ESAT Mathematics Mock Module 2 Worked Solutions

Correct Answer: B. A decrease of $4\%$

Fastest Approach (🚀):
Mass $\times 1.5$; speed squared $\times 0.8^{2} = 0.64$.
$1.5\times 0.64 = 0.96 \implies -4\%$.

Matches Option B.

Step-by-Step Breakdown:

1. Write the multipliers

Mass: $1.5$. Speed: $0.8$.

2. Respect the powers

$E = \tfrac12 mv^{2}$, so the speed multiplier is squared and the mass multiplier is not:
$\dfrac{E_{\text{new}}}{E_{\text{old}}} = 1.5\times(0.8)^{2}$

The constant $\tfrac12$ cancels — constants never affect a percentage change.

3. Evaluate and interpret

$(0.8)^{2} = 0.64$ and $1.5\times 0.64 = 0.96$, a $4\%$ decrease.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. An increase of $20\%$ — Power Ignored
  Forgetting to square the speed multiplier.
• C. A decrease of $10\%$ — Additive Error
  Adding the percentage changes.
• D. An increase of $4\%$ — Sign Error
  Correct size, wrong direction.
• E. A decrease of $30\%$ — Power Misapplied
  Squaring the mass multiplier instead of the speed multiplier.

Common Mistake (⚠️):
Applying the speed change once instead of squaring it, giving $1.2$ and a $20\%$ increase.

Takeaway (📌):
In $y = kx^{a}z^{b}$, scaling $x$ by $f$ scales $y$ by $f^{a}$. Percentage change on any formula is one product of powers.

Correct Answer: E. $252$

Fastest Approach (🚀):
Total $= 900$.
No sevens $= 8\times 9\times 9 = 648$.
$900-648 = 252$.

Matches Option E.

Step-by-Step Breakdown:

1. Count the whole set

Three-digit numbers run from $100$ to $999$:
$999-100+1 = 900$

2. Count the unwanted case

'No digit is a $7$'. The leading digit cannot be $0$ or $7$, leaving $8$ choices; the other two digits have $9$ each:
$8\times 9\times 9 = 648$

3. Subtract

$900-648 = 252$

Counting directly would mean 'exactly one seven', 'exactly two' and 'exactly three' plus their positional overlaps. The complement has no overlaps at all.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $271$ — Incomplete Case Analysis
  Counting 'exactly one seven' only and missing the multi-seven cases.
• B. $243$ — Boundary Error
  Using $9\times 9\times 9$, allowing a leading zero in the complement.
• C. $300$ — Estimation Error
  Assuming one third of the range contains a seven.
• D. $648$ — Complement Inversion
  Giving the complement itself: the count with no sevens.

Common Mistake (⚠️):
Using $9$ choices for the leading digit too, forgetting that a three-digit number cannot start with $0$.

Takeaway (📌):
'At least one' over independent slots is always total minus none — the same move as at-least-one-red and at-least-one-head.

Correct Answer: E. $200\pi\ \text{m}^{2}$

Fastest Approach (🚀):
$R = 51$, $r = 49$.
$R-r = 2$, $R+r = 100$.
Area $= 200\pi\ \text{m}^{2}$.

Matches Option E.

Step-by-Step Breakdown:

1. Identify the two radii

$r = 49$ and $R = 49+2 = 51$

2. Write the annulus area

$A = \pi R^{2}-\pi r^{2} = \pi\left(R^{2}-r^{2}\right)$

3. Factorise rather than square

$R^{2}-r^{2} = (R-r)(R+r) = 2\times 100 = 200$

The path width is the small factor and the sum of the radii is round — no four-figure squares needed.

$A = 200\pi\ \text{m}^{2}$

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $400\pi\ \text{m}^{2}$ — Double Counting
  Doubling the correct answer.
• B. $4\pi\ \text{m}^{2}$ — Formula Misuse
  Using $\pi(R-r)^{2}$, treating the width as a radius.
• C. $196\pi\ \text{m}^{2}$ — Setup Error
  Using $\pi r^{2}$ scaled, or $R+r$ alone.
• D. $100\pi\ \text{m}^{2}$ — Halving Error
  Halving the correct product.

Common Mistake (⚠️):
Using $\pi(R-r)^{2} = 4\pi$, treating the path width as though it were a radius.

Takeaway (📌):
Every ring, border or frame area is $(\text{outer}-\text{inner})(\text{outer}+\text{inner})$ — the same identity that shortcuts $403^{2}-397^{2}$.

Correct Answer: B. $16$

Fastest Approach (🚀):
$S_\infty = \dfrac{8}{\tfrac12} = 16$.

Matches Option B.

Step-by-Step Breakdown:

1. Identify a and r

$a = 8$, and each term is half the previous one, so $r = \tfrac12$.

2. Check convergence

$|r| < 1$, so the sum to infinity exists.

3. Apply the formula

$S_{\infty} = \dfrac{8}{1-\tfrac12} = \dfrac{8}{\tfrac12} = 16$

The partial sums run $8, 12, 14, 15, 15.5,\ldots$ — climbing towards $16$ without reaching it.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $15$ — Incomplete Answer
  Giving a partial sum rather than the limit.
• C. $32$ — Formula Misuse
  Dividing by $r$ rather than by $1-r$.
• D. $14$ — Incomplete Answer
  Stopping after four terms.
• E. The series diverges — Conceptual Error
  Assuming an infinite sum of positive terms cannot converge.

Common Mistake (⚠️):
Stopping at a partial sum such as $15$, or dividing by $r$ rather than by $1-r$.

Takeaway (📌):
When $r = \tfrac12$ the sum to infinity is exactly twice the first term.

Correct Answer: A. $1$

Fastest Approach (🚀):
$2\left(x^{2}-4x\right)+9 = 2\left[(x-2)^{2}-4\right]+9 = 2(x-2)^{2}+1$.
Minimum $= 1$.

Matches Option A.

Step-by-Step Breakdown:

1. Factor out the leading coefficient

$2x^{2}-8x+9 = 2\left(x^{2}-4x\right)+9$

Only the $x$ terms go inside the bracket — the constant stays outside.

2. Complete the square inside

$x^{2}-4x = (x-2)^{2}-4$

3. Multiply back and tidy

$2\left[(x-2)^{2}-4\right]+9 = 2(x-2)^{2}-8+9 = 2(x-2)^{2}+1$

4. Read off the minimum

$(x-2)^{2}\ge 0$ for all real $x$, with equality at $x=2$. So the least value of the expression is
$2(0)+1 = 1$

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $9$ — Misread Question
  Giving the constant term $9$, the value at $x=0$.
• C. $-1$ — Sign Error
  Sign slip in the final constant: $-8+9$ evaluated as $-1$.
• D. $2$ — Misread Question
  Giving the $x$-coordinate of the minimum rather than the minimum value.
• E. $-7$ — Omitted Term
  Forgetting to add the $9$ back, leaving $-8+1$.

Common Mistake (⚠️):
Forgetting to multiply the $-4$ by the factor of $2$ when expanding back, which gives a minimum of $5$ instead of $1$.

Takeaway (📌):
$a(x-p)^{2}+q$ has minimum $q$ at $x=p$ when $a>0$. The completed square answers minimum, vertex and line of symmetry at once.

Correct Answer: D. $12$ minutes

Fastest Approach (🚀):
$\dfrac{3}{60}+\dfrac{2}{60} = \dfrac{1}{12} \implies 12$ minutes.

Matches Option D.

Step-by-Step Breakdown:

1. Convert to rates

Tap A fills $\dfrac{1}{20}$ of the bath per minute, tap B fills $\dfrac{1}{30}$.

2. Add over a common denominator

$\dfrac{3}{60}+\dfrac{2}{60} = \dfrac{5}{60} = \dfrac{1}{12}$

3. Invert

$T = 12$ minutes.

Two taps must beat the faster tap alone, so any answer above $20$ minutes is impossible on inspection.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $25$ minutes — Averaging Error
  Averaging the two times.
• B. $50$ minutes — Additive Error
  Adding the two times.
• C. $10$ minutes — Estimation Error
  Halving the faster time.
• E. $15$ minutes — Arithmetic Error
  Denominator slip when adding the fractions.

Common Mistake (⚠️):
Averaging the times to get $25$ minutes — slower than one tap alone, and so impossible.

Takeaway (📌):
$\dfrac{1}{T} = \dfrac{1}{t_1}+\dfrac{1}{t_2}$, and the answer is always below the smallest individual time.

Correct Answer: A. $5$

Fastest Approach (🚀):
$25+20-10 = 35$ play at least one.
$40-35 = 5$ play neither.

Matches Option A.

Step-by-Step Breakdown:

1. State the principle

$|T\cup S| = |T|+|S|-|T\cap S|$

The overlap sits inside both counts, so it must be removed once.

2. Substitute

$|T\cup S| = 25+20-10 = 35$

3. Take the complement within the club

$40-35 = 5$

Check the four regions: tennis only $15$, squash only $10$, both $10$, neither $5$. They total $40$.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $0$ — Assumption Error
  Assuming every member plays at least one sport.
• C. $15$ — Double Subtraction
  Subtracting the overlap twice.
• D. $10$ — Misread Question
  Giving the size of the overlap rather than the outside region.
• E. $45$ — Double Counting
  Adding the two groups without removing the overlap.

Common Mistake (⚠️):
Adding $25+20 = 45$ and concluding the numbers are impossible, or subtracting the overlap twice to get $25$.

Takeaway (📌):
Two sets: add, subtract the overlap once, then check the four regions sum to the whole group.

Correct Answer: A. $\left(0,\tfrac12\right)$

Fastest Approach (🚀):
$m = -\dfrac34$.
$c = y-mx = -1-\left(-\tfrac34\right)(2) = -1+\tfrac32 = \tfrac12$.

Matches Option A.

Step-by-Step Breakdown:

1. Find the gradient of the given line

$3x+4y = 12 \implies 4y = -3x+12 \implies y = -\dfrac34 x+3$

So $m = -\dfrac34$. (The $+3$ belongs to the original line and is not carried over.)

2. Use the same gradient

A parallel line has the same gradient, so it has the form
$y = -\dfrac34 x + c$

3. Substitute the point

$(2,-1)$ lies on it:
$-1 = -\dfrac34(2)+c = -\dfrac32+c \implies c = -1+\dfrac32 = \dfrac12$

4. State the intercept

The line crosses the $y$-axis at $\left(0,\tfrac12\right)$.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $\left(0,-\tfrac12\right)$ — Sign Error
  Sign error when rearranging $c = y-mx$.
• C. $(0,3)$ — Conceptual Error
  Copying the intercept from the original line.
• D. $\left(0,\tfrac52\right)$ — Gradient Error
  Using the perpendicular gradient $\tfrac43$ instead of $-\tfrac34$.
• E. $\left(0,-\tfrac52\right)$ — Sign Error
  Using $m = \tfrac34$ (sign dropped) when substituting.

Common Mistake (⚠️):
Copying the intercept $3$ from the original line, or using the negative reciprocal $\tfrac43$ — that is the perpendicular gradient, not the parallel one.

Takeaway (📌):
Parallel: same $m$, new $c$. Perpendicular: negative reciprocal $m$. Mixing the two is the single most common error in this topic.

Correct Answer: B. $930$

Fastest Approach (🚀):
$2\times\dfrac{30\times 31}{2} = 30\times 31 = 930$.

Matches Option B.

Step-by-Step Breakdown:

1. Write the series and factor

$2+4+\cdots+60 = 2(1+2+\cdots+30)$

2. Use the triangular number

$1+2+\cdots+30 = \dfrac{30\times 31}{2} = 465$

3. Multiply back

$2\times 465 = 930$

Equivalently the sum of the first $n$ even numbers is always $n(n+1) = 30\times 31$.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $465$ — Omitted Factor
  Forgetting to multiply back by 2.
• C. $900$ — Wrong Standard Sum
  Using $30^{2}$, the sum of the first 30 odd numbers.
• D. $870$ — Off-by-one Error
  Off-by-one in the count of terms.
• E. $1860$ — Double Counting
  Doubling the correct answer.

Common Mistake (⚠️):
Answering $465$, having factored the $2$ out and forgotten to put it back.

Takeaway (📌):
Odds sum to $n^{2}$, evens to $n(n+1)$, integers to $\tfrac{n(n+1)}{2}$ — all three from the same pairing argument.

Correct Answer: C. $6\sqrt{2}$

Fastest Approach (🚀):
$\sqrt{50} = 5\sqrt2,\quad \sqrt{18} = 3\sqrt2,\quad \sqrt{8} = 2\sqrt2$.
$5+3-2 = 6$, so the result is $6\sqrt2$.

Matches Option C.

Step-by-Step Breakdown:

1. Extract the square factors

$\sqrt{50} = \sqrt{25\times 2} = 5\sqrt{2}$
$\sqrt{18} = \sqrt{9\times 2} = 3\sqrt{2}$
$\sqrt{8} = \sqrt{4\times 2} = 2\sqrt{2}$

2. Collect like terms

Every term is now a multiple of $\sqrt2$, so they combine like algebra:
$5\sqrt2+3\sqrt2-2\sqrt2 = (5+3-2)\sqrt2 = 6\sqrt2$

3. Check the size

$6\sqrt2 \approx 8.49$, and $\sqrt{50}+\sqrt{18}-\sqrt{8} \approx 7.07+4.24-2.83 = 8.48$. Consistent.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $\sqrt{60}$ — Root Distribution Error
  Adding and subtracting under the root: $\sqrt{50+18-8}$.
• B. $4\sqrt{2}$ — Arithmetic Error
  Coefficient slip: $5+3-4$ or subtracting $\sqrt8$ as $4\sqrt2$.
• D. $10\sqrt{2}$ — Sign Error
  Adding all three coefficients: $5+3+2$.
• E. $8\sqrt{2}$ — Sign Error
  Adding the first two coefficients and forgetting to subtract the third: $5+3$.

Common Mistake (⚠️):
Adding under the root sign: $\sqrt{50}+\sqrt{18} \ne \sqrt{68}$. Roots do not distribute over addition.

Takeaway (📌):
Simplify every surd to $k\sqrt{m}$ first. Only then can terms be collected, and the whole question becomes integer arithmetic.

Correct Answer: E. $30$

Fastest Approach (🚀):
$\dfrac{5!}{2!\,2!} = \dfrac{120}{4} = 30$.

Matches Option E.

Step-by-Step Breakdown:

1. Count as if every letter were distinct

$5! = 120$

2. Identify the repeats

L appears twice, E appears twice, V once.

3. Divide out the duplicates

Swapping the two Ls gives the same word, and likewise the two Es:
$\dfrac{5!}{2!\times 2!} = \dfrac{120}{4} = 30$

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $120$ — Repeats Ignored
  Ignoring the repeated letters entirely.
• B. $60$ — Incomplete Correction
  Dividing by 2! once instead of twice.
• C. $24$ — Counting Error
  Using $4!$, dropping a letter.
• D. $20$ — Factorial Error
  Dividing by $3!$ or an incorrect repeat count.

Common Mistake (⚠️):
Dividing by $2$ only once, giving $60$.

Takeaway (📌):
Arrangements with repeats is $\dfrac{n!}{n_1!\,n_2!\cdots}$ — the same over-count correction as dividing by $r!$ for combinations.

Correct Answer: B. $6\pi\ \text{cm}^{2}$

Fastest Approach (🚀):
Fraction $= \dfrac{60}{360} = \dfrac16$.
Full circle $= 36\pi$.
Sector $= \dfrac{36\pi}{6} = 6\pi\ \text{cm}^{2}$.

Matches Option B.

Step-by-Step Breakdown:

1. Find the fraction of the circle

$\dfrac{\theta}{360^{\circ}} = \dfrac{60}{360} = \dfrac{1}{6}$

2. Find the area of the whole circle

$A = \pi r^{2} = \pi(6)^{2} = 36\pi\ \text{cm}^{2}$

3. Take the fraction

$A_{\text{sector}} = \dfrac16\times 36\pi = 6\pi\ \text{cm}^{2}$

4. Note the arc length shortcut is identical

The same fraction gives the arc length: $\tfrac16\times 2\pi(6) = 2\pi$ cm. One fraction answers both parts of a typical sector question.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $3\pi\ \text{cm}^{2}$ — Fraction Error
  Halving the correct area, or using $r=6$ with a $\tfrac{1}{12}$ fraction.
• C. $12\pi\ \text{cm}^{2}$ — Fraction Error
  Using $\tfrac13$ of the circle instead of $\tfrac16$.
• D. $2\pi\ \text{cm}^{2}$ — Misread Question
  Giving the arc length rather than the area.
• E. $36\pi\ \text{cm}^{2}$ — Incomplete Answer
  Giving the area of the whole circle.

Common Mistake (⚠️):
Using the arc-length fraction of the circumference instead of the area, giving $2\pi$, or forgetting to square the radius.

Takeaway (📌):
Sector area $= \dfrac{\theta}{360}\pi r^{2}$ and arc length $= \dfrac{\theta}{360}2\pi r$. Compute the fraction once and reuse it.

Correct Answer: B. the 6th

Fastest Approach (🚀):
$486\div 2 = 243 = 3^{5}$.
$n-1 = 5 \implies n = 6$.

Matches Option B.

Step-by-Step Breakdown:

1. Identify a and r

$a = 2$ and each term is 3 times the previous one, so $r = 3$.

2. Write the nth term

$u_{n} = ar^{\,n-1} = 2\times 3^{\,n-1}$

3. Set it equal to the target

$2\times 3^{\,n-1} = 486 \implies 3^{\,n-1} = 243$

4. Match the powers

$243 = 3^{5}$, so
$n-1 = 5 \implies n = 6$

Check: $2,6,18,54,162,486$ — the sixth term. Correct.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. the 5th — Off-by-one Error
  Solving $3^{n} = 243$ and forgetting the $n-1$ offset.
• C. the 7th — Off-by-one Error
  Adding one too many to the exponent.
• D. the 8th — Counting Error
  Using $r=2$ or listing terms and miscounting.
• E. the 9th — Model Error
  Treating the sequence as arithmetic and dividing by the difference.

Common Mistake (⚠️):
Solving $3^{n} = 243$ and answering the 5th term, forgetting the $n-1$ in the exponent.

Takeaway (📌):
$u_n = ar^{n-1}$: the exponent is one less than the term number. Divide by $a$ first, then match powers.

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

Fastest Approach (🚀):
$27^{2/3} = 3^{2} = 9$.
$16^{-1/4} = \dfrac{1}{2}$.
$9\times\dfrac12 = \dfrac92$.

Matches Option B.

Step-by-Step Breakdown:

1. Split each fractional index

$a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^{m}$

Taking the root before the power keeps the arithmetic small: $\sqrt[3]{27} = 3$, then $3^{2} = 9$. (Going the other way means finding $\sqrt[3]{729}$ — the same answer, three times the work.)

2. Handle the negative index

$a^{-k} = \dfrac{1}{a^{k}}$, so
$16^{-\frac14} = \dfrac{1}{16^{\frac14}} = \dfrac{1}{\sqrt[4]{16}} = \dfrac{1}{2}$

3. Multiply

$9\times\dfrac12 = \dfrac{9}{2}$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $18$ — Index Sign Error
  Multiplying $9$ by $2$ instead of by $\tfrac12$ — the negative index inverted the wrong way.
• C. $\dfrac{2}{9}$ — Inversion Error
  Inverting the whole product rather than just the second factor.
• D. $36$ — Formula Misuse
  Using $27^{2/3} = 18$ (multiplying by $\tfrac23$) and $16^{-1/4} = 2$.
• E. $\dfrac{1}{18}$ — Inversion Error
  Inverting both factors.

Common Mistake (⚠️):
Treating the negative index as a negative number and answering $-18$, or multiplying $27$ by $\tfrac23$ instead of taking a root.

Takeaway (📌):
Denominator of the index is the root, numerator is the power, minus sign is a reciprocal. Three independent decisions, applied in that order.

Correct Answer: D. $30$

Fastest Approach (🚀):
At least one language: $120-20 = 100$.
Spanish only $= 100-70 = 30$.

Matches Option D.

Step-by-Step Breakdown:

1. Work out the four regions

Neither $= 20$, so at least one $= 120-20 = 100$.
French (total) $= 70$, of which both $= 45$, so French only $= 25$.

2. The rest of the 100 must be Spanish only

$100-70 = 30$

3. Check every region

French only $25$, both $45$, Spanish only $30$, neither $20$. Total $= 120$. Correct, and Spanish total is $45+30 = 75$.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $25$ — Wrong Region
  Giving French only rather than Spanish only.
• B. $45$ — Wrong Region
  Giving the overlap.
• C. $50$ — Complement Ignored
  Using $120-70$ and ignoring the 20 who study neither.
• E. $75$ — Wrong Region
  Giving the Spanish total, including those who also study French.

Common Mistake (⚠️):
Giving the Spanish total $75$ rather than Spanish-only, or forgetting the $20$ who study neither and working from $120$.

Takeaway (📌):
Draw the four regions before answering. Whichever one the question wants, the totals pin it down.

Correct Answer: E. $10$

Fastest Approach (🚀):
$\Delta x = 5-(-3) = 8,\quad \Delta y = -2-4 = -6$.
$(6,8,10)$ triple $\implies d = 10$.

Matches Option E.

Step-by-Step Breakdown:

1. Find the coordinate differences

$\Delta x = 5-(-3) = 8$
$\Delta y = -2-4 = -6$

Signs do not matter — both get squared.

2. Apply Pythagoras

$d = \sqrt{(\Delta x)^{2}+(\Delta y)^{2}} = \sqrt{64+36} = \sqrt{100}$

3. Evaluate

$d = 10$

4. The triple shortcut

$(6,8,10)$ is the $(3,4,5)$ triple doubled. Spotting it removes the squaring entirely.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\sqrt{14}$ — Formula Misuse
  Adding the differences $8+6$ and taking the root.
• B. $14$ — Formula Misuse
  Adding the two differences without squaring.
• C. $2\sqrt{7}$ — Arithmetic Error
  Mis-simplifying $\sqrt{28}$ after a subtraction error.
• D. $8$ — Incomplete Answer
  Using $\Delta x$ alone.

Common Mistake (⚠️):
Subtracting the coordinates in the wrong pairing, for example $\Delta x = 5-4$, or adding the differences instead of their squares.

Takeaway (📌):
Distance is Pythagoras on the coordinate differences. Check for $(3,4,5)$, $(6,8,10)$ and $(5,12,13)$ before computing anything.

Correct Answer: D. £288

Fastest Approach (🚀):
$5-3 = 2$ parts $= 48 \implies$ one part $= 24$.
Total $= 12\times 24 = £288$.

Matches Option D.

Step-by-Step Breakdown:

1. Express the difference in parts

Largest share $= 5$ parts, smallest $= 3$ parts, so the stated gap is
$5-3 = 2\ \text{parts}$

2. Find one part

$2\ \text{parts} = £48 \implies 1\ \text{part} = £24$

3. Total the parts

$3+4+5 = 12\ \text{parts}$

4. Multiply

$12\times £24 = £288$

Check: the shares are £72, £96 and £120, and $120-72 = 48$. Correct.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. £96 — Parts Error
  Using $2$ parts as the total, or doubling £48.
• B. £144 — Parts Error
  Totalling only $6$ parts, or using a $1:2:3$ split.
• C. £192 — Omitted Term
  Using $8$ parts, dropping the middle share.
• E. £576 — Parts Error
  Treating £48 as one part rather than as two.

Common Mistake (⚠️):
Treating £48 as the value of one part, which gives a total of £576, or as the largest share itself.

Takeaway (📌):
Always translate the given information into parts first. One part is the only number you need; everything else is multiplication.

Correct Answer: E. $5 \le x \lt 10$

Fastest Approach (🚀):
Divide each frequency by its class width.
$1.4,\ 2.4,\ 1.6,\ 0.9$ — the largest is the second class.

Matches Option E.

Step-by-Step Breakdown:

1. Note the widths differ

Widths are $5$, $5$, $10$ and $20$. The class with the biggest frequency is also the widest, which should immediately make you suspicious.

2. Convert to frequency density

$\dfrac{7}{5} = 1.4, \qquad \dfrac{12}{5} = 2.4, \qquad \dfrac{16}{10} = 1.6, \qquad \dfrac{18}{20} = 0.9$

3. Read the mode from the density

The greatest density is $2.4$, in $5 \le x \lt 10$. Data is most concentrated there — the last class holds more values only because it is four times as wide.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $0 \le x \lt 5$ — Density Ignored
  Choosing the narrowest class without computing its density.
• B. All classes are equally modal — Conceptual Error
  Assuming unequal widths make the classes incomparable.
• C. $10 \le x \lt 20$ — Estimation Error
  Choosing the middle class by an eyeballed compromise.
• D. $20 \le x \lt 40$ — Density Ignored
  Choosing the largest raw frequency — the intended trap.

Common Mistake (⚠️):
Choosing $20 \le x \lt 40$ because it has the largest frequency. Frequency alone only identifies the mode when every class has the same width.

Takeaway (📌):
Equal widths: compare frequencies. Unequal widths: compare densities. Check the widths before you compare anything.

Correct Answer: E. $24$

Fastest Approach (🚀):
$30\%$ of $40\% = 12\%$.
$12\%$ of $200 = 24$.

Matches Option E.

Step-by-Step Breakdown:

1. Translate 'of' into multiplication

$30\%\ \text{of}\ 40\%\ \text{of}\ 200 = 0.3\times 0.4\times 200$

2. Combine the two percentages first

$0.3\times 0.4 = 0.12$, that is $12\%$.

Combining the percentages before touching the 200 keeps the arithmetic to a single easy multiplication.

3. Apply to 200

$0.12\times 200 = 24$

4. Check the order does not matter

$40\%$ of $200 = 80$, and $30\%$ of $80 = 24$. Same answer — multiplication is commutative.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $140$ — Additive Error
  Adding the percentages: 70% of 200.
• B. $60$ — Omitted Term
  Applying 30% to 200 and ignoring the 40%.
• C. $80$ — Omitted Term
  Applying 40% to 200 and ignoring the 30%.
• D. $12$ — Arithmetic Error
  Halving the correct answer, or using 6% instead of 12%.

Common Mistake (⚠️):
Adding the percentages to get 70% and answering 140. Successive percentages multiply; they never add.

Takeaway (📌):
Percentages of percentages multiply. $30\%$ of $40\%$ is $12\%$, not $70\%$ and not $10\%$.

Correct Answer: D. $\dfrac{1}{9}$

Fastest Approach (🚀):
Pairs totalling 9: $(3,6),(4,5),(5,4),(6,3)$ — four of them.
$P = \dfrac{4}{36} = \dfrac19$.

Matches Option D.

Step-by-Step Breakdown:

1. Size of the sample space

Each die has 6 outcomes and they are independent:
$6\times 6 = 36$ equally likely ordered pairs.

2. List the favourable outcomes

Totals of 9 require the first die to be at least 3:
$(3,6),\ (4,5),\ (5,4),\ (6,3)$

That is 4 outcomes. Order matters because the dice are distinguishable.

3. Divide

$P(\text{total}=9) = \dfrac{4}{36} = \dfrac{1}{9}$

4. Sanity check against the distribution

Totals run from 2 to 12 with 7 the most likely (6 ways). A total of 9 sits two steps out from 7, so 4 ways is exactly what the symmetric pattern $1,2,3,4,5,6,5,4,3,2,1$ predicts.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\dfrac{1}{6}$ — Wrong Total
  Using $\tfrac{6}{36}$, the count for a total of 7.
• B. $\dfrac{1}{12}$ — Ordering Error
  Counting unordered pairs only: $\tfrac{2}{36}$ simplified incorrectly.
• C. $\dfrac{1}{4}$ — Conceptual Error
  Using $\tfrac{9}{36}$ from the target total rather than a count.
• E. $\dfrac{5}{36}$ — Counting Error
  Using $5$ ways, the count for a total of 8 or 6.

Common Mistake (⚠️):
Counting $(3,6)$ and $(6,3)$ as a single outcome, giving $\tfrac{2}{36}$, or using 21 unordered pairs as the sample space.

Takeaway (📌):
With two distinguishable dice the sample space is 36 ordered pairs, and the number of ways to make each total follows $1,2,3,4,5,6,5,4,3,2,1$.

Correct Answer: C. $22$

Fastest Approach (🚀):
Middle $= \dfrac{100}{5} = 20$.
Largest $= 20+2 = 22$.

Matches Option C.

Step-by-Step Breakdown:

1. Use the symmetry

Write the five integers as
$n-2,\ n-1,\ n,\ n+1,\ n+2$

2. Sum them

The offsets cancel in pairs:
$(n-2)+(n-1)+n+(n+1)+(n+2) = 5n$

3. Solve

$5n = 100 \implies n = 20$

$n$ is the middle term, so the five integers are $18, 19, 20, 21, 22$.

4. Answer the question asked

The largest is $n+2 = 22$.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $20$ — Misread Question
  Giving the middle term, which is the mean of the five.
• B. $21$ — Off-by-one Error
  Off by one when stepping from the middle to the largest.
• D. $24$ — Setup Error
  Starting the run at $n$ rather than centring it, then adding 4.
• E. $96$ — Conceptual Error
  Subtracting 4 from 100 rather than using the structure.

Common Mistake (⚠️):
Giving the middle value $20$, which is the mean rather than the largest term.

Takeaway (📌):
Centre a run of consecutive numbers on $n$ rather than starting at $n$. The offsets cancel and the algebra disappears.

Correct Answer: C. $56$

Fastest Approach (🚀):
$\dfrac{8\times 7\times 6}{6} = 8\times 7 = 56$.

Matches Option C.

Step-by-Step Breakdown:

1. Decide whether order matters

A team of 3 is a set, not a ranking — choosing Ana, Ben, Cara is the same team in any order. So this is a combination, not a permutation.

2. Write the combination

$\binom{8}{3} = \dfrac{8!}{3!\,5!}$

3. Cancel before multiplying

The $5!$ cancels the tail of $8!$:
$\binom{8}{3} = \dfrac{8\times 7\times 6}{3\times 2\times 1}$

4. Evaluate

$\dfrac{336}{6} = 56$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $24$ — Formula Misuse
  Computing $8\times 3$ or dividing $\binom{8}{3}$ by an extra factor.
• B. $336$ — Order Error
  Giving the permutation count, without dividing by 3!.
• D. $112$ — Factorial Error
  Dividing $336$ by $3$ instead of by $3! = 6$.
• E. $512$ — Model Error
  Using $8^{3}$ — selection with repetition and order.

Common Mistake (⚠️):
Using the permutation $8\times 7\times 6 = 336$, which counts each team $3! = 6$ times over — once for every ordering.

Takeaway (📌):
Order matters: permutation. Order does not: divide by $r!$. Write the $r$-term product over $r!$ and cancel first.

Correct Answer: E. $2$ hours

Fastest Approach (🚀):
$90+60 = 150$ km/h.
$\dfrac{300}{150} = 2$ hours.

Matches Option E.

Step-by-Step Breakdown:

1. Work in the closing frame

When two objects move towards each other, the distance between them shrinks at the sum of their speeds:
$v_{\text{closing}} = 90+60 = 150\ \text{km/h}$

2. Divide the gap by the closing speed

$t = \dfrac{300}{150} = 2\ \text{hours}$

3. Check by tracking both trains

In 2 hours the first covers $180$ km and the second $120$ km. Together that is $300$ km — they meet exactly. The meeting point is 180 km from the first train's start.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $1$ hour — Setup Error
  Using $300/300$ or another mismatched pairing.
• B. $1.5$ hours — Omitted Term
  Using only the faster train's speed: $300/90 \approx 3.3$, then adjusting.
• C. $3$ hours — Averaging Error
  Averaging the speeds to $75$ and adjusting, or using $300/100$.
• D. $5$ hours — Omitted Term
  Using only the slower train: $300/60$.

Common Mistake (⚠️):
Using one train's speed alone ($300/90$ or $300/60$), or averaging the two speeds instead of adding them.

Takeaway (📌):
Towards each other: add the speeds. Same direction: subtract them. Either way it becomes a single distance-over-speed division.

Correct Answer: A. $17$

Fastest Approach (🚀):
$168+342 = 510$.
$\dfrac{510}{30} = 17$.

Matches Option A.

Step-by-Step Breakdown:

1. Convert each mean to a total

$\text{boys}: 12\times 14 = 168$
$\text{girls}: 18\times 19 = 342$

2. Combine

$\text{total} = 168+342 = 510, \qquad \text{count} = 12+18 = 30$

3. Divide

$\bar x = \dfrac{510}{30} = 17$

4. Check the position

The simple average of 14 and 19 is 16.5, but there are more girls, so the true mean must be pulled above 16.5 towards 19. It is — 17.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $16.5$ — Weighting Error
  Taking the unweighted average of 14 and 19.
• C. $16$ — Weighting Error
  Weighting towards the smaller group instead of the larger.
• D. $33$ — Additive Error
  Adding the two means.
• E. $15$ — Weighting Error
  Weighting by the wrong group sizes entirely.

Common Mistake (⚠️):
Averaging the two means to get 16.5. That would only be correct if the groups were the same size.

Takeaway (📌):
A combined mean is a weighted mean. Convert to totals, add, divide by the combined count — and check the answer leans towards the bigger group.

Correct Answer: D. $1.35\ \text{kg}$

Fastest Approach (🚀):
$2.7\times 500 = 1350$ g.
$1350\ \text{g} = 1.35$ kg.

Matches Option D.

Step-by-Step Breakdown:

1. Choose the relationship

$\text{density} = \dfrac{\text{mass}}{\text{volume}} \implies \text{mass} = \text{density}\times\text{volume}$

2. Multiply, keeping track of the units

$m = 2.7\ \text{g/cm}^{3}\times 500\ \text{cm}^{3} = 1350\ \text{g}$

The cm³ cancels, leaving grams — the units confirm the operation was the right way round.

3. Convert to kilograms

$1\ \text{kg} = 1000\ \text{g}$, so
$1350\ \text{g} = 1.35\ \text{kg}$

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $185\ \text{kg}$ — Operation Error
  Dividing $500$ by $2.7$ instead of multiplying.
• B. $13.5\ \text{kg}$ — Conversion Error
  Converting with a factor of 100 rather than 1000.
• C. $0.135\ \text{kg}$ — Conversion Error
  Dividing by $10\,000$ — an extra factor of ten in the conversion.
• E. $1350\ \text{kg}$ — Units Error
  Leaving the answer in grams but labelling it kilograms.

Common Mistake (⚠️):
Dividing instead of multiplying, or leaving the answer in grams. A quick unit check catches both: g/cm³ × cm³ must give grams.

Takeaway (📌):
Carry the units through the calculation. If they do not cancel to the unit you want, the operation is the wrong way round.

Correct Answer: C. $24x$

Fastest Approach (🚀):
$A-B = 6,\quad A+B = 4x$.
Product $= 6\times 4x = 24x$.

Matches Option C.

Step-by-Step Breakdown:

1. Recognise the structure

The expression is $A^{2}-B^{2}$ with
$A = 2x+3, \qquad B = 2x-3$

2. Factorise rather than expand

$A^{2}-B^{2} = (A-B)(A+B)$

$A-B = (2x+3)-(2x-3) = 6$
$A+B = (2x+3)+(2x-3) = 4x$

3. Multiply

$(6)(4x) = 24x$

4. Confirm by expanding

$(4x^{2}+12x+9)-(4x^{2}-12x+9) = 24x$. Same answer, twice the writing — and the subtraction of the second bracket is where sign errors usually appear.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $0$ — Conceptual Error
  Assuming the two squares cancel because the brackets look similar.
• B. $12x$ — Arithmetic Error
  Halving the correct result, or using $A+B = 2x$.
• D. $18$ — Bracket Error
  Sign error when subtracting the second bracket, leaving only the constants.
• E. $8x^{2}+18$ — Sign Error
  Adding the two expansions instead of subtracting.

Common Mistake (⚠️):
Dropping the brackets when subtracting: $4x^{2}+12x+9-4x^{2}-12x+9$ gives $18$, because only the first term of the second bracket was negated.

Takeaway (📌):
The difference of two squares applies to any two expressions, not just numbers. Factorising avoids the sign trap of subtracting a whole bracket.

Correct Answer: B. $x > -5$

Fastest Approach (🚀):
$3x-6 < 5x+4$.
Subtract $3x$ and $4$: $-10 < 2x$.
$x > -5$.

Matches Option B.

Step-by-Step Breakdown:

1. Expand

$3(x-2) < 5x+4 \implies 3x-6 < 5x+4$

2. Collect x on the side with the larger coefficient

Subtracting $3x$ from both sides keeps the $x$ coefficient positive:
$-6 < 2x+4$

3. Isolate x

$-6-4 < 2x \implies -10 < 2x \implies -5 < x$

4. State it the usual way round

$-5 < x$ is the same as $x > -5$.

Check $x=0$: $3(-2) = -6 < 4$. True, and $0 > -5$. Consistent.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $x < -5$ — Inequality Reversal
  Dividing by a negative coefficient without reversing the inequality.
• C. $x > 5$ — Sign Error
  Sign error on the constant, giving the mirrored answer.
• D. $x < 5$ — Inequality Reversal
  Both a sign error and a missed reversal.
• E. $x > -1$ — Bracket Error
  Expanding $3(x-2)$ as $3x-2$.

Common Mistake (⚠️):
Collecting $x$ on the left to get $-2x < 10$ and then dividing by $-2$ without reversing the inequality, which gives $x < -5$.

Takeaway (📌):
Move the $x$ terms towards the larger coefficient. It costs nothing and removes every opportunity to forget the reversal rule.