g. treating $WM:MX = 1:2$ as length $1/2$ instead of $1/3$).

Takeaway (📌):
A ratio $a:b$ on a line segment of length $L$ yields sub-lengths $\frac{a}{a+b}L$ and $\frac{b}{a+b}L$.

Correct Answer: E. $\frac{1}{2} = \frac{1}{x} + \frac{1}{y}$

Fastest Approach (🚀):
Divide $xy = 2(x+y)$ by $2xy$: $\frac{1}{2} = \frac{x+y}{xy} = \frac{1}{x} + \frac{1}{y}$.
Matches Option E.

Step-by-Step Breakdown:

1. Formulating the Given Relation

The statement 'product is double their sum' translates to:
$xy = 2(x + y)$

2. Dividing by $xy$

Assuming non-zero $x$ and $y$, divide both sides by $xy$:
$1 = \frac{2(x + y)}{xy}$
$1 = 2 \left(\frac{x}{xy} + \frac{y}{xy}\right)$
$1 = 2 \left(\frac{1}{y} + \frac{1}{x}\right)$

3. Isolating the Reciprocal Sum

$\frac{1}{x} + \frac{1}{y} = \frac{1}{2}$

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\frac{1}{x} + \frac{1}{y} = 2$ — Reciprocal Flipped
  Dividing $xy=2(x+y)$ by $xy$ gives $1=2\left(\tfrac1x+\tfrac1y\right)$, so the sum of reciprocals is $\tfrac12$.
• B. $\frac{1}{x+y} = 2xy$ — Inverted
  Inverts one side only. From $xy=2(x+y)$, $\tfrac{1}{x+y}=\tfrac{2}{xy}$.
• C. $y = \frac{2x}{2-y}$ — Variable Mixed
  Leaves $y$ on both sides. Solving for $y$ gives $y=\tfrac{2x}{x-2}$.
• D. $x = \frac{2y}{2-y}$ — Denominator Sign
  Sign in the denominator: $xy=2x+2y$ rearranges to $x=\tfrac{2y}{y-2}$.

Common Mistake (⚠️):
Failing to divide by $xy$ and instead trying to solve for $y$ as $y = \frac{2x}{x-2}$.

Takeaway (📌):
Whenever an equation relates $xy$ and $x+y$, divide by $xy$ to convert to sum of reciprocals $\frac{1}{x} + \frac{1}{y}$.

Correct Answer: A. $2^{16}$

Fastest Approach (🚀):
$2^{4x-y} = 2^{16}$.
Matches Option A.

Step-by-Step Breakdown:

1. Base Conversion to Base 2

Express $16^x$ as a power of $2$:
$16^x = (2^4)^x = 2^{4x}$

2. Quotient Law of Indices

$\frac{16^x}{2^y} = \frac{2^{4x}}{2^y} = 2^{4x - y}$

3. Substituting the Given Condition

We are given $4x - y = 16$:
$2^{4x - y} = 2^{16}$

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $4^6$ — Wrong Exponent
  $4^{6}=2^{12}$, which would need $4x-y=12$.
• C. $8^5$ — Wrong Exponent
  $8^{5}=2^{15}$ - one power short.
• D. $2^{14}$ — Wrong Exponent
  Two powers short. $\tfrac{16^{x}}{2^{y}}=2^{4x-y}$, and $4x-y$ is given as $16$.

Common Mistake (⚠️):
Writing $16^x = 4^{2x}$ and mixing up bases 4 and 2.

Takeaway (📌):
Convert all numbers in index questions to the smallest common prime base (here base 2).

Correct Answer: A. $4 \times 10^3$

Fastest Approach (🚀):
$2^{12} = 4096 \approx 4 \times 10^3$.
Matches Option A.

Step-by-Step Breakdown:

1. Factoring Powers of 2

Recall that $2^{10} = 1024 \approx 10^3$.
Break down $2^{12}$:
$2^{12} = 2^2 \times 2^{10} = 4 \times 1024 = 4096$

2. Expressing in Standard Form

$4096 \approx 4000 = 4 \times 10^3$

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $4 \times 10^4$ — Order of Magnitude
  $40{,}000$, roughly ten times $2^{12}=4096$.
• C. $4 \times 10^5$ — Order of Magnitude
  $400{,}000$, which is about $2^{18.6}$.
• D. $4 \times 10^6$ — Order of Magnitude
  $4\times10^{6}$ is around $2^{22}$.
• E. $4 \times 10^7$ — Order of Magnitude
  $4\times10^{7}$ is around $2^{25}$. The anchor worth memorising is $2^{10}\approx1000$.

Common Mistake (⚠️):
Confusing exponent powers of 10, e.g. selecting $4 \times 10^4 = 40,000$.

Takeaway (📌):
Use $2^{10} = 1024 \approx 10^3$ as an instant benchmark for large powers of 2.

Correct Answer: A. $17+12\sqrt{2}$

Fastest Approach (🚀):
$\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)^2 = (3+2\sqrt{2})^2 = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2}$.
Matches Option A.

Step-by-Step Breakdown:

1. Factoring Common Radicals

Rewrite $\sqrt{12} = \sqrt{6}\sqrt{2}$:
$\frac{\sqrt{12}+\sqrt{6}}{\sqrt{12}-\sqrt{6}} = \frac{\sqrt{6}(\sqrt{2}+1)}{\sqrt{6}(\sqrt{2}-1)} = \frac{\sqrt{2}+1}{\sqrt{2}-1}$

2. Rationalizing the Base

Multiply by the conjugate $(\sqrt{2}+1)$:
$\frac{(\sqrt{2}+1)(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)} = \frac{(\sqrt{2}+1)^2}{2 - 1} = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2}$

3. Squaring the Result

$(3 + 2\sqrt{2})^2 = 3^2 + 2(3)(2\sqrt{2}) + (2\sqrt{2})^2$
$= 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2}$

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $17-12\sqrt{2}$ — Fraction Inverted
  The reciprocal of the answer, and about $0.03$. The larger sum sits on top, so the value must exceed $1$.
• C. $17+12\sqrt{3}$ — Wrong Surd
  $\sqrt3$ cannot appear: factoring $\sqrt6$ from both parts leaves $\tfrac{\sqrt2+1}{\sqrt2-1}$.
• D. $17-12\sqrt{3}$ — Negative Square
  Negative, at about $-3.8$, and a squared quantity cannot be negative.

Common Mistake (⚠️):
Expanding the squares of top and bottom immediately without simplifying $\sqrt{6}$ first.

Takeaway (📌):
Always simplify the interior fraction before applying high powers.

Correct Answer: E. Drawing 2 consecutive diamonds with replacement

Fastest Approach (🚀):
$P(E) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} = 0.0625$, which is smaller than $P(B) = 1/13 \approx 0.0769$.
Matches Option E.

Step-by-Step Breakdown:

1. Calculating Probabilities for All Options

• Option A (Two dice score $\ge 11$): Outcomes are $(5,6), (6,5), (6,6) \implies 3/36 = 1/12 \approx 0.0833$.
• Option B (Ace from 52 cards): $4/52 = 1/13 \approx 0.0769$.
• Option C (3 dice even): $(1/2)^3 = 1/8 = 0.125$.
• Option D (Same suit without replacement): $12/51 \approx 0.2353$.
• Option E (2 consecutive diamonds with replacement): $(13/52)^2 = (1/4)^2 = 1/16 = 0.0625$.

2. Comparison

$0.0625 < 0.0769 < 0.0833 < 0.125 < 0.2353$

The least likely event is Option E.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. Rolling two regular, 6-sided dice and scoring $\ge 11$ — Not Smallest
  Scoring $11$ or more means $(5,6)$, $(6,5)$ or $(6,6)$ - three of thirty-six, about $0.083$.
• B. Drawing an Ace from a deck of 52 cards — Not Smallest
  $\tfrac{4}{52}=\tfrac{1}{13}\approx0.077$, just above E's $0.0625$.
• C. Rolling 3 dice all landing on even numbers — Not Smallest
  $\left(\tfrac12\right)^{3}=0.125$ - the likeliest of the low-probability options.
• D. Drawing two cards of the same suit, without replacement — Most Likely
  The second card matches the first's suit with probability $\tfrac{12}{51}\approx0.235$, much the likeliest here.

Common Mistake (⚠️):
Confusing replacement vs non-replacement in card probabilities.

Takeaway (📌):
Convert all probabilities to simple fractions with small denominators to compare quickly.

Correct Answer: B. $9\sqrt{2}$

Fastest Approach (🚀):
$k = \frac{3\sqrt{6}}{\sqrt{3}} = 3\sqrt{2} \implies \text{Area} = \frac{1}{2} (3\sqrt{2}) (3\sqrt{2} \times \sqrt{2}) = 9\sqrt{2}$.
Matches Option B.

Step-by-Step Breakdown:

1. Ratio of Triangle Sides

Since $\tan\theta = \sqrt{2} = \frac{\text{Opposite}}{\text{Adjacent}}$, let adjacent side $= x$ and opposite side $= x\sqrt{2}$.
By Pythagoras, hypotenuse $H = \sqrt{x^2 + (x\sqrt{2})^2} = \sqrt{3x^2} = x\sqrt{3}$.

2. Finding Side Lengths

Given $H = 3\sqrt{6}\ \text{cm}$:
$x\sqrt{3} = 3\sqrt{6} \implies x = \frac{3\sqrt{6}}{\sqrt{3}} = 3\sqrt{2}\ \text{cm}$

The two perpendicular legs are:

• Base $= x = 3\sqrt{2}\ \text{cm}$
• Height $= x\sqrt{2} = 3\sqrt{2} \times \sqrt{2} = 6\ \text{cm}$

3. Calculating Area

$\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times (3\sqrt{2}) \times 6 = 9\sqrt{2}\ \text{cm}^2$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\frac{9\sqrt{2}}{2}$ — Halved
  Half the answer - the factor $\tfrac12$ applied twice.
• C. $9\sqrt{3}$ — Wrong Surd
  $\sqrt3$ belongs to the hypotenuse relation $k\sqrt3$, not to the area $\tfrac12(3\sqrt2)(6)$.
• D. $9\sqrt{6}$ — Hypotenuse As Leg
  Uses the hypotenuse $3\sqrt6$ as though it were a leg.
• E. $18\sqrt{2}$ — Factor of Two
  Twice the answer - the $\tfrac12$ omitted from $\tfrac12\times\text{base}\times\text{height}$.

Common Mistake (⚠️):
Assuming the hypotenuse is $x\sqrt{2}$ instead of $x\sqrt{3}$.

Takeaway (📌):
If $\tan\theta = m$, the sides of a right triangle are in ratio $1 : m : \sqrt{1+m^2}$.

Correct Answer: D. 0.3

Fastest Approach (🚀):
$x^2 = 180/80 = 9/4 \implies x = 3/2 = 1.5 \implies z = 0.2(1.5) = 0.3$.
Matches Option D.

Step-by-Step Breakdown:

1. Finding Relationship for $y$ and $x$

Given $y \propto \frac{1}{x^2} \implies y = \frac{k_1}{x^2}$.
When $y = 20, x = 3 \implies 20 = \frac{k_1}{9} \implies k_1 = 180$.
Thus $y = \frac{180}{x^2}$.

2. Finding Relationship for $z$ and $x$

Given $z \propto x \implies z = k_2 x$.
When $z = 1.2, x = 6 \implies 1.2 = 6 k_2 \implies k_2 = 0.2$.
Thus $z = 0.2 x$.

3. Evaluating $z$ when $y = 80$

$80 = \frac{180}{x^2} \implies x^2 = \frac{180}{80} = \frac{9}{4} \implies x = \frac{3}{2} = 1.5$
$z = 0.2 \times 1.5 = 0.3$

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. 0.0225 — Squared
  $0.15^{2}$. $z$ is directly proportional to $x$, so it is linear, never squared.
• B. 0.03 — Decimal Slip
  A tenth of the answer, from $x=0.15$ instead of $1.5$.
• C. 0.15 — Constant Halved
  Half the answer, from $z=0.1x$. The constant is $\tfrac{1.2}{6}=0.2$.
• E. 0.36 — Root Slip
  Uses $x=1.8$. From $x^{2}=\tfrac{180}{80}=2.25$ the positive root is exactly $1.5$.
• F. 0.6 — Old Value Used
  Twice the answer, from $x=3$ - the original value rather than the new one.

Common Mistake (⚠️):
Taking $x = -1.5$ despite the prompt stating $x, y, z$ are positive variables.

Takeaway (📌):
Determine constant of proportionality $k$ immediately from given initial conditions before solving target values.

Correct Answer: B. $a=2$ or $a=-3$

Fastest Approach (🚀):
$2(a^2+a) = 12 \implies a^2+a-6=0 \implies (a+3)(a-2)=0 \implies a=2, -3$.
Matches Option B.

Step-by-Step Breakdown:

1. Formulating Equations from Substitutions

Let $P(x) = x^3 + ax^2 + bx + a^2$.

1. For $x = 1$:

$P(1) = 1^3 + a(1)^2 + b(1) + a^2 = 5$
$a^2 + a + b = 4 \quad \text{--- (Equation 1)}$

1. For $x = -1$:

$P(-1) = (-1)^3 + a(-1)^2 + b(-1) + a^2 = 7$
$-1 + a - b + a^2 = 7 \implies a^2 + a - b = 8 \quad \text{--- (Equation 2)}$

2. Eliminating $b$ and Solving for $a$

Add Equation 1 and Equation 2:
$(a^2 + a + b) + (a^2 + a - b) = 4 + 8$
$2a^2 + 2a = 12$
$a^2 + a - 6 = 0$

Factor the quadratic:
$(a + 3)(a - 2) = 0 \implies a = 2 \quad \text{or} \quad a = -3$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $a=-2$ — Not a Root
  $-2$ is not a root of $a^{2}+a-6$: it gives $4-2-6=-4$.
• C. $a=1$ or $a=0$ — Not a Root
  Neither satisfies $a^{2}+a-6=0$; they give $-4$ and $-6$.
• D. $a=-3$ — Root Missed
  One of the two roots. $(a+3)(a-2)=0$ also allows $a=2$.
• E. No possible values — Premature Rejection
  Adding the two evaluations gives $2a^{2}+2a=12$, which factorises cleanly into two real roots.

Common Mistake (⚠️):
Miscalculating $(-1)^3 = -1$ or $(-1)^2 = 1$ when evaluating $P(-1)$.

Takeaway (📌):
Add and subtract symmetric equations $P(1)$ and $P(-1)$ to separate even powers ($a^2+a$) from odd powers ($b$).