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.
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.
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
| Question 1[1 mark] | |
|---|---|
| Answer or working | Marks |
| 4.5 x 10^4 | B1oe |
| Question 2[2 marks] | |
|---|---|
| Answer or working | Marks |
| correct outline drawn: a row of 4 unit squares (rectangle 4 cm by 1 cm) | B1 |
| correct dimensions stated, 4 cm by 1 cm | B1oe |
| Final answer: Rectangle 4 cm long by 1 cm high (a row of four 1 cm squares). | |
| Question 3[2 marks] | |
|---|---|
| Answer or working | Marks |
| -2 | B1cao |
| (0, 7) | B1cao |
| Final answer: -2 | (0, 7) | |
| Question 4[3 marks] | |
|---|---|
| Answer or working | Marks |
| 4 x 1.20 | M1oe |
| 4.80 | A1cao |
| correct conclusion that the day ticket is cheaper, with correct saving stated | C1 |
| Final answer: The day ticket is cheaper, saving 30p (£4.50 compared with £4.80 for 4 single tickets) | |
| Question 5[1 mark] | |
|---|---|
| Answer or working | Marks |
| 8k cao (oe k x 8) | B1 |
| Final answer: 8k | |
| Question 6[1 mark] | |
|---|---|
| Answer or working | Marks |
| -2 | B1cao |
| Question 7[1 mark] | |
|---|---|
| Answer or working | Marks |
| 7 | B1cao |
| Question 8[1 mark] | |
|---|---|
| Answer or working | Marks |
| A | B1 |
| Question 9[2 marks] | |
|---|---|
| Answer or working | Marks |
| at least two points plotted correctly and a straight line drawn through them | M1oe |
| all three points plotted accurately and correct straight line | A1cao |
| Final answer: Line with equation y = 1/2 x - 1, passing through (0,-1), (2,0), (4,1). | |
| Question 10[3 marks] | |
|---|---|
| Answer or working | Marks |
| 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). | |
| Question 11[2 marks] | |
|---|---|
| Answer or working | Marks |
| 15 < p oe or p <= 40 oe, one boundary correctly represented | M1 |
| 15 < p <= 40 oe cao, both boundaries correct with correct strict/non-strict signs | A1 |
| Final answer: 15 < p <= 40 | |
| Question 12[3 marks] | |
|---|---|
| Answer or working | Marks |
| 360 / 120 = 3 | M1oe |
| 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 360 | A1 |
| Final answer: Hexagons tessellate (360/120 = 3 exactly); nonagons do not (360/140 is not a whole number). | |
| Question 13[2 marks] | |
|---|---|
| Answer or working | Marks |
| substitute 2 into f, e.g. f(2) = 2 x 2 - 7 = -3 | M1 |
| 0 cao (g(-3) = -3 + 3 = 0) | A1 |
| Final answer: 0 (f(2) = -3, g(-3) = 0). | |
| Question 14[4 marks] | |
|---|---|
| Answer or working | Marks |
| use relationship A = 1/2 r s (from Q6) or area = 1/2 * r * arc | M1 |
| substitute A = 90 and s = 15 to get 90 = 1/2 * r * 15 | M1 |
| solve for r: r = 90 * 2 / 15 = 12 | M1 |
| 12.0 cm | A1cao |
| Question 15[4 marks] | |
|---|---|
| Answer or working | Marks |
| multiplies both sides by 2, e.g. 2A = (a + b)h | M1 |
| h = 2A/(a + b) oe | A1cao |
| multiplies by 2 and divides by h, e.g. 2A/h = a + b | M1 |
| a = 2A/h - b oe | A1cao |
| Final answer: h = 2A/(a + b) | a = 2A/h - b | |
| Question 16[4 marks] | |
|---|---|
| Answer or working | Marks |
| sets each factor equal to 0 | M1 |
| 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) | |
| Question 17[5 marks] | |
|---|---|
| Answer or working | Marks |
| lambda = v/f | B1cao |
| substitute values and divide: lambda = 330/440 | M1 |
| 0.75 | A1cao |
| units m stated or implied | B1 |
| answer correct to 3 s.f. (0.75) award | SC1 |
| Final answer: lambda = v/f | 0.75 m | |
| Question 18[5 marks] | |
|---|---|
| Answer or working | Marks |
| sqrt(5^2 - 3^2) or 5^2 - 3^2 (= 16) seen | M1 |
| 4 cm | A1cso |
| triangle area 0.5x6x4 (= 12), doubled (= 24) | M1 |
| three rectangles: 6x12, 5x12, 5x12 (= 72, 60, 60) summed | M1 |
| 216 cm^2 | A1cao |
| Final answer: 4 cm | 216 cm^2 | |
| Question 19[6 marks] | |
|---|---|
| Answer or working | Marks |
| (5x - 12) + (3x + 8) = 180 oe, using co-interior angles sum to 180 for parallel lines | M1 |
| 8x - 4 = 180 (oe simplification) | M1 |
| x = 23 | A1cao |
| substitutes x = 11 into both expressions: 5(11) - 12 = 43 and 3(11) + 8 = 41 | M1 |
| 43 + 41 = 84, compared with 180 | M1 |
| correct conclusion: since 84 is not equal to 180, AB is not parallel to CD, so Priya is incorrect | A1 |
| 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 | |
| Question 20[8 marks] | |
|---|---|
| Answer or working | Marks |
| Note that in R^2 any three vectors are coplanar; state this or show they lie in R^2 | M1 |
| Set up equations for linear dependence: 2x + 5y + 7z = 0 and -x + 2y + z = 0 | M1 |
| Solve system: eliminate x using x = 2y + z from second equation or solve by elimination | M1 |
| Substitute x = 2y + z into first: 2(2y+z) + 5y + 7z = 0 => 4y + 2z + 5y + 7z = 0 | M1 |
| Combine to get 9y + 9z = 0 so y = -z | M1 |
| Choose z = 1 then y = -1 and x = 2y + z = 2( -1) + 1 = -1 | M1 |
| Provide nontrivial solution x = -1, y = -1, z = 1 and conclude vectors are linearly | A1dependent |
| State that the vectors are coplanar (in R^2) and linearly dependent since a nonzero solution exists | A1 |
| 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). | |