ESAT Mathematics Mock Module 6 Worked Solutions

Correct Answer: E. $\tfrac16$

Fastest Approach (🚀):
$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ — six ways.
$\dfrac{6}{36} = \dfrac16$.

Matches Option E.

Step-by-Step Breakdown:

1. Count the total outcomes

Six faces on each die, and the two are independent:
$6\times 6 = 36$ equally likely outcomes.

2. List the favourable ones systematically

Work up through the first die:
$(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$

Six outcomes. Going in order guarantees none is missed and none is counted twice.

3. Divide

$P(\text{total }7) = \dfrac{6}{36} = \dfrac16$

4. Why order matters here

$(2,5)$ and $(5,2)$ are different outcomes, because the dice are distinguishable — imagine one red and one blue. Counting unordered pairs gives three and halves the probability to $\tfrac{1}{12}$, which is Option B.

5. Why 7 is the most likely total

It has more ways of being made than any other, because every face value on the first die has exactly one partner. Totals of $2$ and $12$ have only one way each. That is why $7$ is the pivot of the distribution — and why it matters in games built on two dice.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\tfrac{5}{36}$ — Wrong Total
  The probability of a total of $6$ or of $8$.
• B. $\tfrac{1}{12}$ — Order Ignored
  Counting unordered pairs, halving the true count.
• C. $\tfrac{7}{36}$ — Numerator Confused
  Using the total $7$ as the numerator.
• D. $\tfrac{1}{7}$ — Sample Space Wrong
  Assuming one outcome in seven.

Common Mistake (⚠️):
Treating $(2,5)$ and $(5,2)$ as the same outcome. The dice are distinguishable, so both count separately in the sample space of $36$.

Takeaway (📌):
Two dice give $36$ equally likely ordered outcomes. List favourable pairs in order of the first die so none is missed.

Correct Answer: C. The dashed sides are equal in length; the arrowed lines are parallel

Fastest Approach (🚀):
Dashes $\Rightarrow$ equal lengths. Arrows $\Rightarrow$ parallel.

Matches Option C.

Step-by-Step Breakdown:

1. The standard markings

Dashes on sides — sides carrying the same number of dashes are equal in length. One dash matches one dash, two match two, and so on.

Arrowheads on lines — lines carrying the same arrowhead are parallel.

Arcs on angles — angles with the same number of arcs are equal.

A small square at a vertex — that angle is a right angle.

2. Apply them

Two sides with one dash each are equal in length, so the triangle is at least isosceles.

Two lines with one arrowhead each are parallel.

3. Why the conventions matter

A diagram is not drawn to scale, and nothing may be assumed from how it looks. The markings are the only statements of fact on it — which is why reading them correctly is the first step in any geometry question, and why examiners mark diagrams rather than annotate them in words.

4. Why perpendicular is not among them

Perpendicularity is shown by the small square, never by arrowheads. Option D attaches the wrong property to the arrows.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. The dashed sides are parallel; the arrowed lines are equal in length — Conventions Swapped
  The two conventions swapped.
• B. Both markings indicate right angles — Wrong Convention
  A right angle is shown by a small square.
• D. The dashed sides are equal in length; the arrowed lines are perpendicular — Wrong Property
  Arrowheads indicate parallel, not perpendicular.
• E. The dashes indicate the shortest sides; the arrows indicate the direction of travel — Wrong Meaning
  Dashes indicate equality, not relative size.

Common Mistake (⚠️):
Swapping the two conventions. Dashes measure length and arrows show direction, which is why arrows indicate parallelism.

Takeaway (📌):
Dashes equal lengths, arrows parallel, arcs equal angles, small square a right angle. Never infer anything else from a diagram.

Correct Answer: A. $(1,-1)$

Fastest Approach (🚀):
$x: \dfrac{-3+5}{2} = 1$. $y: \dfrac{2-4}{2} = -1$.

Matches Option A.

Step-by-Step Breakdown:

1. The midpoint formula

$M = \left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$

It is simply the average of each coordinate, taken separately.

2. Substitute

$x$: $\dfrac{-3+5}{2} = \dfrac{2}{2} = 1$
$y$: $\dfrac{2+(-4)}{2} = \dfrac{-2}{2} = -1$

$M = (1,-1)$

3. Check it sits between the points

$1$ lies between $-3$ and $5$; $-1$ lies between $2$ and $-4$. ✓ A midpoint must always fall between its endpoints in both coordinates, which rules out Option E immediately.

4. The negatives

$A$ is in the second quadrant and $B$ in the fourth, so the segment crosses the axes. Adding a negative is the same as subtracting, so $2+(-4) = -2$; the halving then gives $-1$.

Subtracting instead of adding gives the difference, not the average — that is Option D's origin, and it is the vector from one point to the other rather than the midpoint.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $(2,-2)$ — Arithmetic Error
  Halving only one coordinate, or an arithmetic slip.
• C. $(-1,1)$ — Sign Error
  Signs of both coordinates reversed.
• D. $(4,-3)$ — Wrong Operation
  Subtracting rather than averaging — this is the displacement.
• E. $(8,-6)$ — Halving Omitted
  Adding without halving.

Common Mistake (⚠️):
Subtracting the coordinates instead of adding them. Subtraction gives the displacement between the points; the midpoint needs their average.

Takeaway (📌):
Midpoint is the average of each coordinate. Check the result lies between the endpoints in both directions.

Correct Answer: D. $12$

Fastest Approach (🚀):
$36 = 2^{2}3^{2}$, $48 = 2^{4}3$.
Lower power each: $2^{2}\times 3 = 12$.

Matches Option D.

Step-by-Step Breakdown:

1. Prime factorise

$36 = 2\times 2\times 3\times 3 = 2^{2}\times 3^{2}$
$48 = 2\times 2\times 2\times 2\times 3 = 2^{4}\times 3^{1}$

2. Take the lower power of each shared prime

$2$: lower of $2^{2}$ and $2^{4}$ is $2^{2} = 4$
$3$: lower of $3^{2}$ and $3^{1}$ is $3^{1} = 3$

$\text{HCF} = 4\times 3 = 12$

3. Check

$36\div 12 = 3$ and $48\div 12 = 4$, both whole. And $3$ and $4$ share no factor, which confirms $12$ is the highest — if they did, more could still be extracted.

4. The LCM from the same working

Take the higher power of each prime instead:
$\text{LCM} = 2^{4}\times 3^{2} = 144$

That is Option D, and it is the answer to the opposite question. One factorisation gives both.

5. The relationship worth knowing

$\text{HCF}\times\text{LCM} = $ the product of the two numbers.
$12\times 144 = 1728 = 36\times 48$ ✓ — a complete check on both answers at once.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $6$ — Not Highest
  A common factor, but not the highest.
• B. $144$ — LCM Not HCF
  The lowest common multiple.
• C. $4$ — Prime Omitted
  Only the shared power of $2$; the factor of $3$ is missing.
• E. $18$ — Not Common
  A factor of $36$ but not of $48$.

Common Mistake (⚠️):
Giving $6$ — a common factor, but not the highest. After finding one, always check whether the quotients still share a factor.

Takeaway (📌):
HCF takes the lower power of each shared prime; LCM takes the higher. Their product equals the product of the two numbers.

Correct Answer: B. $30$

Fastest Approach (🚀):
$\dfrac{60}{2} = 30$.

Matches Option B.

Step-by-Step Breakdown:

1. Where the median sits

The median splits the data in half, so on a cumulative frequency graph you read across from
$\dfrac{n}{2} = \dfrac{60}{2} = 30$

then down to the horizontal axis for the estimated median value.

2. Why not $\tfrac{n+1}{2}$

For a listed set of $60$ values, the median is the mean of the $30$th and $31$st — position $30.5$. On a cumulative frequency graph the data are treated as continuous, so $\dfrac{n}{2}$ is used and the distinction disappears.

Option B applies the discrete rule to a continuous graph, which is the standard confusion between the two contexts.

3. The quartiles, from the same graph

lower quartile at $\dfrac{n}{4} = 15$
upper quartile at $\dfrac{3n}{4} = 45$

Both appear here as distractors, and both are read the same way — across from the cumulative frequency axis, then down.

The interquartile range is the difference between the two values read off.

4. Why it is an estimate

The original values are lost once the data are grouped, so any statistic read from the graph is an estimate rather than an exact figure.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $30.5$ — Wrong Context
  The discrete-list rule applied to a continuous graph.
• C. $15$ — Wrong Statistic
  The lower quartile position.
• D. $45$ — Wrong Statistic
  The upper quartile position.
• E. $60$ — Wrong Value
  The total frequency, not the halfway point.

Common Mistake (⚠️):
Reading across at a frequency rather than a cumulative frequency, or halving the class width instead of the total.

Takeaway (📌):
Median at $\dfrac{n}{2}$, quartiles at $\dfrac{n}{4}$ and $\dfrac{3n}{4}$. Read across, then down.

Correct Answer: B. $1$

Fastest Approach (🚀):
Sum $= 40$, so the mean is $8$.
The list is already ordered, so the median is the third value, $7$.
$8-7 = 1$.

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

The list is already in order and has $5$ values, so the median sits at position $\dfrac{5+1}{2} = 3$:
median $= 7$

3. Subtract

$8-7 = 1$

The mean exceeds the median because $13$ pulls it upwards — the signature of a set skewed to the right.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $0$ — Conceptual Error
  Assuming mean and median always coincide.
• C. $2$ — Average Confusion
  Using the mode $7$ against a mis-computed mean.
• D. $-1$ — Order Error
  Subtracting in the wrong order.
• E. $4$ — Misread Question
  Giving the difference between the mean and the smallest value.

Common Mistake (⚠️):
Taking the median as the middle of the unordered list, or dividing the sum by $4$ instead of $5$.

Takeaway (📌):
Mean above median means the data is skewed high; mean below means skewed low. The gap itself tells you about the shape.

Correct Answer: D. A segment

Fastest Approach (🚀):
Chord $+$ arc $=$ segment.
Two radii $+$ arc $=$ sector.

Matches Option D.

Step-by-Step Breakdown:

1. Define each term by its boundaries

Radius — centre to circumference.
Diameter — a chord through the centre; twice the radius.
Chord — any straight line joining two points on the circumference.
Tangent — a straight line touching the circle at exactly one point, perpendicular to the radius there.
Arc — part of the circumference itself, a curve rather than a region.
Sector — the region bounded by two radii and an arc, the pizza-slice shape.
Segment — the region bounded by a chord and an arc.

2. Apply

A chord with its arc bounds a segment.

3. The distinction that is always tested

Sector and segment are the pair examiners rely on. A sector reaches the centre; a segment does not. Cutting a circle along a chord leaves a smaller segment and a larger one; cutting along two radii gives two sectors.

4. Why an arc is not the answer

An arc is a curve, not a region. It forms part of the boundary of both a sector and a segment, but encloses nothing by itself.

5. Where the terms are used

Segment area is sector minus triangle — the calculation that appears in the radian work in Mathematics 2. Getting the vocabulary right is what makes that formula make sense.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. A sector — Wrong Region
  Bounded by two radii and an arc, and it reaches the centre.
• B. An arc — Not A Region
  A curve, not a region.
• C. A quadrant — Special Case
  A sector of exactly a quarter circle.
• E. A tangent — Not A Region
  A line touching the circle once, not a region.

Common Mistake (⚠️):
Confusing sector with segment. The sector reaches the centre by two radii; the segment is cut off by a chord and never touches the centre.

Takeaway (📌):
Chord and arc bound a segment; two radii and an arc bound a sector. An arc is a curve, not a region.

Correct Answer: E. $x^{2}+2x-15$

Fastest Approach (🚀):
$x^{2}+(5-3)x+(5)(-3) = x^{2}+2x-15$.

Matches Option E.

Step-by-Step Breakdown:

1. Multiply every pair

$x\times x = x^{2}$
$x\times(-3) = -3x$
$5\times x = +5x$
$5\times(-3) = -15$

2. Collect the middle terms

$-3x+5x = +2x$

$(x+5)(x-3) = x^{2}+2x-15$

3. The shortcut for two simple brackets

For $(x+a)(x+b)$ the result is $x^{2}+(a+b)x+ab$. Here $a = 5$ and $b = -3$:
sum $= 2$, product $= -15$

That reads the answer off in one line, and it is the same relationship used in reverse when factorising.

4. Where the signs go wrong

$5+(-3) = +2$, not $-2$ — Option D reverses it.
$5\times(-3) = -15$, not $+15$ — Option E.

Both come from dropping a minus sign, which is why writing all four products before collecting is worth the extra line.

5. Why Option A is tempting

$(x+5)(x-5)$ would give $x^{2}-25$ with no middle term, because the two middle products cancel. That only happens when the numbers are equal and opposite, which they are not here.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $x^{2}-15$ — Middle Term Lost
  Assumes the middle terms cancel, which needs equal and opposite constants.
• B. $x^{2}+2x+15$ — Sign Error
  Sign of the constant term reversed.
• C. $x^{2}+8x-15$ — Sign Error
  Middle terms added without their signs.
• D. $x^{2}-2x-15$ — Sign Error
  Sign of the middle term reversed.

Common Mistake (⚠️):
Assuming the middle terms cancel. They cancel only in a difference of two squares, where the two constants are equal and opposite.

Takeaway (📌):
$(x+a)(x+b) = x^{2}+(a+b)x+ab$. Sum in the middle, product at the end, signs included.

Correct Answer: C. $32$

Fastest Approach (🚀):
$2^{8-3} = 2^{5} = 32$.

Matches Option C.

Step-by-Step Breakdown:

1. Apply the division law

$\dfrac{a^{m}}{a^{n}} = a^{m-n}$, valid because the bases match:
$\dfrac{2^{8}}{2^{3}} = 2^{5}$

2. Evaluate

$2^{5} = 32$

The question says 'evaluate', so a numerical answer is wanted rather than a power.

3. Why subtraction works

$2^{8}$ is eight twos multiplied; $2^{3}$ is three of them. Dividing cancels three, leaving five. The index law is bookkeeping for that cancellation, not a separate rule to memorise.

4. The neighbouring laws

$a^{m}\times a^{n} = a^{m+n}$ — indices add
$\dfrac{a^{m}}{a^{n}} = a^{m-n}$ — indices subtract
$\left(a^{m}\right)^{n} = a^{mn}$ — indices multiply

Adding when the operation is division gives $2^{11}$, and multiplying gives $2^{24}$. Both are offered, and both come from reaching for the wrong law.

5. The condition

All three require the same base. $\dfrac{2^{8}}{4^{3}}$ needs $4$ rewriting as $2^{2}$ first.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $2^{11}$ — Wrong Law
  Adding the indices, which is the law for multiplication.
• B. $2^{24}$ — Wrong Law
  Multiplying the indices, which is the law for a power of a power.
• D. $\tfrac{8}{3}$ — Wrong Operation
  Dividing the indices.
• E. $64$ — Arithmetic Error
  Computing $2^{6}$.

Common Mistake (⚠️):
Dividing the indices to get $\tfrac83$. Division of the powers means subtraction of the indices, never division of them.

Takeaway (📌):
Multiplying adds indices, dividing subtracts them, a power of a power multiplies them — and all three need the same base.

Correct Answer: A. $8.5$

Fastest Approach (🚀):
Position $= \dfrac{9}{2} = 4.5$.
4th and 5th values are $8$ and $9$, so the median is $8.5$.

Matches Option A.

Step-by-Step Breakdown:

1. Check the list is ordered

$3, 5, 6, 8, 9, 11, 14, 20$ — already ascending, with $n = 8$.

2. Locate the median by position

$\dfrac{n+1}{2} = \dfrac{9}{2} = 4.5$

Position $4.5$ means halfway between the 4th and 5th values.

3. Average the two middle values

$\dfrac{8+9}{2} = 8.5$

The median need not be a member of the data set, and here it is not.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $9$ — Position Error
  Taking the 5th value.
• C. $8$ — Position Error
  Taking the 4th value, from $\tfrac{n}{2} = 4$.
• D. $9.5$ — Position Error
  Averaging the 5th and 6th values.
• E. $10$ — Average Confusion
  Giving the mean of the whole set, or averaging the extremes.

Common Mistake (⚠️):
Taking the 4th value alone because $\tfrac{8}{2} = 4$. Dividing $n$ by 2 gives a count, not the median's position.

Takeaway (📌):
$\dfrac{n+1}{2}$ is a position, not a value. A fractional position means averaging the two values either side.

Correct Answer: D. $60\ \text{cm}^{2}$

Fastest Approach (🚀):
$\tfrac12(20) = 10$, then $10\times 6 = 60\ \text{cm}^{2}$.

Matches Option D.

Step-by-Step Breakdown:

1. Apply the formula

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

where $a$ and $b$ are the parallel sides and $h$ is the perpendicular distance between them.

$A = \dfrac12(7+13)\times 6 = \dfrac12(20)\times 6 = 10\times 6 = 60\ \text{cm}^{2}$

2. Why it works

$\dfrac{7+13}{2} = 10$ is the average width of the shape. A rectangle of that width and the same height has the same area — the trapezium's excess on one side exactly fills its deficit on the other.

Seeing it that way makes the formula reconstructable rather than memorised.

3. Perpendicular height, not a slanted side

$h$ must be the perpendicular distance between the parallel sides. A sloping side is longer and would overstate the area — the same requirement as in a triangle or a parallelogram.

4. The related formulae

triangle: $\dfrac12\times\text{base}\times\text{height}$
parallelogram: $\text{base}\times\text{height}$
trapezium: $\dfrac12(a+b)h$

Setting $a = b$ in the trapezium formula gives the parallelogram, and letting one parallel side shrink to zero gives the triangle. All three are one idea.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $26\ \text{cm}^{2}$ — Operation Error
  Adding the three lengths.
• B. $120\ \text{cm}^{2}$ — Factor Dropped
  Omitting the factor of a half.
• C. $546\ \text{cm}^{2}$ — Formula Misuse
  Computing $7\times 13\times 6$.
• E. $39\ \text{cm}^{2}$ — Side Omitted
  Computing $\tfrac12\times 13\times 6$, using only one parallel side.

Common Mistake (⚠️):
Omitting the factor of a half, giving $120\ \text{cm}^{2}$. That is the area of the surrounding parallelogram, not the trapezium.

Takeaway (📌):
$A = \tfrac12(a+b)h$ — average the parallel sides, multiply by the perpendicular height.

Correct Answer: C. $(x-3)(x-4)$

Fastest Approach (🚀):
Need product $+12$, sum $-7$: $-3$ and $-4$.
$(x-3)(x-4)$.

Matches Option C.

Step-by-Step Breakdown:

1. Read the signs first

The constant is positive, so the two numbers share a sign.
The $x$ coefficient is negative, so that shared sign is negative.

Both numbers are therefore negative, which eliminates Options B and E without trying any pairs.

2. Find the pair

Negative pairs multiplying to $12$: $(-1,-12)$, $(-2,-6)$, $(-3,-4)$.
Their sums: $-13$, $-8$, $-7$.

The one that sums to $-7$ is $-3$ and $-4$.

3. Write the factorisation

$x^{2}-7x+12 = (x-3)(x-4)$

4. Check by expanding

$x^{2}-4x-3x+12 = x^{2}-7x+12$ ✓ — the reverse of the previous question, and always worth the ten seconds.

5. What it gives you

Setting each bracket to zero solves $x^{2}-7x+12 = 0$: the roots are $x = 3$ and $x = 4$. Factorising is the fastest route to the roots when the numbers are this friendly.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $(x-2)(x-6)$ — Sum Wrong
  Sums to $-8$, not $-7$.
• B. $(x+3)(x+4)$ — Sign Error
  Expands to $x^{2}+7x+12$ — the middle sign is wrong.
• D. $(x-1)(x-12)$ — Sum Wrong
  Sums to $-13$.
• E. $(x+6)(x-2)$ — Product Wrong
  Product is $-12$, not $+12$.

Common Mistake (⚠️):
Choosing $(x+3)(x+4)$, which expands to $x^{2}+7x+12$. A positive constant with a negative middle term needs two negatives.

Takeaway (📌):
Product gives the constant, sum gives the $x$ coefficient. Read both signs first to halve the search.

Correct Answer: A. $\pounds 35$

Fastest Approach (🚀):
$12$ parts, so one part $= \pounds 5$.
$7\times 5 = \pounds 35$.

Matches Option A.

Step-by-Step Breakdown:

1. Total the parts

$2+3+7 = 12$

2. Find the value of one part

$\dfrac{60}{12} = \pounds 5$

3. Scale to the share wanted

The largest share is $7$ parts:
$7\times 5 = \pounds 35$

4. Check by totalling

$2\times 5 = \pounds 10$, $3\times 5 = \pounds 15$, $7\times 5 = \pounds 35$.
$10+15+35 = \pounds 60$ ✓

Adding the three shares back to the original total is a complete check and takes seconds.

5. The error the numbers are chosen to catch

Dividing $60$ by $7$ rather than by the total of $12$ gives about $\pounds 8.57$, and dividing by $3$ because there are three people gives $\pounds 20$ — Option E. The divisor is always the number of parts, never the number of shares.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $\pounds 30$ — Ratio Error
  Halving the total, or taking $6$ parts.
• C. $\pounds 42$ — Wrong Part Total
  Computing $\dfrac{60}{10}\times 7$.
• D. $\pounds 7$ — Wrong Quantity
  Quoting the number of parts as an amount of money.
• E. $\pounds 20$ — Wrong Divisor
  Dividing by the number of people rather than the parts.

Common Mistake (⚠️):
Dividing by the number of people rather than the number of parts. Three people share $12$ parts here, and it is the parts that divide the money.

Takeaway (📌):
Add the parts, divide the total by that, multiply by the share wanted, then check the shares sum back to the original.

Correct Answer: B. $1.45$

Fastest Approach (🚀):
$\sum fx = (0)(4)+(1)(6)+(2)(7)+(3)(3) = 0+6+14+9 = 29$.
$\bar x = \dfrac{29}{20} = 1.45$.

Matches Option B.

Step-by-Step Breakdown:

1. Multiply each value by its frequency

$0\times 4 = 0$
$1\times 6 = 6$
$2\times 7 = 14$
$3\times 3 = 9$

2. Total both columns

$\sum fx = 0+6+14+9 = 29$, and $\sum f = 4+6+7+3 = 20$

3. Divide

$\bar x = \dfrac{29}{20} = 1.45$

4. Sanity check

The values run from $0$ to $3$ and the bulk sit at $1$ and $2$, so a mean just under $1.5$ is exactly where it should be.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $1.5$ — Rounding Error
  Rounding, or averaging the four values $0,1,2,3$ and adjusting.
• C. $2$ — Average Confusion
  Giving the modal value rather than the mean.
• D. $7.25$ — Divisor Error
  Dividing $29$ by the $4$ distinct values instead of the $20$ matches.
• E. $1.2$ — Arithmetic Error
  Mis-computing $\sum fx$ as $24$.

Common Mistake (⚠️):
Dividing by the number of distinct values ($4$) rather than the total frequency ($20$), giving $7.25$.

Takeaway (📌):
Divide by $\sum f$, never by the number of rows. The table lists $4$ values but describes $20$ matches.

Correct Answer: D. $90\pi\ \text{cm}^{3}$

Fastest Approach (🚀):
$\pi(9)(10) = 90\pi\ \text{cm}^{3}$.

Matches Option D.

Step-by-Step Breakdown:

1. A cylinder is a prism

Its volume is the cross-sectional area times the length. The cross-section is a circle of area $\pi r^{2}$:
$V = \pi r^{2}h$

2. Substitute

$V = \pi(3)^{2}(10) = \pi(9)(10) = 90\pi\ \text{cm}^{3}$

3. Square, do not double

$3^{2} = 9$, not $6$. Doubling gives $60\pi$ — Option B — and it is the commonest error in every circle formula.

4. Leaving it in terms of π

The question asks for an answer in terms of $\pi$, so $90\pi$ is the finished answer. Multiplying out to about $282.7\ \text{cm}^{3}$ would be less exact and is not what was requested.

An answer in terms of $\pi$ is exact; a decimal is a rounding of it.

5. The related formulae

curved surface area: $2\pi rh$
total surface area: $2\pi rh+2\pi r^{2}$

Both come from unrolling the curved surface into a rectangle of width $2\pi r$ and adding the two circular ends.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $900\pi\ \text{cm}^{3}$ — Extra Square
  Squaring the height as well.
• B. $60\pi\ \text{cm}^{3}$ — Radius Doubled
  Doubling the radius instead of squaring.
• C. $30\pi\ \text{cm}^{3}$ — Not Squared
  Using $r$ rather than $r^{2}$.
• E. $180\pi\ \text{cm}^{3}$ — Wrong Formula
  Computing $2\pi r h$, the curved surface area.

Common Mistake (⚠️):
Doubling the radius instead of squaring it. The circle's area formula has $r^{2}$, and that is what a cylinder's volume inherits.

Takeaway (📌):
$V = \pi r^{2}h$ — circle area times height. Square the radius, and leave $\pi$ in place when asked for an exact answer.

Correct Answer: A. $2$

Fastest Approach (🚀):
Add: $4x = 16 \implies x = 4$.
$4-2y = 0 \implies y = 2$.

Matches Option A.

Step-by-Step Breakdown:

1. Spot the elimination

$3x+2y = 16$
$x-2y = 0$

The $y$ coefficients are $+2$ and $-2$ — equal and opposite. Adding the equations removes $y$ with no multiplication required.

2. Add

$(3x+x)+(2y-2y) = 16+0$
$4x = 16$
$x = 4$

3. Substitute back

Use the simpler equation:
$4-2y = 0 \implies 2y = 4 \implies y = 2$

4. Check in the equation you did not use

$3(4)+2(2) = 12+4 = 16$ ✓

Substituting into the other equation is the real check — the one you solved from will agree automatically.

5. Add or subtract

Coefficients equal and opposite: add.
Coefficients equal and the same: subtract.
Neither: multiply one or both equations first to make them match.

Getting that choice wrong here would give $2x+4y = 16$, which eliminates nothing.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $4$ — Wrong Variable
  The value of $x$, not $y$.
• C. $8$ — Arithmetic Error
  Doubling $x$, or a slip in the substitution.
• D. $0$ — Misread
  Reading the right-hand side of the second equation as $y$.
• E. $6$ — Arithmetic Error
  Solving $x-2y = 0$ incorrectly.

Common Mistake (⚠️):
Answering $4$ — the value of $x$. Read which variable the question asks for before choosing.

Takeaway (📌):
Add for equal and opposite coefficients, subtract for equal ones. Check in the equation you did not substitute into.

Correct Answer: C. $15\%$

Fastest Approach (🚀):
Increase $= 6$. $\dfrac{6}{40} = 0.15 = 15\%$.

Matches Option C.

Step-by-Step Breakdown:

1. Find the change

$46-40 = \pounds 6$

2. Divide by the original

$\text{percentage change} = \dfrac{\text{change}}{\text{original}}\times 100 = \dfrac{6}{40}\times 100 = 15\%$

3. Why the original, not the new value

A percentage change measures the change relative to where it started. Dividing by $46$ gives about $13\%$ — Option B — which answers the different question of what fraction of the new price the rise represents.

The rule holds in both directions: for a decrease, still divide by the original.

4. The multiplier check

A $15\%$ rise means multiplying by $1.15$:
$40\times 1.15 = 46$ ✓

That confirms the answer and gives the multiplier for any further change.

5. Why $115\%$ is offered

That is the multiplier expressed as a percentage — what the new price is as a percentage of the old — not the increase. Reading which of the two the question wants is the last step.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $6\%$ — Wrong Quantity
  Quoting the change in pounds as a percentage.
• B. $13\%$ — Wrong Base
  Dividing by the new price rather than the original.
• D. $115\%$ — Multiplier Not Change
  The multiplier as a percentage, not the increase.
• E. $87\%$ — Inverted
  Computing $\dfrac{40}{46}$.

Common Mistake (⚠️):
Dividing by the new price instead of the original. Percentage change is always measured against the starting value.

Takeaway (📌):
$\dfrac{\text{change}}{\text{original}}\times 100$. Check with the multiplier: a $15\%$ rise is $\times 1.15$.

Correct Answer: E. $18$ minutes

Fastest Approach (🚀):
Midpoints $5, 15, 25$.
$\sum fx = 20+90+250 = 360$; $\sum f = 20$.
$\bar x \approx \dfrac{360}{20} = 18$ minutes.

Matches Option E.

Step-by-Step Breakdown:

1. Find each class midpoint

$\dfrac{0+10}{2} = 5$, $\quad\dfrac{10+20}{2} = 15$, $\quad\dfrac{20+30}{2} = 25$

2. Weight by frequency

$5\times 4 = 20$, $\quad 15\times 6 = 90$, $\quad 25\times 10 = 250$

3. Divide by the total frequency

$\bar x \approx \dfrac{20+90+250}{4+6+10} = \dfrac{360}{20} = 18$

4. Why 'estimated'

The individual times are unknown, so replacing each by its class midpoint is an approximation. Only if the data were evenly spread within each class would $18$ be exact.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $15$ minutes — Weighting Error
  Averaging the midpoints without weighting them.
• B. $20$ minutes — Midpoint Error
  Using class upper bounds instead of midpoints.
• C. $12$ minutes — Midpoint Error
  Using lower class bounds $0, 10, 20$ as the values.
• D. $16.5$ minutes — Weighting Error
  Using midpoints $5, 15, 25$ with equal weights and adjusting.

Common Mistake (⚠️):
Averaging the three midpoints to get $15$, which ignores that the top class holds half the people.

Takeaway (📌):
Grouped mean $= \dfrac{\sum fx}{\sum f}$ with $x$ the midpoint — and it is always an estimate, never exact.

Correct Answer: B. $3\pi\ \text{cm}$

Fastest Approach (🚀):
$\dfrac{60}{360} = \dfrac16$; circumference $= 18\pi$.
$\dfrac{18\pi}{6} = 3\pi\ \text{cm}$.

Matches Option B.

Step-by-Step Breakdown:

1. Find the fraction of the circle

$\dfrac{60}{360} = \dfrac{1}{6}$

2. Find the whole circumference

$C = 2\pi r = 2\pi(9) = 18\pi\ \text{cm}$

3. Take the fraction

$\dfrac16\times 18\pi = 3\pi\ \text{cm}$

4. The companion formula

Sector area works the same way, applied to the circle's area instead:
$\dfrac16\times\pi(9)^{2} = \dfrac{81\pi}{6} = 13.5\pi\ \text{cm}^{2}$

That is Option D — the right answer to the area question, and wrong here because an arc length is a length, not an area. Check the units the answer should carry before choosing.

5. Why the fraction comes first

Simplifying $\dfrac{60}{360}$ to $\dfrac16$ before multiplying keeps every number small. Substituting into $\dfrac{\theta}{360}\times 2\pi r$ directly means handling $\dfrac{60\times 18\pi}{360}$, which is the same answer by a slower and more error-prone route.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $18\pi\ \text{cm}$ — Angle Ignored
  The whole circumference, with the angle ignored.
• C. $1.5\pi\ \text{cm}$ — Formula Misuse
  Using $\pi r$ rather than $2\pi r$.
• D. $13.5\pi\ \text{cm}$ — Wrong Quantity
  The sector area, not the arc length.
• E. $6\pi\ \text{cm}$ — Fraction Error
  Using a fraction of $\dfrac13$.

Common Mistake (⚠️):
Using the area formula and giving $13.5\pi$. An arc is a length and uses the circumference; a sector is an area and uses $\pi r^{2}$.

Takeaway (📌):
Arc length is $\dfrac{\theta}{360}\times 2\pi r$; sector area is $\dfrac{\theta}{360}\times\pi r^{2}$. Simplify the fraction first.

Correct Answer: A. $x = 2$ and $x = 3$

Fastest Approach (🚀):
Product $6$, sum $-5$: $-2$ and $-3$.
$(x-2)(x-3) = 0 \implies x = 2, 3$.

Matches Option A.

Step-by-Step Breakdown:

1. Factorise

Two numbers multiplying to $+6$ and adding to $-5$: both negative, so $-2$ and $-3$.
$(x-2)(x-3) = 0$

2. Use the zero product rule

If two things multiply to zero, at least one of them is zero:
$x-2 = 0 \implies x = 2$
$x-3 = 0 \implies x = 3$

3. Why the signs flip

The bracket $(x-2)$ vanishes when $x$ is $+2$, not $-2$. Reading the roots straight out of the brackets without changing the sign gives $x = -2$ and $x = -3$ — Option B, and the commonest error in the topic.

4. Check both

$2^{2}-5(2)+6 = 4-10+6 = 0$ ✓
$3^{2}-5(3)+6 = 9-15+6 = 0$ ✓

Substituting both roots back is quick and catches any sign slip.

5. Why the equation must equal zero first

The zero product rule needs a zero on the right. For $x^{2}-5x = -6$, rearranging to $x^{2}-5x+6 = 0$ comes first; factorising the left-hand side as it stands would prove nothing.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $x = -2$ and $x = -3$ — Sign Reversed
  Signs taken straight from the brackets without reversing.
• C. $x = 1$ and $x = 6$ — Sum Wrong
  A pair with product $6$ but sum $7$.
• D. $x = 5$ and $x = 6$ — Coefficients As Roots
  Reading $b$ and $c$ as the roots.
• E. $x = -1$ and $x = -6$ — Sign Reversed
  Both errors combined.

Common Mistake (⚠️):
Reading the roots directly from the brackets without reversing the signs. $(x-2)$ is zero at $x = +2$.

Takeaway (📌):
Rearrange to zero, factorise, set each bracket to zero, and reverse the sign inside each. Check both roots by substitution.

Correct Answer: C. $20 \le x \lt 30$

Fastest Approach (🚀):
Median position $= 20.5$.
Cumulative frequency reaches $15$ at $20$ and $28$ at $30$.
So the median falls in $20 \le x \lt 30$.

Matches Option C.

Step-by-Step Breakdown:

1. Locate the median by position

$\dfrac{n+1}{2} = \dfrac{41}{2} = 20.5$

The $20$th and $21$st observations straddle the median.

2. Read across the running totals

Up to $10$: $6$ observations — not yet.
Up to $20$: $15$ observations — still short of $20.5$.
Up to $30$: $28$ observations — this passes $20.5$.

3. Identify the class

The $16$th to $28$th observations all lie in $20 \le x \lt 30$, so the median is in that class.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $0 \le x \lt 10$ — Position Ignored
  Reading the first class listed.
• B. $10 \le x \lt 20$ — Off-by-one Error
  Stopping at the running total of $15$, which is just short of $20.5$.
• D. $30 \le x \lt 40$ — Position Ignored
  Using the final class because it completes the total.
• E. It cannot be determined — Conceptual Error
  Assuming grouped data cannot locate a median class.

Common Mistake (⚠️):
Reading the class whose own frequency is largest, or treating the cumulative figures as ordinary frequencies.

Takeaway (📌):
Cumulative frequency answers 'how many so far'. Find the position first, then read across for the first total that reaches it.

Correct Answer: B. $\sin\theta = \tfrac{7}{25}$, third side $24\ \text{cm}$

Fastest Approach (🚀):
Opposite and hypotenuse $\Rightarrow$ sine.
$7$-$24$-$25$ is a Pythagorean triple.

Matches Option B.

Step-by-Step Breakdown:

1. Label relative to the angle

Hypotenuse — always opposite the right angle, and always the longest side.
Opposite — across from $\theta$.
Adjacent — the remaining side, next to $\theta$.

Here the opposite is $7$ and the hypotenuse is $25$.

2. Choose the ratio

$\text{SOH}$ — $\sin = \dfrac{\text{opposite}}{\text{hypotenuse}}$
$\text{CAH}$ — $\cos = \dfrac{\text{adjacent}}{\text{hypotenuse}}$
$\text{TOA}$ — $\tan = \dfrac{\text{opposite}}{\text{adjacent}}$

Opposite with hypotenuse is sine:
$\sin\theta = \dfrac{7}{25}$

3. Find the third side

$a^{2}+7^{2} = 25^{2}$
$a^{2} = 625-49 = 576$
$a = 24\ \text{cm}$

$7$–$24$–$25$ is a Pythagorean triple, alongside $3$–$4$–$5$ and $5$–$12$–$13$. Recognising it removes the arithmetic entirely.

4. Why sine can never exceed one

The opposite side is always shorter than the hypotenuse, so $\sin\theta < 1$ for any angle in a right-angled triangle. $\dfrac{25}{7}$ is greater than one and therefore impossible — Option E can be rejected on that alone.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\cos\theta = \tfrac{7}{25}$, third side $24\ \text{cm}$ — Wrong Ratio
  Cosine uses the adjacent, not the opposite.
• C. $\tan\theta = \tfrac{7}{25}$, third side $24\ \text{cm}$ — Wrong Ratio
  Tangent uses the adjacent, not the hypotenuse.
• D. $\sin\theta = \tfrac{7}{25}$, third side $18\ \text{cm}$ — Arithmetic Error
  Pythagoras applied incorrectly; $\sqrt{576} = 24$.
• E. $\sin\theta = \tfrac{25}{7}$, third side $24\ \text{cm}$ — Inverted
  Ratio inverted — sine cannot exceed one.

Common Mistake (⚠️):
Inverting the ratio. The hypotenuse is the longest side, so it belongs on the bottom and sine can never come out above one.

Takeaway (📌):
SOH CAH TOA, chosen by which two sides you have. Sine and cosine never exceed one; recognise the common triples.

Correct Answer: D. $13$

Fastest Approach (🚀):
$n = 11$, median is the 6th value $10$.
$Q_1$ is the middle of the lower five: $5$. $Q_3$ is the middle of the upper five: $18$.
IQR $= 18-5 = 13$.

Matches Option D.

Step-by-Step Breakdown:

1. Find the median

$n = 11$, so the median is at position $\dfrac{12}{2} = 6$: the value $10$.

2. Split into halves

Lower half: $2, 4, 5, 7, 8$. Upper half: $12, 15, 18, 21, 30$.

The median itself is excluded from both halves.

3. Take the middle of each half

$Q_1 = 5$ (3rd of five), $\quad Q_3 = 18$ (3rd of five)

4. Subtract

$\text{IQR} = Q_3-Q_1 = 18-5 = 13$

The full range is $30-2 = 28$, more than twice the IQR — the value $30$ is stretching it, which is exactly why the IQR is the more robust measure.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $28$ — Measure Confusion
  Giving the full range rather than the interquartile range.
• B. $10$ — Misread Question
  Giving the median rather than the IQR.
• C. $11$ — Position Error
  Including the median in both halves, shifting the quartiles.
• E. $16$ — Counting Error
  Using $Q_3 = 21$ from a miscount of the upper half.

Common Mistake (⚠️):
Giving the range $28$ instead of the interquartile range, or including the median in both halves when locating the quartiles.

Takeaway (📌):
The IQR is the spread of the middle half. It ignores the extremes, which is what makes it resistant to outliers.

Correct Answer: B. mean $64$, range $36$

Fastest Approach (🚀):
Mean: $3(20)+4 = 64$.
Range: $3\times 12 = 36$ (the $+4$ cancels in any difference).

Matches Option B.

Step-by-Step Breakdown:

1. Transform the mean

The mean is an average of values, so it undergoes the same transformation:
$\bar y = 3\bar x+4 = 3(20)+4 = 64$

2. Transform the range

The range is a difference between two values:
$y_{\max}-y_{\min} = (3x_{\max}+4)-(3x_{\min}+4) = 3\left(x_{\max}-x_{\min}\right)$

The $+4$ appears in both terms and cancels.
$\text{range} = 3\times 12 = 36$

3. The general rule

For $y = ax+b$: every measure of location (mean, median, quartiles) becomes $a(\text{value})+b$, while every measure of spread (range, IQR, standard deviation) becomes $|a|\times(\text{spread})$.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. mean $64$, range $40$ — Spread Shift Error
  Adding $4$ to the range as well as multiplying by $3$.
• C. mean $60$, range $36$ — Omitted Term
  Forgetting the $+4$ in the mean.
• D. mean $24$, range $16$ — Omitted Factor
  Adding $4$ to the mean without multiplying by $3$.
• E. mean $64$, range $12$ — Scaling Ignored
  Leaving the range unchanged, as though the multiplication did not affect it either.

Common Mistake (⚠️):
Adding the $4$ to the range as well, giving $40$. Shifting every value moves the data along the scale without stretching it.

Takeaway (📌):
Location follows $ax+b$; spread follows $|a|$ alone. Sliding a data set never changes how spread out it is.

Correct Answer: E. $31.5$

Fastest Approach (🚀):
$\text{IQR} = 15-4 = 11$.
$1.5\times 11 = 16.5$.
Upper limit $= 15+16.5 = 31.5$.

Matches Option E.

Step-by-Step Breakdown:

1. Compute the IQR

$\text{IQR} = Q_3-Q_1 = 15-4 = 11$

2. Scale it

$1.5\times 11 = 16.5$

3. Add to the upper quartile

$Q_3+1.5\times\text{IQR} = 15+16.5 = 31.5$

Anything above $31.5$ is an outlier, so $31.5$ itself is the largest value that is not.

4. The other boundary

$Q_1-1.5\times\text{IQR} = 4-16.5 = -12.5$. If the data cannot be negative, no low outliers are possible at all.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $15$ — Boundary Error
  Giving $Q_3$ itself.
• B. $22.5$ — Formula Misuse
  Using $1.5\times Q_3$ rather than $1.5\times\text{IQR}$.
• C. $30$ — Arithmetic Error
  Rounding the boundary, or using an IQR of $10$.
• D. $26$ — Factor Error
  Adding the IQR once rather than $1.5$ times.

Common Mistake (⚠️):
Adding $1.5\times\text{IQR}$ to the median or to $Q_1$, or using $1.5\times Q_3$ instead of $1.5\times\text{IQR}$.

Takeaway (📌):
Outlier fences are $Q_1-1.5\,\text{IQR}$ and $Q_3+1.5\,\text{IQR}$. Compute the IQR first; everything else is one addition.

Correct Answer: B. $15$

Fastest Approach (🚀):
$80-65 = 15$.

Matches Option B.

Step-by-Step Breakdown:

1. Read what cumulative frequency means

'Up to $70$ kg: $65$' means $65$ of the $80$ people weigh $70$ kg or less.

2. Subtract from the total

$80-65 = 15$ weigh more than $70$ kg.

3. What the other figure is for

The $18$ up to $50$ kg is not needed here — it would answer 'how many between $50$ and $70$ kg', namely $65-18 = 47$.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $47$ — Misread Question
  Giving the count between $50$ and $70$ kg.
• C. $65$ — Misread Question
  Quoting the cumulative frequency unchanged.
• D. $62$ — Wrong Subtraction
  Subtracting $18$ from $80$.
• E. $18$ — Misread Question
  Quoting the first cumulative frequency.

Common Mistake (⚠️):
Answering $47$, the count between the two stated weights, or quoting the cumulative figure $65$ unchanged.

Takeaway (📌):
Cumulative frequency answers 'at most'. For 'more than', subtract from the total; for 'between', subtract one running total from another.

Correct Answer: A. $8$

Fastest Approach (🚀):
$5\times 12 = 60$; $4\times 13 = 52$.
$60-52 = 8$.

Matches Option A.

Step-by-Step Breakdown:

1. Convert each mean to a total

Before: $5\times 12 = 60$
After: $4\times 13 = 52$

2. The removed number is the difference

$60-52 = 8$

3. Sanity check the direction

Removing a value raised the mean from $12$ to $13$, so the removed value must have been below the original mean. $8 < 12$ — consistent.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $1$ — Conceptual Error
  Taking the change in the mean as the removed value.
• C. $12$ — Conceptual Error
  Assuming the removed value equals the original mean.
• D. $13$ — Conceptual Error
  Assuming the removed value equals the new mean.
• E. $25$ — Arithmetic Error
  Adding the two means, or using $60-35$.

Common Mistake (⚠️):
Answering $1$ because the mean rose by $1$. The mean of four numbers rising by one requires a shift of four units in the total, not one.

Takeaway (📌):
Mean $\times$ count $=$ total. Every 'mean changes when a value is added or removed' question is a subtraction once both totals are written.