ESAT Mathematics Mock Module 3 Worked Solutions

Correct Answer: A. $0.02$

Fastest Approach (🚀):
$0.0004 = \dfrac{4}{10\,000}$.
$\sqrt{\ } = \dfrac{2}{100} = 0.02$.

Matches Option A.

Step-by-Step Breakdown:

1. Write as a fraction

$0.0004 = \dfrac{4}{10\,000} = \dfrac{4}{10^{4}}$

2. Root numerator and denominator separately

$\sqrt{\dfrac{4}{10^{4}}} = \dfrac{\sqrt{4}}{\sqrt{10^{4}}} = \dfrac{2}{10^{2}} = \dfrac{2}{100}$

Note the index halves: $10^{4}\to 10^{2}$.

3. Convert back

$\dfrac{2}{100} = 0.02$

4. Check by squaring

$0.02^{2} = 0.0004$. Correct — and squaring your answer is always the fastest check here.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $0.002$ — Place Value Error
  Counting decimal places instead of halving the index.
• C. $0.2$ — Place Value Error
  Rooting $0.04$ rather than $0.0004$.
• D. $0.0002$ — Operation Error
  Halving the original number rather than taking its root.
• E. $0.00002$ — Place Value Error
  Over-shifting the decimal point by two places.

Common Mistake (⚠️):
Halving the number of decimal places by counting rather than halving the index, which usually produces $0.002$. The root of a decimal below 1 is always larger than the decimal itself.

Takeaway (📌):
Convert the decimal to $\dfrac{a}{10^{2k}}$, then the root is $\dfrac{\sqrt a}{10^{k}}$. Confirm by squaring the answer.

Correct Answer: C. $\dfrac{8}{15}$

Fastest Approach (🚀):
$P(\text{no faulty}) = \dfrac{7}{10}\times\dfrac{6}{9} = \dfrac{42}{90} = \dfrac{7}{15}$.
$1-\dfrac{7}{15} = \dfrac{8}{15}$.

Matches Option C.

Step-by-Step Breakdown:

1. Take the complement

'At least one faulty' is the complement of 'both working'. One product replaces a three-case sum.

2. Probability both are working

There are 7 working components out of 10; after one is removed, 6 of the remaining 9 work:
$P(\text{both working}) = \dfrac{7}{10}\times\dfrac{6}{9} = \dfrac{42}{90} = \dfrac{7}{15}$

3. Subtract from 1

$P(\text{at least one faulty}) = 1-\dfrac{7}{15} = \dfrac{8}{15}$

4. Verify by counting

$\binom{10}{2} = 45$ pairs; $\binom{7}{2} = 21$ are all-working; $45-21 = 24$ contain a faulty one, and $\dfrac{24}{45} = \dfrac{8}{15}$. Confirmed.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{3}{10}$ — Misread Question
  Giving the probability that a single component is faulty.
• B. $\dfrac{7}{15}$ — Complement Inversion
  Giving the complement itself — the probability that neither is faulty.
• D. $\dfrac{3}{5}$ — Independence Error
  Treating the draws as independent: $1-\left(\tfrac{7}{10}\right)^{2}$ mis-evaluated.
• E. $\dfrac{1}{15}$ — Wrong Case
  Computing the probability that both are faulty.

Common Mistake (⚠️):
Using $\tfrac{7}{10}\times\tfrac{7}{10}$ and treating the draws as independent. Without replacement, both the numerator and denominator drop for the second draw.

Takeaway (📌):
'At least one' means $1-P(\text{none})$. Without replacement, reduce both the favourable count and the total for the second draw.

Correct Answer: C. $720^{\circ}$

Fastest Approach (🚀):
$(6-2)\times 180 = 4\times 180 = 720^{\circ}$.

Matches Option C.

Step-by-Step Breakdown:

1. Split into triangles

Drawing every diagonal from a single vertex of an $n$-gon produces $n-2$ triangles. For a hexagon that is $4$ triangles.

2. Sum the triangles

Each triangle's angles sum to $180^{\circ}$, and together they account for exactly the polygon's interior angles:
$(n-2)\times 180^{\circ} = 4\times 180^{\circ} = 720^{\circ}$

3. Cross-check via exterior angles

A regular hexagon has exterior angles of $\tfrac{360}{6} = 60^{\circ}$, so each interior angle is $120^{\circ}$, and $6\times 120 = 720^{\circ}$. Consistent — and note the sum holds for any hexagon, regular or not.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $1080^{\circ}$ — Formula Misuse
  Using $n\times 180$ instead of $(n-2)\times 180$.
• B. $540^{\circ}$ — Substitution Error
  Using $n=5$ — the pentagon's sum.
• D. $360^{\circ}$ — Conceptual Error
  Giving the exterior angle sum, which is 360° for every polygon.
• E. $900^{\circ}$ — Off-by-one Error
  Using $(n-1)\times 180$.

Common Mistake (⚠️):
Using $n\times 180$ and answering $1080^\circ$, or confusing the interior sum with the exterior sum of $360^\circ$.

Takeaway (📌):
Interior sum $= (n-2)180^\circ$ for any polygon; exterior sum $= 360^\circ$ always. Only the interior sum depends on $n$.

Correct Answer: A. $2n^{2}+n$

Fastest Approach (🚀):
Second differences $= 4 \implies a = 2$.
$u_n-2n^{2}: 1, 2, 3, 4, 5 = n$.
$u_n = 2n^{2}+n$.

Matches Option A.

Step-by-Step Breakdown:

1. Take differences

First differences: $7,\ 11,\ 15,\ 19$. Second differences: $4,\ 4,\ 4$.

2. Fix the leading coefficient

The second difference is $2a$, so $a = 2$.

3. Subtract and read off the rest

$u_{n}-2n^{2}$: $3-2 = 1$, $10-8 = 2$, $21-18 = 3$, $36-32 = 4$ — the sequence $n$.

$u_{n} = 2n^{2}+n$

Check $n=5$: $50+5 = 55$. Correct.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $n^{2}+2n$ — Coefficient Error
  Using $a=1$, which gives $3, 8, 15$ rather than $3, 10, 21$.
• C. $3n^{2}$ — Insufficient Checking
  Matching the first term only.
• D. $2n^{2}+1$ — Term Error
  Correct leading coefficient but a constant instead of the linear term.
• E. $4n^{2}-n$ — Coefficient Error
  Doubling the leading coefficient and patching the linear term.

Common Mistake (⚠️):
Using the second difference $4$ as the leading coefficient, or accepting a formula that matches only the first term.

Takeaway (📌):
Second difference $= 2a$ every time. Subtract $an^{2}$ and the remainder is always linear.

Correct Answer: B. $\dfrac{10}{3}$

Fastest Approach (🚀):
$\alpha+\beta = -\dfrac{b}{a} = \dfrac{10}{3}$.

Matches Option B.

Step-by-Step Breakdown:

1. Identify the coefficients

$a = 3,\quad b = -10,\quad c = 8$

2. Apply Vieta

$\alpha+\beta = -\dfrac{b}{a} = -\dfrac{-10}{3} = \dfrac{10}{3}$

Note the double negative — that is where the sign errors live.

3. Verify by factorising

$3x^{2}-10x+8 = (3x-4)(x-2)$, so the roots are $\tfrac43$ and $2$.
$\dfrac43+2 = \dfrac{4+6}{3} = \dfrac{10}{3}$. Confirmed.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{8}{3}$ — Formula Confusion
  Giving the product of the roots instead of the sum.
• C. $10$ — Omitted Divisor
  Using $-b$ without dividing by $a$.
• D. $-\dfrac{10}{3}$ — Sign Error
  Sign error: the double negative in $-b/a$ was missed.
• E. $\dfrac{3}{10}$ — Inversion Error
  Inverting the ratio.

Common Mistake (⚠️):
Dropping the leading minus in $-b/a$ and answering $-\tfrac{10}{3}$, or giving $c/a = \tfrac83$ (the product) instead of the sum.

Takeaway (📌):
Sum $=-b/a$, product $=c/a$. Both are two-second reads once the coefficients are written down.

Correct Answer: D. $2$

Fastest Approach (🚀):
$\dfrac13+\dfrac16 = \dfrac26+\dfrac16 = \dfrac36 = \dfrac12$.
$\dfrac{1}{1/2} = 2$.

Matches Option D.

Step-by-Step Breakdown:

1. Add the fractions in the denominator

Common denominator 6:
$\dfrac13+\dfrac16 = \dfrac26+\dfrac16 = \dfrac36 = \dfrac12$

2. Invert once

$\dfrac{1}{\frac12} = 2$

3. Why term-by-term inversion fails

$\dfrac{1}{\frac13}+\dfrac{1}{\frac16} = 3+6 = 9$, which is not the answer. The reciprocal of a sum is not the sum of the reciprocals.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $9$ — Reciprocal Error
  Inverting each fraction and adding: 3 + 6.
• B. $\dfrac{1}{2}$ — Incomplete Answer
  Giving the value of the denominator rather than the whole expression.
• C. $\dfrac{2}{9}$ — Inversion Error
  Inverting the correct answer.
• E. $\dfrac{9}{2}$ — Fraction Error
  Adding the denominators to get $\tfrac29$ and inverting.

Common Mistake (⚠️):
Inverting each fraction separately and adding, giving $3+6 = 9$ — the classic error, and the reason this structure appears in resistor and work-rate questions.

Takeaway (📌):
$\dfrac{1}{a+b} \ne \dfrac1a+\dfrac1b$. Combine the denominator into a single fraction, then take one reciprocal.

Correct Answer: A. $6$

Fastest Approach (🚀):
At least one language $= 32-5 = 27$.
$18+15-27 = 6$ study both.

Matches Option A.

Step-by-Step Breakdown:

1. Find the union from the complement

$5$ study neither, so
$|F\cup G| = 32-5 = 27$

2. Rearrange the two-set identity

$|F\cup G| = |F|+|G|-|F\cap G| \implies |F\cap G| = |F|+|G|-|F\cup G|$

3. Substitute

$|F\cap G| = 18+15-27 = 6$

Check the regions: French only $12$, German only $9$, both $6$, neither $5$ — totalling $32$.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $3$ — Arithmetic Error
  Using $33-30$ or another mis-substitution.
• C. $8$ — Setup Error
  Ignoring the five students who study neither in a different way.
• D. $1$ — Complement Ignored
  Using the class total as the union: $18+15-32$.
• E. $11$ — Misread Question
  Giving the size of a single-language region.

Common Mistake (⚠️):
Using $32$ as the union and getting $1$, forgetting that five students are outside both sets.

Takeaway (📌):
The same identity solves for whichever term is missing. Deduce the union from the 'neither' count first.

Correct Answer: A. $053^{\circ}$

Fastest Approach (🚀):
$\tan\theta = \dfrac{4}{3} \implies \theta \approx 53^{\circ}$.
Measured clockwise from north: $053^{\circ}$.

Matches Option A.

Step-by-Step Breakdown:

1. Sketch the displacement

North 3 km, then east 4 km — a right-angled triangle with the north leg adjacent to the bearing angle and the east leg opposite it. (The hypotenuse is 5 km, the familiar 3-4-5 triangle.)

2. Choose the right ratio

The bearing angle $\theta$ is measured at the start, from north, turning clockwise towards the destination:
$\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{\text{east}}{\text{north}} = \dfrac{4}{3}$

3. Evaluate

$\theta = \arctan\left(\tfrac43\right) \approx 53^{\circ}$

4. Write it as a bearing

Bearings take three figures, so the answer is $053^{\circ}$. It lies between $000^\circ$ and $090^\circ$, as it must for a north-east displacement.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $037^{\circ}$ — Reference Direction Error
  Using $\tan\theta = \tfrac34$, the angle measured from east.
• C. $127^{\circ}$ — Reference Direction Error
  Measuring from south, or adding $90^\circ$ to the angle from east.
• D. $045^{\circ}$ — Estimation Error
  Assuming the displacement is at 45° because both legs are 'similar'.
• E. $233^{\circ}$ — Direction Reversed
  Giving the back bearing, from the destination to the start.

Common Mistake (⚠️):
Using $\tan\theta = \tfrac34$ and answering $037^\circ$ — that is the angle from east, not from north.

Takeaway (📌):
Bearings: clockwise from north, three figures. Set the ratio up as east over north and the angle comes out directly.

Correct Answer: D. $203$

Fastest Approach (🚀):
$d = 4$; zeroth term $= 7-4 = 3$.
$u_{n} = 4n+3 \implies u_{50} = 200+3 = 203$.

Matches Option D.

Step-by-Step Breakdown:

1. Find the common difference

$11-7 = 4,\quad 15-11 = 4,\quad 19-15 = 4 \implies d = 4$

2. Find the constant by stepping back

The term before the first (the 'zeroth' term) is
$7-4 = 3$

3. Write the nth term

$u_{n} = 4n+3$

Check: $u_{1} = 7$, $u_{2} = 11$. Correct.

4. Substitute

$u_{50} = 4(50)+3 = 203$

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $199$ — Off by One
  Using $n = 49$ rather than $50$, an off-by-one on the term number.
• B. $207$ — Constant Error
  Using $u_n = 4n+7$, taking the first term as the constant.
• C. $197$ — Sign Error
  Using $u_n = 4n-3$.
• E. $4\times 50$ — Incomplete Answer
  Giving the multiplication rather than completing it.

Common Mistake (⚠️):
Writing $u_n = 4n$ and answering 200, or using $u_n = 4n+7$ by taking the first term as the constant.

Takeaway (📌):
$u_n = dn + u_0$. Stepping back one term to get $u_0$ is quicker and safer than substituting and solving.

Correct Answer: E. $25\%$

Fastest Approach (🚀):
Profit $= 100-80 = 20$.
$\dfrac{20}{80} = \dfrac14 = 25\%$.

Matches Option E.

Step-by-Step Breakdown:

1. Find the actual profit

$£100-£80 = £20$

2. Choose the correct base

Percentage profit is measured against what was paid:
$\text{percentage profit} = \dfrac{\text{profit}}{\text{cost price}}\times 100$

3. Evaluate

$\dfrac{20}{80}\times 100 = 25\%$

4. Contrast with the wrong base

Dividing by the selling price gives $\dfrac{20}{100} = 20\%$ — that is the profit margin, a different measure and the intended trap.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $20\%$ — Wrong Base
  Dividing by the selling price — that is the margin, not the profit percentage.
• B. $80\%$ — Wrong Base
  Giving the cost as a percentage of the sale price.
• C. $125\%$ — Misread Question
  Giving the selling price as a percentage of the cost price.
• D. $5\%$ — Arithmetic Error
  Dividing the profit by 400 or misplacing a factor of 5.

Common Mistake (⚠️):
Dividing the profit by the selling price and answering 20%. That is margin, not percentage profit.

Takeaway (📌):
Percentage change $= \dfrac{\text{change}}{\text{original}}$. For profit the 'original' is always the cost price.

Correct Answer: B. $56$

Fastest Approach (🚀):
$\dfrac{84}{360}\times 240 = \dfrac{84}{3}\times 2 = 56$.

Matches Option B.

Step-by-Step Breakdown:

1. Turn the angle into a fraction

$\dfrac{84}{360} = \dfrac{7}{30}$

2. Apply it to the total

$\dfrac{7}{30}\times 240 = 7\times 8 = 56$

3. Sanity check

$84^{\circ}$ is a little under a quarter of the circle, and a quarter of $240$ is $60$. An answer of $56$ sits just below that, as it should.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $84$ — Angle vs Frequency
  Reading the angle as the number of people.
• C. $70$ — Denominator Error
  Using $\tfrac{84}{288}$ or another wrong denominator.
• D. $28$ — Incomplete Answer
  Dividing $84$ by $3$ and stopping.
• E. $120$ — Estimation Error
  Taking half the total, treating $84^{\circ}$ as a semicircle.

Common Mistake (⚠️):
Reading the angle as the frequency and answering $84$ — the angle only equals the count when the total happens to be $360$.

Takeaway (📌):
Frequency $= \dfrac{\text{angle}}{360}\times\text{total}$. Going the other way, $\text{angle} = \dfrac{\text{frequency}}{\text{total}}\times 360$.

Correct Answer: C. $14\pi$ cm

Fastest Approach (🚀):
$r^{2} = 49 \implies r = 7$.
$C = 2\pi r = 14\pi$ cm.

Matches Option C.

Step-by-Step Breakdown:

1. Recover the radius

$A = \pi r^{2} = 49\pi \implies r^{2} = 49 \implies r = 7\ \text{cm}$

The $\pi$ cancels immediately — leaving it symbolic keeps the arithmetic exact.

2. Substitute into the circumference formula

$C = 2\pi r = 2\pi(7) = 14\pi\ \text{cm}$

3. Check the units

Area was in cm², so the radius is in cm and the circumference is in cm. Consistent.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $49\pi$ cm — Misread Question
  Repeating the area as though it were the circumference.
• B. $7\pi$ cm — Formula Misuse
  Using $C = \pi r$ instead of $2\pi r$.
• D. $98\pi$ cm — Formula Misuse
  Using $C = 2\pi r^{2}$.
• E. $28\pi$ cm — Arithmetic Error
  Using $r = 14$ from $49\pi$ mis-rooted.

Common Mistake (⚠️):
Using $C = \pi r$ or $C = 2\pi r^{2}$, or treating the given $49\pi$ as the radius squared including the $\pi$.

Takeaway (📌):
The radius is the bridge between every circle formula. Extract it first and the rest is one substitution.

Correct Answer: B. $12.55$ cm

Fastest Approach (🚀):
Half of $0.1$ is $0.05$.
Upper bound $= 12.5+0.05 = 12.55$ cm.

Matches Option B.

Step-by-Step Breakdown:

1. Identify the rounding unit

The measurement is to the nearest $0.1$ cm, so $u = 0.1$.

2. Take half of it

$\dfrac{u}{2} = 0.05$

3. Form the bounds

$\text{lower bound} = 12.5-0.05 = 12.45$
$\text{upper bound} = 12.5+0.05 = 12.55$

4. Note the convention

Any true value in $12.45 \le x < 12.55$ rounds to $12.5$. The upper bound is quoted as $12.55$ even though that exact value would round up — writing $12.549$ or $12.5499$ is the classic over-thinking error.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $12.6$ cm — Bound Error
  Adding the whole rounding unit rather than half of it.
• C. $12.51$ cm — Bound Error
  Using a rounding unit of $0.02$ or adding $0.01$.
• D. $12.549$ cm — Convention Error
  Trying to write the largest value strictly below the bound.
• E. $13.0$ cm — Accuracy Error
  Treating the measurement as correct to the nearest centimetre.

Common Mistake (⚠️):
Adding the full rounding unit and answering $12.6$, or trying to write the largest number strictly below the bound, such as $12.549$.

Takeaway (📌):
Bounds are always the stated value $\pm$ half the rounding unit. Quote the half-unit boundary itself as the upper bound.

Correct Answer: B. Class A scored higher and was more consistent

Fastest Approach (🚀):
Median: $62 > 58 \implies$ A higher.
IQR: $8 < 20 \implies$ A more consistent.

Matches Option B.

Step-by-Step Breakdown:

1. Compare location

The medians are $62$ and $58$, so Class A is typically higher. Comparing medians with medians is the like-for-like comparison the question invites.

2. Compare spread

The interquartile ranges are $8$ and $20$. A smaller IQR means the middle half of the data is packed into a narrower band, so Class A is the more consistent.

3. What cannot be concluded

Neither summary says anything about the single highest score — the IQR deliberately ignores the extremes. Option E is unsupported for that reason.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Class B scored higher and was more consistent — Comparison Reversed
  Reversing both comparisons.
• C. Class A scored higher but was less consistent — Spread Misread
  Reading the smaller IQR as less consistent.
• D. The two classes performed identically — Comparison Ignored
  Ignoring both differences.
• E. Class B contained the single highest score — Over-reach
  The IQR excludes the extremes, so it cannot support a claim about the maximum.

Common Mistake (⚠️):
Reading a larger IQR as 'better'. A large interquartile range means more variability, not higher achievement.

Takeaway (📌):
One measure of location, one of spread, and never mix the two. A smaller IQR is more consistent, not higher.

Correct Answer: C. $216\ \text{cm}^{2}$

Fastest Approach (🚀):
$s = 6$.
$SA = 6\times 6^{2} = 216\ \text{cm}^{2}$.

Matches Option C.

Step-by-Step Breakdown:

1. Find the side length

$V = s^{3} = 216 \implies s = \sqrt[3]{216} = 6\ \text{cm}$

2. Apply the surface area formula

A cube has 6 identical square faces:
$SA = 6s^{2} = 6\times 6^{2} = 6\times 36 = 216\ \text{cm}^{2}$

3. Note the coincidence

The volume and surface area are numerically equal here — a quirk of $s=6$ only, since $s^{3} = 6s^{2}$ exactly when $s=6$. Do not read anything general into it.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $36\ \text{cm}^{2}$ — Omitted Factor
  Giving the area of one face only.
• B. $1296\ \text{cm}^{2}$ — Formula Misuse
  Using $6s^{3}$ or squaring the volume-derived value twice.
• D. $72\ \text{cm}^{2}$ — Face Count Error
  Using $2s^{2}$ — only two faces.
• E. $144\ \text{cm}^{2}$ — Face Count Error
  Using $4s^{2}$ — the lateral faces only.

Common Mistake (⚠️):
Using $s^{2} = 36$ as the surface area (one face only), or forgetting the cube root and working from $s = 216$.

Takeaway (📌):
Cube: $V = s^{3}$, $SA = 6s^{2}$. Extract $s$ first and both formulas are one step away.

Correct Answer: A. $42$

Fastest Approach (🚀):
$12 = 2^{2}\times 3,\quad 18 = 2\times 3^{2}$.
HCF $= 2\times 3 = 6$; LCM $= 2^{2}\times 3^{2} = 36$.
$6+36 = 42$.

Matches Option A.

Step-by-Step Breakdown:

1. Prime factorise both numbers

$12 = 2^{2}\times 3$
$18 = 2\times 3^{2}$

2. Build the HCF from the lowest powers

Take the smaller index of each shared prime:
$\text{HCF} = 2^{1}\times 3^{1} = 6$

3. Build the LCM from the highest powers

Take the larger index of every prime that appears:
$\text{LCM} = 2^{2}\times 3^{2} = 36$

4. Add, and check with the product rule

$6+36 = 42$

Check: $\text{HCF}\times\text{LCM} = 6\times 36 = 216 = 12\times 18$. The identity holds, confirming both values.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $36$ — Incomplete Answer
  Giving the LCM alone.
• C. $30$ — LCM Error
  Using $\text{HCF}=6$ and $\text{LCM}=24$.
• D. $48$ — HCF Error
  Using $\text{HCF}=12$ (the smaller number) and $\text{LCM}=36$.
• E. $216$ — Misread Question
  Giving $\text{HCF}\times\text{LCM}$ rather than the sum.

Common Mistake (⚠️):
Swapping the two rules — taking the highest powers for the HCF and the lowest for the LCM. The HCF can never exceed the smaller number, which catches this instantly.

Takeaway (📌):
HCF: lowest powers. LCM: highest powers. Verify with $\text{HCF}\times\text{LCM} = $ the product of the numbers.

Correct Answer: A. $8$

Fastest Approach (🚀):
At least one: $30-5 = 25$.
$18+15-25 = 8$.

Matches Option A.

Step-by-Step Breakdown:

1. Find how many study at least one subject

$30-5 = 25$

The 5 who study neither sit outside both circles.

2. Apply inclusion-exclusion

$n(M\cup P) = n(M)+n(P)-n(M\cap P)$
$25 = 18+15-n(M\cap P)$

3. Solve

$n(M\cap P) = 33-25 = 8$

4. Check the regions

Maths only: $18-8 = 10$. Physics only: $15-8 = 7$. Both: $8$. Neither: $5$.
$10+7+8+5 = 30$. The four regions account for the whole class.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $3$ — Omitted Group
  Ignoring the 5 who study neither.
• C. $25$ — Misread Question
  Giving the number who study at least one subject.
• D. $13$ — Region Error
  Computing $18-5$ or another partial region.
• E. $5$ — Misread Question
  Repeating the 'neither' figure.

Common Mistake (⚠️):
Forgetting the 5 who study neither and using $18+15-30 = 3$, which under-counts the overlap.

Takeaway (📌):
Remove the 'neither' group first, then apply inclusion-exclusion. Finish by checking the four regions sum to the total.

Correct Answer: D. $\dfrac{5}{11}$

Fastest Approach (🚀):
Two recurring digits $\implies \dfrac{45}{99}$.
$\div 9$: $\dfrac{5}{11}$.

Matches Option D.

Step-by-Step Breakdown:

1. Set up the algebra

Let $x = 0.454545\ldots$

The recurring block has length 2, so multiply by $10^{2}$:
$100x = 45.454545\ldots$

2. Subtract

$100x-x = 45.4545\ldots-0.4545\ldots$
$99x = 45$

The recurring tails cancel exactly — that is the whole point of choosing $10^{2}$.

3. Solve

$x = \dfrac{45}{99}$

4. Simplify

$\dfrac{45}{99} = \dfrac{5}{11}$ (dividing top and bottom by 9).

The question asks for simplest form, so $\tfrac{45}{99}$ is not the final answer.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{45}{100}$ — Conceptual Error
  Treating the decimal as terminating at two places.
• B. $\dfrac{1}{2}$ — Denominator Error
  Dividing by $90$ rather than $99$ for a two-digit recurring block.
• C. $\dfrac{45}{99}$ — Incomplete Answer
  Correct fraction, but not in simplest form as the question requires.
• E. $\dfrac{4}{9}$ — Misread Question
  Treating only the first digit as recurring.

Common Mistake (⚠️):
Stopping at $\tfrac{45}{99}$ when the question demands simplest form, or treating the decimal as terminating and writing $\tfrac{45}{100}$.

Takeaway (📌):
$k$ recurring digits over $k$ nines, then simplify. $0.\dot3 = \tfrac39 = \tfrac13$ and $0.\dot{1}\dot{2} = \tfrac{12}{99} = \tfrac{4}{33}$ follow the same rule.

Correct Answer: C. $15$ km/h

Fastest Approach (🚀):
$\dfrac{600}{40} = 15$ km/h.

Matches Option C.

Step-by-Step Breakdown:

1. Time each leg

Up: $\dfrac{5}{10} = 0.5$ h. Down: $\dfrac{5}{30} = \dfrac16$ h.

2. Total distance over total time

$\dfrac{10}{0.5+\tfrac16} = \dfrac{10}{\tfrac23} = 15$ km/h

3. Note how far below the midpoint it falls

The arithmetic mean is $20$, but three times as long is spent climbing as descending, so the true average is $15$ — a quarter below.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $20$ km/h — Averaging Error
  Taking the arithmetic mean.
• B. $18$ km/h — Estimation Error
  Partially correcting from the midpoint without computing.
• D. $12$ km/h — Estimation Error
  Over-correcting towards the slow leg.
• E. $25$ km/h — Weighting Error
  Weighting towards the descent.

Common Mistake (⚠️):
Answering $20$ km/h. The wider the speed ratio, the further the harmonic mean sits below the arithmetic one.

Takeaway (📌):
The harmonic mean never exceeds the slower speed doubled, and approaches the slow speed as the ratio widens.

Correct Answer: B. $48$ km/h

Fastest Approach (🚀):
$\dfrac{2\times 40\times 60}{40+60} = \dfrac{4800}{100} = 48$ km/h.

Matches Option B.

Step-by-Step Breakdown:

1. Use the definition

Average speed is total distance over total time, not the average of the speeds.

2. Time each leg

Out: $\dfrac{12}{40} = 0.3$ h. Back: $\dfrac{12}{60} = 0.2$ h.

3. Combine

$\dfrac{24}{0.5} = 48$ km/h

The distance never mattered: for equal legs the result is always $\dfrac{2uv}{u+v}$.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $50$ km/h — Averaging Error
  Taking the arithmetic mean of the two speeds.
• C. $45$ km/h — Estimation Error
  Over-correcting downwards without computing the times.
• D. $52$ km/h — Weighting Error
  Weighting towards the faster leg.
• E. $100$ km/h — Additive Error
  Adding the two speeds.

Common Mistake (⚠️):
Averaging $40$ and $60$ to get $50$. More time is spent at the slower speed, so the true average must fall below the midpoint.

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

Correct Answer: E. It increases by a factor of 4

Fastest Approach (🚀):
$(2)^{2} = 4$, so $y$ is multiplied by $4$.

Matches Option E.

Step-by-Step Breakdown:

1. Write the relationship

$y \propto x^{2} \implies y = kx^{2}$ for some constant $k$.

2. Substitute the doubled input

$y_{\text{new}} = k(2x)^{2} = k\times 4x^{2} = 4kx^{2}$

3. Compare

$\dfrac{y_{\text{new}}}{y_{\text{old}}} = \dfrac{4kx^{2}}{kx^{2}} = 4$

The constant $k$ cancels — which is why proportion questions never require finding it.

4. General rule

For $y \propto x^{n}$, scaling $x$ by $f$ scales $y$ by $f^{n}$. Doubling with $n=3$ would give a factor of 8; with $n=\tfrac12$ it would give $\sqrt2$.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. It doubles — Power Ignored
  Applying direct proportion to $x$ rather than to $x^{2}$.
• B. It is halved — Proportion Inversion
  Treating the relationship as inverse proportion.
• C. It increases by a factor of 8 — Power Error
  Using $x^{3}$ instead of $x^{2}$.
• D. It stays the same — Conceptual Error
  Assuming the constant absorbs the change.

Common Mistake (⚠️):
Doubling $y$ as well, which would only be right for $y \propto x$. The square applies to the scale factor too.

Takeaway (📌):
The whole bracket gets raised to the power, scale factor included. $f^{n}$ is the only calculation needed.

Correct Answer: B. $16x^{12}$

Fastest Approach (🚀):
$2^{4} = 16;\ x^{3\times 4} = x^{12}$.
$16x^{12}$.

Matches Option B.

Step-by-Step Breakdown:

1. Distribute the outer index

$\left(2x^{3}\right)^{4} = 2^{4}\times\left(x^{3}\right)^{4}$

Every factor inside the bracket takes the outer power, the coefficient included.

2. Evaluate each part

$2^{4} = 16$
$\left(x^{3}\right)^{4} = x^{3\times 4} = x^{12}$

Powers of powers multiply; they do not add.

3. Combine

$16x^{12}$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $2x^{12}$ — Coefficient Ignored
  Forgetting to raise the coefficient to the fourth power.
• C. $8x^{12}$ — Index Error
  Using $2^{3}$ instead of $2^{4}$.
• D. $16x^{7}$ — Index Law Error
  Adding the indices: $3+4 = 7$.
• E. $2x^{7}$ — Index Law Error
  Both errors together.

Common Mistake (⚠️):
Leaving the coefficient untouched ($2x^{12}$) or adding the indices ($x^{7}$) instead of multiplying them.

Takeaway (📌):
$(ab)^n = a^n b^n$ and $(a^m)^n = a^{mn}$. Coefficient raised, indices multiplied.

Correct Answer: A. $77\ \text{cm}^{2}$

Fastest Approach (🚀):
$l+w = \dfrac{36}{2} = 18,\quad l-w = 4$.
Adding: $2l = 22 \implies l = 11,\ w = 7$.
Area $= 11\times 7 = 77\ \text{cm}^{2}$.

Matches Option A.

Step-by-Step Breakdown:

1. Halve the perimeter

$P = 2(l+w) = 36 \implies l+w = 18$

Working with the semi-perimeter avoids carrying a factor of 2 through the algebra.

2. Write the second relationship

$l = w+4 \implies l-w = 4$

3. Solve the sum-and-difference pair

Adding the two equations:
$2l = 22 \implies l = 11$
Subtracting:
$2w = 14 \implies w = 7$

4. Compute the area

$A = 11\times 7 = 77\ \text{cm}^{2}$

Check: perimeter $= 2(11+7) = 36$. Correct.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $80\ \text{cm}^{2}$ — Solving Error
  Using $l=10$, $w=8$ from a mis-solved pair.
• C. $72\ \text{cm}^{2}$ — Setup Error
  Using $l=12$, $w=6$ — a difference of 6 rather than 4.
• D. $96\ \text{cm}^{2}$ — Perimeter Error
  Forgetting to halve the perimeter and scaling down partially.
• E. $64\ \text{cm}^{2}$ — Constraint Ignored
  Assuming a square with side 8.

Common Mistake (⚠️):
Using $l+w = 36$ instead of $18$, which gives a rectangle of 20 by 16 and an area far too large.

Takeaway (📌):
Given a sum and a difference, add and subtract the equations. Halving the perimeter first is what makes them a clean sum-and-difference pair.

Correct Answer: E. $70$

Fastest Approach (🚀):
$42\div 3 = 14$ is one fifth.
$14\times 5 = 70$.

Matches Option E.

Step-by-Step Breakdown:

1. Set up the relationship

$\dfrac35 \times N = 42$

2. Find one fifth

If three fifths is 42, then one fifth is
$42\div 3 = 14$

3. Scale to the whole

$N = 14\times 5 = 70$

4. Equivalent method

$N = 42\div\dfrac35 = 42\times\dfrac53 = 70$. Dividing by a fraction is multiplying by its reciprocal.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $25.2$ — Direction Error
  Multiplying by $\tfrac35$ rather than dividing.
• B. $14$ — Incomplete Answer
  Giving one fifth of the number.
• C. $105$ — Fraction Error
  Dividing by $\tfrac25$ instead of $\tfrac35$.
• D. $63$ — Fraction Error
  Adding half of 42, or using $\tfrac23$.

Common Mistake (⚠️):
Multiplying by the fraction instead of dividing, giving $42\times\tfrac35 = 25.2$. The answer must be larger than 42, since 42 is only part of it.

Takeaway (📌):
Find one part, then scale. Sanity-check the direction: taking a proper fraction shrinks a number, so reversing it must grow.

Correct Answer: E. $54$ km/h

Fastest Approach (🚀):
$\times 3600$ then $\div 1000$ is $\times 3.6$.
$15\times 3.6 = 54$ km/h.

Matches Option E.

Step-by-Step Breakdown:

1. Convert the time unit

In one hour the cyclist covers
$15\ \text{m/s}\times 3600\ \text{s} = 54\,000\ \text{m}$

2. Convert the distance unit

$54\,000\ \text{m} = \dfrac{54\,000}{1000} = 54\ \text{km}$

3. State the speed

$54$ km/h.

4. The single factor

$\dfrac{3600}{1000} = 3.6$, so m/s to km/h is always $\times 3.6$, and km/h to m/s is always $\div 3.6$.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $4.17$ km/h — Direction Error
  Dividing by 3.6 rather than multiplying.
• B. $90$ km/h — Conversion Error
  Multiplying by 6, or converting only the minutes.
• C. $900$ km/h — Conversion Error
  Multiplying by 60 rather than 3.6.
• D. $150$ km/h — Conversion Error
  Multiplying by 10.

Common Mistake (⚠️):
Dividing by 3.6 instead of multiplying, giving $4.17$. A speed in km/h is numerically larger than the same speed in m/s.

Takeaway (📌):
$\times 3.6$ for m/s $\to$ km/h; $\div 3.6$ the other way. The km/h figure is always the bigger number.

Correct Answer: D. $2.4\times 10^{4}$

Fastest Approach (🚀):
$24\times 10^{3}$.
Renormalise: $2.4\times 10^{4}$.

Matches Option D.

Step-by-Step Breakdown:

1. Separate mantissas and powers

$\left(6\times 4\right)\times\left(10^{5}\times 10^{-2}\right)$

2. Multiply and add

$6\times 4 = 24$
$10^{5}\times 10^{-2} = 10^{\,5+(-2)} = 10^{3}$

3. Renormalise

$24\times 10^{3}$ is not standard form, since the mantissa must lie in $[1,10)$:
$24\times 10^{3} = 2.4\times 10^{1}\times 10^{3} = 2.4\times 10^{4}$

Moving the decimal point one place left raises the index by one.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $24\times 10^{3}$ — Normalisation Error
  Correct value but not written in standard form.
• B. $2.4\times 10^{3}$ — Index Error
  Renormalising the mantissa without adjusting the index.
• C. $2.4\times 10^{-10}$ — Index Error
  Multiplying the indices instead of adding them.
• E. $10\times 10^{3}$ — Arithmetic Error
  Mis-multiplying $6\times 4$ as $10$.

Common Mistake (⚠️):
Leaving the answer as $24\times 10^{3}$. It is numerically right but not in standard form, and it is offered as a distractor for exactly that reason.

Takeaway (📌):
Multiply, add indices, then check the mantissa is between 1 and 10. Renormalising changes the index by the number of places moved.

Correct Answer: B. $53$

Fastest Approach (🚀):
$(x+y)^{2} = 81$ and $2xy = 28$.
$81-28 = 53$.

Matches Option B.

Step-by-Step Breakdown:

1. Choose the identity

$(x+y)^{2} = x^{2}+2xy+y^{2}$, so
$x^{2}+y^{2} = (x+y)^{2}-2xy$

2. Substitute

$= 9^{2}-2(14) = 81-28 = 53$

3. Check

$x$ and $y$ are the roots of $t^{2}-9t+14 = 0$, namely $2$ and $7$. Indeed $4+49 = 53$.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $81$ — Incomplete Answer
  Giving $(x+y)^{2}$ without subtracting $2xy$.
• C. $67$ — Omitted Factor
  Subtracting $xy$ once instead of twice.
• D. $95$ — Sign Error
  Adding $2xy$ rather than subtracting it.
• E. $25$ — Formula Misuse
  Using $(x-y)^{2}$ by mistake.

Common Mistake (⚠️):
Answering $81$, which is $(x+y)^{2}$ — the identity's starting point, not its result.

Takeaway (📌):
$x^{2}+y^{2} = (x+y)^{2}-2xy$ and $x^{3}+y^{3} = (x+y)^{3}-3xy(x+y)$. Both convert a symmetric target into given data.