Higher Tier - Year 10

Year 10 Paper 4: Ratio and Geometry

Covers linear algebra, ratio and proportion, indices and standard form, graphs and coordinates, angles and geometrical reasoning, and area, volume and measures.

20 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.

Download printable PDF

Questions

Question 1 [1 marks]

Indices and Standard Form

Write the number 45000 in standard form.

Question 2 [2 marks]

Angles and Geometrical Reasoning

The diagram shows a solid made from four identical 1 cm cubes joined edge-to-edge in a single straight row. Draw the front elevation of the solid, as viewed from arrow F, and state its dimensions.

Question 3 [2 marks]

Graphs and Coordinates

A line has equation y = -2x + 7.

Write down the gradient of the line.

Write down the coordinates of the point where the line crosses the y-axis.

Question 4 [3 marks]

Ratio and Proportion

Declan is deciding how to pay for his bus travel for one day. A day ticket costs £4.50 and allows unlimited journeys. A single ticket costs £1.20 and Declan plans to make 4 journeys that day.

Question 5 [1 marks]

Linear Algebra

A box contains 8 chocolates. Write down an expression for the number of chocolates in k boxes.

Question 6 [1 marks]

Indices and Standard Form

Work out the cube root of -8.

Question 7 [1 marks]

Graphs and Coordinates

Write down the gradient of the line with equation y = 7x + 2.

Question 8 [1 marks]

Linear Algebra

Given that y = 2x + 3, which of the following correctly makes x the subject of the formula?

Question 9 [2 marks]

Graphs and Coordinates

Plot the three points from the table on graph paper and draw the straight line joining them. x: 0, 2, 4 y: -1, 0, 1

Question 10 [3 marks]

Angles and Geometrical Reasoning

Enlarge triangle T with vertices (2, 1), (4, 1) and (2, 3) by a scale factor of -2 centre (1, 1). Find the coordinates of the image vertices.

Question 11 [2 marks]

Linear Algebra

Anjali has p pounds. She has more than £15 but no more than £40. Write down an inequality, in terms of p, to show this information.

Question 12 [3 marks]

Angles and Geometrical Reasoning

A regular hexagon has an interior angle of 120 degrees. A regular nonagon (a 9-sided polygon) has an interior angle of 140 degrees. Using calculations, explain why regular hexagons tessellate but regular nonagons do not.

Question 13 [2 marks]

Linear Algebra

Let f(x) = 2x - 7 and g(x) = x + 3. Work out g(f(2)).

Question 14 [4 marks]

Area, Volume and Measures

A sector has arc length 15 cm and area 90 cm^2. Find the radius of the circle. Give your answer correct to 3 significant figures.

Question 15 [4 marks]

Linear Algebra

The area of a trapezium with parallel sides a and b and height h is given by A = (1/2)(a + b)h

Make h the subject of the formula.

Make a the subject of the formula.

Question 16 [4 marks]

Graphs and Coordinates

A cubic graph has the equation y = (x - 1)(x + 3)(x - 2).

Find the three x-coordinates of the points where the graph crosses the x-axis.

Find the coordinates of the point where the graph crosses the y-axis.

Question 17 [5 marks]

Linear Algebra

Rearrange the formula for the frequency f of a wave: v = f * lambda, to make lambda the subject. Then a wave has speed v = 330 m/s and frequency f = 440 Hz. Calculate the wavelength lambda to 3 significant figures.

Make lambda the subject of v = f * lambda.

Calculate lambda for v = 330 m/s and f = 440 Hz and give the answer to 3 significant figures.

Question 18 [5 marks]

Area, Volume and Measures

An isosceles triangular prism has a cross-section with base 6 cm and two equal sides of 5 cm. The prism has length 12 cm.

Show that the perpendicular height of the triangular cross-section is 4 cm.

Work out the total surface area of the prism.

Question 19 [6 marks]

Angles and Geometrical Reasoning

Lines AB and CD are crossed by a straight transversal EF at G (on AB) and H (on CD). Angle EGB = (5x - 12) degrees and angle GHD = (3x + 8) degrees; these two angles are co-interior.

Find the value of x for which AB would be parallel to CD.

Priya says that, in fact, x = 11, and that AB is parallel to CD. Determine, with full reasons, whether Priya is correct.

Question 20 [8 marks]

Angles and Geometrical Reasoning

Show that the vectors (2, -1), (5, 2) and (7, 1) are coplanar in R^2 and then determine whether they are linearly dependent by finding constants x, y, z, not all zero, such that x(2, -1) + y(5, 2) + z(7, 1) = (0, 0). Give one nontrivial solution.

Model solutions

Mark scheme for Question 1 [1 mark]
Question 1[1 mark]
Answer or workingMarks
4.5 x 10^4B1oe
Mark scheme for Question 2 [2 marks]
Question 2[2 marks]
Answer or workingMarks
correct outline drawn: a row of 4 unit squares (rectangle 4 cm by 1 cm)B1
correct dimensions stated, 4 cm by 1 cmB1oe
Final answer: Rectangle 4 cm long by 1 cm high (a row of four 1 cm squares).
Mark scheme for Question 3 [2 marks]
Question 3[2 marks]
Answer or workingMarks
-2B1cao
(0, 7)B1cao
Final answer: -2 | (0, 7)
Mark scheme for Question 4 [3 marks]
Question 4[3 marks]
Answer or workingMarks
4 x 1.20M1oe
4.80A1cao
correct conclusion that the day ticket is cheaper, with correct saving statedC1
Final answer: The day ticket is cheaper, saving 30p (£4.50 compared with £4.80 for 4 single tickets)
Mark scheme for Question 5 [1 mark]
Question 5[1 mark]
Answer or workingMarks
8k cao (oe k x 8)B1
Final answer: 8k
Mark scheme for Question 6 [1 mark]
Question 6[1 mark]
Answer or workingMarks
-2B1cao
Mark scheme for Question 7 [1 mark]
Question 7[1 mark]
Answer or workingMarks
7B1cao
Mark scheme for Question 8 [1 mark]
Question 8[1 mark]
Answer or workingMarks
AB1
Mark scheme for Question 9 [2 marks]
Question 9[2 marks]
Answer or workingMarks
at least two points plotted correctly and a straight line drawn through themM1oe
all three points plotted accurately and correct straight lineA1cao
Final answer: Line with equation y = 1/2 x - 1, passing through (0,-1), (2,0), (4,1).
Mark scheme for Question 10 [3 marks]
Question 10[3 marks]
Answer or workingMarks
subtract centre (1,1) then multiply by -2 for each coordinate (method shown)M1oe
image of (2,1) is (-1,1)A1cao
images of other vertices: (4,1) -> (-5,1) and (2,3) -> (-1,-3)A1cao
Final answer: Images: (2,1) -> (-1,1); (4,1) -> (-5,1); (2,3) -> (-1,-3).
Mark scheme for Question 11 [2 marks]
Question 11[2 marks]
Answer or workingMarks
15 < p oe or p <= 40 oe, one boundary correctly representedM1
15 < p <= 40 oe cao, both boundaries correct with correct strict/non-strict signsA1
Final answer: 15 < p <= 40
Mark scheme for Question 12 [3 marks]
Question 12[3 marks]
Answer or workingMarks
360 / 120 = 3M1oe
360 / 140 = 2.57... (not a whole number)M1oe
correct conclusion linking calculations: hexagons tessellate because exactly 3 hexagons meet at a point (3 x 120 = 360); nonagons do not because a whole number of 140 degree angles cannot make exactly 360A1
Final answer: Hexagons tessellate (360/120 = 3 exactly); nonagons do not (360/140 is not a whole number).
Mark scheme for Question 13 [2 marks]
Question 13[2 marks]
Answer or workingMarks
substitute 2 into f, e.g. f(2) = 2 x 2 - 7 = -3M1
0 cao (g(-3) = -3 + 3 = 0)A1
Final answer: 0 (f(2) = -3, g(-3) = 0).
Mark scheme for Question 14 [4 marks]
Question 14[4 marks]
Answer or workingMarks
use relationship A = 1/2 r s (from Q6) or area = 1/2 * r * arcM1
substitute A = 90 and s = 15 to get 90 = 1/2 * r * 15M1
solve for r: r = 90 * 2 / 15 = 12M1
12.0 cmA1cao
Mark scheme for Question 15 [4 marks]
Question 15[4 marks]
Answer or workingMarks
multiplies both sides by 2, e.g. 2A = (a + b)hM1
h = 2A/(a + b) oeA1cao
multiplies by 2 and divides by h, e.g. 2A/h = a + bM1
a = 2A/h - b oeA1cao
Final answer: h = 2A/(a + b) | a = 2A/h - b
Mark scheme for Question 16 [4 marks]
Question 16[4 marks]
Answer or workingMarks
sets each factor equal to 0M1
x = 1, x = -3, x = 2, all correct (any order)A1
substitutes x = 0 into y = (x-1)(x+3)(x-2)M1
(0, 6)A1cao
Final answer: x = 1, x = -3, x = 2 | (0, 6)
Mark scheme for Question 17 [5 marks]
Question 17[5 marks]
Answer or workingMarks
lambda = v/fB1cao
substitute values and divide: lambda = 330/440M1
0.75A1cao
units m stated or impliedB1
answer correct to 3 s.f. (0.75) awardSC1
Final answer: lambda = v/f | 0.75 m
Mark scheme for Question 18 [5 marks]
Question 18[5 marks]
Answer or workingMarks
sqrt(5^2 - 3^2) or 5^2 - 3^2 (= 16) seenM1
4 cmA1cso
triangle area 0.5x6x4 (= 12), doubled (= 24)M1
three rectangles: 6x12, 5x12, 5x12 (= 72, 60, 60) summedM1
216 cm^2A1cao
Final answer: 4 cm | 216 cm^2
Mark scheme for Question 19 [6 marks]
Question 19[6 marks]
Answer or workingMarks
(5x - 12) + (3x + 8) = 180 oe, using co-interior angles sum to 180 for parallel linesM1
8x - 4 = 180 (oe simplification)M1
x = 23A1cao
substitutes x = 11 into both expressions: 5(11) - 12 = 43 and 3(11) + 8 = 41M1
43 + 41 = 84, compared with 180M1
correct conclusion: since 84 is not equal to 180, AB is not parallel to CD, so Priya is incorrectA1
Final answer: x = 23 | Priya is incorrect; when x = 11 the co-interior angles sum to 84 degrees, not 180 degrees, so AB is not parallel to CD
Mark scheme for Question 20 [8 marks]
Question 20[8 marks]
Answer or workingMarks
Note that in R^2 any three vectors are coplanar; state this or show they lie in R^2M1
Set up equations for linear dependence: 2x + 5y + 7z = 0 and -x + 2y + z = 0M1
Solve system: eliminate x using x = 2y + z from second equation or solve by eliminationM1
Substitute x = 2y + z into first: 2(2y+z) + 5y + 7z = 0 => 4y + 2z + 5y + 7z = 0M1
Combine to get 9y + 9z = 0 so y = -zM1
Choose z = 1 then y = -1 and x = 2y + z = 2( -1) + 1 = -1M1
Provide nontrivial solution x = -1, y = -1, z = 1 and conclude vectors are linearlyA1dependent
State that the vectors are coplanar (in R^2) and linearly dependent since a nonzero solution existsA1
Final answer: Coplanar in R^2. One nontrivial solution is x = -1, y = -1, z = 1 since -1*(2,-1) + -1*(5,2) + 1*(7,1) = (0,0).