Higher Tier - Year 10

Year 10 Paper 5: Number and Shape

Covers number and calculation, fractions, decimals and percentages, sequences, indices and standard form, graphs and coordinates, angles and geometrical reasoning, and area, volume and measures.

23 questions - 60 marks - calculator allowed

Year 10 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 10, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.

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Questions

Question 1 [1 marks]

Area, Volume and Measures

The diameter of a circle is 12 cm. Write down the radius of the circle.

Question 2 [1 marks]

Graphs and Coordinates

Which of these points lies on the line with equation y = 2x + 1?

Question 3 [3 marks]

Indices and Standard Form

Work out the value of:

64^(1/2)

8^(1/3)

81^(1/4)

Question 4 [2 marks]

Angles and Geometrical Reasoning

A cuboid has length 6 cm, width 4 cm and height 3 cm. It stands on a table. Arrow F points towards the front face of the cuboid.

Write down the dimensions of the front elevation of the cuboid.

Write down the dimensions of the plan of the cuboid.

Question 5 [1 marks]

Number and Calculation

Priya pours 250 ml of squash equally into 4 cups. Work out the volume in each cup in millilitres.

Question 6 [1 marks]

Indices and Standard Form

Which is larger, 6.4 x 10^3 or 5.9 x 10^4? Write down the larger number, in standard form.

Question 7 [2 marks]

Sequences

Write down the first four terms of the sequence defined by T(n) = 2n^2 - n + 3.

Question 8 [1 marks]

Number and Calculation

Write down the bounds for a number given as 5, correct to 1 significant figure, using inequality notation.

Question 9 [1 marks]

Angles and Geometrical Reasoning

A wooden box measures length 12 cm, width 5 cm and height 3 cm. State the two measurements that are shown on the plan (top view).

Question 10 [2 marks]

Number and Calculation

A padlock code is made from 3 digits, each chosen from 0 to 9. Digits may be repeated. Work out how many different codes are possible.

Question 11 [3 marks]

Fractions, Decimals and Percentages

A shop increases the price of a coat by 20%. In a later sale, it reduces the new price by 20%. A customer says: "The price is now back to what it started at, because a 20% increase and a 20% decrease cancel each other out."

Using a starting price of £50, show that the customer is wrong and explain why.

Question 12 [1 marks]

Number and Calculation

Two offcuts of skirting board have lengths a = 14 cm and b = 9 cm, each measured correct to the nearest cm. Find the upper bound of a + b.

Question 13 [3 marks]

Fractions, Decimals and Percentages

In 2024, a school had 240 pupils in Year 7. In 2025, the school had 258 pupils in Year 7. Work out the percentage increase in the number of Year 7 pupils.

Question 14 [3 marks]

Sequences

The first three terms of a sequence are 3, 8, 15. (a) Work out the 4th term. (b) Suggest an nth term formula for this sequence.

Question 15 [3 marks]

Fractions, Decimals and Percentages

Tom invests £3000 in an account that pays compound interest at a rate of 3% per year. Use trial and improvement to find the number of complete years after which his investment first exceeds £3500.

Question 16 [2 marks]

Graphs and Coordinates

A feasible region is defined by the inequalities x >= 0, y >= 0 and y <= 2x + 1. Which of the following points are vertices (corner points) of this region: (0,0), (0,1), (1,3)? Explain your answer.

Question 17 [3 marks]

Fractions, Decimals and Percentages

The price of a laptop is £750 now. Due to inflation, the price increases by 2% each year. Work out the price of the laptop in 4 years' time. Give your answer to the nearest penny.

Question 18 [3 marks]

Number and Calculation

A recipe uses 350 g of flour to make 4 portions. Work out how much flour is needed for 7 portions.

Question 19 [3 marks]

Fractions, Decimals and Percentages

Sam paints 3/5 of a fence in the morning. In the afternoon he paints 1/4 of the remaining part. What fraction of the whole fence has Sam painted by the end of the day?

Question 20 [6 marks]

Indices and Standard Form

Simplify each expression fully.

(8x^6)^(1/3)

(9x^4)^(1/2)

(x^3)^-2 * x^4

Question 21 [2 marks]

Area, Volume and Measures

A right-angled isosceles triangle has its two perpendicular sides each of length 8 cm. Find the exact perimeter of the triangle, giving your answer in the form (a + b sqrt(2)) cm, where a and b are integers.

Question 22 [5 marks]

Graphs and Coordinates

Two savings accounts are modelled by A = 500 x 1.03^t and B = 400 x 1.045^t, where t is the number of complete years the money has been invested. Find the first whole number of years, t, after which the value of account B first exceeds the value of account A.

Question 23 [8 marks]

Area, Volume and Measures

A solid dome is made from a cone of radius 8 cm and perpendicular height 15 cm fixed on top of a hemisphere of radius 8 cm, with the flat faces joined together.

Calculate the slant height of the cone.

Calculate the total area of the dome's surface that needs to be painted: the curved surface of the cone plus the curved surface of the hemisphere only (do not include the flat circle where the two solids join). Give your answer as a multiple of pi.

Paint costs 2p per cm^2. Work out the total cost of painting 50 of these dome solids. Give your answer in pounds, to the nearest penny.

Model solutions

Mark scheme for Question 1 [1 mark]
Question 1[1 mark]
Answer or workingMarks
6 cmB1cao
Mark scheme for Question 2 [1 mark]
Question 2[1 mark]
Answer or workingMarks
CB1cao
Final answer: C) (3, 7)
Mark scheme for Question 3 [3 marks]
Question 3[3 marks]
Answer or workingMarks
8B1cao
2B1cao
3B1cao
Final answer: 8 | 2 | 3
Mark scheme for Question 4 [2 marks]
Question 4[2 marks]
Answer or workingMarks
6 cm by 3 cmB1oe
6 cm by 4 cmB1oe
Final answer: 6 cm x 3 cm | 6 cm x 4 cm
Mark scheme for Question 5 [1 mark]
Question 5[1 mark]
Answer or workingMarks
62.5 mlB1cao
Final answer: 62.5 ml. Working check: 250 ÷ 4 = 62.5.
Mark scheme for Question 6 [1 mark]
Question 6[1 mark]
Answer or workingMarks
5.9 x 10^4 oe (59000)B1
Final answer: 5.9 x 10^4
Mark scheme for Question 7 [2 marks]
Question 7[2 marks]
Answer or workingMarks
substitute n = 1, 2, 3, 4 into 2n^2 - n + 3M1oe
4, 9, 18, 31A1cao
Mark scheme for Question 8 [1 mark]
Question 8[1 mark]
Answer or workingMarks
4.5 <= x < 5.5B1cao
Mark scheme for Question 9 [1 mark]
Question 9[1 mark]
Answer or workingMarks
12 cm by 5 cmB1cao
Final answer: 12 cm by 5 cm (length and width). 12 cm, 5 cm are shown on the plan.
Mark scheme for Question 10 [2 marks]
Question 10[2 marks]
Answer or workingMarks
recognise 10 choices for each of 3 positions and set up 10^3M1oe
1000A1cao
Mark scheme for Question 11 [3 marks]
Question 11[3 marks]
Answer or workingMarks
after increase: 50 x 1.2 = 60 (pounds)M1
after decrease: 60 x 0.8 = 48 (pounds)M1
final price is £48, not 50; the 20% decrease is taken from the larger £60, not the original £50C1
Final answer: Final price £48 (not 50); the decrease is of a larger amount
Mark scheme for Question 12 [1 mark]
Question 12[1 mark]
Answer or workingMarks
24 (cm)B1cao
Final answer: 24 cm
Mark scheme for Question 13 [3 marks]
Question 13[3 marks]
Answer or workingMarks
correct difference 258 - 240 = 18M1
divide by the original value and multiply by 100: (18 / 240) x 100M1
7.5%A1cao
Mark scheme for Question 14 [3 marks]
Question 14[3 marks]
Answer or workingMarks
24B1cao
recognise quadratic form with second difference 2 and write an^2 + bn + c with a = 1M1
n^2 + 2nA1cao
Final answer: 24 | n^2 + 2n
Mark scheme for Question 15 [3 marks]
Question 15[3 marks]
Answer or workingMarks
trial at n = 5: 3000 x 1.03^5 = 3477.82 (too low)M1
trial at n = 6: 3000 x 1.03^6 = 3582.16 (exceeds 3500)M1
6 (years)A1cao
Final answer: 6 years
Mark scheme for Question 16 [2 marks]
Question 16[2 marks]
Answer or workingMarks
identify that (0,0) and (0,1) are intersections of two boundary lines so are vertices, but (1,3) lies on the sloping boundary onlyM1
(0,0) and (0,1) are the corner points; (1,3) is not a vertex since it only lies on y = 2x + 1 and not on another boundaryA1cao
Final answer: Vertices: (0,0) and (0,1). Reason: (0,0) is intersection of x >= 0 and y >= 0; (0,1) is intersection of x = 0 and y = 2x + 1. Point (1,3) lies on the sloping line only, not at an intersection of two boundaries.
Mark scheme for Question 17 [3 marks]
Question 17[3 marks]
Answer or workingMarks
1.02^4 (= 1.08243216)M1oe
750 x their 1.08243216M1
£811.82 (awrt)A1cao
Final answer: £811.82
Mark scheme for Question 18 [3 marks]
Question 18[3 marks]
Answer or workingMarks
finds flour per portion: 350 ÷ 4M1
multiplies by 7 to scale up (method)M1
612.5 gA1cao
Final answer: 612.5 g. Working check: 350 ÷ 4 = 87.5 g per portion; 87.5 × 7 = 612.5 g.
Mark scheme for Question 19 [3 marks]
Question 19[3 marks]
Answer or workingMarks
calculate remaining after morning, 1 - 3/5 = 2/5 and find 1/4 of this, 1/4 × 2/5M1
add morning and afternoon amounts, e.g. 3/5 + 1/10 shownM1
7/10A1cao
Mark scheme for Question 20 [6 marks]
Question 20[6 marks]
Answer or workingMarks
8^(1/3) = 2 and x^6 raised to power 1/3 gives x^2, both foundM1
2x^2A1cao
9^(1/2) = 3 and x^4 raised to power 1/2 gives x^2, both foundM1
3x^2A1cao
(x^3)^-2 = x^-6 foundM1
1/x^2 oe (x^-2)A1
Final answer: 2x^2 | 3x^2 | x^-2 (= 1/x^2)
Mark scheme for Question 21 [2 marks]
Question 21[2 marks]
Answer or workingMarks
Use Pythagoras to find the hypotenuse: sqrt(8^2 + 8^2) = sqrt(128) = 8sqrt(2)M1
Perimeter = 8 + 8 + 8sqrt(2) = (16 + 8sqrt(2)) cmA1cao
Final answer: (16 + 8sqrt(2)) cm
Mark scheme for Question 22 [5 marks]
Question 22[5 marks]
Answer or workingMarks
attempts trial values of A and B at t=15M1
A15 awrt 779, B15 awrt 774 (B < A)A1
attempts trial values of A and B at t=16M1
A16 awrt 802, B16 awrt 809 (B > A)A1
t = 16 stated as conclusionB1
Final answer: t = 16
Mark scheme for Question 23 [8 marks]
Question 23[8 marks]
Answer or workingMarks
sqrt(8^2 + 15^2) oe (64 + 225 = 289 seen)M1
17 cmA1cao
pi x 8 x their 17 oe (cone curved surface area = 136 piM1ft
2 x pi x 8^2 oe (hemisphere curved surface area = 128 pi)M1
264 pi cm^2 oeA1cao
264 pi x 50 oe (total area for 50 domes, ft from part b, awrt 41500)M1
their total area x 0.02 (converts 2p per cm^2 correctly to poundsM1dep
awrt 829.38 (pounds)A1ft
Final answer: 17 cm | 264 pi cm^2 | GBP 829.38