ESAT Mathematics Mock Module 1 Worked Solutions

Correct Answer: C. $\sqrt{5}$

Fastest Approach (🚀):
Common denominator $= 7-5 = 2$.
Numerator $= (\sqrt{7}+\sqrt{5})-(\sqrt{7}-\sqrt{5}) = 2\sqrt{5}$.
So the value is $\dfrac{2\sqrt{5}}{2} = \sqrt{5}$.

Matches Option C.

Step-by-Step Breakdown:

1. Recognise the conjugate pair

The two denominators $\sqrt{7}-\sqrt{5}$ and $\sqrt{7}+\sqrt{5}$ are conjugates, so their product is a difference of two squares:
$(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5}) = (\sqrt{7})^2-(\sqrt{5})^2 = 7-5 = 2$

2. Combine over the common denominator

$\dfrac{1}{\sqrt{7}-\sqrt{5}}-\dfrac{1}{\sqrt{7}+\sqrt{5}} = \dfrac{(\sqrt{7}+\sqrt{5})-(\sqrt{7}-\sqrt{5})}{2}$

3. Simplify the numerator

The $\sqrt{7}$ terms cancel and the $\sqrt{5}$ terms add:
$(\sqrt{7}+\sqrt{5})-(\sqrt{7}-\sqrt{5}) = 2\sqrt{5}$

4. Final value

$\dfrac{2\sqrt{5}}{2} = \sqrt{5}$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $\sqrt{7}$ — Sign Error
  Cancelling the $\sqrt{5}$ terms instead of the $\sqrt{7}$ terms in the numerator.
• B. $2\sqrt{5}$ — Incomplete Simplification
  Forgetting to divide the numerator by the common denominator of 2.
• D. $\dfrac{\sqrt{5}}{2}$ — Arithmetic Error
  Dividing $\sqrt{5}$ by 2 as well — the factor 2 has already been used once.
• E. $2\sqrt{7}$ — Sign Error
  Combining the numerators by adding rather than subtracting the second fraction.

Common Mistake (⚠️):
Rationalising each fraction separately. That works, but it doubles the algebra and invites a sign slip when the second numerator $\sqrt{7}-\sqrt{5}$ is subtracted.

Takeaway (📌):
When conjugate surds appear in separate denominators, combine first: the denominator becomes an integer in one line.

Correct Answer: A. $100$

Fastest Approach (🚀):
Centre $(3,-4)$, radius $5$.
$d = \sqrt{3^2+4^2} = 5$.
Maximum distance $= d + r = 10$, so maximum of $x^2+y^2$ is $100$.

Matches Option A.

Step-by-Step Breakdown:

1. Complete the square

$x^2-6x+y^2+8y = 0$
$(x-3)^2-9+(y+4)^2-16 = 0$
$(x-3)^2+(y+4)^2 = 25$

So the centre is $C(3,-4)$ and the radius is $r = 5$.

2. Interpret the target expression

For a point $P(x,y)$, $x^2+y^2 = OP^2$, the square of its distance from the origin $O$.

3. Maximum distance from a point to a circle

$OC = \sqrt{3^2+(-4)^2} = 5$, so the origin lies on the circle itself. The farthest point of a circle from any point is along the line through the centre:
$OP_{\max} = OC + r = 5+5 = 10$

4. Square the maximum distance

$\left(x^2+y^2\right)_{\max} = 10^2 = 100$

The maximising point is the end of the diameter through the origin, namely $(6,-8)$.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $25$ — Conceptual Error
  Quoting $r^2 = 25$ — the radius squared, not the maximum squared distance from the origin.
• C. $50$ — Formula Misuse
  Using $d^2 + r^2 = 25+25$ instead of $(d+r)^2$.
• D. $10$ — Incomplete Answer
  Giving the maximum distance 10 rather than its square.
• E. $75$ — Arithmetic Error
  Computing $(d+r)^2$ but with $d+r$ mis-evaluated, or mixing $3r^2$.

Common Mistake (⚠️):
Reporting the maximum distance $10$ instead of its square, or stopping at $r^2 = 25$ having confused the radius with the distance from the origin.

Takeaway (📌):
Whenever $x^2+y^2$ appears, ask whether it is a squared distance. Constrained maxima on circles then need no calculus at all.

Correct Answer: D. $\dfrac{9}{14}$

Fastest Approach (🚀):
$P(\text{both blue}) = \dfrac58\times\dfrac47 = \dfrac{20}{56} = \dfrac{5}{14}$.
$P(\text{at least one red}) = 1-\dfrac{5}{14} = \dfrac{9}{14}$.

Matches Option D.

Step-by-Step Breakdown:

1. Choose the complement

The event 'at least one red' is the complement of 'no reds', that is 'both counters blue'. One calculation replaces three.

2. Probability that both are blue

Without replacement, the second probability is conditional on the first:
$P(\text{blue then blue}) = \dfrac{5}{8}\times\dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$

3. Subtract from 1

$P(\text{at least one red}) = 1-\dfrac{5}{14} = \dfrac{9}{14}$

4. Check by direct counting

$\binom{8}{2}=28$ equally likely pairs; $\binom{5}{2}=10$ are all-blue, so $28-10=18$ contain a red, and $\dfrac{18}{28}=\dfrac{9}{14}$. The two routes agree.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{3}{8}$ — Misread Question
  Giving the probability that a single counter drawn is red.
• B. $\dfrac{5}{14}$ — Complement Inversion
  Giving the complement itself: the probability that both counters are blue.
• C. $\dfrac{15}{28}$ — Incomplete Case Analysis
  Treating the draws as with replacement, or summing the two 'exactly one red' orderings only.
• E. $\dfrac{13}{28}$ — Arithmetic Error
  Arithmetic slip when converting $\tfrac{20}{56}$ before subtracting from 1.

Common Mistake (⚠️):
Using $\tfrac58\times\tfrac57$ and forgetting that the counter is not replaced, or adding $P(\text{exactly one red})$ and $P(\text{two red})$ but missing one of the two orderings for 'exactly one'.

Takeaway (📌):
'At least one' almost always means $1-P(\text{none})$. It also removes the risk of double-counting orderings.

Correct Answer: B. $3367$

Fastest Approach (🚀):
$\dfrac{100\times 101}{2} = 5050$.
Multiples of 3 up to 100: $3+6+\cdots+99 = 3\times\dfrac{33\times 34}{2} = 3\times 561 = 1683$.
$5050-1683 = 3367$.

Matches Option B.

Step-by-Step Breakdown:

1. Sum the full range

$\sum_{k=1}^{100} k = \dfrac{100\times 101}{2} = 5050$

2. Identify the unwanted subset

The multiples of 3 in the range are $3,6,\ldots,99$ — that is $3\times 1$ up to $3\times 33$, so there are $33$ of them.

3. Sum the unwanted subset

Factor the 3 out first:
$3(1+2+\cdots+33) = 3\times\dfrac{33\times 34}{2} = 3\times 561 = 1683$

4. Subtract

$5050-1683 = 3367$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $1683$ — Complement Inversion
  Giving the sum of the multiples of 3 instead of the sum of the rest.
• C. $3350$ — Arithmetic Error
  Using $33$ multiples but averaging incorrectly, e.g. $3\times 550$.
• D. $3400$ — Off-by-one Error
  Rounding or using 34 multiples of 3 (including 102, which is out of range).
• E. $5050$ — Incomplete Answer
  Giving the total sum from 1 to 100 without removing the multiples of 3.

Common Mistake (⚠️):
Answering $1683$ — the sum that was supposed to be removed — or counting $34$ multiples of 3 by including $102$.

Takeaway (📌):
'Not', 'at least', and 'excluding' are all signals for complementary counting: total minus the easy bad case.

Correct Answer: B. $5$

Fastest Approach (🚀):
$8 = 2^3$, $4 = 2^2$.
$2^{3x+3} = 2^{4x-2}$.
Equate indices: $3x+3 = 4x-2 \implies x = 5$.

Matches Option B.

Step-by-Step Breakdown:

1. Write both sides in base 2

$8 = 2^3$ and $4 = 2^2$, so
$8^{\,x+1} = (2^3)^{x+1} = 2^{3x+3}$
$4^{\,2x-1} = (2^2)^{2x-1} = 2^{4x-2}$

2. Equate the indices

An exponential function is one-to-one, so equal powers of the same base force equal exponents:
$3x+3 = 4x-2$

3. Solve

$3 + 2 = 4x-3x \implies x = 5$

4. Check

$8^{6} = 2^{18}$ and $4^{9} = 2^{18}$. Both sides agree.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $1$ — Index Law Error
  Dropping the multiplication in one index, solving $3x+1 = 4x-2$ style variants.
• C. $-5$ — Sign Error
  Sign slip when collecting terms, giving the negative of the correct root.
• D. $2$ — Bracket Error
  Equating bases instead of indices, or solving $3(x+1)=2(2x-1)$ with a dropped bracket.
• E. $3$ — Conceptual Error
  Guessing from the base values 8 and 4 rather than solving the linear equation.

Common Mistake (⚠️):
Multiplying the outer power into only one bracket, for example writing $8^{x+1}=2^{3x+1}$, which loses the $3\times 1$ term.

Takeaway (📌):
Any equation mixing $2,4,8,16$ (or $3,9,27$) is a common-base question. Convert first, then it is linear.

Correct Answer: D. $\dfrac{7}{3}$

Fastest Approach (🚀):
For $ax^2+bx+c=0$: $\alpha+\beta = -b/a = 7/2$ and $\alpha\beta = c/a = 3/2$.
$\dfrac{1}{\alpha}+\dfrac{1}{\beta} = \dfrac{\alpha+\beta}{\alpha\beta} = \dfrac{7/2}{3/2} = \dfrac{7}{3}$.

Matches Option D.

Step-by-Step Breakdown:

1. Read the symmetric functions off the coefficients

For $ax^2+bx+c=0$ with roots $\alpha,\beta$:
$\alpha+\beta = -\dfrac{b}{a}, \qquad \alpha\beta = \dfrac{c}{a}$

Here $a=2$, $b=-7$, $c=3$, so
$\alpha+\beta = \dfrac{7}{2}, \qquad \alpha\beta = \dfrac{3}{2}$

2. Rewrite the target expression

$\dfrac{1}{\alpha}+\dfrac{1}{\beta} = \dfrac{\beta+\alpha}{\alpha\beta}$

3. Substitute

$\dfrac{\alpha+\beta}{\alpha\beta} = \dfrac{7/2}{3/2} = \dfrac{7}{2}\times\dfrac{2}{3} = \dfrac{7}{3}$

4. Sanity check

The roots are $3$ and $\tfrac{1}{2}$, so $\dfrac{1}{3}+2 = \dfrac{7}{3}$. The shortcut agrees with the direct calculation.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{3}{7}$ — Inversion Error
  Inverting the ratio: computing $\alpha\beta/(\alpha+\beta)$ instead.
• B. $\dfrac{2}{3}$ — Formula Misuse
  Using $c/b$ rather than forming the ratio of the two Vieta results.
• C. $\dfrac{7}{2}$ — Incomplete Answer
  Stopping at $\alpha+\beta = 7/2$ and never dividing by $\alpha\beta$.
• E. $\dfrac{14}{3}$ — Coefficient Error
  Dropping the factor $a=2$ from $\alpha\beta$, using $\alpha\beta = 3$ then doubling.

Common Mistake (⚠️):
Factorising to find $\alpha=3$, $\beta=\tfrac12$ first. It gives the right answer here, but on the ESAT the quadratic often does not factorise and the Vieta route is the only fast one.

Takeaway (📌):
Any symmetric expression in the roots — $\tfrac1\alpha+\tfrac1\beta$, $\alpha^2+\beta^2$, $\alpha^3+\beta^3$ — should be converted to $\alpha+\beta$ and $\alpha\beta$ and read straight off the coefficients.

Correct Answer: C. $-\dfrac{\sqrt{6}}{12}$

Fastest Approach (🚀):
$\sin 150^\circ = \tfrac12$ (second quadrant, sine positive).
$\cos 225^\circ = -\tfrac{\sqrt2}{2}$ (third quadrant, cosine negative).
$\tan 240^\circ = \tan 60^\circ = \sqrt3$ (third quadrant, tangent positive).
$\dfrac{\tfrac12 \times \left(-\tfrac{\sqrt2}{2}\right)}{\sqrt3} = \dfrac{-\tfrac{\sqrt2}{4}}{\sqrt3} = -\dfrac{\sqrt2}{4\sqrt3} = -\dfrac{\sqrt6}{12}$.

Matches Option C.

Step-by-Step Breakdown:

1. Reduce each angle to its reference angle

$150^\circ = 180^\circ-30^\circ$ (second quadrant, reference $30^\circ$)
$225^\circ = 180^\circ+45^\circ$ (third quadrant, reference $45^\circ$)
$240^\circ = 180^\circ+60^\circ$ (third quadrant, reference $60^\circ$)

2. Apply the quadrant signs (ASTC)

In the second quadrant only sine is positive; in the third quadrant only tangent is positive.
$\sin 150^\circ = +\sin 30^\circ = \tfrac12$
$\cos 225^\circ = -\cos 45^\circ = -\tfrac{\sqrt2}{2}$
$\tan 240^\circ = +\tan 60^\circ = \sqrt3$

3. Substitute

$\dfrac{\tfrac12 \times \left(-\tfrac{\sqrt2}{2}\right)}{\sqrt3} = \dfrac{-\tfrac{\sqrt2}{4}}{\sqrt3} = -\dfrac{\sqrt2}{4\sqrt3}$

4. Rationalise

$-\dfrac{\sqrt2}{4\sqrt3} \times \dfrac{\sqrt3}{\sqrt3} = -\dfrac{\sqrt6}{12}$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $-\dfrac{\sqrt{2}}{4}$ — Incomplete Answer
  Leaving the answer as $-\tfrac{\sqrt2}{4}$ — the numerator only, with the division by $\sqrt3$ forgotten.
• B. $\dfrac{\sqrt{6}}{12}$ — Sign Error
  Correct magnitude but $\tan 240^\circ$ or $\cos 225^\circ$ given the wrong sign.
• D. $-\dfrac{\sqrt{6}}{6}$ — Arithmetic Error
  Rationalising incorrectly: $\tfrac{\sqrt2}{4\sqrt3}$ simplified to $\tfrac{\sqrt6}{6}$ rather than $\tfrac{\sqrt6}{12}$.
• E. $-\dfrac{1}{4}$ — Value Error
  Treating $\cos 225^\circ$ as $-\tfrac12$ and $\tan 240^\circ$ as $1$.

Common Mistake (⚠️):
Taking $\tan 240^\circ$ as negative. $240^\circ$ is in the third quadrant, where sine and cosine are both negative and their ratio is therefore positive.

Takeaway (📌):
Reference angle first, sign second. Splitting the two decisions removes almost every exact-value error.

Correct Answer: A. $400$

Fastest Approach (🚀):
$a-b = 6$, $a+b = 800$.
$403^2-397^2 = 6\times 800 = 4800$.
$\dfrac{4800}{12} = 400$.

Matches Option A.

Step-by-Step Breakdown:

1. Apply the difference of two squares

For any $a$ and $b$,
$a^2-b^2 = (a-b)(a+b)$

With $a = 403$ and $b = 397$:
$a-b = 6, \qquad a+b = 800$

2. Evaluate the numerator

$403^2-397^2 = 6\times 800 = 4800$

3. Divide

$\dfrac{4800}{12} = 400$

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $200$ — Double Counting
  Halving the correct result, usually by using 6 as the divisor as well as the difference.
• C. $800$ — Incomplete Simplification
  Using the sum 800 alone and dividing by 1 instead of completing the division by 12.
• D. $300$ — Arithmetic Error
  Mis-computing $a+b$ as 600, giving $6\times 600/12$.
• E. $4800$ — Incomplete Answer
  Stopping at the numerator and forgetting to divide by 12.

Common Mistake (⚠️):
Squaring $403$ and $397$ in full. Both squares are six-figure numbers, the subtraction is error-prone, and it costs well over a minute.

Takeaway (📌):
Any time you see one square subtracted from another, factorise. The gap $(a-b)$ is usually small and the sum $(a+b)$ is usually round.

Correct Answer: E. $\dfrac{1}{4}$

Fastest Approach (🚀):
Midpoint $= \left(\tfrac{-2+4}{2},\tfrac{5-3}{2}\right) = (1,1)$.
$m_{AB} = \dfrac{-3-5}{4-(-2)} = \dfrac{-8}{6} = -\dfrac43 \implies m_\perp = \dfrac34$.
$y = 1+\tfrac34(x-1)$; at $x=0$, $y = 1-\tfrac34 = \tfrac14$.

Matches Option E.

Step-by-Step Breakdown:

1. Midpoint of AB

$M = \left(\dfrac{-2+4}{2},\ \dfrac{5+(-3)}{2}\right) = (1,1)$

2. Gradient of AB

$m_{AB} = \dfrac{-3-5}{4-(-2)} = \dfrac{-8}{6} = -\dfrac{4}{3}$

3. Gradient of the perpendicular bisector

Perpendicular gradients multiply to $-1$:
$m_\perp = -\dfrac{1}{m_{AB}} = \dfrac{3}{4}$

4. Equation and y-intercept

Through $M(1,1)$ with gradient $\tfrac34$:
$y-1 = \dfrac34(x-1)$
At $x=0$: $y = 1-\dfrac34 = \dfrac14$

So $c = \tfrac14$.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $-\dfrac{1}{4}$ — Sign Error
  Sign slip when evaluating the intercept, giving the negative of the correct value.
• B. $\dfrac{3}{4}$ — Misread Question
  Reporting the gradient $\tfrac34$ rather than the intercept.
• C. $\dfrac{7}{4}$ — Sign Error
  Adding $\tfrac34$ to 1 instead of subtracting when moving from $x=1$ to $x=0$.
• D. $-\dfrac{3}{4}$ — Formula Misuse
  Using $m_\perp = -\tfrac34$ (negating instead of taking the negative reciprocal correctly).

Common Mistake (⚠️):
Using $-m_{AB} = \tfrac43$ instead of the negative reciprocal $\tfrac34$, or taking the midpoint as the difference rather than the average of the coordinates.

Takeaway (📌):
Perpendicular bisector = midpoint + negative reciprocal. Two quick calculations replace any attempt to solve for equidistant points algebraically.

Correct Answer: B. $72$

Fastest Approach (🚀):
Total: $5! = 120$.
Adjacent: glue Ana and Ben into one block $\implies 4! = 24$ arrangements, doubled for their internal order $\implies 48$.
$120-48 = 72$.

Matches Option B.

Step-by-Step Breakdown:

1. Count the unrestricted arrangements

Five distinct students in five chairs:
$5! = 120$

2. Count the arrangements that break the rule

Treat Ana and Ben as a single glued block. That leaves 4 objects to arrange:
$4! = 24$

3. Account for the block's internal order

Within the block, Ana–Ben and Ben–Ana are different seatings:
$24\times 2 = 48$

4. Subtract

$120-48 = 72$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $48$ — Complement Inversion
  Giving the number of arrangements where the two are adjacent, rather than the number where they are not.
• C. $96$ — Omitted Factor
  Forgetting to double for the internal order of the block: $120-24$.
• D. $60$ — Symmetry Fallacy
  Dividing 120 by 2 on the assumption that half the arrangements are adjacent.
• E. $120$ — Omitted Constraint
  Ignoring the restriction entirely.

Common Mistake (⚠️):
Forgetting the factor of 2 for the internal order of the block, giving $120-24 = 96$.

Takeaway (📌):
'Not adjacent' is the complement of 'glued together'. Glue, count, double for the internal order, subtract.

Correct Answer: A. $1010$

Fastest Approach (🚀):
$a = 3,\ l = 98$.
$S_{20} = \dfrac{20}{2}(3+98) = 10\times 101 = 1010$.

Matches Option A.

Step-by-Step Breakdown:

1. Recognise the sequence type

$u_n = 5n-2$ is linear in $n$, so the sequence is arithmetic with common difference $5$.

2. Find the first and last terms

$u_1 = 5(1)-2 = 3$
$u_{20} = 5(20)-2 = 98$

3. Use the pairing formula

Pairing the first term with the last, the second with the second-to-last, and so on gives 10 pairs each summing to 101:
$S_n = \dfrac{n}{2}(a+l) = \dfrac{20}{2}(3+98) = 10\times 101 = 1010$

4. Cross-check with the other form

$S_n = \dfrac{n}{2}\left[2a+(n-1)d\right] = 10\left[6+19(5)\right] = 10(101) = 1010$. Both agree.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $1030$ — Off-by-one Error
  Using $l = u_{21} = 103$ by mis-indexing the final term.
• C. $990$ — Off-by-one Error
  Using $a = u_0 = -2$ in the pairing formula.
• D. $98$ — Misread Question
  Giving the 20th term rather than the sum.
• E. $2020$ — Formula Misuse
  Forgetting the factor $\tfrac{n}{2}$ and computing $20\times 101$ instead.

Common Mistake (⚠️):
Giving the 20th term, $98$, instead of the sum, or using $n=19$ pairs by mis-indexing the last term.

Takeaway (📌):
For any linear $n$th term, the average of the first and last term times the number of terms is the sum. That is the fastest arithmetic-series route on the paper.

Correct Answer: E. An increase of 10%

Fastest Approach (🚀):
$1.25 \times 0.80 = 1$ (the first two changes cancel exactly).
$1 \times 1.10 = 1.10$.
So the net effect is a 10% increase.

Matches Option E.

Step-by-Step Breakdown:

1. Convert each change to a multiplier

A rise of $p\%$ is a multiplier of $1+\dfrac{p}{100}$; a fall of $p\%$ is $1-\dfrac{p}{100}$.
$+25\% \to 1.25, \qquad -20\% \to 0.80, \qquad +10\% \to 1.10$

2. Multiply in any order

Multiplication is commutative, so pair the friendly numbers first:
$1.25 \times 0.80 = 1$

3. Apply the remaining multiplier

$1 \times 1.10 = 1.10$

4. Interpret the result

A net multiplier of $1.10$ is an increase of $10\%$.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. An increase of 5% — Additive Error
  Adding and mis-collecting the percentages, or applying $-20\%$ to the original price.
• B. An increase of 15% — Additive Error
  Adding the percentages directly: 25 − 20 + 10.
• C. No change — Incomplete Answer
  Cancelling the first two changes correctly but forgetting the final $+10\%$.
• D. A decrease of 10% — Sign Error
  Treating the 20% fall as acting last on the inflated price and mis-signing the result.

Common Mistake (⚠️):
Adding the percentages: $+25-20+10 = +15\%$. Percentage changes act on different bases, so they compound rather than add.

Takeaway (📌):
A rise of $25\%$ is exactly undone by a fall of $20\%$ because $\tfrac54 \times \tfrac45 = 1$. Recognising cancelling multiplier pairs removes most of the arithmetic.

Correct Answer: A. $1:7$

Fastest Approach (🚀):
Small cone $= \tfrac18$ of the whole; frustum $= \tfrac78$ of the whole.
Ratio $= 1:7$.

Matches Option A.

Step-by-Step Breakdown:

1. Establish similarity

A plane parallel to the base cuts off a cone similar to the original, with every length halved:
$k = \dfrac{1}{2}$

2. Convert the length ratio to a volume ratio

For similar solids, volumes scale as $k^3$:
$\dfrac{V_{\text{small}}}{V_{\text{whole}}} = \left(\dfrac12\right)^3 = \dfrac18$

3. Find the frustum

The frustum is the whole minus the small cone:
$\dfrac{V_{\text{frustum}}}{V_{\text{whole}}} = 1-\dfrac18 = \dfrac78$

4. Form the required ratio

$V_{\text{small}} : V_{\text{frustum}} = \dfrac18 : \dfrac78 = 1:7$

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $1:8$ — Misread Question
  Comparing the small cone with the whole cone instead of with the frustum.
• C. $1:2$ — Scale Factor Error
  Using the linear scale factor 1:2 without cubing it.
• D. $1:3$ — Scale Factor Error
  Halving the volume and comparing $\tfrac12:\tfrac12$ after a partial correction.
• E. $1:4$ — Scale Factor Error
  Using the area scale factor 1:4 rather than the volume scale factor.

Common Mistake (⚠️):
Answering $1:8$, which is the small cone compared with the whole cone rather than with the frustum. Read carefully which two parts are being compared.

Takeaway (📌):
$k,\ k^2,\ k^3$ for lengths, areas and volumes — then check whether the question wants part-to-whole or part-to-remainder.

Correct Answer: A. $0.5$

Fastest Approach (🚀):
$A$ but not $B$: $0.6\times 0.5 = 0.3$.
$B$ but not $A$: $0.4\times 0.5 = 0.2$.
Total $= 0.3+0.2 = 0.5$.

Matches Option A.

Step-by-Step Breakdown:

1. List the two favourable cases

'Exactly one' excludes both the case where neither occurs and the case where both occur:
Case 1: $A$ occurs, $B$ does not.
Case 2: $B$ occurs, $A$ does not.

These are mutually exclusive, so their probabilities add.

2. Use independence

$P(A\cap B') = P(A)P(B') = 0.6\times 0.5 = 0.3$
$P(A'\cap B) = P(A')P(B) = 0.4\times 0.5 = 0.2$

3. Add

$P(\text{exactly one}) = 0.3+0.2 = 0.5$

4. Check the full partition

$P(\text{neither}) = 0.4\times 0.5 = 0.2$ and $P(\text{both}) = 0.6\times 0.5 = 0.3$.
$0.2+0.5+0.3 = 1$. The four outcomes account for everything.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $0.3$ — Incomplete Case Analysis
  Giving only one of the two cases, usually A but not B.
• C. $0.8$ — Misread Question
  Computing $P(A\cup B)$, which is 'at least one'.
• D. $0.2$ — Wrong Case
  Giving the probability that neither event occurs.
• E. $0.1$ — Operation Error
  Subtracting the two case probabilities instead of adding them.

Common Mistake (⚠️):
Computing $P(A\cup B) = 0.8$, which is 'at least one' and wrongly includes the case where both occur.

Takeaway (📌):
Build the full 2 by 2 table of outcomes when events are independent. Every question of this type is then a matter of adding the right cells.

Correct Answer: A. $9$ minutes

Fastest Approach (🚀):
Use 36 as the common denominator: $\dfrac{3}{36}+\dfrac{2}{36}-\dfrac{1}{36} = \dfrac{4}{36} = \dfrac19$ of a tank per minute.
Time $=$ reciprocal of the rate $= 9$ minutes.

Matches Option A.

Step-by-Step Breakdown:

1. Convert each time to a rate

Rate $=$ fraction of the tank handled per minute.
$A: \dfrac{1}{12}, \qquad B: \dfrac{1}{18}, \qquad \text{drain}: \dfrac{1}{36}\ \text{(outflow)}$

2. Combine the rates

Inflows add, the outflow subtracts. The lowest common denominator of 12, 18 and 36 is 36:
$\dfrac{1}{12}+\dfrac{1}{18}-\dfrac{1}{36} = \dfrac{3}{36}+\dfrac{2}{36}-\dfrac{1}{36} = \dfrac{4}{36} = \dfrac{1}{9}$

3. Invert the net rate

A net rate of $\tfrac19$ of a tank per minute means a full tank takes
$T = \dfrac{1}{1/9} = 9\ \text{minutes}$

4. Sanity check

The drain is the weakest of the three flows, so the answer must lie between the 12 minutes for A alone and the roughly 7.2 minutes for A and B together — and 9 does.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $6$ minutes — Omitted Term
  Ignoring the drain: $\tfrac1{12}+\tfrac1{18} = \tfrac{5}{36}$ gives 7.2 minutes, often rounded to 6 or 7.
• C. $12$ minutes — Conceptual Error
  Quoting pipe A's time alone, treating B and the drain as cancelling.
• D. $8$ minutes — Sign Error
  Adding the drain instead of subtracting it: $\tfrac{6}{36} = \tfrac16$.
• E. $10$ minutes — Averaging Error
  Averaging the times of the two filling pipes and adjusting by eye.

Common Mistake (⚠️):
Averaging or adding the times (for example $12+18-36$), which has no physical meaning. Only rates are additive.

Takeaway (📌):
Every 'how long together' question is one line: $\dfrac{1}{T} = \sum \dfrac{1}{t_i}$, with a minus sign for anything working in the opposite direction.

Correct Answer: D. $15$

Fastest Approach (🚀):
Exterior angle $= 180-156 = 24^\circ$.
$n = \dfrac{360}{24} = 15$.

Matches Option D.

Step-by-Step Breakdown:

1. Convert to the exterior angle

Interior and exterior angles at each vertex are supplementary:
$\text{exterior} = 180^\circ-156^\circ = 24^\circ$

2. Use the exterior angle sum

For any convex polygon the exterior angles sum to $360^\circ$, and for a regular polygon they are all equal:
$n\times 24^\circ = 360^\circ$

3. Solve

$n = \dfrac{360}{24} = 15$

4. Check with the interior formula

$\dfrac{(15-2)\times 180}{15} = \dfrac{2340}{15} = 156^\circ$. Correct, but note how much slower that route is.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $12$ — Arithmetic Error
  Using $360/30$ after taking the exterior angle as $30^\circ$.
• B. $13$ — Rounding Error
  Rounding an inexact division rather than checking $24$ divides $360$ exactly.
• C. $14$ — Off-by-one Error
  Solving the interior-angle equation and mis-rearranging by one.
• E. $24$ — Misread Question
  Giving the exterior angle in degrees rather than the number of sides.

Common Mistake (⚠️):
Starting from $\dfrac{(n-2)\times 180}{n} = 156$ and solving the resulting equation. It is correct but involves rearranging a fraction under time pressure for no gain.

Takeaway (📌):
Always convert interior to exterior. The $360^\circ$ sum is independent of $n$, which is what makes it a one-step question.

Correct Answer: D. $24$

Fastest Approach (🚀):
$9\times 16 = 144$.
$8\times 15 = 120$.
$144-120 = 24$.

Matches Option D.

Step-by-Step Breakdown:

1. Convert the first mean to a total

$\text{total} = \text{mean}\times\text{count} = 8\times 15 = 120$

2. Convert the second mean to a total

$9\times 16 = 144$

3. The new value is the difference

$144-120 = 24$

4. Sanity check with the shift argument

Adding the ninth number lifted the mean of nine values by 1, which requires $9\times 1 = 9$ extra on top of the new mean's own share: $15+9 = 24$. Both arguments agree.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $16$ — Conceptual Error
  Assuming the new number equals the new mean.
• B. $17$ — Conceptual Error
  Adding 1 to the new mean, as if the mean shift equalled the value shift.
• C. $20$ — Count Error
  Using $8\times 16-120 = 8$ style totals with the wrong count.
• E. $31$ — Arithmetic Error
  Using $9\times 16-8\times 14$ or another mis-multiplication.

Common Mistake (⚠️):
Answering 17 or 16 by reasoning that the mean rose by 1 so the new number is 'one more'. Lifting the mean of nine numbers by 1 requires nine extra units, not one.

Takeaway (📌):
Mean $\times$ count $=$ total. Once every mean is a total, these questions become subtraction.

Correct Answer: C. $12$ km/h

Fastest Approach (🚀):
Equal distances $\implies$ harmonic mean: $\dfrac{2uv}{u+v} = \dfrac{2(15)(10)}{25} = 12$ km/h.

Matches Option C.

Step-by-Step Breakdown:

1. Use the definition

$\text{average speed} = \dfrac{\text{total distance}}{\text{total time}}$

2. Compute each leg's time

Out: $t_1 = \dfrac{30}{15} = 2$ h
Back: $t_2 = \dfrac{30}{10} = 3$ h

3. Combine

Total distance $= 60$ km, total time $= 5$ h:
$\text{average speed} = \dfrac{60}{5} = 12\ \text{km/h}$

4. The shortcut

For two equal distances at speeds $u$ and $v$, the distance cancels and the result is always the harmonic mean:
$\bar v = \dfrac{2uv}{u+v} = \dfrac{2(15)(10)}{15+10} = \dfrac{300}{25} = 12$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $11$ km/h — Estimation Error
  Over-correcting downwards from 12.5 without computing the times.
• B. $12.5$ km/h — Averaging Error
  Taking the arithmetic mean of 15 and 10 — the classic trap.
• D. $13$ km/h — Weighting Error
  Weighting the speeds by distance rather than by time.
• E. $25$ km/h — Additive Error
  Adding the two speeds instead of averaging them.

Common Mistake (⚠️):
Averaging the two speeds to get $12.5$ km/h. More time is spent at the slower speed, so the true average must be below the arithmetic mean.

Takeaway (📌):
Equal distances: harmonic mean. Equal times: arithmetic mean. Identify which is equal before averaging anything.

Correct Answer: E. $13$ cm

Fastest Approach (🚀):
$3^2+4^2+12^2 = 9+16+144 = 169$.
$d = \sqrt{169} = 13$ cm.

Matches Option E.

Step-by-Step Breakdown:

1. Diagonal of the base

The base is $3\times 4$, so its diagonal is
$\sqrt{3^2+4^2} = 5$

2. Rise to the opposite vertex

That base diagonal and the vertical edge of 12 form a right-angled triangle:
$d = \sqrt{5^2+12^2} = \sqrt{25+144} = \sqrt{169} = 13$

3. The one-step form

Combining the two stages gives the general result for a cuboid:
$d = \sqrt{a^2+b^2+c^2} = \sqrt{9+16+144} = 13\ \text{cm}$

Note the two Pythagorean triples doing the work: $(3,4,5)$ and $(5,12,13)$.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $5$ cm — Incomplete Answer
  Giving the diagonal of the 3 by 4 base only.
• B. $\sqrt{19}$ cm — Formula Misuse
  Adding the edge lengths $3+4+12=19$ and then square-rooting.
• C. $19$ cm — Formula Misuse
  Adding the three edge lengths.
• D. $12.5$ cm — Estimation Error
  Estimating from $12$ and $\sqrt{25}$ without completing the calculation.

Common Mistake (⚠️):
Stopping at the base diagonal of $5$ cm, or adding the lengths rather than their squares.

Takeaway (📌):
Space diagonal: square, add all three, square-root. Recognising $(3,4,5)$ and $(5,12,13)$ removes the arithmetic entirely.

Correct Answer: D. $133$

Fastest Approach (🚀):
$(x+y)^3 = 343$.
$3xy(x+y) = 3\times 10\times 7 = 210$.
$x^3+y^3 = 343-210 = 133$.

Matches Option D.

Step-by-Step Breakdown:

1. Choose the identity

$(x+y)^3 = x^3+3x^2y+3xy^2+y^3 = x^3+y^3+3xy(x+y)$

Rearranged:
$x^3+y^3 = (x+y)^3-3xy(x+y)$

2. Substitute the given values

$(x+y)^3 = 7^3 = 343$
$3xy(x+y) = 3\times 10\times 7 = 210$

3. Evaluate

$x^3+y^3 = 343-210 = 133$

4. Check

$x$ and $y$ are the roots of $t^2-7t+10=0$, namely $2$ and $5$. Indeed $8+125 = 133$.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $63$ — Formula Misuse
  Using $x^3+y^3 = (x+y)^3-3xy$ with the factor $(x+y)$ omitted, or $343-280$.
• B. $91$ — Formula Misuse
  Using $(x+y)(x^2+y^2)$ instead of the correct factorisation $(x+y)(x^2-xy+y^2)$.
• C. $210$ — Incomplete Answer
  Giving the subtracted term $3xy(x+y) = 210$ rather than the difference.
• E. $343$ — Incomplete Answer
  Giving $(x+y)^3 = 343$ and forgetting to subtract.

Common Mistake (⚠️):
Solving for $x=2$, $y=5$ first. It works here, but as soon as the roots are irrational the identity route is the only practical one — and it is faster anyway.

Takeaway (📌):
Learn the two workhorses: $x^2+y^2=(x+y)^2-2xy$ and $x^3+y^3=(x+y)^3-3xy(x+y)$. They convert symmetric targets into given data.

Correct Answer: B. $\dfrac{x+3}{x+4}$

Fastest Approach (🚀):
Numerator: $(x-3)(x+3)$.
Denominator: $(x+4)(x-3)$.
Cancel $(x-3)$: $\dfrac{x+3}{x+4}$.

Matches Option B.

Step-by-Step Breakdown:

1. Factorise the numerator

A difference of two squares:
$x^2-9 = (x-3)(x+3)$

2. Factorise the denominator

Two numbers multiplying to $-12$ and adding to $+1$ are $+4$ and $-3$:
$x^2+x-12 = (x+4)(x-3)$

3. Cancel the common factor

$\dfrac{(x-3)(x+3)}{(x+4)(x-3)} = \dfrac{x+3}{x+4}$

valid for $x \ne 3$ (and $x \ne -4$).

4. Check with a value

At $x=0$: the original is $\dfrac{-9}{-12} = \dfrac34$, and the simplified form gives $\dfrac{3}{4}$. They agree.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{x-3}{x-4}$ — Cancellation Error
  Cancelling $(x+3)$ instead of $(x-3)$, which is not a common factor.
• C. $\dfrac{x+3}{x-4}$ — Pairing Error
  Pairing the wrong brackets after factorising correctly.
• D. $\dfrac{x-3}{x+4}$ — Factorisation Error
  Factorising the denominator as $(x-4)(x+3)$.
• E. $\dfrac{3}{4}$ — Misread Question
  Substituting $x=0$ into the original expression and giving the numerical value.

Common Mistake (⚠️):
Cancelling the $x^2$ terms or the $9$ and $12$ directly. Only whole factors of the numerator and denominator can be cancelled, never individual terms.

Takeaway (📌):
Factorise first, cancel second. A shared root between numerator and denominator is exactly what the question is built around.

Correct Answer: E. £205

Fastest Approach (🚀):
$1.05^{2} = 1.1025$.
$2000\times 1.1025 = 2205$.
Interest $= 2205-2000 = £205$.

Matches Option E.

Step-by-Step Breakdown:

1. Write the multiplier

A 5% annual increase is a multiplier of $1.05$ per year.

2. Apply it for two years

$A = 2000\times 1.05^{2}$

$1.05^{2} = 1.1025$, so
$A = 2000\times 1.1025 = £2205$

3. Subtract the principal

The question asks for the interest earned, not the balance:
$I = 2205-2000 = £205$

4. Why not £200

Simple interest would give $2\times 5\% \times 2000 = £200$. The extra £5 is the second year's interest on the first year's £100 — that £5 is precisely what 'compound' means.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. £200 — Model Error
  Using simple interest: 5% of £2000 twice.
• B. £2205 — Misread Question
  Giving the final balance rather than the interest earned.
• C. £210 — Arithmetic Error
  Applying 5% to £2100 but mis-rounding the second year's interest.
• D. £100 — Incomplete Answer
  Giving one year of interest only.

Common Mistake (⚠️):
Giving the final balance £2205 instead of the interest, or using simple interest and answering £200.

Takeaway (📌):
Balance $= P\times(\text{multiplier})^{n}$; interest $=$ balance $- P$. Read which of the two the question wants.

Correct Answer: B. $2$

Fastest Approach (🚀):
$x$ is multiplied by $4 \implies \sqrt{x}$ is multiplied by $2$.
Inverse proportion $\implies y$ is divided by $2$.
$y = 4 \div 2 = 2$.

Matches Option B.

Step-by-Step Breakdown:

1. Write the relationship

$y \propto \dfrac{1}{\sqrt{x}} \implies y = \dfrac{k}{\sqrt{x}}$

2. The scaling shortcut

The constant $k$ never has to be found. Since $y\sqrt{x} = k$ is fixed, multiplying $\sqrt{x}$ by a factor divides $y$ by that same factor.
$x: 9 \to 36$ is a factor of $4$, so $\sqrt{x}: 3 \to 6$ is a factor of $2$.
Therefore $y: 4 \to 2$.

3. Confirm with the constant

$k = y\sqrt{x} = 4\times 3 = 12$, so at $x=36$:
$y = \dfrac{12}{\sqrt{36}} = \dfrac{12}{6} = 2$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $1$ — Scale Factor Error
  Dividing $y$ by $4$ (the factor for $x$) instead of by $2$ (the factor for $\sqrt{x}$).
• C. $3$ — Misread Question
  Reporting $\sqrt{x}=3$ or the value of $\sqrt{9}$, not $y$.
• D. $8$ — Proportion Inversion
  Treating the relationship as direct proportion and doubling.
• E. $16$ — Proportion Inversion
  Direct proportion to $x$: multiplying $y$ by $4$.

Common Mistake (⚠️):
Applying the factor of $4$ to $y$ directly and answering $1$. The proportionality is to $\sqrt{x}$, not to $x$, so the factor that matters is $2$.

Takeaway (📌):
Track the factor, not the constant. 'Inversely proportional to the square root' means quadruple the input, halve the output.

Correct Answer: D. £80.00

Fastest Approach (🚀):
$68 \div 0.85 = \dfrac{6800}{85} = 80$.
Check: $15\%$ of $80$ is $12$, and $80-12 = 68$.

Matches Option D.

Step-by-Step Breakdown:

1. Identify the multiplier

A 15% discount is a multiplier of
$1-0.15 = 0.85$

2. Set up the equation

The multiplier acts on the original price $P$:
$0.85P = 68$

3. Divide, do not multiply

$P = \dfrac{68}{0.85} = \dfrac{6800}{85} = 80$

4. Verify

15% of £80 is £12, and $80-12 = 68$. Correct.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. £57.80 — Direction Error
  Applying the discount a second time: 68 × 0.85.
• B. £78.20 — Wrong Base
  Adding 15% to the sale price: 68 × 1.15 — the intended trap.
• C. £79.50 — Estimation Error
  Estimating between the trap value and the correct answer.
• E. £83.00 — Units Error
  Adding 15 pounds rather than 15 per cent.

Common Mistake (⚠️):
Adding 15% back to £68 to get £78.20. That applies the percentage to the wrong base — the discount was 15% of the original price, not of the sale price.

Takeaway (📌):
Forward change: multiply. Reverse change: divide. Always check by applying the change forwards to your answer.

Correct Answer: E. $3\times 10^{11}$

Fastest Approach (🚀):
$3.6\div 1.2 = 3$.
$10^{8}\div 10^{-3} = 10^{8+3} = 10^{11}$.
$3\times 10^{11}$.

Matches Option E.

Step-by-Step Breakdown:

1. Separate the two parts

$\dfrac{3.6\times 10^{8}}{1.2\times 10^{-3}} = \dfrac{3.6}{1.2}\times\dfrac{10^{8}}{10^{-3}}$

2. Divide the mantissas

$\dfrac{3.6}{1.2} = 3$

3. Subtract the indices

$\dfrac{10^{8}}{10^{-3}} = 10^{\,8-(-3)} = 10^{11}$

Subtracting a negative index adds — this is the step the question is testing.

4. Recombine

$3\times 10^{11}$

The mantissa already lies between 1 and 10, so no normalisation is needed.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $3\times 10^{5}$ — Sign Error
  Using $10^{8-3} = 10^{5}$ — the sign of the negative index dropped.
• B. $3\times 10^{-11}$ — Sign Error
  Subtracting in the wrong order: $10^{-3-8}$.
• C. $4.32\times 10^{5}$ — Operation Error
  Multiplying the mantissas ($3.6\times 1.2$) instead of dividing, and mishandling the index.
• D. $3\times 10^{10}$ — Arithmetic Error
  Index arithmetic off by one: $8+3 = 10$.

Common Mistake (⚠️):
Computing $10^{8-3} = 10^{5}$. The index in the denominator is $-3$, so the subtraction becomes an addition.

Takeaway (📌):
Dividing by a negative power of ten makes a number larger. If your answer shrank, the sign of the index went the wrong way.

Correct Answer: C. $8$

Fastest Approach (🚀):
Add the equations: $8x+8y = 64$.
Divide by $8$: $x+y = 8$.

Matches Option C.

Step-by-Step Breakdown:

1. Spot the swapped coefficients

$3x+5y = 31$
$5x+3y = 33$

The coefficient pair $(3,5)$ appears in both equations, reversed. That is the signal to add and subtract rather than to eliminate.

2. Add the equations

$(3x+5x)+(5y+3y) = 31+33$
$8x+8y = 64 \implies 8(x+y) = 64 \implies x+y = 8$

3. Check against the full solution

Subtracting gives $2x-2y = 2$, so $x-y = 1$. With $x+y=8$ that gives $x = 4.5$, $y = 3.5$, which does satisfy both original equations — but finding them was unnecessary.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $6$ — Arithmetic Error
  Arithmetic slip in $31+33$, or dividing $64$ by a wrong factor.
• B. $7$ — Rounding Error
  Rounding $x=4.5$ and $y=3.5$ to integers before adding.
• D. $16$ — Arithmetic Error
  Dividing $64$ by $4$ instead of by $8$.
• E. $64$ — Incomplete Answer
  Giving $8x+8y = 64$ without dividing by $8$.

Common Mistake (⚠️):
Solving the system by elimination or substitution. It works, but it produces fractions and takes three times as long for a question that only asks for the sum.

Takeaway (📌):
When the coefficients are mirrored, add for $x+y$ and subtract for $x-y$. Check what the question actually asks before solving for the variables separately.

Correct Answer: C. $108$ km/h

Fastest Approach (🚀):
$3600 \div 40 = 90$ intervals per hour.
$1.2\times 90 = 108$ km/h.

Matches Option C.

Step-by-Step Breakdown:

1. Check the units already given

The distance is in kilometres, which is what the answer needs. Only the time has to change from seconds to hours.

2. Scale the time to one hour

$1\ \text{hour} = 3600\ \text{s}, \qquad \dfrac{3600}{40} = 90$

So the train covers its 1.2 km ninety times in an hour.

3. Scale the distance by the same factor

$1.2\times 90 = 108\ \text{km/h}$

4. Confirm with the formula

$v = \dfrac{d}{t} = \dfrac{1.2}{40}\ \text{km/s} = 0.03\ \text{km/s}$, and $0.03\times 3600 = 108\ \text{km/h}$.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $30$ km/h — Direction Error
  Dividing rather than multiplying somewhere in the conversion chain.
• B. $48$ km/h — Conversion Error
  Using $1.2\times 40$ and rescaling incorrectly.
• D. $120$ km/h — Rounding Error
  Multiplying by 100 instead of 90 as a rounded shortcut.
• E. $180$ km/h — Conversion Error
  Using 60 intervals per hour rather than 90.

Common Mistake (⚠️):
Multiplying by 60 instead of 3600, giving 1.8 km/min misread as a km/h figure, or converting the distance to metres unnecessarily and then losing a factor of 1000.

Takeaway (📌):
Convert only the unit that is wrong. Scaling the time interval up to an hour keeps every number small and mental.