Exam Structure Set B: Higher Paper 2
An 80-mark, 90-minute calculator paper using the Edexcel 1MA1 paper structure as its reference.
Questions
Question 1 [1 marks]
Statistics and Probability
State one feature of a good line of best fit drawn on a scatter graph.
Question 2 [1 marks]
Angles and Geometrical Reasoning
A bearing of 315 degrees represents which compass direction?
Question 3 [2 marks]
Sequences
Here are the first four terms of a sequence. 2, 6, 10, 14
Write down the next two terms of the sequence.
Describe, in words, the term-to-term rule for the sequence.
Question 4 [2 marks]
Area, Volume and Measures
The diagram shows a rectangle.
Question 5 [3 marks]
Graphs and Coordinates
Complete the table of values for y = x^3 - 1. x : -2 , -1 , 0 , 1 , 2 y : -9 , _ , _ , _ , 7
Question 6 [1 marks]
Indices and Standard Form
Simplify sqrt(32) fully.
Question 7 [1 marks]
Ratio and Proportion
Which is cheaper per litre: a 750 ml bottle for £1.20 or a 1.5 litre bottle for £2.50? Give the cheaper option and the price per litre.
Question 8 [1 marks]
Fractions, Decimals and Percentages
Write 175% as a fraction in its simplest form.
Question 9 [1 marks]
Indices and Standard Form
Write the number 0.000000083 in standard form.
Question 10 [2 marks]
Fractions, Decimals and Percentages
Priya ate 2/5 of a cake. The part she ate weighed 160 g. Work out the weight of the whole cake in grams.
Question 11 [2 marks]
Area, Volume and Measures
A cylindrical candle has radius 3.5 cm and height 9 cm. Calculate the volume of wax needed to make the candle. Give your answer correct to 3 significant figures.
Question 12 [2 marks]
Graphs and Coordinates
Find the gradient of the line joining the points A(2, 5) and B(6, 13).
Question 13 [2 marks]
Quadratics
Factorise 5x^2 - 13x - 6
Question 14 [3 marks]
Number and Calculation
A combination lock has 3 dials. Each dial can be set to any digit from 0 to 9, and digits may repeat across dials. Work out the number of different combinations possible.
Question 15 [3 marks]
Quadratics
The graph of y = x^2 - 4x + 3 is a parabola. On graph paper, sketch this parabola. On your sketch, clearly mark the vertex and the x- and y-intercepts.
Question 16 [3 marks]
Angles and Geometrical Reasoning
A chord PQ of a circle with centre O has length 24 cm. The radius of the circle is 13 cm. M is the midpoint of PQ, and OM is drawn perpendicular to PQ. Diagram: circle, centre O, chord PQ = 24 cm, M marked as the midpoint of PQ, OM drawn perpendicular to PQ from the centre, radius OP = 13 cm marked.
Explain why PM = 12 cm.
Work out the length OM.
Question 17 [4 marks]
Ratio and Proportion
Salma and Tom share some money in the ratio 3:4. Salma receives £27.
Write Tom's share as a fraction of Salma's share.
Work out the total amount of money shared between Salma and Tom.
Question 18 [4 marks]
Quadratics
Write 5x^2 + 20x + 3 in the form a(x + p)^2 + q, stating the values of a, p and q.
Question 19 [5 marks]
Linear Algebra
Solve the simultaneous equations: 3x + 2y = 16 2x + 3y = 19
Question 20 [5 marks]
Pythagoras and Trigonometry
A circle has centre O and a chord AB of length 14 cm. The perpendicular from O to AB meets AB at M and OM = 6 cm. Find the radius of the circle.
Question 21 [5 marks]
Linear Algebra
Two mobile phone companies charge for calls as follows. Company A: a fixed monthly charge of £15 plus 5p per minute of calls. Company B: a fixed monthly charge of £9 plus 8p per minute of calls. Let C = the total monthly cost in pounds and m = the number of minutes of calls used in a month.
Write down an equation for C in terms of m for each company.
Find the number of minutes of calls, m, for which both companies charge the same monthly amount, and state this common cost.
Question 22 [6 marks]
Quadratics
A rectangle has length (x + 5) cm and width x cm. The area of the rectangle is 84 cm^2.
Show that x^2 + 5x - 84 = 0
Solve the equation to find the value of x, given that x > 0
Hence find the length of the rectangle
Question 23 [1 marks]
Pythagoras and Trigonometry
In triangle ABC, angle A and sides a and b are given (side a is opposite angle A). Which condition guarantees that the ambiguous case (two possible triangles) cannot occur?
Question 24 [2 marks]
Indices and Standard Form
Show that sqrt(3^5) = 3^(5/2).
Question 25 [5 marks]
Area, Volume and Measures
A factory makes solid rubber doorstops in the shape of a triangular prism. Each doorstop has a cross-section that is a right-angled triangle with base 6 cm and perpendicular height 4 cm, and the prism length is 11 cm.
Show that the volume of one doorstop is 132 cm^3.
The factory has a block of rubber with volume 3.168 m^3. Work out the maximum number of complete doorstops that can be made from this block of rubber, assuming there is no wastage.
Question 26 [6 marks]
Angles and Geometrical Reasoning
PAQ is a straight tangent line touching a circle at A. B and C are points on the circle, with chords AB and AC drawn, and BC also drawn to complete triangle ABC. Angle QAB = 47 degrees and angle PAC = 65 degrees (the angles between the tangent and each chord). Diagram: circle with straight tangent line PAQ touching at A, chords AB and AC drawn to points B and C on either side, chord BC also drawn, angle QAB = 47 degrees and angle PAC = 65 degrees marked between tangent and each chord.
State the size of angle ACB, giving a reason for your answer.
State the size of angle ABC, giving a reason for your answer.
Calculate the size of angle BAC.
Question 27 [7 marks]
Area, Volume and Measures
A sector-shaped garden ornament is cut from a sheet of metal. The sector has radius 18 cm. Its perimeter (the two straight edges plus the curved arc) is 54 cm. The metal costs £0.09 per cm^2.
Show that the arc length of the sector is 18 cm.
Find the angle of the sector, giving your answer correct to 1 decimal place.
Find the area of the sector, giving your answer correct to 3 significant figures.
Calculate the cost of the metal needed for the ornament.
Model solutions
| Question 1[1 mark] | |
|---|---|
| Answer or working | Marks |
| any correct feature, e.g. it follows the trend of the points with roughly equal numbers of points above and below it (oe), or it passes through the mean point | B1 |
| Final answer: It should follow the trend of the data with roughly equal numbers of points above and below the line. | |
| Question 2[1 mark] | |
|---|---|
| Answer or working | Marks |
| C selected | B1 |
| Final answer: C) North-West | |
| Question 3[2 marks] | |
|---|---|
| Answer or working | Marks |
| 18 and 22 both correct | B1oe |
| start at 2 and add 4 each time | B1oe |
| Final answer: 18, 22 | Start at 2 and add 4 each time (oe). | |
| Question 4[2 marks] | |
|---|---|
| Answer or working | Marks |
| 8 x 5 | M1 |
| 40 (cm^2) | A1cao |
| Final answer: 40 cm^2 | |
| Question 5[3 marks] | |
|---|---|
| Answer or working | Marks |
| y = -2 at x = -1 | B1 |
| y = -1 at x = 0 | B1 |
| y = 0 at x = 1 | B1 |
| Final answer: x = -1: y = -2; x = 0: y = -1; x = 1: y = 0 | |
| Question 6[1 mark] | |
|---|---|
| Answer or working | Marks |
| 4sqrt(2) oe | B1cao |
| Final answer: 4sqrt(2) | |
| Question 7[1 mark] | |
|---|---|
| Answer or working | Marks |
| 750 ml bottle cheaper, 120p/0.75 = 160p per litre | B1cao |
| Final answer: 750 ml bottle, 160p per litre (cheaper than 166.67p per litre for 1.5 L). | |
| Question 8[1 mark] | |
|---|---|
| Answer or working | Marks |
| 7/4 oe (e.g. 1 3/4) | B1cao |
| Final answer: 7/4 | |
| Question 9[1 mark] | |
|---|---|
| Answer or working | Marks |
| 8.3 x 10^-8 | B1oe |
| Question 10[2 marks] | |
|---|---|
| Answer or working | Marks |
| method: divide 160 by 2/5 or multiply by 5/2 | M1 |
| 400 g | A1cao |
| Question 11[2 marks] | |
|---|---|
| Answer or working | Marks |
| correct substitution into V = pi x r^2 x h, e.g. pi x 3.5^2 x 9 | M1oe |
| awrt 346 cm^3 | A1cao |
| Final answer: 346 cm^3 (3 s.f.) | |
| Question 12[2 marks] | |
|---|---|
| Answer or working | Marks |
| correct substitution into (y2 - y1)/(x2 - x1), e.g. (13-5)/(6-2) | M1 |
| 2 | A1cao |
| Question 13[2 marks] | |
|---|---|
| Answer or working | Marks |
| identifies a pair of numbers with product -30 and sum -13 (-15 and 2), or one correct bracket seen | M1 |
| (5x + 2)(x - 3) | A1cao |
| Question 14[3 marks] | |
|---|---|
| Answer or working | Marks |
| 10 choices identified for each dial | B1 |
| (dep) 10 * 10 * 10 | M1oe |
| 1000 | A1cao |
| Question 15[3 marks] | |
|---|---|
| Answer or working | Marks |
| complete sketch showing a parabola with correct shape and axis of symmetry x = 2 (vertex visible) | M1ft |
| vertex marked at (2, -1) | A1cao |
| x-intercepts at x = 1 and x = 3 and y-intercept at (0, 3) all marked | A1cao |
| Final answer: Vertex (2, -1); x-intercepts (1, 0) and (3, 0); y-intercept (0, 3). | |
| Question 16[3 marks] | |
|---|---|
| Answer or working | Marks |
| the perpendicular from the centre to a chord bisects the chord, so PM = 24 / 2 = 12 cm | B1oe |
| sqrt(13^2 - 12^2) | M1oe |
| 5 cm | A1cao |
| Final answer: The perpendicular from the centre to a chord bisects the chord, so PM = 24 / 2 = 12 cm. | 5 cm | |
| Question 17[4 marks] | |
|---|---|
| Answer or working | Marks |
| 4/3 | B1oe |
| 27 / 3 (= 9), the value of one part | M1 |
| 9 x (3 + 4), ft their value of one part | M1 |
| £63 | A1cao |
| Final answer: 4/3 | £63 | |
| Question 18[4 marks] | |
|---|---|
| Answer or working | Marks |
| 5(x^2 + 4x) + 3 seen, factor of 5 taken out correctly | M1 |
| (x + 2)^2 seen within the bracket | M1 |
| 5(x + 2)^2 - 20 + 3 correctly simplified (dep. on both M marks) | A1 |
| a = 5, p = 2, q = -17 all correct, fully simplified 5(x + 2)^2 - 17 | A1cao |
| Final answer: 5(x + 2)^2 - 17 (a = 5, p = 2, q = -17) | |
| Question 19[5 marks] | |
|---|---|
| Answer or working | Marks |
| multiplies one equation to match a coefficient, e.g. first equation x3 | M1 |
| multiplies the other equation to match the same coefficient, e.g. second equation x2 | M1 |
| subtracts or adds correctly to eliminate one variable | M1 |
| x = 2 | A1 |
| y = 5 | A1cao |
| Final answer: x = 2, y = 5 | |
| Question 20[5 marks] | |
|---|---|
| Answer or working | Marks |
| half-chord AM = 14/2 = 7 cm and right triangle OMA used | M1 |
| use Pythagoras: radius^2 = OM^2 + AM^2 | M1oe |
| compute radius^2 = 6^2 + 7^2 = 36 + 49 = 85 | M1 |
| radius = sqrt(85) | A1oe |
| give decimal to 3 s.f. 9.22 cm (sqrt(85)=9.2195...) | A1awrt |
| Final answer: 9.22 cm | |
| Question 21[5 marks] | |
|---|---|
| Answer or working | Marks |
| C = 15 + 0.05m | B1oe |
| C = 9 + 0.08m | B1oe |
| sets 15 + 0.05m = 9 + 0.08m | M1 |
| rearranges correctly to solve for m | M1 |
| m = 200 minutes and C = £25 (both required | A1cao |
| Final answer: C = 15 + 0.05m and C = 9 + 0.08m | m = 200 minutes, common cost = £25 | |
| Question 22[6 marks] | |
|---|---|
| Answer or working | Marks |
| forms the equation x(x + 5) = 84 | M1 |
| cso, rearranges correctly to x^2 + 5x - 84 = 0 | A1 |
| correctly factorises to (x + 12)(x - 7) = 0 | M1 |
| identifies both roots x = 7 and x = -12 | A1 |
| (dep) selects x = 7, rejecting x = -12 since x > 0 | A1 |
| 12 cm, ft from x = 7 in part (b) | B1 |
| Final answer: x^2 + 5x - 84 = 0 (shown) | x = 7 | 12 cm | |
| Question 23[1 mark] | |
|---|---|
| Answer or working | Marks |
| B | B1oe |
| Final answer: B) angle A is acute and a >= b | |
| Question 24[2 marks] | |
|---|---|
| Answer or working | Marks |
| rewrite sqrt as power 1/2: sqrt(3^5) = (3^5)^(1/2) | M1 |
| 3^(5/2) | A1cso |
| Final answer: 3^(5/2) ((3^5)^(1/2) = 3^(5 x 1/2) = 3^(5/2)). | |
| Question 25[5 marks] | |
|---|---|
| Answer or working | Marks |
| area of triangle = (1/2) x 6 x 4 = 12 cm^2 | M1 |
| 12 x 11 = 132 cm^3 fully shown with no errors | A1cso |
| converts 3.168 m^3 to cm^3, e.g. 3.168 x 1,000,000 (oe), = 3,168,000 cm^3 | M1 |
| 3,168,000 / 132 (oe, ft their cm^3 volume and the value from part (a)) | M1 |
| 24000 doorstops | A1cao |
| Final answer: 132 cm^3 (shown) | 24000 doorstops | |
| Question 26[6 marks] | |
|---|---|
| Answer or working | Marks |
| 47 degrees | B1 |
| reason: alternate segment theorem | B1 |
| 65 degrees | B1 |
| reason: alternate segment theorem | B1 |
| 180 - 47 - 65 | M1oe |
| 68 degrees | A1cao |
| Final answer: 47 degrees | 65 degrees | 68 degrees | |
| Question 27[7 marks] | |
|---|---|
| Answer or working | Marks |
| perimeter = 2 x 18 + arc length, i.e. 54 = 36 + arc length | M1oe |
| arc length = 54 - 36 = 18 cm, correctly shown | A1 |
| sets up 18 = (theta/360) x 2 x pi x 18 | M1oe |
| theta = 57.3 degrees awrt | A1cao |
| (their theta / 360) x pi x 18^2 oe, ft their part (b) | M1 |
| 162 cm^2 awrt | A1cao |
| their part (c) area x 0.09, cao = £14.58 (accept £14.58 or £14.58) | B1ft |
| Final answer: a) 18 cm (shown) b) 57.3 degrees c) 162 cm^2 d) £14.58 | |