Question 1

\xi = \{odd numbers between 10 and 40\}

\textit{A} = \{11,13,27,39\}

\textit{B} = \{15,19,21,27,35\}

\textit{C} = \{13,19,27,31\}

 

 

Question 1(a) [4 marks]

Which of the following Venn diagrams is correct?

 

Answer type: Multiple choice

A:

B:

C:

D:

 

ANSWER: A

 

 

Question 1(b) [2 marks]

A number is chosen at random from \sigma.

Find the probability that the number is a member of A \cup C

 

 

Answer type: Fraction

 

ANSWER: \dfrac{6}{15}

WORKING:

A \cup C = A or C

The numbers that satisfy A or C are 11,13,19,27,31,39

There are a total of 15 numbers in \sigma

Probability =\dfrac{6}{15}

 

 


Question 2 [2 marks]

Michael has information about how many hours 10 students revised for their maths exam and the results they achieved.

He is asked to draw a scatter graph and a line of best fit for this information.

Here is his answer.

Michael has plotted the points accurately.

What is wrong with his answer?

 

Answer type: Multiple choice

A: The line of best fit is inaccurate.

B: The labels on the y axis are not linear.

C: There is no correlation between the time spent revising and the result achieved.

D: The time should start at 140 hours.

 

ANSWER: A

 

 

 


Question 3 [4 marks]

\textit{BEG} is a triangle.

\textit{ABC} and \textit{DEF} are parallel lines.

What is the size of angle x?

 

Answer type: Simple text answer

 

 

ANSWER: 52 \degree

WORKING:

\angle\textit{BEG} = 180 - \angle\textit{DEB} - \angle\textit{GEF} = 180 - 120 - 110 = 50 \degree

\angle\textit{ABE} + \angle\textit{DEB} = 180 \degree so = \angle\textit{ABE} = 180 - 120 = 60 \degree

\angle\textit{EBG} = 180 -\angle\textit{ABE} - \angle\textit{CBG} = 180 - 60 - 42 = 78 \degree

\angle\textit{BGE} = 180 - \angle\textit{BEG} - \angle\textit{ABE} = 180 - 50 - 78 =52 \degree

So, x = 52 \degree


Question 4

Southern Bank has two types of account.

Both accounts pay compound interest.

Savings account: Interest 3\% per annum

Shares account: Interest 4.5\% per annum

 

Amy invests £3500 in the savings account.

Barney invests £2100 in the shares account.

 

Question 4(a) [4 marks]

How much more interest will Amy get than Barney by the end of 3 years?

 

Answer type: Multiple choice

A: £28.05

B: £82.50

C: £25.08

D: £52.80

 

ANSWER: A

WORKING:

Amy:

3500 \times 1.03^3 = £3824.54 (nearest penny)

Interest = £3824.54 - £3500 = £324.54

 

Barney:

2100 \times 1.045^3 = £2396.45 (nearest penny)

Interest = £2396.45 - £2100 = £296.45

 

Difference = £324.54 - £296.45 = £28.09

 

 

Question 4(b) [1 mark]

In the 3rd year, the rate of interest for the savings account is changed to 3.5\% per annum.

Would Amy still get the most interest by the end of 3 years?

 

Answer type: Multiple choice

A: Yes

B: No

 

ANSWER: A

WORKING:

Amy already gets the most interest and so increasing her interest to 3.5\% will only make her get even more.

 

 


Question 5 [5 marks]

The diagram shows a floor in the shape of a trapezium.

James is going to varnish the floor.

Each 8 litre tin of varnish costs £21.49

1 litre of varnish covers an area of 5 m^2

James has £130 to spend on varnish.

How much more money would James need to buy all the varnish required to varnish the whole floor?

 

Answer type: Simple text answer

 

ANSWER: £20.43

WORKING:

Area of the trapezium =\dfrac{1}{2}(18+25) \times 12 = 258 m^2

As 1 litre of varnish is 5 m^2, 1 tin of varnish covers an area of 8 \times 5 = 40 m^2

\dfrac{258}{40} = 6.45, so James needs 7 tins

 

Total cost = 7 \times 21.49 = £150.43

Money needed = 150.43 - 130 = £20.43

 

 


Question 6 [3 marks]

\textit{A} is the point with coordinates (4, 19)

\textit{B} is the point with coordinates (7,z)

The gradient of the line \textit{AB} is 5

What is the value of z?

 

Answer type: Simple text value

 

ANSWER: 34

WORKING:

Gradient of \textit{AB} = \dfrac{\text{Change in } y}{\text{Change in } x} = \dfrac{z - 19}{7 - 4} = \dfrac{z - 19}{3} = 5

 

Rearrange to get

z - 19 = 15

z = 34

 

 


Question 7(a) [1 mark]

What is the number 0.0000031897 in standard form?

 

Answer type: Multiple choice

A: 3.1897 \times 10^{-6}

B: 31.897 \times 10^{-6}

C: 3.1897 \times 10^{-5}

D: 31.897 \times 10^{-7}

 

ANSWER: A

 

 

Question 7(b) [2 marks]

What is the number \dfrac{2.4 \times 10^{4} - 1.5 \times 10^{-3}}{3.1 \times 10^{-2}} in standard form, correct to 4 significant figures?

 

Answer type: Multiple choice

A: 7.742 \times 10^5

B: 77.72 \times 10^3

C: 7.742 \times 10^7

D: 777.4 \times 10^4

 

ANSWER: A

WORKING:

\dfrac{2.4 \times 10^{4} - 1.5 \times 10^{-3}}{3.1 \times 10^{-2}} = 774193.5

774193.5 = 774200 (4 sf)

774200 = 7.742 \times 10^5 in standard form

 

 


Question 8 [3 marks]

Triangle A is reflected in the line y = 1 to give triangle B

Triangle B is reflected in the line y = -x to give triangle C

Where will triangle C be located on the graph?

 

Answer type: Multiple choice

A:

B:

C:

D:

 

ANSWER: A

WORKING:

 

 

 

 


Question 9 [2 marks]

Laura truncates the number \textit{N} to 2 digits.

The result is 3.4

What is the error interval for \textit{N}?

 

Answer type: Multiple choice

A: 3.4 \leq \textit{N} < 3.5

B: 3.4 \leq \textit{N} < 4.4

C: 3.4 < \textit{N} < 4

D: 3.4 < \textit{N} < 3.5

 

ANSWER: A

WORKING:

Truncate means we limit the number of digits right of the decimal point without rounding.

 

 


Question 10 [5 marks] – here

Eleanor makes 80 litres of orange paint by mixing litres of red paint and litres of yellow paint.

The ratio of red paint to yellow paint is 3:7

Red paint is sold in 6 litre tins and each tin of red paint costs £32

Yellow paint is sold in 14 litre tins and each tin of yellow paint costs £60

Eleanor sells all the orange paint she makes in 10 litre tins.

She sells each tin of orange paint for £74.25

What will be Eleanor’s percentage profit on each tin of orange paint she sells, rounded to the nearest whole number?

 

Answer: Simple text answer

 

ANSWER: 61\%

WORKING:

Profit = Income - Cost

 

Cost:

Number of parts in ratio = 3 + 7 = 10

1 part = \dfrac{80}{10} = 8 litres

 

8 \times 3 = 24 litres of red paint

\dfrac{24}{6} = 4 tins of red paint

Red paint cost = 4 \times £32 = £128

 

8 \times 7 = 56 litres of yellow paint

\dfrac{56}{14} = 4 tins of yellow paint

Yellow paint cost = 4 \times £60 = £240

 

Total cost = £128 + £240 = £368

 

 

Income:

\dfrac{80}{10} = 8 tins of orange paint

Income = 8 \times £74.25= £594

 

 

Profit = 594 - 368 = £226

 

Percentage profit = \dfrac{226}{368} \times 100 = 61.4130435 = 61 \% (nearest whole number)

 

 


Question 11 [3 marks]

In a coffee shop there are

  • 6 types of coffee
  • 14 types of panini
  • 12 types of cake

Charlotte is going to order one of the following combinations

  • A coffee and a panini
  • A panini and a cake
  • A coffee, a panini and a cake

How many different ways are there for Charlotte to choose her order?

 

Answer type: Simple text answer

 

ANSWER: 1240 ways

WORKING:

6 \times 14 = 84 ways for Charlotte to order a coffee and a panini

14 \times 12 = 168 ways for Charlotte to order a panini and a cake

6 \times 14 \times 12 = 1008 ways for Charlotte to order a coffee, a panini and a cake

Total number of ways = 84 + 168 + 1008 = 1240

 

 


Question 12(a) [3 marks]

What is \dfrac{9x^2 - 25}{6x + 10} \times \dfrac{3x}{x^2 + 5x} in the form \dfrac{ax+b}{cx+d} where a,b,c and d are integers?

 

Answer type: Multiple choice

A: \dfrac{9x-15}{2x+10}

B: \dfrac{2x+10}{9x+15}

C: \dfrac{3x+5}{2x+10}

D: \dfrac{3x}{6x+5}

 

 

ANSWER: A

WORKING:

\dfrac{(3x + 5)(3x - 5)}{2(3x + 5)} \times \dfrac{3x}{x(x+5)}

= \dfrac{3x - 5}{2} \times \dfrac{3}{x+5}

= \dfrac{9x-15}{2x+10}

 

Question 12(b) [3 marks]

What is \dfrac{5}{x + 3} - \dfrac{1}{x + 4} - \dfrac{2}{x} as a single fraction?

 

Answer type: Multiple choice

A: \dfrac{2x^2 + 3x - 24}{x(x+3)(x+4)}

B: \dfrac{8x^2 + 37x + 24}{x(x+3)(x+4)}

C: \dfrac{4x^2 + 9x - 24}{x(x+3)(x+4)}

D: \dfrac{6x^2 + 31x + 24}{x(x+3)(x+4)}

 

ANSWER: A

WORKING:

\dfrac{5(x+4)(x)}{(x+3)(x+4)(x)} - \dfrac{1(x+3)(x)}{(x+3)(x+4)(x)} - \dfrac{2(x+3)(x+4)}{(x+3)(x+4)(x)}

=\dfrac{(5x^2 + 20x) - (x^2 + 3x) - (2x^2 + 14x + 24)}{x(x+3)(x+4)}

=\dfrac{2x^2 + 3x - 24}{x(x+3)(x+4)}

 


Question 13 [3 marks]

The diagram shows a circle and an equilateral triangle.

One side of the equilateral triangle is the diameter of the circle.

The circle has a circumference of 32 cm

 

What is the area of the triangle?

Give your answer correct to 3 significant figures.

 

Answer type: Simple text answer

 

ANSWER: 44.9 cm^2

WORKING:

c = 32 = \pi  d, so  d = \dfrac{32}{\pi}

All the angles of the triangle are 60^\circ because it is an equilateral triangle

Area = \dfrac{1}{2} \times \dfrac{32}{\pi} \times \dfrac{32}{\pi} \times \sin 60 = 44.926 = 44.9 cm^2 (3 sf)

 


Question 14 [2 marks]

What is the graph with equation x + y = 3?

 

 

Answer type: Multiple choice

A:

B:

C:

D:

 

ANSWER: A

 

 


Question 15 [1 mark]

The equation of a circle is x^2 + y^2 = 110.25

What is the diameter of the circle?

 

Answer type: Simple text answer

 

ANSWER: 21

WORKING:

r^2 = 110.25

r = \sqrt{110.25} = 10.5 (r is positive)

d = 2r = 2 \times 10.5 = 21

 

 


Question 16

There are only black pens and red pens in a box.

Jack takes a random pen from the box.

The probability that the pen is black is 0.45

Jack puts the pen back in the box.

Molly takes a random pen from the box.

She puts the pen back in the box.

 

 

Question 16(a) [2 marks]

What is the probability that the pen Jack and Molly took out was the same colour?

 

Answer type: Simple text answer

 

ANSWER: 0.505

WORKING:

P_{(B,B)} = 0.45 \times 0.45 = 0.2025

P_{(R,R)}  = 0.55 \times 0.55 = 0.3025

P_{(B,B)} + P_{(R,R)} = 0.2025 + 0.3025 = 0.505

 

 

Question 16(b) [2 marks]

There are 77 red pens in the box.

How many black pens are there in the box?

 

Answer type: Simple text answer

 

ANSWER: 63

WORKING:

\dfrac{77}{0.55} = 140 total pens

140 - 77 = 63 black pens

 

 


Question 17 [5 marks]

a and b are two numbers such that a > b

When you add 4 to a and add 4 to b the answers are in the ratio 2:1

When you subtract 6 from a and subtract 6 from b the answers are in the ratio 5:2

What is the ratio a:b in its simplest form?

 

Answer type: Multiple answers

 

ANSWER: a = 28, b = 13

WORKING:

a + 4:b + 4 = 2:1

\dfrac{a+4}{b+4} = \dfrac{2}{1}

1(a+4) = 2(b+4)

a + 4 = 2b + 8 , so a - 2b = 4

 

a - 6:b - 6 = 5:2

\dfrac{a-6}{b-6} = \dfrac{5}{2}

2(a-6) = 5(b-6)

2a - 12 = 5b - 30 so 2a - 5b = -18

 

Treat a - 2b = 4 and 2a - 5b = -18 as simultaneous equations and solve

2a - 4b = 8 \quad \quad\,\, (1)

2a - 5b = -18 \quad (2)

Subtract (2) from (1) to get b = 26

Substitute b = 26 into (1):

a - 2(26) = 4 so a = 56

 

The ratio is 56:26 = 28:13

So, a = 28, b = 13

 

 


Question 18 [4 marks]

The straight line L_{1} passes through the points with coordinates (4,2) and (6,7)

The straight line L_{2} passes through the origin and has gradient \dfrac{3}{2}

The lines L_1 and L_2 intersect at the point P.

What are the coordinates of P ?

 

Answer type: Multiple answers

 

ANSWER: x = 8, y = 12

WORKING:

L_{1} : M = \dfrac{\text{Change in } y}{\text{Change in } x} = \dfrac{7-2}{6-4} = \dfrac{5}{2}

y = \dfrac{5}{2}x + c

2 = \dfrac{5}{2}(4) + c

c = 2 - 10 = -8

So, y = \dfrac{5}{2}x - 8

 

L_{2} : y = \dfrac{3}{2}x

 

As L_{1} and L_{2} intersect at point \textit{P}, the values of x and y are the same.

So, we have two simultaneous equations, that we can solve as follows

\dfrac{5}{2}x - 8 = \dfrac{3}{2}x

x = 8

Substitute x=8 into the equation of L_2 to get

y = \dfrac{3}{2}x = \dfrac{3}{2}(8) = 12

 

Hence, x = 8, y = 12

 

 


Question 19 [5 marks]

Solve 5 < \dfrac{m^4 + 9}{5} < 18

 

Answer type: Multiple choice

A: 2 < m < 3 and -3 < m < -2

B: 2 < m < 3

C: 16 < m < 81

D: 16 < m < 81 and -81 < m < -16

 

ANSWER: A

WORKING:

25 < m^4 + 9 < 90

16 < m^4 < 81

m^4 > 16 so m > 2 and m < -2

m^4 < 81 so m < 3 and m > -3

 

Combine into two inequalities to get

2 < m < 3 and -3 < m < -2

 

 


Question 20 [5 marks]

Solid A, shown below, is a frustum of a cone.

The diagram above shows that the frustum is made by removing a cone with height 3.7 cm from a solid cone with height 7.4 cm and a base diameter of 5.4 cm.

The frustum is joined to a solid cylinder with diameter 5.4 cm and a height of 1.4 cm, to form solid B, as shown below.

The density of the frustum is 3.3 g/cm^3

The density of the cylinder is 2.8 g/cm^3

 

Volume of cone = \dfrac{1}{3} \pi r^2 h

 

Calculate the average density of solid B, correct to 2 decimal places.

 

Answer type: Multiple choice type 1

A: 3.10 g/cm^3

B: 6.10 g/cm^3

C: 3.05 g/cm^3

D: 3.16 g/cm^3

 

ANSWER: A

WORKING:

Frustum:

Radius of big cone = 5.4 \div 2 = 2.7 cm

Volume of big cone = \dfrac{1}{3} \times \pi \times 2.7^2 \times 7.4 = 56.4921... cm^3

 

Radius of small cone = 2.7 \div 2 = 1.35 cm, since the height of the big cone is twice the height of the small cone.

Volume of small cone = \dfrac{1}{3} \times \pi \times 1.35^2 \times 3.7 = 7.0615... cm^3

 

Volume of frustum = 56.4921... - 7.0615... = 49.4306... cm^3

 

\text{M} = \text{D} \times \text{V} = 3.3 \times 49.4306... = 163.1209... g

 

 

Cylinder:

Radius = 2.7 cm

Volume of cylinder = \pi r^2 h = \pi \times 2.7^2 \times 1.4 = 32.0630... cm^3

\text{M} = \text{D} \times \text{V} = 2.8 \times 32.0630... = 89.7766... g

 

 

Total volume = 49.4306... + 32.0630... = 81.4936... cm^3

Total mass = 163.1209... + 89.7766... = 252.8976... g

\text{Average density} = \dfrac{\text{Total mass}}{\text{Total volume}} = \dfrac{252.8976...}{81.4936...} = 3.10 g/cm^3

 


Question 21 [4 marks]

\textit{A.B,Q} and \textit{P} are four points on a circle with centre \textit{O}.

\textit{A,O,Q} and \textit{C} are four points on a different circle.

The two circles intersect at the points \textit{A} and \textit{Q}.

\textit{CPA}, \textit{CQB} and \textit{AOB} are straight lines.

Angle \textit{CAB} = 78 \degree

What is angle \textit{ACB} ?

 

Answer type: Simple text answer

 

ANSWER: 24 \degree

WORKING:

\angle\textit{CQO} = 180 - \angle\textit{CAB} = 180 - 78 = 102 \degree (Cyclic quadrilaterals)

\angle\textit{OQB} = 180 - \angle\textit{CQO} = 180 - 102 = 78 \degree (Angles on a straight line)

OQ = OB, hence \angle\textit{OQB} = \angle\textit{OBQ} = 78 \degree

\angle\textit{ACB} = 180 - \angle\textit{CAB} - \angle\textit{OQB} = 180 - 78 - 78 = 24 \degree (Angles in a triangle)