ESAT Mathematics Mock Module 7 Worked Solutions

Correct Answer: A. $16$

Fastest Approach (🚀):
$3.2\times 5 = 16$.

Matches Option A.

Step-by-Step Breakdown:

1. State the relationship

$\text{frequency density} = \dfrac{\text{frequency}}{\text{class width}} \implies \text{frequency} = \text{density}\times\text{width}$

2. Substitute

$3.2\times 5 = 16$

3. Why density exists

When classes have different widths, a taller bar does not necessarily hold more data. Dividing by the width makes the area proportional to frequency, so bars of unequal width can be compared fairly.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $0.64$ — Operation Error
  Dividing $3.2$ by $5$ rather than multiplying.
• C. $3.2$ — Height vs Area
  Reading the density as the frequency.
• D. $8$ — Width Error
  Halving the class width before multiplying.
• E. $5$ — Misread Question
  Giving the class width.

Common Mistake (⚠️):
Reading the height as the frequency, or dividing by the width instead of multiplying — density is already the divided quantity.

Takeaway (📌):
Histogram: area is frequency. Bar chart: height is frequency. Confusing the two is the most common histogram error there is.

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

Fastest Approach (🚀):
Each order gives $\dfrac{4}{10}\times\dfrac{6}{9} = \dfrac{24}{90}$.
Two orders: $\dfrac{48}{90} = \dfrac{8}{15}$.

Matches Option D.

Step-by-Step Breakdown:

1. List the ways it can happen

Red then blue, or blue then red. Both give two different colours.

2. Multiply along each branch

Red then blue: $\dfrac{4}{10}\times\dfrac{6}{9} = \dfrac{24}{90}$
Blue then red: $\dfrac{6}{10}\times\dfrac{4}{9} = \dfrac{24}{90}$

The denominators fall from $10$ to $9$ because the first counter is not replaced.

3. Add across the branches

$\dfrac{24}{90}+\dfrac{24}{90} = \dfrac{48}{90} = \dfrac{8}{15}$

4. Check with the complement

Same colour: $\dfrac{4}{10}\cdot\dfrac39+\dfrac{6}{10}\cdot\dfrac59 = \dfrac{12+30}{90} = \dfrac{42}{90} = \dfrac{7}{15}$, and $\dfrac{8}{15}+\dfrac{7}{15} = 1$. Correct.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{4}{15}$ — Ordering Omitted
  Counting one ordering only.
• B. $\dfrac{1}{3}$ — Independence Error
  Rounding, or treating the two draws as independent.
• C. $\dfrac{12}{25}$ — Replacement Assumed
  Treating the two draws as being with replacement: $2\times\tfrac{4}{10}\times\tfrac{6}{10}$.
• E. $\dfrac{24}{100}$ — Independence Error
  Treating the draws as independent: $\tfrac{4}{10}\times\tfrac{6}{10}$ doubled.

Common Mistake (⚠️):
Counting only one order and answering $\tfrac{24}{90} = \tfrac{4}{15}$. Two different colours can arrive in either sequence, and both must be counted.

Takeaway (📌):
Along a branch, multiply. Across branches, add. Missing the second ordering is the most common error in the whole topic.

Correct Answer: E. $4320$

Fastest Approach (🚀):
$6$ objects (block + T, R, N, G, L): $6! = 720$.
Vowels within the block: $3! = 6$.
$720\times 6 = 4320$.

Matches Option E.

Step-by-Step Breakdown:

1. Check the letters

TRIANGLE has eight distinct letters: T, R, I, A, N, G, L, E. The vowels are I, A and E.

2. Glue the vowels into one block

Treat [IAE] as a single object. The objects to arrange are then the block plus T, R, N, G, L — six objects:
$6! = 720$

3. Arrange within the block

The three vowels can be ordered among themselves in $3! = 6$ ways.

4. Multiply

$720\times 6 = 4320$

For comparison, the unrestricted total is $8! = 40\,320$, so a little over one arrangement in ten has the vowels together.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $720$ — Omitted Factor
  Forgetting the internal arrangements of the three vowels.
• B. $120$ — Counting Error
  Using $5!$ for the other letters and dropping the block.
• C. $2160$ — Factorial Error
  Using $2!$ for the block instead of $3!$.
• D. $40\,320$ — Constraint Ignored
  Ignoring the constraint entirely.

Common Mistake (⚠️):
Forgetting the internal $3!$ and answering $720$, or treating the block as occupying only one of the eight original positions.

Takeaway (📌):
Glue a block of $k$ objects: arrange $(n-k+1)$ objects, then multiply by $k!$ for the internal order.

Correct Answer: A. $12$

Fastest Approach (🚀):
Ends: $2$ vowels in $2! = 2$ orders.
Middle: $3! = 6$.
$2\times 6 = 12$.

Matches Option A.

Step-by-Step Breakdown:

1. Identify the constrained positions

EXAMS has five distinct letters, of which E and A are vowels. Both ends must be vowels, and there are exactly two vowels — so both are used at the ends and neither can appear in the middle.

2. Fill the ends first

E at the front and A at the back, or A at the front and E at the back: $2! = 2$ ways.

3. Arrange the remainder

X, M and S fill the three middle positions in $3! = 6$ ways.

4. Multiply

$2\times 6 = 12$

The order matters: satisfying the constraint first means the remaining count is unrestricted. Arranging freely and then trying to discard invalid cases is far slower.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $24$ — Counting Error
  Using $4!$ for the remaining letters, or forgetting that both vowels are used up.
• C. $6$ — Omitted Factor
  Giving $3!$ alone and forgetting the two ways to place the vowels.
• D. $48$ — Over-count
  Doubling twice, or treating the ends as having $4$ choices.
• E. $120$ — Constraint Ignored
  Ignoring the constraint entirely: $5!$.

Common Mistake (⚠️):
Computing $5! = 120$ and then trying to subtract the invalid arrangements. With a positional constraint it is nearly always faster to place the constrained letters first.

Takeaway (📌):
Constrained positions first, free positions second, then multiply. The same method handles 'must start with', 'must end with' and 'must be in the middle'.

Correct Answer: C. $60$

Fastest Approach (🚀):
$6! = 720$; repeats are $3$ As and $2$ Ns.
$\dfrac{720}{3!\times 2!} = \dfrac{720}{12} = 60$.

Matches Option C.

Step-by-Step Breakdown:

1. Count as if every letter were distinct

Six letters give $6! = 720$ orderings.

2. Identify the repeats

B appears once, A three times, N twice.

3. Divide out each repeat

The three As can be permuted among themselves in $3! = 6$ ways without changing the word, and the two Ns in $2! = 2$ ways:
$\dfrac{6!}{3!\times 2!} = \dfrac{720}{6\times 2} = 60$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $720$ — Repeats Ignored
  Ignoring the repeated letters entirely.
• B. $120$ — Factorial Error
  Dividing by $(3+2)!$ instead of $3!\times 2!$.
• D. $180$ — Repeat Count Error
  Dividing by $2!\times 2!$, missing one of the As.
• E. $36$ — Factorial Error
  Dividing by $4!$ or another mis-counted repeat.

Common Mistake (⚠️):
Dividing by $(3+2)! = 120$ rather than by $3!\times 2!$. Each repeated letter contributes its own factorial, and those factorials multiply.

Takeaway (📌):
Arrangements of $n$ letters with repeats is $\dfrac{n!}{n_1!\,n_2!\cdots}$ — one factorial per repeated letter, multiplied together.

Correct Answer: D. $48$

Fastest Approach (🚀):
$4! = 24$ arrangements of the block plus M, A, S.
$\times 2$ for TH and HT: $48$.

Matches Option D.

Step-by-Step Breakdown:

1. Glue the pair together

Treat TH as a single object. The objects to arrange are then [TH], M, A and S — four objects.

2. Arrange the objects

$4! = 24$

3. Account for the internal order

Inside the block, TH and HT are different arrangements:
$24\times 2 = 48$

4. Sanity check against the complement

Total arrangements are $5! = 120$, so $120-48 = 72$ have T and H apart. Both counts are plausible and they sum correctly.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $24$ — Omitted Factor
  Forgetting to double for the block's internal order.
• B. $12$ — Counting Error
  Using $3!$ objects, losing one letter.
• C. $120$ — Constraint Ignored
  Ignoring the constraint entirely.
• E. $72$ — Complement Inversion
  Giving the number of arrangements where T and H are apart.

Common Mistake (⚠️):
Forgetting the factor of $2$ for the block's internal order, giving $24$.

Takeaway (📌):
Glue for 'together', and subtract the glued count from the total for 'apart'. The internal-order factor is what most answers miss.

Correct Answer: A. $4$

Fastest Approach (🚀):
Deviations already squared and summed: $\sigma^{2} = \dfrac{32}{8} = 4$.

Matches Option A.

Step-by-Step Breakdown:

1. Identify which sum you have

Two equivalent formulas:
$\sigma^{2} = \dfrac{\sum(x-\bar x)^{2}}{n} \qquad\text{or}\qquad \sigma^{2} = \dfrac{\sum x^{2}}{n}-\bar x^{2}$

The question gives $\sum(x-\bar x)^{2}$ — deviations, already squared and summed — so the first formula applies.

2. Divide

$\sigma^{2} = \dfrac{32}{8} = 4$

3. Do not subtract again

The mean has already been removed inside the bracket. Subtracting $\bar x^{2} = 25$ as well would give $4-25 = -21$, a negative variance and therefore impossible.

4. Cross-check

If $\sigma^{2} = 4$ and $\bar x = 5$, then $\sum x^{2} = n\left(\sigma^{2}+\bar x^{2}\right) = 8(29) = 232$. Feeding that into the second formula: $\dfrac{232}{8}-25 = 29-25 = 4$. The two agree, as they must.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $32$ — Omitted Division
  Giving $\sum(x-\bar x)^{2}$ without dividing by $n$.
• C. $25$ — Misread Question
  Giving $\bar x^{2}$.
• D. $5$ — Misread Question
  Giving the mean, or the standard deviation of a different figure.
• E. $-21$ — Double Subtraction
  Subtracting $\bar x^{2}$ from an already-centred sum — a negative variance.

Common Mistake (⚠️):
Applying the sums formula to the deviations sum and subtracting $\bar x^{2}$ a second time. A negative answer is the immediate signal, since variance can never be below zero.

Takeaway (📌):
$\sum(x-\bar x)^{2}$ needs only a division. $\sum x^{2}$ needs the extra $-\bar x^{2}$. Read which sum you were given before choosing.

Correct Answer: D. The median, because it is unaffected by the extreme value

Fastest Approach (🚀):
The mean is pulled to roughly $\pounds 47{,}000$ — above nine of the ten salaries.
The median stays inside the $\pounds 22{,}000$ to $\pounds 30{,}000$ cluster.

Matches Option D.

Step-by-Step Breakdown:

1. Estimate the mean

Nine salaries near $\pounds 26{,}000$ total about $\pounds 234{,}000$; adding $\pounds 250{,}000$ gives roughly $\pounds 484{,}000$ over ten employees:
$\bar x \approx \pounds 48{,}000$

That figure is higher than nine of the ten actual salaries, so it describes nobody.

2. Locate the median

With ten values, the median is the average of the 5th and 6th — both inside the $\pounds 22{,}000$–$\pounds 30{,}000$ band. It is unmoved by how extreme the tenth value is.

3. Rule out the others

The mode may not exist or may be an accident of rounding. The range is a measure of spread, not an average at all.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. The mean, because it uses every value — Robustness Ignored
  Using every value is exactly what makes the mean vulnerable here.
• B. The range, because it shows the spread — Measure Confusion
  The range measures spread, not a typical value.
• C. The mode, because it is the most common salary — Measure Confusion
  With salaries spread across a band there may be no meaningful mode.
• E. The mean, because the data set is small — Wrong Reason
  Sample size is not the issue; the extreme value is.

Common Mistake (⚠️):
Choosing the mean because it 'uses all the data'. Using every value is precisely why one extreme value can distort it.

Takeaway (📌):
Skewed data: median. Symmetric data: mean. Categorical data: mode. The right average depends on the shape, not on preference.

Correct Answer: B. $\dfrac{5}{12}$

Fastest Approach (🚀):
Given a car owner, the population is the $60$ car owners.
$\dfrac{25}{60} = \dfrac{5}{12}$.

Matches Option B.

Step-by-Step Breakdown:

1. Fill the four regions

Neither: $20$, so at least one: $80$.
Car total: $60$, of which both: $25$, so car only: $35$.
Bicycle only: $80-60 = 20$.

Check: $35+25+20+20 = 100$.

2. Identify the conditioning

The person is already known to own a car, so the relevant population is the $60$ car owners — not all $100$ people.

3. Divide

$P(\text{bike}\mid\text{car}) = \dfrac{25}{60} = \dfrac{5}{12}$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{1}{4}$ — Wrong Denominator
  Dividing by the whole survey: $P(\text{car and bike})$, not the conditional probability.
• C. $\dfrac{25}{45}$ — Wrong Denominator
  Using bicycle owners ($45$) as the denominator, which conditions on the wrong event.
• D. $\dfrac{3}{5}$ — Misread Question
  Giving $P(\text{car})$.
• E. $\dfrac{9}{20}$ — Wrong Denominator
  Using bicycle owners over the whole survey.

Common Mistake (⚠️):
Dividing by $100$ to get $\tfrac{25}{100} = \tfrac14$. That is $P(\text{car and bike})$, the probability for a randomly chosen person from the whole survey — a different question.

Takeaway (📌):
Conditional probability changes the denominator, not the numerator. 'Given that…' tells you which group to divide by.

Correct Answer: C. $10$

Fastest Approach (🚀):
$\dfrac{n(n-1)}{2} = 45 \implies n(n-1) = 90$.
$10\times 9 = 90 \implies n = 10$.

Matches Option C.

Step-by-Step Breakdown:

1. Recognise the structure

Each handshake involves two people and order does not matter, so the number of handshakes among $n$ people is
$\binom{n}{2} = \dfrac{n(n-1)}{2}$

2. Set it equal to 45

$\dfrac{n(n-1)}{2} = 45 \implies n(n-1) = 90$

3. Solve

Two consecutive integers multiplying to $90$: $10\times 9$. So $n = 10$.

Formally $n^{2}-n-90 = 0$ gives $(n-10)(n+9) = 0$, and the negative root is rejected.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $9$ — Off-by-one Error
  Giving $n-1$, the number of handshakes one person makes.
• B. $45$ — Misread Question
  Giving the number of handshakes.
• D. $15$ — Omitted Factor
  Solving $n(n-1) = 45$ without the factor of $2$, then rounding.
• E. $90$ — Incomplete Answer
  Giving $n(n-1)$ rather than $n$.

Common Mistake (⚠️):
Forgetting the division by $2$ and solving $n(n-1) = 45$, or answering $9$ by giving the number of handshakes each person makes rather than the number of people.

Takeaway (📌):
Handshakes, chords, matches in a round-robin and lines through pairs of points are all $\binom{n}{2}$ — the same count wearing different clothes.

Correct Answer: A. $676\,000$

Fastest Approach (🚀):
$26\times 26 = 676$; $10^{3} = 1000$.
$676\times 1000 = 676\,000$.

Matches Option A.

Step-by-Step Breakdown:

1. Count each position independently

Repetition is allowed, so the first letter does not reduce the choices for the second:
letters: $26\times 26 = 676$
digits: $10\times 10\times 10 = 1000$

2. Multiply the two blocks

$676\times 1000 = 676\,000$

3. Contrast with no repetition

Without repetition the count would be $26\times 25\times 10\times 9\times 8 = 468\,000$ — noticeably smaller, and the reason the question states the rule explicitly.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $67\,600$ — Index Error
  Using $10^{2}$ instead of $10^{3}$ for the digits.
• C. $468\,000$ — Constraint Misread
  Assuming no repetition despite the question allowing it.
• D. $6\,760\,000$ — Index Error
  Using $26^{2}\times 10^{4}$.
• E. $15\,600$ — Counting Error
  Using $26\times 25\times 10\times 2$ or another partial product.

Common Mistake (⚠️):
Using $26\times 25$ and $10\times 9\times 8$ despite the question permitting repeats, giving $468\,000$.

Takeaway (📌):
Repetition allowed: each position keeps its full range. Repetition forbidden: the range falls by one each time.

Correct Answer: E. $0.25$

Fastest Approach (🚀):
$0.3+0.45 = 0.75$.
$1-0.75 = 0.25$.

Matches Option E.

Step-by-Step Breakdown:

1. Check the events are exhaustive

The spinner can land only on those three colours, and it cannot land on two at once. So the three are mutually exclusive and exhaustive.

2. Apply the total

$P(\text{red})+P(\text{blue})+P(\text{green}) = 1$

3. Solve

$0.3+0.45+P(\text{green}) = 1 \implies P(\text{green}) = 1-0.75 = 0.25$

4. Why 'exhaustive' matters

If a fourth colour were possible, the three would not sum to $1$ and the question could not be answered. The word 'only' in the stem is doing real work.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $0.75$ — Incomplete Answer
  Giving the sum of the two known probabilities.
• B. $0.15$ — Operation Error
  Computing $0.45-0.3$.
• C. $0.3$ — Assumption Error
  Assuming green matches red.
• D. $0.55$ — Omitted Term
  Subtracting only $P(\text{red})$ from 1.

Common Mistake (⚠️):
Giving $0.75$, the total of the two known probabilities, rather than what remains.

Takeaway (📌):
Mutually exclusive and exhaustive means the probabilities sum to exactly $1$. Check both conditions before relying on it.

Correct Answer: B. $61$

Fastest Approach (🚀):
Have it: $200\times 0.3 = 60 \implies 0.9\times 60 = 54$ positive.
Do not: $140 \implies 0.05\times 140 = 7$ positive.
$54+7 = 61$.

Matches Option B.

Step-by-Step Breakdown:

1. Split the group

$30\%$ of $200 = 60$ have the condition; the other $140$ do not.

2. Apply each branch rate

Of the $60$ with the condition, $90\%$ test positive: $54$.
Of the $140$ without, $5\%$ test positive: $7$.

3. Add the two positive branches

$54+7 = 61$

4. Why the false positives matter

Only $54$ of the $61$ positives actually have the condition, so $P(\text{condition}\mid\text{positive}) = \dfrac{54}{61} \approx 0.89$. A test that is $90\%$ accurate on those who have a condition does not make a positive result $90\%$ reliable — the size of the healthy group matters too.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $54$ — Omitted Branch
  Counting only the true positives.
• C. $60$ — Misread Question
  Giving the number with the condition rather than the number testing positive.
• D. $67$ — Wrong Base
  Using $5\%$ of $200$ rather than of the $140$ without the condition.
• E. $95$ — Rate Misapplied
  Adding $90\%$ and $5\%$ as percentages of the whole group.

Common Mistake (⚠️):
Giving $54$ and forgetting the false positives from the much larger healthy group. When the condition is rare, that second branch can dominate.

Takeaway (📌):
Convert percentages to people before branching. The arithmetic stays in whole numbers and the false-positive branch is impossible to overlook.

Correct Answer: C. $93$

Fastest Approach (🚀):
French: $0.6\times 300 = 180 \implies 0.25\times 180 = 45$.
Not French: $120 \implies 0.4\times 120 = 48$.
$45+48 = 93$.

Matches Option C.

Step-by-Step Breakdown:

1. Split by the first stage

$60\%$ of $300 = 180$ study French; the remaining $120$ do not.

2. Apply the two German rates

French and German: $25\%$ of $180 = 45$
German but not French: $40\%$ of $120 = 48$

3. Add both routes to German

$45+48 = 93$

4. What the counts then give free

Since $93$ study German and $45$ of them study French,
$P(\text{French}\mid\text{German}) = \dfrac{45}{93} = \dfrac{15}{31}$
Once the tree is in whole numbers, every conditional in the question is a single division.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $45$ — Omitted Branch
  Counting only those who study both.
• B. $48$ — Omitted Branch
  Counting only the German-without-French branch.
• D. $105$ — Wrong Base
  Using $40\%$ of the whole $300$ plus a partial branch.
• E. $120$ — Misread Question
  Giving the number who do not study French.

Common Mistake (⚠️):
Applying the $40\%$ to the whole $300$, or stopping at $45$ because the French branch was the one the question opened with.

Takeaway (📌):
Each branch rate applies to its own group only. Sweep every path that ends in the outcome before adding.

Correct Answer: A. $30$

Fastest Approach (🚀):
Ways to total at least $10$: $6$ of $36 \implies P = \dfrac16$.
$180\times\dfrac16 = 30$.

Matches Option A.

Step-by-Step Breakdown:

1. Count the favourable outcomes

Total $10$: $(4,6),(5,5),(6,4)$ — three ways.
Total $11$: $(5,6),(6,5)$ — two ways.
Total $12$: $(6,6)$ — one way.

$3+2+1 = 6$ favourable outcomes out of $36$.

2. Find the probability

$P = \dfrac{6}{36} = \dfrac16$

3. Scale by the number of trials

$\text{expected frequency} = 180\times\dfrac16 = 30$

4. What 'expected' means

This is a long-run average, not a guarantee. An actual run of $180$ rolls would give a number near $30$, not exactly $30$ — which is precisely the distinction between expected and observed frequency.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $18$ — Counting Error
  Using only the totals of $11$ and $12$, or $P = \tfrac{1}{10}$.
• C. $45$ — Probability Error
  Using $P = \tfrac14$.
• D. $60$ — Probability Error
  Using $P = \tfrac13$.
• E. $36$ — Ordering Error
  Counting unordered pairs, or using $\tfrac{36}{180}$ inverted.

Common Mistake (⚠️):
Counting only the total of $10$, or forgetting that $(4,6)$ and $(6,4)$ are different outcomes on distinguishable dice.

Takeaway (📌):
Expected frequency $= P\times n$. Get the probability from the possibility space, then multiply once.

Correct Answer: B. $14.3$

Fastest Approach (🚀):
Midpoints: $\tfrac{0+10}{2} = 5$, $\tfrac{11+20}{2} = 15.5$, $\tfrac{21+30}{2} = 25.5$.
$\sum fx = 30+155+102 = 287$; $\sum f = 20$.
$\bar x \approx \dfrac{287}{20} = 14.35$.

Matches Option B.

Step-by-Step Breakdown:

1. Take each midpoint from its own interval

$\dfrac{0+10}{2} = 5, \qquad \dfrac{11+20}{2} = 15.5, \qquad \dfrac{21+30}{2} = 25.5$

There is no need to invent a shared boundary at $10.5$. The classes are written as they are because the marks are whole numbers, so $11$–$20$ genuinely begins at $11$.

2. Weight by frequency

$5\times 6 = 30, \qquad 15.5\times 10 = 155, \qquad 25.5\times 4 = 102$

3. Divide by the total frequency

$\bar x \approx \dfrac{30+155+102}{6+10+4} = \dfrac{287}{20} = 14.35$

To one decimal place, $14.3$ — recognising that the whole figure is an estimate in the first place, because the individual marks within each class are unknown.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $15$ — Weighting Error
  Averaging the three midpoints without weighting by frequency.
• C. $13.75$ — Boundary Convention
  Using continuous boundaries $10.5$ and $20.5$ to build the midpoints.
• D. $14.05$ — Midpoint Error
  Using lower bounds, or a midpoint slip in the middle class.
• E. $16$ — Midpoint Error
  Using upper class limits as the representative values.

Common Mistake (⚠️):
Treating the classes as continuous and using boundaries of $10.5$ and $20.5$, so the midpoints become $5.25$, $15.5$ and $25.5$. That convention belongs to continuous data such as time or mass, where a value of $10.4$ is possible and has to fall somewhere.

Takeaway (📌):
Discrete classes written $0$–$10$, $11$–$20$: midpoint of each interval as written. Continuous classes written $0 \le t \lt 10$: the boundaries already meet, so nothing is lost.

Correct Answer: C. Students who did more past papers tended to score higher

Fastest Approach (🚀):
Correlation gives direction and strength, not cause and not certainty.
Only the statement about a tendency is supported.

Matches Option C.

Step-by-Step Breakdown:

1. What correlation states

A strong positive correlation means that as one variable increases the other tends to increase, and that the points lie close to a straight-line trend.

2. What it does not state

It does not identify a cause. A third factor — motivation, available study time, prior attainment — could drive both variables. Nor does it claim every individual follows the trend; 'strong' still permits exceptions.

3. Test each option

Options A and B both assert a direction of causation the data cannot establish. Option D claims a rule without exceptions, which correlation never provides. Option E contradicts the stated correlation. Option C claims a tendency, which is exactly what a correlation supports.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Doing past papers causes higher scores — Causation Fallacy
  Inferring causation from an observed association.
• B. Higher scores cause students to do more past papers — Causation Fallacy
  Inferring causation in the reverse direction, equally unsupported.
• D. Every student who did more past papers scored higher — Over-generalisation
  Treating a strong correlation as an exceptionless rule.
• E. There is no relationship between the two variables — Misread Question
  Contradicting the stated correlation.

Common Mistake (⚠️):
Reading a strong correlation as proof of cause. The association is real; the explanation for it is not supplied by the data.

Takeaway (📌):
Correlation licenses 'tended to', never 'caused' and never 'every'. Watch for the absolute word in the options — it is usually the trap.

Correct Answer: D. $\dfrac{3}{4}$

Fastest Approach (🚀):
Product is odd only if both dice are odd: $\dfrac12\times\dfrac12 = \dfrac14$.
$1-\dfrac14 = \dfrac34$.

Matches Option D.

Step-by-Step Breakdown:

1. Find when the product is odd

A product is odd only when every factor is odd. Each die is odd with probability $\dfrac{3}{6} = \dfrac12$, and the dice are independent:
$P(\text{both odd}) = \dfrac12\times\dfrac12 = \dfrac14$

2. Take the complement

$P(\text{product even}) = 1-\dfrac14 = \dfrac34$

3. Check by counting

Of the $36$ outcomes, $3\times 3 = 9$ have both dice odd, so $36-9 = 27$ give an even product, and $\dfrac{27}{36} = \dfrac34$. Confirmed.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{1}{2}$ — Analogy Error
  Assuming products behave like sums, which are even half the time.
• B. $\dfrac{1}{4}$ — Complement Inversion
  Giving the probability that the product is odd.
• C. $\dfrac{2}{3}$ — Counting Error
  Using $\tfrac{4}{6}$ from a mis-counted sample space.
• E. $\dfrac{5}{6}$ — Counting Error
  Over-counting the even outcomes.

Common Mistake (⚠️):
Assuming a product is even 'about half the time' by analogy with a sum. Sums and products behave quite differently: a sum is even when both are odd or both are even.

Takeaway (📌):
Odd product needs all odd factors, which makes the complement tiny and easy. Even sums, by contrast, split evenly.

Correct Answer: B. $\dfrac{5}{14}$

Fastest Approach (🚀):
$\dfrac58\times\dfrac47 = \dfrac{20}{56} = \dfrac{5}{14}$.

Matches Option B.

Step-by-Step Breakdown:

1. Route one: count the selections

Total ways to choose $2$ from $8$: $\binom{8}{2} = 28$.
Ways to choose $2$ reds from $5$: $\binom{5}{2} = 10$.
$P = \dfrac{10}{28} = \dfrac{5}{14}$

2. Route two: multiply along the branch

First counter red: $\dfrac58$. Second red, given one red is gone: $\dfrac47$.
$P = \dfrac58\times\dfrac47 = \dfrac{20}{56} = \dfrac{5}{14}$

3. Note the agreement

Both routes must agree, which makes each a check on the other. Option D is the same value unsimplified — correct arithmetic, but the question asks for a probability in its simplest form.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{25}{64}$ — Independence Error
  Treating the draws as independent: $\left(\tfrac58\right)^{2}$.
• C. $\dfrac{5}{8}$ — Misread Question
  Giving the probability that a single counter is red.
• D. $\dfrac{5}{16}$ — Denominator Error
  Reducing the number of red counters for the second draw but forgetting to reduce the total, giving $\tfrac{5}{8}\times\tfrac{4}{8}$.
• E. $\dfrac{1}{2}$ — Estimation Error
  Estimating from the majority colour.

Common Mistake (⚠️):
Treating the draws as independent and computing $\left(\tfrac58\right)^{2} = \tfrac{25}{64}$. Without replacement, both the numerator and the denominator fall for the second draw.

Takeaway (📌):
Without replacement, either count combinations or multiply conditional probabilities. Both work; the independent-events product does not.

Correct Answer: B. $8:7$

Fastest Approach (🚀):
Mean $= \dfrac{40}{5} = 8$; median $= 7$.
Ratio $= 8:7$.

Matches Option B.

Step-by-Step Breakdown:

1. Find the mean

$\dfrac{4+7+7+9+13}{5} = \dfrac{40}{5} = 8$

2. Find the median

Five values already in order, so the median is the third: $7$.

3. Form the ratio in the stated order

'The ratio of the mean to the median' means mean first:
$8:7$

The ratio exceeds $1$, which is the signature of data skewed to the right — here by the value $13$.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $7:8$ — Order Error
  Reversing the order of the ratio.
• C. $1:1$ — Conceptual Error
  Assuming mean and median always agree.
• D. $8:5$ — Value Error
  Using the mode or the smallest value as the second term.
• E. $13:4$ — Misread Question
  Using the largest and smallest values.

Common Mistake (⚠️):
Writing the ratio the wrong way round as $7:8$. 'The ratio of A to B' always means $A:B$, in that order.

Takeaway (📌):
Mean to median above $1$ means right-skewed; below $1$ means left-skewed; exactly $1$ suggests symmetry. The ratio carries information the difference does not.

Correct Answer: C. $\dfrac{54}{61}$

Fastest Approach (🚀):
Positives: $54+7 = 61$.
Of those, $54$ have the condition.
$P = \dfrac{54}{61}$.

Matches Option C.

Step-by-Step Breakdown:

1. Identify the conditioning group

We are told the person tested positive. Everyone who tested negative is now irrelevant.

2. Count that group

$54$ true positives $+\ 7$ false positives $= 61$ people.

3. Count the favourable cases within it

Of those $61$, exactly $54$ actually have the condition.

$P(\text{condition}\mid\text{positive}) = \dfrac{54}{61} \approx 0.885$

4. Why this is not 90%

$\dfrac{54}{60} = 90\%$ is $P(\text{positive}\mid\text{condition})$ — the test's accuracy on people who have it. That is a different question, and reversing the two is the classic error. The two coincide only when the false-positive branch is empty.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{9}{10}$ — Conditional Reversed
  Quoting the test's accuracy, which is $P(\text{positive}\mid\text{condition})$.
• B. $\dfrac{54}{200}$ — Wrong Denominator
  Dividing by the whole screening rather than by the positives.
• D. $\dfrac{1}{20}$ — Wrong Conditional
  The false-positive rate, $\tfrac{7}{140}$ - the chance of testing positive given that you do not have the condition, which is a different conditional again.
• E. $\dfrac{61}{200}$ — Misread Question
  Giving the probability of testing positive at all.

Common Mistake (⚠️):
Answering $\tfrac{9}{10}$ by quoting the test's accuracy. $P(A\mid B)$ and $P(B\mid A)$ are different quantities and swapping them is the single most consequential mistake in conditional probability.

Takeaway (📌):
Conditioning sets the denominator. 'Given positive' means divide by everyone who tested positive, however they got there.

Correct Answer: B. $\dfrac{9}{40}$

Fastest Approach (🚀):
$\dfrac{45}{200} = \dfrac{9}{40} = 0.225$.

Matches Option B.

Step-by-Step Breakdown:

1. Use relative frequency

With no information about how the spinner is constructed, the best available estimate is
$\dfrac{\text{successes}}{\text{trials}} = \dfrac{45}{200}$

2. Simplify

$\dfrac{45}{200} = \dfrac{9}{40} = 0.225$

3. What this estimate is and is not

It is an estimate, not the true probability. Repeating the experiment would give a slightly different figure, and more spins would narrow the variation. A theoretical probability would require knowing the spinner's construction.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{1}{4}$ — Rounding Error
  Rounding $0.225$ to a convenient quarter.
• C. $\dfrac{45}{155}$ — Wrong Denominator
  Dividing by the non-red spins instead of the total trials.
• D. $\dfrac{1}{45}$ — Inversion Error
  Inverting the fraction.
• E. It cannot be estimated — Conceptual Error
  Assuming a probability cannot be estimated from data.

Common Mistake (⚠️):
Dividing by the $155$ non-red spins rather than by the $200$ trials. The denominator is always the total number of trials.

Takeaway (📌):
Relative frequency $= \dfrac{\text{successes}}{\text{trials}}$, and it approaches the true probability as the number of trials grows.

Correct Answer: E. $74$

Fastest Approach (🚀):
All committees: $\binom{9}{3} = 84$.
All-male committees: $\binom{5}{3} = 10$.
$84-10 = 74$.

Matches Option E.

Step-by-Step Breakdown:

1. Count every committee

There are $9$ people in total:
$\binom{9}{3} = \dfrac{9\times 8\times 7}{3\times 2\times 1} = 84$

2. Count the committees that break the requirement

'No women' means all three chosen from the five men:
$\binom{5}{3} = \dfrac{5\times 4\times 3}{6} = 10$

3. Subtract

$84-10 = 74$

4. Why not count directly

Counting one, two and three women separately gives $\binom41\binom52+\binom42\binom51+\binom43 = 40+30+4 = 74$ — the same answer from three calculations instead of two.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $84$ — Constraint Ignored
  Giving the total number of committees without the restriction.
• B. $64$ — Arithmetic Error
  Subtracting an incorrect complement, such as $\binom{5}{2}\times 2$.
• C. $40$ — Incomplete Case Analysis
  Counting only the committees with exactly one woman.
• D. $10$ — Complement Inversion
  Giving the all-male count — the complement itself.

Common Mistake (⚠️):
Computing $\binom{4}{1}\binom{5}{2} = 40$ and stopping, which counts only the committees with exactly one woman.

Takeaway (📌):
'At least one' over selections is total minus none, exactly as it is over trials. One subtraction beats three cases.

Correct Answer: A. $90$

Fastest Approach (🚀):
Girls: $\binom42 = 6$. Boys: $\binom62 = 15$.
$6\times 15 = 90$.

Matches Option A.

Step-by-Step Breakdown:

1. Fix the split

Exactly $2$ girls in a team of $4$ means exactly $2$ boys as well.

2. Count each group

$\binom{4}{2} = \dfrac{4\times 3}{2} = 6$ ways to choose the girls
$\binom{6}{2} = \dfrac{6\times 5}{2} = 15$ ways to choose the boys

3. Multiply

Each choice of girls can pair with each choice of boys:
$6\times 15 = 90$

4. Compare with the unrestricted total

$\binom{10}{4} = 210$ teams in total, so $90$ of them — a little under half — have exactly two girls.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $210$ — Constraint Ignored
  Giving the total number of teams with no restriction.
• C. $60$ — Arithmetic Error
  Using $\binom42\times\binom62$ with a mis-computed $\binom62 = 10$.
• D. $36$ — Split Error
  Using $\binom42\times\binom61$ or another wrong split.
• E. $120$ — Order Error
  Using permutations rather than combinations for one group.

Common Mistake (⚠️):
Adding $6+15 = 21$ instead of multiplying, or forgetting that fixing the number of girls also fixes the number of boys.

Takeaway (📌):
'Exactly $k$ from one group' means choose from each group and multiply. 'At least $k$' means total minus the complement.

Correct Answer: E. $18$

Fastest Approach (🚀):
$\dfrac{60}{1200} = \dfrac{1}{20}$.
$\dfrac{360}{20} = 18$.

Matches Option E.

Step-by-Step Breakdown:

1. Find the sampling fraction

$\dfrac{\text{sample size}}{\text{population}} = \dfrac{60}{1200} = \dfrac{1}{20}$

One student in every twenty is sampled.

2. Apply it to the stratum

$\dfrac{1}{20}\times 360 = 18$

3. Check the whole sample

Year 12: $\tfrac{240}{20} = 12$. Year 13: $18$. Earlier years: $\tfrac{600}{20} = 30$. Total $= 60$. Correct.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $20$ — Equal Split Error
  Dividing the sample equally between the three groups.
• B. $12$ — Wrong Stratum
  Giving the Year 12 figure.
• C. $30$ — Wrong Stratum
  Giving the figure for the earlier years.
• D. $36$ — Fraction Error
  Using a sampling fraction of $\tfrac{1}{10}$.

Common Mistake (⚠️):
Splitting the sample equally between the three groups, giving $20$ each — which over-represents Year 12 and under-represents the largest group.

Takeaway (📌):
One sampling fraction, applied to every stratum. Check the parts sum to the intended sample size before answering.

Correct Answer: D. $60$

Fastest Approach (🚀):
$4\times 5\times 3 = 60$.

Matches Option D.

Step-by-Step Breakdown:

1. Identify the independent choices

Choosing a starter does not restrict the main course, and neither restricts the dessert. The three decisions are independent.

2. Apply the product rule

If one task can be done in $m$ ways and a second in $n$ ways, the pair can be done in $mn$ ways. Extending to three:
$4\times 5\times 3 = 60$

3. Why not add

Adding, $4+5+3 = 12$, would count the number of individual dishes on the menu — not the number of meals. Addition counts alternatives; multiplication counts combinations of independent choices.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $12$ — Additive Error
  Adding the three counts instead of multiplying.
• B. $120$ — Over-count
  Multiplying by an extra factor, such as treating the order of courses as variable.
• C. $35$ — Operation Error
  Adding pairwise products, or $4\times 5+3\times 5$.
• E. $45$ — Operation Error
  Computing $4\times 5+5\times 3+4\times 3$ style combinations.

Common Mistake (⚠️):
Adding the three numbers to get $12$. Add when the choices are alternatives to each other; multiply when they are made together.

Takeaway (📌):
Independent choices multiply. Every permutation and combination formula is this rule applied repeatedly.

Correct Answer: C. $y = x^{2}$

Fastest Approach (🚀):
$y = x^{2}$: mean of squares $= \dfrac{14}{3} \approx 4.67$, square of mean $= 4$. Different.
All the linear maps agree, so squaring is the odd one out.

Matches Option C.

Step-by-Step Breakdown:

1. Test a linear transformation

$y = 3x+4$ gives $7, 10, 13$ with mean $10$. Transforming the mean gives $3(2)+4 = 10$. They agree.

2. Test the squaring

$y = x^{2}$ gives $1, 4, 9$ with mean $\dfrac{14}{3} \approx 4.67$. Squaring the mean gives $2^{2} = 4$. They do not agree.

3. Why linearity is the dividing line

For $y = ax+b$:
$\bar y = \dfrac{\sum(ax_i+b)}{n} = \dfrac{a\sum x_i+nb}{n} = a\bar x+b$

The sum passes straight through a multiplication and an addition. It cannot pass through a square, because $\sum x_i^{2} \ne \left(\sum x_i\right)^{2}$.

4. The gap has a name

The difference between the mean of the squares and the square of the mean is exactly the variance: $\dfrac{14}{3}-4 = \dfrac23$. It is zero only when every value is identical.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $y = 3x+4$ — Linearity Holds
  Linear: mean of the transformed values is $10$, and $3(2)+4 = 10$.
• B. $y = x-7$ — Linearity Holds
  Linear: a shift passes through the mean unchanged.
• D. $y = \tfrac12 x$ — Linearity Holds
  Linear: scaling passes through the mean unchanged.
• E. None of them differ — Conceptual Error
  Squaring genuinely does differ, by exactly the variance.

Common Mistake (⚠️):
Assuming that because the shortcut works for coding, it works for any transformation. It is linearity that makes it work, and the moment a power, root or reciprocal appears the shortcut fails.

Takeaway (📌):
Transform the summary statistic only for linear maps. For anything else, transform every value first and then average.