Cambridge IGCSE · thinka-original Practice Paper

2023 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

Thinka Jun 2023 (V2) Cambridge IGCSE-Style Mock — Mathematics (0580)

200 marks240 mins2023

An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V2) Cambridge IGCSE Mathematics (0580) paper. Not affiliated with or reproduced from Cambridge.

Paper 2 (Extended)

Answer all questions. Use of a calculator is allowed where appropriate. You must show all necessary working.

23 Question · 69.6 marks

Question 1 · Short Answer

3 marks

A cylinder has a radius of $4.5\text{ cm}$ and a height of $12\text{ cm}$. A cone has a radius of $3\text{ cm}$ and the same volume as the cylinder. Calculate the perpendicular height of the cone.

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Worked solution

The volume of a cylinder is given by the formula:
\[V_{\text{cylinder}} = \pi r^2 h\]
Substituting $r = 4.5\text{ cm}$ and $h = 12\text{ cm}$:
\[V_{\text{cylinder}} = \pi \times 4.5^2 \times 12 = 243\pi\text{ cm}^3\]

The volume of a cone is given by the formula:
\[V_{\text{cone}} = \frac{1}{3} \pi R^2 H\]
Substituting $R = 3\text{ cm}$ and equating the two volumes:
\[\frac{1}{3} \pi \times 3^2 \times H = 243\pi\]
\[3\pi H = 243\pi\]
\[H = \frac{243}{3} = 81\text{ cm}\]

Marking scheme

**M1** for calculating the volume of the cylinder: $\pi \times 4.5^2 \times 12$ (or $243\pi$ or $763.4$)
**M1** for equating their volume to the formula for the volume of the cone: $\frac{1}{3} \pi \times 3^2 \times H = 243\pi$
**A1** for $81$

Question 2 · Short Answer

3 marks

Solve the simultaneous equations. You must show all your working.
\[3x + 2y = 4\]
\[4x - 3y = 11\]

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Worked solution

Multiply the first equation by 3:
\[9x + 6y = 12\]

Multiply the second equation by 2:
\[8x - 6y = 22\]

Add the two resulting equations to eliminate $y$:
\[(9x + 8x) + (6y - 6y) = 12 + 22\]
\[17x = 34\]
\[x = 2\]

Substitute $x = 2$ back into the first equation:
\[3(2) + 2y = 4\]
\[6 + 2y = 4\]
\[2y = -2\]
\[y = -1\]

So, the solution is $x = 2$, $y = -1$.

Marking scheme

**M1** for a correct method to eliminate one variable (e.g., multiplying equations to align coefficients of $x$ or $y$)
**A1** for $x = 2$
**A1** for $y = -1$

Question 3 · Short Answer

3 marks

Make $t$ the subject of the formula:
\[w = \frac{3t + 5}{2t - 1}\]

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Worked solution

First, multiply both sides by $2t - 1$ to clear the fraction:
\[w(2t - 1) = 3t + 5\]
\[2wt - w = 3t + 5\]

Next, collect all terms containing $t$ on one side and constant terms on the other:
\[2wt - 3t = w + 5\]

Factorise $t$ out of the left-hand side:
\[t(2w - 3) = w + 5\]

Finally, divide by $2w - 3$ to make $t$ the subject:
\[t = \frac{w + 5}{2w - 3}\]

Marking scheme

**M1** for multiplying by $2t - 1$ to clear the fraction: $w(2t - 1) = 3t + 5$
**M1** for isolating the terms in $t$ on one side: $2wt - 3t = w + 5$ (or equivalent)
**A1** for $t = \frac{w + 5}{2w - 3}$ or $t = \frac{-w - 5}{3 - 2w}$

Question 4 · Short Answer

3 marks

A regular polygon has an interior angle of $162^\circ$. Find the number of sides of this polygon.

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Worked solution

An interior angle and an exterior angle on the same vertex add up to $180^\circ$.
\[\text{Exterior angle} = 180^\circ - 162^\circ = 18^\circ\]

The sum of the exterior angles of any convex polygon is $360^\circ$.
\[\text{Number of sides } (n) = \frac{360^\circ}{\text{Exterior angle}} = \frac{360^\circ}{18^\circ} = 20\]

Marking scheme

**M1** for finding the exterior angle: $180 - 162 = 18$
**M1** for calculating the number of sides: $360 \div 18$ or setting up the formula $\frac{(n-2) \times 180}{n} = 162$
**A1** for $20$

Question 5 · Short Answer

3 marks

A bag contains 4 red counters and 6 black counters. Two counters are taken out of the bag at random, without replacement. Find the probability that both counters are of the same colour.

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Worked solution

There are $4 + 6 = 10$ counters in total.

The probability of choosing two red counters is:
\[P(\text{Red, Red}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}\]

The probability of choosing two black counters is:
\[P(\text{Black, Black}) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{5}{15}\]

The probability that both counters are the same colour is the sum of these two probabilities:
\[P(\text{Same Colour}) = P(\text{Red, Red}) + P(\text{Black, Black}) = \frac{12}{90} + \frac{30}{90} = \frac{42}{90} = \frac{7}{15}\]

Marking scheme

**M1** for calculating the probability of choosing two of one colour (either red or black): $\frac{4}{10} \times \frac{3}{9}$ or $\frac{6}{10} \times \frac{5}{9}$
**M1** for adding the two product probabilities: $\frac{12}{90} + \frac{30}{90}$
**A1** for $\frac{7}{15}$ or any equivalent fraction (e.g. $\frac{42}{90}$), decimal (0.467 or 0.4666...)

Question 6 · Short Answer

3 marks

Simplify completely:
\[\left( \frac{125x^6}{8y^{-3}} \right)^{-\frac{2}{3}}\]

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Worked solution

First, simplify the expression inside the parentheses:
\[\frac{125x^6}{8y^{-3}} = \frac{125x^6y^3}{8}\]

Next, apply the negative exponent by taking the reciprocal of the base:
\[\left( \frac{8}{125x^6y^3} \right)^{\frac{2}{3}}\]

Now, apply the power of $\frac{2}{3}$. This involves taking the cube root first, then squaring:
\[\text{Cube root: } \sqrt[3]{\frac{8}{125x^6y^3}} = \frac{2}{5x^2y}\]
\[\text{Squaring the result: } \left( \frac{2}{5x^2y} \right)^2 = \frac{4}{25x^4y^2}\]

Marking scheme

**M1** for taking the reciprocal of the fraction to remove the negative sign in the exponent: $\left( \frac{8}{125x^6y^3} \right)^{\frac{2}{3}}$ or equivalent step
**M1** for taking the cube root of the terms correctly: $\frac{2}{5x^2y}$ or squaring correctly first
**A1** for $\frac{4}{25x^4y^2}$ or $0.16x^{-4}y^{-2}$

Question 7 · Short Answer

3 marks

In a right-angled triangle, the hypotenuse is $13\text{ cm}$ and one of the shorter sides is $5\text{ cm}$. Calculate the size of the smallest angle in this triangle. Give your answer to 1 decimal place.

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Worked solution

The three angles of the triangle are $90^\circ$ and two acute angles. The smallest angle of a triangle is opposite the shortest side, which is $5\text{ cm}$.

Let the smallest angle be $\theta$.
Using the sine ratio:
\[\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{5}{13}\]
\[\theta = \sin^{-1}\left(\frac{5}{13}\right) \approx 22.6198^\circ\]

Rounding to 1 decimal place gives $22.6^\circ$.

Marking scheme

**M1** for identifying the correct trigonometric ratio: $\sin(\theta) = \frac{5}{13}$ (or using Pythagoras first to find the third side as 12 cm and then using $\cos(\theta) = \frac{12}{13}$ or $\tan(\theta) = \frac{5}{12}$)
**M1** for applying the inverse trigonometric function: $\sin^{-1}\left(\frac{5}{13}\right)$
**A1** for $22.6$ or $22.61$ to $22.62$

Question 8 · Short Answer

3 marks

$y$ is inversely proportional to the square root of $x$. When $x = 16$, $y = 5$. Find $y$ when $x = 25$.

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Worked solution

Since $y$ is inversely proportional to the square root of $x$, we can write the formula as:
\[y = \frac{k}{\sqrt{x}}\]
where $k$ is the constant of proportionality.

Substitute the given values $x = 16$ and $y = 5$ into the equation to find $k$:
\[5 = \frac{k}{\sqrt{16}}\]
\[5 = \frac{k}{4}\]
\[k = 20\]

Now, write the complete formula with $k = 20$:
\[y = \frac{20}{\sqrt{x}}\]

Substitute $x = 25$ into the equation to find the new value of $y$:
\[y = \frac{20}{\sqrt{25}} = \frac{20}{5} = 4\]

Marking scheme

**M1** for translating the proportionality statement into an algebraic equation: $y = \frac{k}{\sqrt{x}}$
**M1** for substituting $x=16$ and $y=5$ to find $k = 20$
**A1** for $4$

Question 9 · short_answer

3.04 marks

Solve the equation $\frac{10}{x} - \frac{6}{x+1} = 1$.

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Worked solution

Multiply each term by $x(x+1)$:
$10(x+1) - 6x = x(x+1)$
$10x + 10 - 6x = x^2 + x$
$4x + 10 = x^2 + x$

Rearrange into a quadratic equation:
$x^2 - 3x - 10 = 0$

Factorise:
$(x-5)(x+2) = 0$

So, $x = 5$ or $x = -2$.

Marking scheme

M1 for clearing the fractions: $10(x+1) - 6x = x(x+1)$ or equivalent
M1 for forming the quadratic equation $x^2 - 3x - 10 = 0$
A1 for $x = 5$ and $x = -2$

Question 10 · short_answer

3.04 marks

Simplify $\frac{2x^2 - 7x - 4}{x^2 - 16}$.

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Worked solution

Factorise the numerator:
$2x^2 - 7x - 4 = (2x+1)(x-4)$

Factorise the denominator:
$x^2 - 16 = (x-4)(x+4)$

Simplify by cancelling the common factor $(x-4)$:
$\frac{(2x+1)(x-4)}{(x-4)(x+4)} = \frac{2x+1}{x+4}$

Marking scheme

M1 for factorising the numerator: $(2x+1)(x-4)$
M1 for factorising the denominator: $(x-4)(x+4)$
A1 for $\frac{2x+1}{x+4}$

Question 11 · short_answer

3.04 marks

A solid cylinder has radius 3 cm and height 8 cm. A solid cone has radius 4 cm and has the same volume as the cylinder. Calculate the perpendicular height of the cone.

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Worked solution

Volume of cylinder:
$V = \pi r^2 h = \pi \times 3^2 \times 8 = 72\pi\text{ cm}^3$

Volume of cone:
$V = \frac{1}{3} \pi r^2 h_{\text{cone}} = \frac{1}{3} \pi \times 4^2 \times h_{\text{cone}} = \frac{16}{3}\pi h_{\text{cone}}$

Set the volumes equal:
$\frac{16}{3}\pi h_{\text{cone}} = 72\pi$

Multiply both sides by 3 and divide by $\pi$:
$16 h_{\text{cone}} = 216$

Divide by 16:
$h_{\text{cone}} = 13.5\text{ cm}$

Marking scheme

M1 for Volume of cylinder = $72\pi$ or 226.2...
M1 for setting up the equation: $\frac{1}{3} \pi \times 4^2 \times h = 72\pi$
A1 for 13.5

Question 12 · short_answer

3.04 marks

A regular polygon has $n$ sides. The interior angle of this polygon is 11 times the size of its exterior angle. Find the value of $n$.

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Worked solution

Let the exterior angle be $e$.
The interior angle is $11e$.

The sum of the interior angle and exterior angle is $180^{\circ}$:
$e + 11e = 180^{\circ}$
$12e = 180^{\circ}$
$e = 15^{\circ}$

The number of sides $n$ is given by:
$n = \frac{360^{\circ}}{e} = \frac{360}{15} = 24$

Marking scheme

M1 for setting up the angle equation: $e + 11e = 180$ or equivalent
M1 for finding the exterior angle $e = 15$
A1 for 24

Question 13 · short_answer

3.04 marks

Calculate $\frac{4.2 \times 10^5 + 8.1 \times 10^4}{3.0 \times 10^{-2}}$, giving your answer in standard form.

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Worked solution

First, add the numbers in the numerator:
$4.2 \times 10^5 + 8.1 \times 10^4 = 4.2 \times 10^5 + 0.81 \times 10^5 = 5.01 \times 10^5$

Next, divide by the denominator:
$\frac{5.01 \times 10^5}{3.0 \times 10^{-2}} = \left(\frac{5.01}{3.0}\right) \times 10^{5 - (-2)} = 1.67 \times 10^7$

Marking scheme

M1 for simplifying the numerator to $5.01 \times 10^5$ (or equivalent)
M1 for division of coefficients and correct subtraction of indices: $\frac{5.01}{3} = 1.67$ and $10^7$
A1 for $1.67 \times 10^7$

Question 14 · short_answer

3.04 marks

Find an expression for the $n$th term of the sequence: $3, 10, 21, 36, \dots$

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Worked solution

First four terms: $3, 10, 21, 36$
First differences: $7, 11, 15$
Second differences: $4, 4$

The second difference is constant, so the sequence is quadratic: $an^2 + bn + c$
$2a = 4 \implies a = 2$

Subtract $2n^2$ from each term:
- $n=1$: $3 - 2(1^2) = 1$
- $n=2$: $10 - 2(2^2) = 2$
- $n=3$: $21 - 2(3^2) = 3$
- $n=4$: $36 - 2(4^2) = 4$

The sequence of differences is $1, 2, 3, 4$, which has the $n$th term $n$.

Therefore, the $n$th term of the original sequence is $2n^2 + n$.

Marking scheme

M1 for finding the second difference is 4
M1 for subtracting $2n^2$ from the terms to get $1, 2, 3, 4$ (or equivalent system of equations)
A1 for $2n^2 + n$

Question 15 · short_answer

3.04 marks

Simplify $\left(\frac{64y^{12}}{x^6}\right)^{-\frac{2}{3}}$, giving your answer with positive indices.

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Worked solution

Apply the negative index to find the reciprocal:
$\left(\frac{64y^{12}}{x^6}\right)^{-\frac{2}{3}} = \left(\frac{x^6}{64y^{12}}\right)^{\frac{2}{3}}$

Apply the fractional power to each part of the fraction:
$\frac{(x^6)^{\frac{2}{3}}}{(64)^{\frac{2}{3}}(y^{12})^{\frac{2}{3}}} = \frac{x^4}{16y^8}$

Marking scheme

M1 for removing negative index: $\left(\frac{x^6}{64y^{12}}\right)^{\frac{2}{3}}$
M1 for applying power to elements, e.g., finding $64^{-2/3} = \frac{1}{16}$ or $(x^6)^{-2/3} = x^{-4}$ or $(y^{12})^{-2/3} = y^{-8}$
A1 for $\frac{x^4}{16y^8}$

Question 16 · short_answer

3.04 marks

The area of a sector of a circle with radius 12 cm is $24\pi\text{ cm}^2$. Calculate the perimeter of this sector, giving your answer correct to 3 significant figures.

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Worked solution

The area of a sector is given by:
$\text{Area} = \frac{\theta}{360} \pi r^2$

Substitute the given values:
$24\pi = \frac{\theta}{360} \pi (12)^2$
$24\pi = \frac{\theta}{360} \pi \times 144$
$24 = 144 \times \frac{\theta}{360}$
$\frac{\theta}{360} = \frac{24}{144} = \frac{1}{6}$

The arc length of the sector is:
$\text{Arc length} = \frac{\theta}{360} \times 2\pi r = \frac{1}{6} \times 2\pi \times 12 = 4\pi\text{ cm}$

The perimeter of the sector is the arc length plus two radii:
$\text{Perimeter} = 4\pi + 2 \times 12 = 4\pi + 24 \approx 36.566\text{ cm}$

Correct to 3 significant figures, the perimeter is $36.6\text{ cm}$.

Marking scheme

M1 for setting up the area equation to find the sector fraction $\frac{1}{6}$ or angle $\theta = 60^{\circ}$
M1 for arc length = $4\pi$ or 12.566...
A1 for 36.6

Question 17 · Short Answer

3.04 marks

Rearrange the formula to make $x$ the subject: $ w = \frac{4x - 3}{2x + 5} $

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Worked solution

Multiply both sides by the denominator:
$ w(2x + 5) = 4x - 3 $
Expand the bracket:
$ 2xw + 5w = 4x - 3 $
Gather all terms containing $x$ on one side and the other terms on the opposite side:
$ 5w + 3 = 4x - 2xw $
Factor out $x$:
$ 5w + 3 = x(4 - 2w) $
Divide by $ 4 - 2w $:
$ x = \frac{5w + 3}{4 - 2w} $

Marking scheme

M1 for correctly multiplying both sides by the denominator.
M1 for correctly gathering all $x$ terms on one side.
A1 for the final correct expression.

Question 18 · Short Answer

3.04 marks

A solid metal cylinder of radius 6 cm and height 15 cm is melted down and recast to form a solid cone of radius 8 cm. Calculate the height of the cone.

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Worked solution

First, calculate the volume of the cylinder:
$ V_{\text{cylinder}} = \pi r^2 h = \pi \times 6^2 \times 15 = 540\pi \text{ cm}^3 $
The volume of the recast cone must equal the volume of the cylinder:
$ V_{\text{cone}} = \frac{1}{3} \pi R^2 H = \frac{1}{3} \pi \times 8^2 \times H = 540\pi $
Solve for the height of the cone, $H$:
$ \frac{64}{3} H = 540 $
$ H = \frac{540 \times 3}{64} = 25.3125 \text{ cm} $
To 3 significant figures, the height is 25.3 cm.

Marking scheme

M1 for calculating the correct volume of the cylinder: $540\pi$ or approx 1696.
M1 for setting the cone volume formula equal to their cylinder volume.
A1 for 25.3 or 25.3125.

Question 19 · Short Answer

3.04 marks

The interior angle of a regular polygon is $ 165^\circ $. Find the number of sides of this polygon.

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Worked solution

The sum of the interior and exterior angle is $ 180^\circ $.
Therefore, the exterior angle is:
$ 180^\circ - 165^\circ = 15^\circ $
The sum of all exterior angles in any regular polygon is $ 360^\circ $.
Number of sides, $n$:
$ n = \frac{360^\circ}{15^\circ} = 24 $

Marking scheme

M1 for finding the exterior angle $180 - 165$.
M1 for $360$ divided by their exterior angle.
A1 for 24.

Question 20 · Short Answer

3.04 marks

The table shows the time, $ t $ minutes, spent by 40 students on their homework.

\begin{array}{|c|c|}
\hline
\text{Time } (t \text{ minutes}) & \text{Frequency} \\
\hline
0 < t \le 10 & 5 \\
10 < t \le 20 & 8 \\
20 < t \le 30 & 12 \\
30 < t \le 40 & 15 \\
\hline
\end{array}

Calculate an estimate of the mean time.

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Worked solution

Find the midpoint of each interval:
- $ 0 < t \le 10 $: midpoint is 5
- $ 10 < t \le 20 $: midpoint is 15
- $ 20 < t \le 30 $: midpoint is 25
- $ 30 < t \le 40 $: midpoint is 35

Calculate the sum of $ \text{frequency} \times \text{midpoint} $:
$ \sum f \cdot x = (5 \times 5) + (8 \times 15) + (12 \times 25) + (15 \times 35) $
$ \sum f \cdot x = 25 + 120 + 300 + 525 = 970 $

Calculate the estimate of the mean:
$ \text{Mean} = \frac{\sum f \cdot x}{\sum f} = \frac{970}{40} = 24.25 \text{ minutes} $

Marking scheme

M1 for correct midpoints identified: 5, 15, 25, 35.
M1 for finding the sum of products of midpoints and frequencies (970).
A1 for 24.25 or 24.3.

Question 21 · Short Answer

3.04 marks

Simplify $ \left( \frac{64y^6}{x^9} \right)^{-\frac{1}{3}} $.

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Worked solution

First, apply the reciprocal to eliminate the negative sign in the exponent:
$ \left( \frac{64y^6}{x^9} \right)^{-\frac{1}{3}} = \left( \frac{x^9}{64y^6} \right)^{\frac{1}{3}} $
Now, apply the power of $ \frac{1}{3} $ (cube root) to each component of the fraction:
$ \frac{(x^9)^{\frac{1}{3}}}{(64)^{\frac{1}{3}} (y^6)^{\frac{1}{3}}} = \frac{x^3}{4y^2} $

Marking scheme

M1 for reciprocating the fraction to address the negative index.
M1 for applying the cube root to at least two elements of the fraction correctly.
A1 for the correct final simplified expression.

Question 22 · Short Answer

3.04 marks

A bag contains 7 blue marbles and 5 red marbles. Two marbles are selected at random from the bag without replacement. Find the probability that both marbles are of the same colour.

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Worked solution

Total number of marbles is $ 7 + 5 = 12 $.
There are two mutually exclusive events for the same colour:
1. Both are blue:
$ P(\text{Blue, Blue}) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132} $

2. Both are red:
$ P(\text{Red, Red}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132} $

Add the probabilities:
$ P(\text{Same colour}) = \frac{42}{132} + \frac{20}{132} = \frac{62}{132} = \frac{31}{66} $

Marking scheme

M1 for calculating $P(\text{Blue, Blue})$ or $P(\text{Red, Red})$ with a denominator of 11 in the second fraction.
M1 for adding their two correct probability products.
A1 for $\frac{31}{66}$ or equivalent decimal (approx 0.470).

Question 23 · Short Answer

3.04 marks

Find the equation of the perpendicular bisector of the line segment joining the points $ A(2, -3) $ and $ B(8, 5) $. Give your answer in the form $ y = m x + c $.

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Worked solution

Find the midpoint of the line segment $ AB $:
$ M = \left( \frac{2 + 8}{2}, \frac{-3 + 5}{2} \right) = (5, 1) $
Find the gradient of line $ AB $:
$ m_{AB} = \frac{5 - (-3)}{8 - 2} = \frac{8}{6} = \frac{4}{3} $
The gradient of the perpendicular line is the negative reciprocal:
$ m_{\perp} = -\frac{3}{4} = -0.75 $
Use the point-slope form with midpoint $ (5, 1) $:
$ y - 1 = -0.75(x - 5) $
$ y - 1 = -0.75x + 3.75 $
$ y = -0.75x + 4.75 $

Marking scheme

M1 for finding the midpoint $(5, 1)$.
M1 for finding the perpendicular gradient $-\frac{3}{4}$ from the gradient of $AB$.
A1 for $ y = -0.75x + 4.75 $ or equivalent fraction form.

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Practice This Topic

Paper 4 (Extended)

Answer all questions. Show all necessary working. Give non-exact numerical answers correct to 3 significant figures.

10 Question · 130 marks

Question 1 · structured

13 marks

A company buys some tablets for a total cost of $7200.
If they had negotiated a discount of $40 per tablet, they would have been able to buy 6 more tablets for the same total cost of $7200.

(a) Let the original price of a tablet be $x. Show that $\frac{7200}{x - 40} - \frac{7200}{x} = 6$ simplifies to $x^2 - 40x - 48000 = 0$.
(b) Solve the equation $x^2 - 40x - 48000 = 0$ to find the original price of a tablet and the number of tablets originally bought.
(c) Solve the simultaneous equations:
$2x + y = 7$
$3x^2 - y^2 = 8$
Give your answers correct to 2 decimal places.

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Worked solution

(a) $\frac{7200}{x - 40} - \frac{7200}{x} = 6 \implies \frac{1200}{x-40} - \frac{1200}{x} = 1 \implies 1200x - 1200(x - 40) = x(x - 40) \implies 1200x - 1200x + 48000 = x^2 - 40x \implies x^2 - 40x - 48000 = 0$.
(b) $x^2 - 40x - 48000 = 0 \implies (x - 240)(x + 200) = 0 \implies x = 240$ (since price must be positive). Original price: $240. Number of tablets: $7200 / 240 = 30$.
(c) From $2x + y = 7$, $y = 7 - 2x$. Substitute into $3x^2 - y^2 = 8$: $3x^2 - (7 - 2x)^2 = 8 \implies 3x^2 - (49 - 28x + 4x^2) = 8 \implies -x^2 + 28x - 57 = 0 \implies x^2 - 28x + 57 = 0$. Using the quadratic formula, $x = \frac{28 \pm \sqrt{(-28)^2 - 4(1)(57)}}{2} = \frac{28 \pm \sqrt{556}}{2}$. Thus, $x \approx 25.79$ or $x \approx 2.21$. Corresponding $y$ values: for $x = 25.79$, $y = 7 - 2(25.79) = -44.58$; for $x = 2.21$, $y = 7 - 2(2.21) = 2.58$.

Marking scheme

(a) M1 for setting up the initial equation, M1 for clearing denominators, M1 for expansion, A1 for fully correct simplification to the target quadratic.
(b) M1 for factorization or formula use, A1 for finding $x = 240$ (rejecting $-200$), A1 for stating price is $240, A1 for number of tablets is 30.
(c) M1 for writing $y = 7 - 2x$ (or equivalent), M1 for substitution into quadratic, A1 for simplifying to $x^2 - 28x + 57 = 0$, A1 for one pair of correct values, A1 for second pair of correct values.

Question 2 · structured

13 marks

(a) Simplify completely: $\frac{2x^2 - 5x - 12}{3x^2 - 12x}$.
(b) Write as a single fraction in its simplest form: $\frac{3}{2a - 1} - \frac{2}{a + 3}$.
(c) Make $t$ the subject of the formula: $w = \frac{3t + 2}{5 - 2t}$.

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Worked solution

(a) Factorise the numerator: $2x^2 - 5x - 12 = (2x + 3)(x - 4)$. Factorise the denominator: $3x^2 - 12x = 3x(x - 4)$. Thus, $\frac{(2x + 3)(x - 4)}{3x(x - 4)} = \frac{2x + 3}{3x}$.
(b) Put over a common denominator: $\frac{3(a + 3) - 2(2a - 1)}{(2a - 1)(a + 3)} = \frac{3a + 9 - 4a + 2}{(2a - 1)(a + 3)} = \frac{11 - a}{(2a - 1)(a + 3)}$.
(c) Multiply by denominator: $w(5 - 2t) = 3t + 2 \implies 5w - 2wt = 3t + 2 \implies 5w - 2 = 3t + 2wt \implies 5w - 2 = t(3 + 2w) \implies t = \frac{5w - 2}{3 + 2w}$.

Marking scheme

(a) M1 for factorising the numerator, M1 for factorising the denominator, A2 for the final simplified fraction.
(b) M1 for common denominator, M1 for expanding the numerator with correct signs, A2 for the final simplified fraction.
(c) M1 for multiplying by denominator, M1 for expanding correctly, M1 for isolating terms containing $t$, M1 for factorising out $t$, A1 for final correct subject form.

Question 3 · structured

13 marks

The diagram shows a toy made from a hemisphere of radius $r$ cm surmounting a cylinder of radius $r$ cm and height $h$ cm.
The total volume of the toy is $792\pi$ $\text{cm}^3$.

(a) Given that the height of the cylinder is 3 times its radius, show that $r = 6$ cm.
(b) Calculate the total surface area of the toy. Give your answer correct to 3 significant figures.
(c) The toy is melted down and recast into a solid cone of radius 8 cm. Calculate the perpendicular height of this cone.

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Worked solution

(a) Total volume: $V = V_{\text{hemi}} + V_{\text{cyl}} = \frac{2}{3}\pi r^3 + \pi r^2 h$. Since $h = 3r$, we have $V = \frac{2}{3}\pi r^3 + 3\pi r^3 = \frac{11}{3}\pi r^3$. Given $V = 792\pi$, so $\frac{11}{3}\pi r^3 = 792\pi \implies r^3 = 792 \times \frac{3}{11} = 72 \times 3 = 216 \implies r = 6$ cm.
(b) Total surface area consists of the curved surface of the hemisphere, the curved surface of the cylinder, and the base of the cylinder: $A = 2\pi r^2 + 2\pi r h + \pi r^2 = 3\pi r^2 + 2\pi r (3r) = 9\pi r^2$. Since $r = 6$, $A = 9\pi (36) = 324\pi \approx 1017.88 \approx 1020$ $\text{cm}^2$.
(c) Volume of cone: $\frac{1}{3}\pi R^2 H = 792\pi \implies \frac{1}{3} (8)^2 H = 792 \implies \frac{64}{3} H = 792 \implies H = \frac{2376}{64} = 37.125$ cm.

Marking scheme

(a) M1 for volume formulas, M1 for substituting $h = 3r$, M1 for setting up $\frac{11}{3}\pi r^3 = 792\pi$, A1 for obtaining $r = 6$.
(b) M1 for surface area components, M1 for correctly identifying the base is included, M1 for substituting $r=6$ and $h=18$, A2 for 1020 or $324\pi$.
(c) M1 for cone volume formula, M1 for equating to $792\pi$, A2 for 37.1 or 37.125.

Question 4 · structured

13 marks

In the diagram, $A$, $B$, $C$ and $D$ are points on a circle, centre $O$.
$AT$ is a tangent to the circle at $A$.
Angle $DAT = 58^\circ$ and angle $ADC = 112^\circ$.

(a) Find, giving a geometric reason for each answer:
(i) angle $ABD$,
(ii) angle $ACD$,
(iii) angle $ABC$,
(iv) angle $AOD$.
(b) A regular polygon has an interior angle of $168^\circ$.
Calculate the number of sides of this polygon.

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Worked solution

(a)(i) Angle $ABD = 58^\circ$. Reason: Alternate segment theorem.
(ii) Angle $ACD = 58^\circ$. Reason: Angles in the same segment are equal.
(iii) Angle $ABC = 180^\circ - 112^\circ = 68^\circ$. Reason: Opposite angles of a cyclic quadrilateral sum to $180^\circ$.
(iv) Angle $AOD = 2 \times 58^\circ = 116^\circ$. Reason: Angle at centre is twice angle at circumference.
(b) Exterior angle $= 180^\circ - 168^\circ = 12^\circ$. Number of sides $n = \frac{360^\circ}{12^\circ} = 30$.

Marking scheme

(a)(i) B1 for $58^\circ$, B1 for 'alternate segment theorem'.
(ii) B1 for $58^\circ$, B1 for 'angles in same segment'.
(iii) B1 for $68^\circ$, B1 for 'opposite angles of cyclic quad sum to $180^\circ$'.
(iv) B1 for $116^\circ$, B1 for 'angle at centre is twice angle at circumference'.
(b) M1 for finding exterior angle $12^\circ$, M1 for $360 / 12$, A1 for 30.

Question 5 · structured

13 marks

A group of 120 students took a science test. The table shows the distribution of their scores.

| Score ($s$) | $0 < s \le 20$ | $20 < s \le 40$ | $40 < s \le 60$ | $60 < s \le 80$ | $80 < s \le 100$ |
| --- | --- | --- | --- | --- | --- |
| Frequency | 12 | 26 | 42 | 28 | 12 |

(a) Complete the cumulative frequency table.

| Score ($s$) | $\le 20$ | $\le 40$ | $\le 60$ | $\le 80$ | $\le 100$ |
| --- | --- | --- | --- | --- | --- |
| Cumulative Frequency | 12 | | | | |

(b) Draw a cumulative frequency diagram to represent this information.
(c) Use your diagram to find an estimate for:
(i) the median score,
(ii) the interquartile range,
(iii) the number of students who scored more than 75 marks.

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Worked solution

(a) Cumulative frequencies: $\le 20$ is 12, $\le 40$ is $12 + 26 = 38$, $\le 60$ is $38 + 42 = 80$, $\le 80$ is $80 + 28 = 108$, $\le 100$ is $108 + 12 = 120$.
(b) Plot points at upper limits: (20, 12), (40, 38), (60, 80), (80, 108), (100, 120) and join with a smooth curve.
(c)(i) Median is at CF = 60. From diagram, score $\approx 50.5$.
(ii) Lower quartile at CF = 30 is $\approx 33.8$. Upper quartile at CF = 90 is $\approx 67.1$. Interquartile range $\approx 67.1 - 33.8 = 33.3$.
(iii) At score 75, the CF is $\approx 101$. Number of students scoring more than 75 is $120 - 101 = 19$.

Marking scheme

(a) B2 for all cumulative frequencies correct (B1 for 3 correct).
(b) B1 for plotting points correctly, B1 for curve starting at origin, B1 for smooth curve through the points.
(c)(i) B2 for estimate 50.5 (accept 49 to 52).
(ii) M1 for finding LQ and UQ, M1 for subtracting their values, A1 for 33.3 (accept 31 to 37).
(iii) M1 for reading CF at 75, M1 for subtracting their reading from 120, A1 for 19 (accept 17 to 21).

Question 6 · structured

13 marks

A surveyor is measuring a triangular plot of land $PQR$.
$PQ = 120$ m and $PR = 150$ m.
The bearing of $Q$ from $P$ is $040^\circ$ and the bearing of $R$ from $P$ is $115^\circ$.

(a) Calculate the distance $QR$.
(b) Calculate the bearing of $R$ from $Q$.
(c) Calculate the area of the plot of land $PQR$.

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Worked solution

(a) The angle $QPR = 115^\circ - 40^\circ = 75^\circ$. By Cosine Rule: $QR^2 = 120^2 + 150^2 - 2(120)(150)\cos(75^\circ) = 14400 + 22500 - 36000(0.258819) = 36900 - 9317.49 = 27582.51 \implies QR \approx 166.08 \approx 166$ m.
(b) Using Sine Rule: $\frac{\sin(PQR)}{150} = \frac{\sin(75^\circ)}{166.08} \implies \sin(PQR) = \frac{150 \cdot \sin(75^\circ)}{166.08} \approx 0.8724 \implies \text{Angle } PQR \approx 60.74^\circ$. The back bearing of $P$ from $Q$ is $180^\circ + 40^\circ = 220^\circ$. The bearing of $R$ from $Q$ is $220^\circ - 60.74^\circ = 159.26^\circ \approx 159.3^\circ$.
(c) Area $= \frac{1}{2} \cdot PQ \cdot PR \cdot \sin(QPR) = \frac{1}{2} \cdot 120 \cdot 150 \cdot \sin(75^\circ) = 9000 \cdot 0.965926 = 8693.33 \approx 8690$ $\text{m}^2$.

Marking scheme

(a) M1 for identifying angle $QPR = 75^\circ$, M2 for correct Cosine Rule substitution, A1 for 166 m.
(b) M1 for Sine Rule setup, M1 for finding angle $PQR = 60.7^\circ$, M1 for back bearing $220^\circ$, M1 for bearing calculation, A1 for $159.3^\circ$.
(c) M1 for correct Area formula setup, M1 for correct values substituted, A2 for 8690.

Question 7 · structured

13 marks

The equation of a curve is $y = 2x^3 - 3x^2 - 12x + 5$.

(a) Find $\frac{dy}{dx}$.
(b) Find the coordinates of the two stationary points on the curve.
(c) Determine the nature of each stationary point. Show your working.
(d) Find the equation of the tangent to the curve at the point where $x = 1$. Give your answer in the form $y = mx + c$.

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Worked solution

(a) $\frac{dy}{dx} = 6x^2 - 6x - 12$.
(b) Setting $\frac{dy}{dx} = 0 \implies 6x^2 - 6x - 12 = 0 \implies x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0$. So $x = 2$ or $x = -1$. At $x = 2$, $y = 2(8) - 3(4) - 12(2) + 5 = -15$. At $x = -1$, $y = 2(-1) - 3(1) - 12(-1) + 5 = 12$. Stationary points are $(2, -15)$ and $(-1, 12)$.
(c) Second derivative: $\frac{d^2y}{dx^2} = 12x - 6$. For $x = 2$, $\frac{d^2y}{dx^2} = 18 > 0$ (minimum). For $x = -1$, $\frac{d^2y}{dx^2} = -18 < 0$ (maximum).
(d) At $x = 1$, $y = 2(1)^3 - 3(1)^2 - 12(1) + 5 = -8$. Gradient $m = 6(1)^2 - 6(1) - 12 = -12$. Tangent equation: $y - (-8) = -12(x - 1) \implies y + 8 = -12x + 12 \implies y = -12x + 4$.

Marking scheme

(a) B2 for $6x^2 - 6x - 12$ (B1 for any one term correct).
(b) M1 for setting their derivative to 0, M1 for solving quadratic to get $x=2, -1$, A1 for one point correct, A1 for second point correct.
(c) M1 for finding second derivative, M1 for substituting $x$ values, A1 for identifying correct nature.
(d) M1 for finding $y = -8$ at $x=1$, M1 for substituting $x=1$ into derivative to get slope $-12$, M1 for straight line equation method, A1 for $y = -12x + 4$.

Question 8 · structured

13 marks

Box A contains 5 red balls and 3 blue balls.
Box B contains 4 red balls and 6 blue balls.
A ball is chosen at random from Box A and placed into Box B.
A ball is then chosen at random from Box B.

(a) Find the probability that the ball chosen from Box A is red.
(b) Find the probability that both balls chosen are red.
(c) Find the probability that the ball chosen from Box B is blue.
(d) Given that the ball chosen from Box B is blue, find the probability that the ball chosen from Box A was red.

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Worked solution

(a) Probability $= \frac{5}{8}$.
(b) Probability both are red $= P(R_A \cap R_B) = P(R_A) \times P(R_B | R_A) = \frac{5}{8} \times \frac{5}{11} = \frac{25}{88}$.
(c) Two cases: (i) Red from A, then Blue from B: $\frac{5}{8} \times \frac{6}{11} = \frac{30}{88}$. (ii) Blue from A, then Blue from B: $\frac{3}{8} \times \frac{7}{11} = \frac{21}{88}$. Total probability $= \frac{30}{88} + \frac{21}{88} = \frac{51}{88}$.
(d) Conditional probability $P(R_A | B_B) = \frac{P(R_A \cap B_B)}{P(B_B)} = \frac{30/88}{51/88} = \frac{30}{51} = \frac{10}{17}$.

Marking scheme

(a) B1 for $\frac{5}{8}$.
(b) M1 for multiplying $P(R_A)$ by conditional probability, M1 for correct fractions used, A1 for $\frac{25}{88}$.
(c) M1 for first branch product, M1 for second branch product, M1 for adding products, A1 for $\frac{51}{88}$.
(d) M1 for conditional probability formula, M2 for substituting correct values, A2 for $\frac{10}{17}$.

Question 9 · structured

13 marks

Part (a)

A closed cylindrical container has radius
$x\text{ cm}$ and height $3x\text{ cm}$.

(i) Write down, in terms of $\pi$ and $x$, an expression for the total surface area of the cylinder.

(ii) The total surface area of the cylinder is $288\pi\text{ cm}^2$. Show that $x = 6$.

(iii) Calculate the volume of this cylinder, giving your answer correct to 3 significant figures.

Part (b)

A solid metal sphere of radius $4.5\text{ cm}$ is melted down and recast into a solid cone of height $12\text{ cm}$.

(i) Calculate the volume of the sphere, leaving your answer in terms of $\pi$.

(ii) Calculate the base radius of the cone.

(iii) Calculate the total surface area of the cone.

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Worked solution

Part (a)
(i) Total surface area of a closed cylinder $A = 2\pi r^2 + 2\pi r h$.
Substituting $r = x$ and $h = 3x$:
$A = 2\pi x^2 + 2\pi x (3x) = 2\pi x^2 + 6\pi x^2 = 8\pi x^2$.

(ii) Set $8\pi x^2 = 288\pi$.
$8x^2 = 288$
$x^2 = 36$
Since $x > 0$, $x = 6$.

(iii) Volume $V = \pi r^2 h = \pi \times 6^2 \times 18 = 648\pi \approx 2040\text{ cm}^3$.

Part (b)
(i) Volume of sphere $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \times 4.5^3 = 121.5\pi\text{ cm}^3$.

(ii) Volume of cone $V = \frac{1}{3} \pi R^2 H$.
$121.5\pi = \frac{1}{3} \pi R^2 \times 12$
$121.5 = 4 R^2$
$R^2 = 30.375 \implies R = \sqrt{30.375} \approx 5.51\text{ cm}$.

(iii) Slant height of cone $L = \sqrt{R^2 + H^2} = \sqrt{30.375 + 12^2} = \sqrt{174.375} \approx 13.205\text{ cm}$.
Total surface area $A = \pi R^2 + \pi R L = \pi \times 30.375 + \pi \times 5.511 \times 13.205 \approx 324\text{ cm}^2$.

Marking scheme

Part (a)
(i) M1 for $2\pi x^2 + 2\pi x(3x)$ or equivalent, A1 for $8\pi x^2$. [2]
(ii) M1 for equating $8\pi x^2 = 288\pi$, M1 for $x^2 = 36$, A1 for $x = 6$. [3]
(iii) M1 for $\pi \times 6^2 \times 18$, A1 for 2040 (accept 2030 to 2040 or $648\pi$). [2]

Part (b)
(i) M1 for $\frac{4}{3}\pi \times 4.5^3$, A1 for $121.5\pi$ (or $\frac{243}{2}\pi$). [2]
(ii) M1 for equating $\frac{1}{3}\pi R^2 \times 12 = 121.5\pi$, A1 for 5.51 (accept 5.51 to 5.52). [2]
(iii) M1 for finding slant height $L = \sqrt{R^2 + 12^2}$ (using their $R$), A1 for 324 (accept 323 to 325). [2]

Question 10 · structured

13 marks

A group of students rent a coach for a trip for a total cost of $480.

Let $x$ be the number of students in the group.

(a) Write down an expression, in terms of $x$, for the cost per student if everyone pays equally.

(b) On the day of the trip, 4 students are unable to go. As a result, the cost per student for the remaining students increases by $4. Write down an equation in terms of $x$ to show this information and show that it simplifies to $x^2 - 4x - 480 = 0$.

(d) Find the number of students who actually went on the trip.

(e) Find the amount that each of these students actually paid.

(f) A second group of $y$ students plans a trip. The coach company offers them a discount, so the total cost is $400. The cost per student for this second group is $4 less than the cost per student in the first group when everyone paid equally. Calculate the value of $y$.

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Worked solution

(a) Cost per student is $\frac{480}{x}$.

(b) With 4 fewer students, the number of students is $x - 4$ and the new cost per student is $\frac{480}{x-4}$.
The equation is: $\frac{480}{x-4} - \frac{480}{x} = 4$.
Divide throughout by 4:
$\frac{120}{x-4} - \frac{120}{x} = 1$.
Multiply by $x(x-4)$:
$120x - 120(x-4) = x(x-4)$
$120x - 120x + 480 = x^2 - 4x$
$x^2 - 4x - 480 = 0$.

(c) $(x - 24)(x + 20) = 0$
$x = 24$ or $x = -20$.

(d) Number of students who actually went = $24 - 4 = 20$.

(e) Amount actually paid = $\frac{480}{20} = \$24$.

(f) Cost per student in the first group initially = $\frac{480}{24} = \$20$.
Cost per student in the second group = $20 - 4 = \$16$.
Number of students $y = \frac{400}{16} = 25$.

Marking scheme

(a) B1 for $\frac{480}{x$}. [1]
(b) M1 for $\frac{480}{x-4} - \frac{480}{x} = 4$ or equivalent,
M1 for $120x - 120(x-4) = x(x-4)$ or equivalent algebraic step to clear fractions,
M1 for expanding brackets: $120x - 120x + 480 = x^2 - 4x$,
A1 for establishing the final form with no errors. [4]
(c) M1 for $(x - a)(x + b) = 0$ where $ab = -480$ or $a - b = 4$,
A1 for $(x-24)(x+20) = 0$,
A1 for $x = 24$ and $x = -20$. [3]
(d) B1 for 20. [1]
(e) M1 for $\frac{480}{\text{their } 20}$, A1 for 24. [2]
(f) M1 for $\frac{480}{24} - 4 = 16$, A1 for $y = 25$. [2]

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