An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 CCEA GCSE Mathematics 2210 paper. Not affiliated with or reproduced from CCEA.
Unit M4: Higher Tier Calculator Paper
Calculator allowed. Answer all 22 questions. Total of 100 marks. 2 hours.
1 mark for correctly evaluating both indices (\( 5^{-2}=\frac{1}{25} \), \( 2^{0}=1 \)); 1 mark for the correct final fraction \( \frac{1}{25} \). Max 2.
Calculate \( (3.2 \times 10^{4}) \times (5 \times 10^{3}) \), giving your answer in standard form.
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Worked solution
\( 3.2 \times 5 = 16 \) and \( 10^{4} \times 10^{3} = 10^{7} \), so the product is \( 16 \times 10^{7} = 1.6 \times 10^{8} \).
Marking scheme
1 mark for correctly multiplying 3.2 by 5 to give 16 (or equivalent correct working); 1 mark for the correct final answer in standard form, \( 1.6 \times 10^{8} \). Max 2.
A cuboid has length 8 cm, width 5 cm and height 4 cm. Calculate the total surface area of the cuboid.
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Worked solution
The cuboid has three pairs of faces with areas \( 8 \times 5 = 40 \), \( 8 \times 4 = 32 \) and \( 5 \times 4 = 20 \) cm\(^2\). Total surface area \( = 2(40+32+20) = 2 \times 92 = 184 \) cm\(^2\).
Marking scheme
1 mark for calculating the three distinct face areas (40, 32, 20 cm²); 1 mark for summing these to 92 cm²; 1 mark for correctly doubling to give the final answer of 184 cm². Max 3.
1 mark for correctly rounding each value to 1 significant figure (60, 4, 10); 1 mark for the correct method \( 60 \times 4 \div 10 \); 1 mark for the correct estimate of 24. Max 3.
Question 9 · Structured Multi-step Problem Solving (4-6 marks)
5 marks
A line passes through the points \( (2, 3) \) and \( (6, 11) \). Find the equation of the line in the form \( y = mx + c \).
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Worked solution
Gradient \( m = \dfrac{11-3}{6-2} = \dfrac{8}{4} = 2 \). Substituting \( (2,3) \): \( 3 = 2(2) + c \Rightarrow 3 = 4 + c \Rightarrow c = -1 \). So the equation is \( y = 2x - 1 \). Check with \( (6,11) \): \( 2(6)-1 = 11 \), correct.
Marking scheme
1 mark for the correct method to find gradient; 1 mark for gradient \( m=2 \); 1 mark for substituting a point into \( y=mx+c \); 1 mark for \( c=-1 \); 1 mark for the correct final equation \( y=2x-1 \). Max 5.
Question 10 · Structured Multi-step Problem Solving (4-6 marks)
5 marks
A line \( L_{1} \) has equation \( y = \dfrac{1}{2}x + 4 \). Line \( L_{2} \) is perpendicular to \( L_{1} \) and passes through the point \( (3, -1) \). Find the equation of \( L_{2} \) in the form \( y = mx + c \).
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Worked solution
Gradients of perpendicular lines multiply to give \( -1 \), so the gradient of \( L_{2} \) is \( -1 \div \frac{1}{2} = -2 \). Substituting \( (3,-1) \): \( -1 = -2(3) + c \Rightarrow -1 = -6 + c \Rightarrow c = 5 \). So \( L_{2} \) is \( y = -2x + 5 \). Check: \( -2(3)+5 = -1 \), correct.
Marking scheme
1 mark for using the perpendicular gradient rule \( m_{1}m_{2}=-1 \); 1 mark for gradient \( m=-2 \); 1 mark for substituting the point \( (3,-1) \); 1 mark for \( c=5 \); 1 mark for the correct final equation \( y=-2x+5 \). Max 5.
Question 11 · Structured Multi-step Problem Solving (4-6 marks)
5 marks
Solve the equation \( 3x^{2} + 5x - 8 = 0 \), giving your answers to 2 decimal places where appropriate.
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Worked solution
Using the quadratic formula with \( a=3, b=5, c=-8 \): \( x = \dfrac{-5 \pm \sqrt{5^{2}-4(3)(-8)}}{2(3)} = \dfrac{-5 \pm \sqrt{25+96}}{6} = \dfrac{-5 \pm \sqrt{121}}{6} = \dfrac{-5 \pm 11}{6} \). So \( x = \dfrac{6}{6} = 1 \) or \( x = \dfrac{-16}{6} = -2.67 \) (2 d.p.). (Equivalently, \( 3x^{2}+5x-8=(3x+8)(x-1) \).)
Marking scheme
1 mark for correct substitution into the quadratic formula (or attempt at factorisation); 1 mark for correct discriminant \( \sqrt{121}=11 \); 1 mark for \( x=1 \); 1 mark for \( x=-2.67 \) (2 d.p.); 1 mark for both solutions clearly stated. Max 5.
Question 12 · Structured Multi-step Problem Solving (4-6 marks)
5 marks
Solve the equation \( \dfrac{2}{x+2} + \dfrac{3}{2x-1} = 1 \).
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Worked solution
Multiplying every term by \( (x+2)(2x-1) \): \( 2(2x-1) + 3(x+2) = (x+2)(2x-1) \). Expanding the left side: \( 4x-2+3x+6 = 7x+4 \). Expanding the right side: \( 2x^{2}+3x-2 \). So \( 7x+4 = 2x^{2}+3x-2 \), giving \( 0 = 2x^{2}-4x-6 \), i.e. \( x^{2}-2x-3=0 \), which factorises as \( (x-3)(x+1)=0 \). So \( x=3 \) or \( x=-1 \). Checking \( x=3 \): \( \frac{2}{5}+\frac{3}{5}=1 \). Checking \( x=-1 \): \( \frac{2}{1}+\frac{3}{-3}=2-1=1 \). Both check.
Marking scheme
1 mark for correctly multiplying through by the common denominator \( (x+2)(2x-1) \); 1 mark for correctly expanding the left-hand side to \( 7x+4 \); 1 mark for correctly expanding the right-hand side to \( 2x^{2}+3x-2 \); 1 mark for forming and factorising \( x^{2}-2x-3=(x-3)(x+1) \); 1 mark for both solutions \( x=3 \) and \( x=-1 \). Max 5.
Question 13 · Structured Multi-step Problem Solving (4-6 marks)
5 marks
From a point P on horizontal ground, the angle of elevation to the top of a vertical tower is 34°. P is 45 m from the base of the tower. Calculate the height of the tower, giving your answer to 3 significant figures.
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Worked solution
Let the height be \( h \). Using \( \tan(34^{\circ}) = \dfrac{h}{45} \), so \( h = 45 \times \tan(34^{\circ}) = 45 \times 0.6745\ldots = 30.35\ldots \), which rounds to 30.4 m (3 s.f.).
Marking scheme
1 mark for identifying \( \tan \) as the correct trigonometric ratio (opposite/adjacent); 1 mark for the correct equation \( h = 45\tan(34^{\circ}) \); 1 mark for correct method/substitution; 1 mark for the unrounded value \( \approx 30.35 \); 1 mark for the final answer 30.4 m to 3 s.f. Max 5.
Question 14 · Structured Multi-step Problem Solving (4-6 marks)
5 marks
Points A, B, C and D lie on a circle, centre O. (a) Angle ACB is an angle at the circumference and equals 52°. Calculate the size of angle AOB, the angle at the centre standing on the same arc AB, giving a reason. (b) ABCD is a cyclic quadrilateral. Angle ABC = 108°. Calculate the size of angle ADC, giving a reason.
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Worked solution
(a) The angle at the centre of a circle is twice the angle at the circumference when both stand on the same arc. So angle AOB \( = 2 \times 52\)° \( = 104\)°. (b) Opposite angles in a cyclic quadrilateral sum to 180°. So angle ADC \( = 180 - 108 = 72\)°.
Marking scheme
(a) 1 mark for stating the correct circle theorem (angle at centre is twice angle at circumference); 1 mark for the correct answer 104°. (b) 1 mark for stating the correct circle theorem (opposite angles in a cyclic quadrilateral sum to 180°); 1 mark for the correct method \( 180-108 \); 1 mark for the correct answer 72°. Max 5.
Question 15 · Structured Multi-step Problem Solving (4-6 marks)
5 marks
The table shows the time, t minutes, spent on homework by 75 students.
Time (t minutes) | Frequency 0 ≤ t < 10 | 8 10 ≤ t < 20 | 22 20 ≤ t < 40 | 30 40 ≤ t < 70 | 15
Calculate the frequency density for each class interval, and state which class interval would be drawn as the tallest bar on a histogram of this data.
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Worked solution
Frequency density = frequency ÷ class width. For 0–10 (width 10): \( 8 \div 10 = 0.8 \). For 10–20 (width 10): \( 22 \div 10 = 2.2 \). For 20–40 (width 20): \( 30 \div 20 = 1.5 \). For 40–70 (width 30): \( 15 \div 30 = 0.5 \). On a histogram, bar height equals frequency density, so the tallest bar is the class with the highest frequency density, which is 10 ≤ t < 20 (frequency density 2.2).
Marking scheme
1 mark for each correctly calculated frequency density (0.8, 2.2, 1.5, 0.5), max 4; 1 mark for correctly identifying the 10 ≤ t < 20 class as the tallest bar. Max 5.
Question 16 · Structured Multi-step Problem Solving (4-6 marks)
4 marks
A box contains a mixture of red and blue counters. The probability that a counter picked at random is red is 0.3. Two counters are picked at random, one after another, with the first counter replaced before the second is picked. (a) Calculate the probability that both counters are red. (b) Calculate the probability that the two counters picked are different colours.
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Worked solution
Since the counter is replaced, the two picks are independent, each with P(red) = 0.3 and P(blue) = 0.7. (a) P(both red) \( = 0.3 \times 0.3 = 0.09 \). (b) P(different colours) = P(red then blue) + P(blue then red) \( = (0.3 \times 0.7) + (0.7 \times 0.3) = 0.21 + 0.21 = 0.42 \).
Marking scheme
(a) 1 mark for the method \( 0.3 \times 0.3 \); 1 mark for the correct answer 0.09. (b) 1 mark for the method \( (0.3\times0.7)+(0.7\times0.3) \); 1 mark for the correct answer 0.42. Max 4.
Question 17 · Structured Multi-step Problem Solving (4-6 marks)
4 marks
A sector of a circle has radius 12 cm and angle 75°. Calculate the arc length and the area of the sector, giving your answers to 3 significant figures.
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Worked solution
Arc length \( = \dfrac{75}{360} \times 2\pi(12) = \dfrac{75}{360} \times 24\pi = 15.70\ldots \), which rounds to 15.7 cm. Sector area \( = \dfrac{75}{360} \times \pi (12)^{2} = \dfrac{75}{360} \times 144\pi = 94.24\ldots \), which rounds to 94.2 cm\(^2\).
Marking scheme
1 mark for the correct arc length method \( \frac{75}{360}\times 2\pi(12) \); 1 mark for arc length 15.7 cm; 1 mark for the correct sector area method \( \frac{75}{360}\times\pi(12)^{2} \); 1 mark for sector area 94.2 cm\(^2\). Max 4.
Question 18 · Structured Multi-step Problem Solving (4-6 marks)
4 marks
Solve the inequality \( 3x - 7 \le 11 \), and describe how the solution would be shown on a number line.
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Worked solution
Adding 7 to both sides: \( 3x \le 18 \). Dividing both sides by 3: \( x \le 6 \). On a number line, this would be shown with a filled (closed) circle at 6, with an arrow extending to the left to represent all values less than or equal to 6.
Marking scheme
1 mark for the correct method of adding 7 to both sides (\( 3x \le 18 \)); 1 mark for dividing by 3; 1 mark for the correct final answer \( x \le 6 \); 1 mark for a correct description of the number line (filled circle at 6, arrow pointing left). Max 4.
Question 19 · Structured Multi-step Problem Solving (4-6 marks)
4 marks
Factorise fully \( 6x^{2} + 7x - 3 \).
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Worked solution
We need two numbers that multiply to give \( 6 \times (-3) = -18 \) and add to give 7: these are 9 and \( -2 \). Rewrite the middle term: \( 6x^{2}+9x-2x-3 \). Factorise by grouping: \( 3x(2x+3) - 1(2x+3) = (3x-1)(2x+3) \). Checking by expanding: \( (3x-1)(2x+3) = 6x^{2}+9x-2x-3 = 6x^{2}+7x-3 \), which matches.
Marking scheme
1 mark for finding the two numbers \( 9 \) and \( -2 \) (product \( -18 \), sum \( 7 \)); 1 mark for correctly splitting the middle term; 1 mark for correct factorisation by grouping; 1 mark for the final answer \( (3x-1)(2x+3) \). Max 4.
A large solid cone has base radius 9 cm and height 24 cm. A smaller, similar cone is removed from the top, leaving a frustum of height 16 cm. (a) Show that the removed small cone has base radius 3 cm. (b) Calculate the volume of the large cone, giving your answer as a multiple of \( \pi \). (c) Calculate the volume of the removed small cone, giving your answer as a multiple of \( \pi \). (d) Hence calculate the volume of the frustum, giving your answer to 3 significant figures.
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Worked solution
(a) The small cone's height is \( 24-16=8 \) cm. Since the small cone is similar to the large cone, its radius scales in the same ratio as its height: \( r = 9 \times \dfrac{8}{24} = 9 \times \dfrac{1}{3} = 3 \) cm, as required. (b) Volume of large cone \( = \dfrac{1}{3}\pi (9)^{2}(24) = \dfrac{1}{3}\pi(1944) = 648\pi \) cm\(^3\). (c) Volume of small cone \( = \dfrac{1}{3}\pi(3)^{2}(8) = \dfrac{1}{3}\pi(72) = 24\pi \) cm\(^3\). (d) Volume of frustum \( = 648\pi - 24\pi = 624\pi = 1960.35\ldots \), which rounds to 1960 cm\(^3\) (3 s.f.).
Marking scheme
(a) 1 mark for finding the small cone's height (8 cm); 1 mark for correctly using the similar-triangle ratio to show radius = 3 cm. (b) 1 mark for correct substitution into the cone volume formula; 1 mark for correct arithmetic; 1 mark for \( 648\pi \) cm\(^3\). (c) 1 mark for correct substitution/method; 1 mark for \( 24\pi \) cm\(^3\). (d) 1 mark for the method (large volume − small volume); 1 mark for the final answer 1960 cm\(^3\) (3 s.f.), following through from (b) and (c). Max 9.
Consider the equation \( \dfrac{3}{x-1} - \dfrac{2}{x+1} = 1 \). (a) Show that this equation can be written as \( x^{2} - x - 6 = 0 \). (b) Hence solve the equation, giving both values of x. (c) Verify that both of your solutions satisfy the original equation.
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Worked solution
(a) Multiplying every term by \( (x-1)(x+1) \): \( 3(x+1) - 2(x-1) = (x-1)(x+1) \). Expanding the left side: \( 3x+3-2x+2 = x+5 \). Expanding the right side: \( x^{2}-1 \). So \( x+5 = x^{2}-1 \), which rearranges to \( x^{2}-x-6=0 \), as required. (b) Factorising: \( x^{2}-x-6 = (x-3)(x+2) = 0 \), so \( x=3 \) or \( x=-2 \). (c) Checking \( x=3 \): \( \dfrac{3}{2} - \dfrac{2}{4} = 1.5 - 0.5 = 1 \), correct. Checking \( x=-2 \): \( \dfrac{3}{-3} - \dfrac{2}{-1} = -1 - (-2) = 1 \), correct. Both solutions satisfy the original equation.
Marking scheme
(a) 1 mark for multiplying through by \( (x-1)(x+1) \); 1 mark for correctly expanding the left side to \( x+5 \); 1 mark for correctly expanding the right side to \( x^{2}-1 \); 1 mark for correctly rearranging to \( x^{2}-x-6=0 \). (b) 1 mark for correct factorisation \( (x-3)(x+2) \); 1 mark for \( x=3 \); 1 mark for \( x=-2 \). (c) 1 mark for correctly verifying \( x=3 \); 1 mark for correctly verifying \( x=-2 \); 1 mark for a clear conclusion that both solutions are valid. Max 10.
A school has 900 students split across three year groups: Year 9 has 300 students, Year 10 has 350 students, and Year 11 has 250 students. A stratified sample of 90 students is to be selected, proportional to year group size, to survey time spent on homework. (a) Calculate the number of students that should be sampled from each year group. The times (in minutes) that the sampled students spent on homework last week are grouped in the table below.
Time (t minutes) | Frequency 0 ≤ t < 10 | 8 10 ≤ t < 20 | 22 20 ≤ t < 40 | 30 40 ≤ t < 70 | 15
(b) Complete a frequency density table for this data (frequency density = frequency ÷ class width). (c) Assuming students are spread evenly within the 10 ≤ t < 20 class, estimate how many of these students spent between 15 and 20 minutes on their homework.
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Worked solution
(a) Sampling fraction \( = \dfrac{90}{900} = \dfrac{1}{10} \). Year 9: \( 300 \times \frac{1}{10} = 30 \). Year 10: \( 350 \times \frac{1}{10} = 35 \). Year 11: \( 250 \times \frac{1}{10} = 25 \). Check: \( 30+35+25=90 \), correct. (b) Frequency density = frequency ÷ class width: 0–10 (width 10): \( 8 \div 10 = 0.8 \); 10–20 (width 10): \( 22\div10=2.2 \); 20–40 (width 20): \( 30\div20=1.5 \); 40–70 (width 30): \( 15\div30=0.5 \). (c) The 10 ≤ t < 20 class has frequency density 2.2. The interval 15–20 has width 5, so, assuming an even spread within the class, the estimated frequency \( = 2.2 \times 5 = 11 \) students.
Marking scheme
(a) 1 mark for finding the sampling fraction \( \frac{1}{10} \); 1 mark for correct Year 9 and Year 10 values; 1 mark for correct Year 11 value with totals checking to 90. (b) 1 mark for each pair of correct frequency densities, up to 4 marks total for all four correct values (0.8, 2.2, 1.5, 0.5). (c) 1 mark for correct method (frequency density \( \times \) sub-interval width); 1 mark for correct working \( 2.2 \times 5 \); 1 mark for the final answer of 11 students. Max 10.
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No calculator permitted. Answer all 14 questions. Total of 50 marks. 1 hour 15 minutes. Show your working out clearly. A solution by trial and improvement will not be accepted.
14 Question · 50 marks
Question 1 · Short Recall & Exact Calculation (1-2 marks)
2 marks
Simplify \( \sqrt{50} \), giving your answer in the form \( a\sqrt{2} \), where a is an integer.
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1 mark for correctly splitting 50 into \( 25 \times 2 \); 1 mark for the final answer \( 5\sqrt{2} \). Max 2.
Question 2 · Short Recall & Exact Calculation (1-2 marks)
2 marks
Change the recurring decimal \( 0.\dot{7} \) (\( 0.7777\ldots \)) to a fraction in its simplest form.
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Worked solution
Let \( x = 0.7777\ldots \). Then \( 10x = 7.7777\ldots \). Subtracting: \( 10x - x = 7.7777\ldots - 0.7777\ldots \), so \( 9x = 7 \), giving \( x = \dfrac{7}{9} \).
Marking scheme
1 mark for correct method (setting up \( x=0.777\ldots \), multiplying by 10 and subtracting); 1 mark for the correct final fraction \( \frac{7}{9} \). Max 2.
Question 3 · Short Recall & Exact Calculation (1-2 marks)
2 marks
Factorise \( x^{2} - 5x + 6 \).
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Worked solution
We need two numbers that multiply to give 6 and add to give \( -5 \): these are \( -2 \) and \( -3 \). So \( x^{2}-5x+6 = (x-2)(x-3) \). Checking: \( (x-2)(x-3) = x^{2}-3x-2x+6 = x^{2}-5x+6 \), which matches.
Marking scheme
1 mark for identifying the correct pair of numbers ( \( -2 \) and \( -3 \) ); 1 mark for the correct final factorisation \( (x-2)(x-3) \). Max 2.
Question 4 · Short Recall & Exact Calculation (1-2 marks)
2 marks
Calculate the size of each exterior angle of a regular nonagon (a 9-sided polygon).
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Worked solution
The exterior angles of any polygon sum to 360°. For a regular nonagon: \( 360 \div 9 = 40\)°.
Marking scheme
1 mark for the correct method \( 360 \div 9 \); 1 mark for the correct answer 40°. Max 2.
Question 5 · Short Recall & Exact Calculation (1-2 marks)
2 marks
The first four terms of a linear sequence are 5, 9, 13, 17. Find an expression for the nth term of the sequence.
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Worked solution
The common difference is \( 9-5=4 \), so the sequence has the form \( 4n+c \). When \( n=1 \), the term is 5, so \( 4(1)+c=5 \), giving \( c=1 \). So the nth term is \( 4n+1 \). Check: \( n=2 \): \( 4(2)+1=9 \), correct.
Marking scheme
1 mark for correctly identifying the common difference as 4; 1 mark for the correct final expression \( 4n+1 \). Max 2.
Question 6 · Short Recall & Exact Calculation (1-2 marks)
2 marks
Make x the subject of the formula \( y = 3x - 7 \).
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Worked solution
Starting with \( y=3x-7 \), add 7 to both sides: \( y+7=3x \). Divide both sides by 3: \( x = \dfrac{y+7}{3} \).
Marking scheme
1 mark for correctly adding 7 to both sides; 1 mark for correctly dividing by 3 to give \( x=\frac{y+7}{3} \). Max 2.
Question 7 · Short Recall & Exact Calculation (1-2 marks)
2 marks
Calculate the value of \( 2^{-3} \), giving your answer as a fraction.
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Worked solution
\( 2^{-3} = \dfrac{1}{2^{3}} = \dfrac{1}{8} \).
Marking scheme
1 mark for correctly applying the negative index rule \( 2^{-3}=\frac{1}{2^{3}} \); 1 mark for the correct final answer \( \frac{1}{8} \). Max 2.
Solve the simultaneous equations \( 3x + 2y = 16 \) and \( x - y = 2 \), giving the values of x and y.
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Worked solution
From the second equation, \( x = y+2 \). Substituting into the first equation: \( 3(y+2)+2y=16 \Rightarrow 3y+6+2y=16 \Rightarrow 5y=10 \Rightarrow y=2 \). Then \( x=y+2=4 \). Check: \( 3(4)+2(2)=12+4=16 \), correct; \( 4-2=2 \), correct.
Marking scheme
1 mark for rearranging one equation to make x or y the subject; 1 mark for correct substitution into the other equation; 1 mark for correctly solving to \( y=2 \); 1 mark for correctly finding \( x=4 \); 1 mark for both values checked/stated clearly. Max 5.
Simplify \( \dfrac{5}{3\sqrt{2}} \) fully, rationalising the denominator.
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Worked solution
Multiply the numerator and denominator by \( \sqrt{2} \): \( \dfrac{5}{3\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{5\sqrt{2}}{3 \times 2} = \dfrac{5\sqrt{2}}{6} \).
Marking scheme
1 mark for the correct method of multiplying numerator and denominator by \( \sqrt{2} \); 1 mark for the correct numerator \( 5\sqrt{2} \); 1 mark for the correct denominator 6; 1 mark for expressing the final fraction correctly; 1 mark for the fraction being fully simplified, \( \frac{5\sqrt{2}}{6} \). Max 5.
Triangle A has vertices \( (1,1) \), \( (3,1) \) and \( (1,4) \). Triangle B has vertices \( (1,-1) \), \( (1,-3) \) and \( (4,-1) \). Describe fully the single transformation that maps triangle A onto triangle B.
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Worked solution
Testing the vertex \( (1,1) \to (1,-1) \), \( (3,1) \to (1,-3) \), and \( (1,4) \to (4,-1) \): each point \( (x,y) \) maps to \( (y,-x) \), which is the rule for a rotation of 90° clockwise about the origin. Checking all three vertices confirms this rule holds consistently, so the single transformation is a rotation of 90° clockwise, centre \( (0,0) \).
Marking scheme
1 mark for correctly identifying the transformation type as a rotation; 2 marks for the correct angle and direction (90° clockwise, or equivalently 270° anticlockwise); 2 marks for correctly identifying the centre of rotation as the origin (0, 0), with evidence from checking at least one vertex mapping. Max 5. All three elements (type, angle/direction, centre) are required for full marks — describing fully requires all three.
y is directly proportional to x. When \( x = 5 \), \( y = 8 \). Find the value of y when \( x = 25 \).
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Worked solution
Since \( y \propto x \), \( y = kx \) for some constant k. Using \( x=5, y=8 \): \( 8 = k(5) \Rightarrow k = \dfrac{8}{5} = 1.6 \). When \( x=25 \): \( y = 1.6 \times 25 = 40 \).
Marking scheme
1 mark for setting up \( y=kx \); 1 mark for finding \( k=1.6 \) (or equivalent, e.g. \( \frac{8}{5} \)); 1 mark for correct substitution of \( x=25 \); 1 mark for the correct final answer \( y=40 \). Max 4.
The cumulative frequency curve for the ages, in complete years, of 60 people attending a coding club shows that at a cumulative frequency of 15, the age is 12 years, and at a cumulative frequency of 45, the age is 27 years. Calculate the interquartile range of the ages.
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Worked solution
With \( n=60 \) people, the lower quartile is at cumulative frequency \( \frac{n}{4}=15 \), giving \( Q_{1}=12 \) years. The upper quartile is at cumulative frequency \( \frac{3n}{4}=45 \), giving \( Q_{3}=27 \) years. Interquartile range \( = Q_{3}-Q_{1} = 27-12 = 15 \) years.
Marking scheme
1 mark for correctly identifying \( Q_{1}=12 \) years; 1 mark for correctly identifying \( Q_{3}=27 \) years; 1 mark for the correct method \( Q_{3}-Q_{1} \); 1 mark for the correct final answer of 15 years. Max 4.
Triangle ABC is similar to triangle DEF, with AB corresponding to DE. \( AB = 6 \) cm, \( BC = 9 \) cm, and \( DE = 15 \) cm. Calculate the length of EF.
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Worked solution
Scale factor from ABC to DEF \( = DE \div AB = 15 \div 6 = 2.5 \). Since BC corresponds to EF, \( EF = BC \times 2.5 = 9 \times 2.5 = 22.5 \) cm.
Marking scheme
1 mark for the correct method to find the scale factor (\( 15 \div 6 \)); 1 mark for scale factor = 2.5; 1 mark for the correct method \( 9 \times 2.5 \); 1 mark for the final answer 22.5 cm. Max 4.
Question 14 · Multi-part Circle / Tangent Problem (9 marks)
9 marks
A circle has centre O, the origin, and equation \( x^{2} + y^{2} = 25 \). The point P has coordinates \( (3, 4) \). (a) Show that P lies on the circle. (b) Find the gradient of the radius OP. (c) Explain why the gradient of the tangent to the circle at P is \( -\dfrac{3}{4} \). (d) Find the equation of the tangent to the circle at P, giving your answer in the form \( y = mx + c \).
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Worked solution
(a) Substituting \( x=3, y=4 \) into \( x^{2}+y^{2} \): \( 3^{2}+4^{2} = 9+16 = 25 \), which equals the right-hand side of the circle's equation, so P lies on the circle. (b) Gradient of OP \( = \dfrac{4-0}{3-0} = \dfrac{4}{3} \). (c) A tangent to a circle is always perpendicular to the radius at the point of contact. Since gradients of perpendicular lines multiply to give \( -1 \), the tangent's gradient is \( -1 \div \dfrac{4}{3} = -\dfrac{3}{4} \). (d) Using \( y-y_{1}=m(x-x_{1}) \) with \( m=-\frac{3}{4} \) and point \( (3,4) \): \( y-4 = -\dfrac{3}{4}(x-3) \Rightarrow y = -\dfrac{3}{4}x + \dfrac{9}{4} + 4 = -\dfrac{3}{4}x + \dfrac{25}{4} \).
Marking scheme
(a) 1 mark for correct substitution; 1 mark for showing \( 9+16=25 \) matches the circle's equation. (b) 1 mark for the correct method \( \frac{4-0}{3-0} \); 1 mark for gradient \( = \frac{4}{3} \). (c) 1 mark for stating that the tangent is perpendicular to the radius at the point of contact; 1 mark for correctly applying \( m_{1}m_{2}=-1 \) to obtain \( -\frac{3}{4} \). (d) 1 mark for correct substitution into \( y-y_{1}=m(x-x_{1}) \); 1 mark for correctly rearranging to \( y=-\frac{3}{4}x+\frac{25}{4} \). Max 9.
Unit M8 Paper 2: Higher Tier Calculator Paper
Calculator allowed. Answer all 14 questions. Total of 50 marks. 1 hour 15 minutes. Show your working out clearly.
14 Question · 50 marks
Question 1 · Short Calculation / Concept (1-2 marks)
2 marks
Find the mean and the range of the data set: 4, 7, 7, 9, 13.
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Worked solution
Mean \( = \dfrac{4+7+7+9+13}{5} = \dfrac{40}{5} = 8 \). Range \( = 13 - 4 = 9 \).
Marking scheme
1 mark for the correct mean of 8; 1 mark for the correct range of 9. Max 2.
Question 2 · Short Calculation / Concept (1-2 marks)
2 marks
A spinner is spun 200 times and lands on red 46 times. Use this data to estimate the probability that the spinner lands on red.
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Worked solution
Estimated probability \( = \dfrac{\text{number of reds}}{\text{total spins}} = \dfrac{46}{200} = 0.23 \).
Marking scheme
1 mark for the correct method \( \frac{46}{200} \); 1 mark for the correct final answer 0.23. Max 2.
Question 3 · Short Calculation / Concept (1-2 marks)
2 marks
A fair six-sided dice is rolled once. Calculate the probability of rolling a number greater than 4.
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Worked solution
The numbers greater than 4 on a dice are 5 and 6, so there are 2 favourable outcomes out of 6 possible outcomes. \( P(\text{greater than } 4) = \dfrac{2}{6} = \dfrac{1}{3} \).
Marking scheme
1 mark for identifying 2 favourable outcomes out of 6; 1 mark for the correct simplified probability \( \frac{1}{3} \). Max 2.
Question 4 · Short Calculation / Concept (1-2 marks)
2 marks
The length of a pencil is given as 12.6 cm, correct to 1 decimal place. State the upper bound of the actual length of the pencil.
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Worked solution
A measurement given as 12.6 cm to 1 decimal place could have been rounded from any value from 12.55 cm up to (but not including) 12.65 cm. The upper bound is 12.65 cm.
Marking scheme
1 mark for correct method (half of 0.1 added to 12.6); 1 mark for the correct upper bound of 12.65 cm. Max 2.
Question 5 · Short Calculation / Concept (1-2 marks)
2 marks
A washing machine costs £84 before VAT is added. VAT is charged at 20%. Calculate the total price including VAT.
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A cylindrical water tank has radius 6 m and height 15 m. Calculate the volume of the tank, giving your answer to the nearest whole number of cubic metres.
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Worked solution
Volume of a cylinder \( = \pi r^{2} h = \pi (6)^{2}(15) = \pi (36)(15) = 540\pi = 1696.46\ldots \), which rounds to 1696 m\(^3\) (nearest whole number).
Marking scheme
1 mark for the correct formula \( \pi r^{2}h \); 1 mark for correct substitution; 1 mark for the unrounded value \( 540\pi \approx 1696.46 \); 1 mark for the correct final answer 1696 m\(^3\). Max 4.
A sector of a circle has radius 8.4 cm and angle 140°. Calculate the area of the sector and the arc length, giving your answers to 3 significant figures.
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Worked solution
Sector area \( = \dfrac{140}{360} \times \pi (8.4)^{2} = \dfrac{140}{360} \times 70.56\pi = 86.20\ldots \), which rounds to 86.2 cm\(^2\). Arc length \( = \dfrac{140}{360} \times 2\pi(8.4) = \dfrac{140}{360} \times 16.8\pi = 20.52\ldots \), which rounds to 20.5 cm.
Marking scheme
1 mark for the correct sector area method; 1 mark for sector area 86.2 cm\(^2\); 1 mark for the correct arc length method; 1 mark for arc length 20.5 cm. Max 4.
Two similar solids have corresponding lengths in the ratio 3 : 5. The smaller solid has a volume of 54 cm\(^3\). Calculate the volume of the larger solid.
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Worked solution
For similar solids, the ratio of volumes is the cube of the ratio of corresponding lengths. Length ratio \( 3:5 \) gives volume ratio \( 3^{3}:5^{3} = 27:125 \). Volume of larger solid \( = 54 \times \dfrac{125}{27} = 250 \) cm\(^3\).
Marking scheme
1 mark for correctly cubing the length ratio to get the volume ratio 27:125; 1 mark for the correct method \( 54 \times \frac{125}{27} \); 1 mark for correct working; 1 mark for the final answer 250 cm\(^3\). Max 4.
y is inversely proportional to x. When \( x = 5 \), \( y = 8 \). Find the value of y when \( x = 10 \).
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Worked solution
Since \( y \propto \frac{1}{x} \), \( y = \dfrac{k}{x} \) for some constant k. Using \( x=5, y=8 \): \( 8 = \dfrac{k}{5} \Rightarrow k = 40 \). When \( x=10 \): \( y = \dfrac{40}{10} = 4 \).
Marking scheme
1 mark for setting up \( y=\frac{k}{x} \); 1 mark for finding \( k=40 \); 1 mark for correct substitution of \( x=10 \); 1 mark for the correct final answer \( y=4 \). Max 4.
A plumber charges a fixed call-out fee of £45 plus £35 for each hour worked. For a particular job, the total bill was £185. Form and solve an equation to find the number of hours the plumber worked.
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Worked solution
Let h be the number of hours worked. The total cost is \( 45 + 35h = 185 \). Subtracting 45 from both sides: \( 35h = 140 \). Dividing both sides by 35: \( h = 4 \).
Marking scheme
1 mark for correctly forming the equation \( 45+35h=185 \); 1 mark for correctly subtracting 45 (\( 35h=140 \)); 1 mark for correctly dividing by 35; 1 mark for the final answer of 4 hours. Max 4.
A car hire company charges a total cost, £y, that increases at a constant rate of £0.45 per mile driven, x. A customer who drives 100 miles is charged a total of £70. (a) Find the fixed charge (the value of y when x = 0). (b) Calculate the total cost for a journey of 220 miles.
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Worked solution
The relationship is \( y = 0.45x + c \), where c is the fixed charge. (a) Using \( x=100, y=70 \): \( 70 = 0.45(100) + c = 45 + c \Rightarrow c = 25 \), so the fixed charge is £25. (b) Total cost for 220 miles \( = 0.45(220) + 25 = 99 + 25 = £124 \).
Marking scheme
(a) 1 mark for correct substitution into \( y=0.45x+c \); 1 mark for the correct fixed charge £25. (b) 1 mark for correct substitution of \( x=220 \); 1 mark for the correct final answer £124. Max 4.
A bag contains 4 red counters and 6 blue counters. A counter is picked at random, its colour noted, and then it is replaced before a second counter is picked at random. Calculate the probability that exactly one of the two counters picked is red.
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Worked solution
P(red) \( = \frac{4}{10} = 0.4 \) and P(blue) \( = \frac{6}{10} = 0.6 \) on each pick, since the counter is replaced. P(exactly one red) = P(red then blue) + P(blue then red) \( = (0.4 \times 0.6) + (0.6 \times 0.4) = 0.24 + 0.24 = 0.48 \).
Marking scheme
1 mark for correct individual probabilities P(red) = 0.4 and P(blue) = 0.6; 1 mark for identifying the two orders (red-then-blue and blue-then-red); 1 mark for correct calculation of each combined probability (0.24 and 0.24); 1 mark for the correct final answer 0.48. Max 4.
Question 13 · Advanced 3D Geometry / Dependent Probability (6 marks)
6 marks
A cuboid has length 8 cm, width 6 cm and height 5 cm. (a) Calculate the length of the diagonal from one bottom corner of the cuboid to the opposite top corner, giving your answer to 3 significant figures. (b) Calculate the angle this diagonal makes with the base of the cuboid, giving your answer to 3 significant figures.
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Worked solution
(a) First find the diagonal of the base rectangle: \( \sqrt{8^{2}+6^{2}} = \sqrt{64+36} = \sqrt{100} = 10 \) cm. The diagonal of the cuboid then forms a right-angled triangle with the base diagonal (10 cm) and the height (5 cm): diagonal \( = \sqrt{10^{2}+5^{2}} = \sqrt{100+25} = \sqrt{125} = 11.18\ldots \), which rounds to 11.2 cm (3 s.f.). (b) The angle \( \theta \) between the diagonal and the base satisfies \( \tan\theta = \dfrac{\text{height}}{\text{base diagonal}} = \dfrac{5}{10} = 0.5 \), so \( \theta = \tan^{-1}(0.5) = 26.56\ldots \), which rounds to 26.6° (3 s.f.).
Marking scheme
(a) 1 mark for correctly finding the base diagonal (10 cm); 1 mark for correct method using Pythagoras with the height; 1 mark for the final answer 11.2 cm (3 s.f.). (b) 1 mark for identifying \( \tan\theta = \frac{5}{10} \); 1 mark for correct method \( \tan^{-1}(0.5) \); 1 mark for the final answer 26.6° (3 s.f.). Max 6.
Question 14 · Advanced 3D Geometry / Dependent Probability (6 marks)
6 marks
A bag contains 5 red counters and 3 blue counters. Two counters are drawn at random, one after another, without replacement. (a) Calculate the probability that both counters drawn are red. (b) Calculate the probability that the two counters drawn are different colours.
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Worked solution
There are 8 counters in total. (a) P(both red) \( = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14} \), since after the first red counter is removed, 4 red counters remain out of 7 total. (b) P(different colours) = P(red then blue) + P(blue then red) \( = \left(\dfrac{5}{8}\times\dfrac{3}{7}\right) + \left(\dfrac{3}{8}\times\dfrac{5}{7}\right) = \dfrac{15}{56}+\dfrac{15}{56} = \dfrac{30}{56} = \dfrac{15}{28} \).
Marking scheme
(a) 1 mark for correct probabilities without replacement \( \frac{5}{8} \) and \( \frac{4}{7} \); 1 mark for correct method (multiplying); 1 mark for the correct simplified answer \( \frac{5}{14} \). (b) 1 mark for identifying both orders (red-then-blue and blue-then-red) with correct without-replacement probabilities; 1 mark for correct addition of both cases; 1 mark for the correct simplified final answer \( \frac{15}{28} \). Max 6.
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