Higher Tier - Year 10

Edexcel-Style Year 10 Higher Paper 3

An Edexcel-style Year 10 Higher Paper 3: 80 marks, 90 minutes, calculator, matching Edexcel 1MA1's published paper format. Covers the same nine Year 10 topics as the board-agnostic set: number and calculation, fractions, decimals and percentages, ratio and proportion, indices and standard form, linear algebra, graphs and coordinates, sequences, angles and geometrical reasoning, and area, volume and measures.

27 questions - 80 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 [2 marks]

Area, Volume and Measures

A sphere has radius 6 cm. Calculate the volume of the sphere. Give your answer correct to 3 significant figures.

Question 2 [2 marks]

Sequences

Here are the first four terms of a sequence. 1, 3, 6, 10

Write down the next two terms of the sequence.

State the name given to this special sequence of numbers.

Question 3 [2 marks]

Number and Calculation

A parcel has mass 3.6 kg, correct to the nearest 100 g. Write down the error interval for the mass, m kilograms.

Question 4 [4 marks]

Angles and Geometrical Reasoning

Triangle OAB has OA = a and OB = b. M is the midpoint of OA.

Find the vector OM in terms of a.

Find the vector MA in terms of a.

Find the vector MB in terms of a and b.

Question 5 [1 marks]

Indices and Standard Form

Write 6^3 as a repeated multiplication (in the form 6 x 6 x ...).

Question 6 [1 marks]

Fractions, Decimals and Percentages

Work out 4/9 of 63.

Question 7 [1 marks]

Linear Algebra

A function machine multiplies the input by 8. The output is 72. Write down the input.

Question 8 [2 marks]

Graphs and Coordinates

Chidi sketches this distance-time graph for a car journey. The graph includes a vertical section: it rises from (10, 20) straight up to (10, 35) before continuing. Explain why this graph cannot show a real journey.

Question 9 [2 marks]

Linear Algebra

Expand and simplify 4(2x + 3) - 9.

Question 10 [2 marks]

Area, Volume and Measures

A rectangle measures 10 cm by 6 cm. A small rectangle of size 3 cm by 2 cm has been removed from one corner. Work out the area of the remaining shape.

Question 11 [2 marks]

Number and Calculation

Grace tosses 3 fair coins, one after another. Work out the number of different sequences of heads and tails possible.

Question 12 [2 marks]

Graphs and Coordinates

A toy car's distance-time graph shows it travels 50 m in the first 10 seconds. Work out the speed in m/s and in km/h.

Question 13 [3 marks]

Ratio and Proportion

The density of a liquid is given by the formula d = m / V, where m is mass and V is volume.

Rearrange the formula to make V the subject.

A liquid has a density of 1.25 g/cm^3. Find the volume of 250 g of the liquid.

Question 14 [3 marks]

Number and Calculation

Two buses leave a bus station at the same time, 08:00. Bus A leaves every 15 minutes. Bus B leaves every 20 minutes. Work out the next time both buses will leave the bus station together.

Question 15 [3 marks]

Fractions, Decimals and Percentages

Write these four numbers in order of size, starting with the smallest: 0.6, 0.6333... (3 recurring), 2/3, 0.65

Question 16 [3 marks]

Number and Calculation

A train journey takes 1 hour 20 minutes and covers 85 miles. Work out the average speed of the train in miles per hour.

Question 17 [3 marks]

Angles and Geometrical Reasoning

ABCD is a cyclic quadrilateral. Angle ABC = 68 degrees. Work out angle ADC and give a reason.

Question 18 [3 marks]

Fractions, Decimals and Percentages

Decrease £340 by 35%.

Question 19 [4 marks]

Angles and Geometrical Reasoning

Points B(8, 0) and C(3, 6) define line BC. (a) Find the equation of line BC in the form y = mx + c. (b) Find the equation of the locus of points equidistant from B and C (the perpendicular bisector). Give the equation in the form y = mx + c and give the midpoint and slope used.

Find the equation of line BC.

Find the equation of the perpendicular bisector of BC (locus equidistant from B and C). Give midpoint and slope used.

Question 20 [4 marks]

Ratio and Proportion

A cylindrical metal rod has length 1.2 m and cross-sectional area 3.5 cm^2. The density of the metal is 7.8 g/cm^3. Calculate the mass of the rod in kilograms. Give your answer to 3 significant figures.

Question 21 [6 marks]

Area, Volume and Measures

A model car is built using a scale of 1:24 compared with the real car (every linear dimension of the real car is 24 times the corresponding dimension of the model). The model car has a volume of 90 cm^3 and a surface area of 220 cm^2.

Work out the volume of the real car. Give your answer in cm^3, in standard form correct to 3 significant figures.

Work out the surface area of the real car in m^2, correct to 3 significant figures.

Question 22 [5 marks]

Linear Algebra

Three angles lie on a straight line. The angles are 2x degrees, 3x degrees and 40 degrees.

Form an equation in x and solve it.

Work out the size of the two unknown angles.

Question 23 [2 marks]

Area, Volume and Measures

Find the exact area of a parallelogram with adjacent sides 10 cm and 7 cm with included angle 60 degrees. Give your answer in exact form.

Question 24 [4 marks]

Indices and Standard Form

Simplify (5 + 2sqrt(2))/(3 - 2sqrt(2)) fully, giving your answer in the form p + qsqrt(2), where p and q are integers.

Question 25 [5 marks]

Area, Volume and Measures

A cone has height 24 cm. It is cut by a plane parallel to the base, 8 cm from the vertex, forming a smaller cone (mathematically similar to the original cone) and a frustum. The whole cone has volume 900 cm^3.

Work out the ratio of the volume of the small cone to the volume of the frustum. Give your answer in the form 1 : n.

Work out the volume of the frustum.

Question 26 [4 marks]

Number and Calculation

a = 2^2 x 3 x 5 and b = 2 x 3^3.

Find the HCF of a and b.

Find the LCM of a and b.

Question 27 [5 marks]

Area, Volume and Measures

Triangle ABC has AB = 12 m, angle A = 60 degrees and area 84 m^2. Work out AC, correct to 2 decimal places.

Model solutions

Mark scheme for Question 1 [2 marks]
Question 1[2 marks]
Answer or working	Marks
4/3 x pi x 6^3 oe (288 pi seen)	M1
awrt 905	A1
Final answer: 905 cm^3 (3 sf)

Mark scheme for Question 2 [2 marks]
Question 2[2 marks]
Answer or working	Marks
15 and 21 both correct	B1cao
triangular numbers	B1cao
Final answer: 15, 21 \| Triangular numbers

Mark scheme for Question 3 [2 marks]
Question 3[2 marks]
Answer or working	Marks
converts 100 g to 0.1 kg and identifies one correct bound, 3.55 or 3.65	M1oe
3.55 <= m < 3.65	A1cao

Mark scheme for Question 4 [4 marks]
Question 4[4 marks]
Answer or working	Marks
(1/2)a (oe)	B1cao
(1/2)a (oe)	B1cao
correct method, e.g. MB = OB - OM (oe MB = MA + AB)	M1
b - (1/2)a (oe)	A1cao
Final answer: OM = (1/2)a \| MA = (1/2)a \| MB = b - (1/2)a

Mark scheme for Question 5 [1 mark]
Question 5[1 mark]
Answer or working	Marks
6 x 6 x 6 oe	B1cao
Final answer: 6 x 6 x 6

Mark scheme for Question 6 [1 mark]
Question 6[1 mark]
Answer or working	Marks
28	B1cao

Mark scheme for Question 7 [1 mark]
Question 7[1 mark]
Answer or working	Marks
9	B1cao
Final answer: 9 (because input = 72 ÷ 8).

Mark scheme for Question 8 [2 marks]
Question 8[2 marks]
Answer or working	Marks
identifies the vertical section as the fault	B1
correct reason, e.g. the distance changes (by 15 km) while the time does not change, meaning the car would travel instantly (infinite speed), which is impossible	B1oe
Final answer: The vertical section shows the distance increasing by 15 km while the time stays the same (t = 10), which would mean travelling 15 km instantly (infinite speed) - impossible for a real car.

Mark scheme for Question 9 [2 marks]
Question 9[2 marks]
Answer or working	Marks
correct expansion 8x + 12 shown	M1
8x + 3	A1cao
Final answer: 8x + 3 (cao)

Mark scheme for Question 10 [2 marks]
Question 10[2 marks]
Answer or working	Marks
60 - 3×2, or equivalent subtraction method	M1
54 cm^2	A1cao

Mark scheme for Question 11 [2 marks]
Question 11[2 marks]
Answer or working	Marks
2 * 2 * 2 oe seen	M1
8	A1cao

Mark scheme for Question 12 [2 marks]
Question 12[2 marks]
Answer or working	Marks
calculate 50/10 = 5 m/s	M1
5 m/s and 18 km/h	A1cao

Mark scheme for Question 13 [3 marks]
Question 13[3 marks]
Answer or working	Marks
V = m / d	B1oe
250 / 1.25, using their (a) formula	M1ft
200 (cm^3)	A1cao
Final answer: V = m / d \| 200 cm^3

Mark scheme for Question 14 [3 marks]
Question 14[3 marks]
Answer or working	Marks
attempt to find the LCM of 15 and 20, e.g. multiples listed or prime factorisation used	M1
60 (minutes) found	A1
ft, 09:00 stated as the final answer	A1
Final answer: 09:00

Mark scheme for Question 15 [3 marks]
Question 15[3 marks]
Answer or working	Marks
converts 2/3 to 0.6666... (or awrt 0.667) so all four values are comparable	M1
correctly compares all four values	M1
correct order: 0.6, 0.6333..., 0.65, 2/3	A1
Final answer: 0.6, 0.6333..., 0.65, 2/3

Mark scheme for Question 16 [3 marks]
Question 16[3 marks]
Answer or working	Marks
converts 1 hour 20 minutes to hours = 4/3 hours	M1
uses 85 ÷ (4/3) or 85 × 3/4 to find speed	M1
63.75 mph	A1cao
Final answer: 63.75 mph. Working check: 1 h 20 min = 1 + 20/60 = 4/3 h; 85 ÷ (4/3) = 85 × 3/4 = 255/4 = 63.75.

Mark scheme for Question 17 [3 marks]
Question 17[3 marks]
Answer or working	Marks
state opposite angles in a cyclic quadrilateral sum to 180 degrees	M1oe
calculate 180 - 68	M1oe
112 degrees	A1cao
Final answer: 112 degrees; opposite angles in a cyclic quadrilateral sum to 180 degrees.

Mark scheme for Question 18 [3 marks]
Question 18[3 marks]
Answer or working	Marks
finds 35% of 340 (= 119), or uses multiplier 0.65	M1
subtracts their 35% from 340	M1oe
221 (pounds)	A1cao
Final answer: £221

Mark scheme for Question 19 [4 marks]
Question 19[4 marks]
Answer or working	Marks
calculate slope m = (6 - 0)/(3 - 8) = 6/(-5) = -6/5	M1
equation y = (-6/5)x + 48/5	A1cao
find midpoint M = ((8+3)/2, (0+6)/2) = (11/2, 3) and perpendicular slope m_perp = 5/6	M1
equation y - 3 = (5/6)(x - 11/2) leading to y = (5/6)x - 55/12 + 36/12 => y = (5/6)x - 19/12	A1cao
Final answer: Slope m = -6/5. Equation y = (-6/5)x + 48/5. (Check: substituting x = 8 gives y = (-6/5)*8 + 48/5 = -48/5 + 48/5 = 0.) \| Midpoint M = (11/2, 3). Perpendicular slope = 5/6. Equation: y - 3 = (5/6)(x - 11/2) which simplifies to y = (5/6)x - 19/12.

Mark scheme for Question 20 [4 marks]
Question 20[4 marks]
Answer or working	Marks
Convert length to cm or area to m^2 consistently, e.g. length 1.2 m = 120 cm then volume = area * length = 3.5 * 120 cm^3	M1
Use mass = density * volume: 7.8 * (3.5*120)	M1
3276 g (or equivalent)	A1cao
3.28 kg awrt (3 s.f.)	A1
Final answer: 3.28 kg

Mark scheme for Question 21 [6 marks]
Question 21[6 marks]
Answer or working	Marks
volume scale factor = 24^3 (=13824)	M1
90 x 13824	M1
1.24 x 10^6 cm^3 awrt (1244160 cm^3)	A1
area scale factor = 24^2 (=576)	M1
220 x 576 (=126720 cm^2), converted by dividing by 10000 to give m^2	M1
12.7 m^2 awrt (12.672 m^2)	A1
Final answer: 1.24 x 10^6 cm^3 (1244160 cm^3) \| 12.7 m^2 (12.672 m^2)

Mark scheme for Question 22 [5 marks]
Question 22[5 marks]
Answer or working	Marks
correct equation, 2x + 3x + 40 = 180 oe, or 5x + 40 = 180	B1
correct method to solve, 5x = 140	M1oe
x = 28	A1cao
substitute x = 28 (ft) into 2x and 3x	M1
56 degrees and 84 degrees	A1cao
Final answer: x = 28 \| 56 degrees and 84 degrees

Mark scheme for Question 23 [2 marks]
Question 23[2 marks]
Answer or working	Marks
Use area = ab sin(theta) = 10 × 7 × sin 60 degrees	M1
35 sqrt(3) cm^2	A1cao

Mark scheme for Question 24 [4 marks]
Question 24[4 marks]
Answer or working	Marks
multiplies numerator and denominator by (3 + 2sqrt(2))	M1
numerator simplifies to 15 + 10sqrt(2) + 6sqrt(2) + 8 (= 23 + 16sqrt(2))	M1
denominator simplifies to 9 - 8 = 1	A1
23 + 16sqrt(2) cao (p = 23, q = 16)	A1
Final answer: 23 + 16sqrt(2) (p = 23, q = 16)

Mark scheme for Question 25 [5 marks]
Question 25[5 marks]
Answer or working	Marks
linear scale factor (small cone : whole cone) = 8/24 (=1/3)	M1
volume ratio small:whole = (1/3)^3 = 1:27, so frustum:whole = 26:27	M1
1 : 26	A1cao
volume of small cone = 900 x 1/27 (=33.33..); frustum = 900 - their small cone volume	M1
866.67 cm^3 awrt (exact 866 2/3)	A1cao
Final answer: 1 : 26 \| 866.67 cm^3 (awrt, exact 866 2/3 cm^3)

Mark scheme for Question 26 [4 marks]
Question 26[4 marks]
Answer or working	Marks
lowest powers of common prime factors identified: 2^1 and 3^1	M1
6	A1cao
highest powers of all prime factors identified: 2^2, 3^3, 5^1	M1
540	A1cao
Final answer: 6 \| 540

Mark scheme for Question 27 [5 marks]
Question 27[5 marks]
Answer or working	Marks
use 1/2 * AB * AC * sin A = area	M1
substitute: 1/2 * 12 * AC * sin60 = 84	M1
rearrange to AC = 84 / (0.512sin60) or equivalent	M1
evaluate denominator correctly (sin60 = sqrt(3)/2) and compute AC	M1
16.17 m	A1cao