Higher Tier - Set B, Paper 2

Exam Structure Set B: Higher Paper 2

An 80-mark, 90-minute calculator paper using the Edexcel 1MA1 paper structure as its reference.

27 questions - 80 marks - calculator allowed

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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

Mark scheme for Question 1 [1 mark]
Question 1[1 mark]
Answer or workingMarks
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 pointB1
Final answer: It should follow the trend of the data with roughly equal numbers of points above and below the line.
Mark scheme for Question 2 [1 mark]
Question 2[1 mark]
Answer or workingMarks
C selectedB1
Final answer: C) North-West
Mark scheme for Question 3 [2 marks]
Question 3[2 marks]
Answer or workingMarks
18 and 22 both correctB1oe
start at 2 and add 4 each timeB1oe
Final answer: 18, 22 | Start at 2 and add 4 each time (oe).
Mark scheme for Question 4 [2 marks]
Question 4[2 marks]
Answer or workingMarks
8 x 5M1
40 (cm^2)A1cao
Final answer: 40 cm^2
Mark scheme for Question 5 [3 marks]
Question 5[3 marks]
Answer or workingMarks
y = -2 at x = -1B1
y = -1 at x = 0B1
y = 0 at x = 1B1
Final answer: x = -1: y = -2; x = 0: y = -1; x = 1: y = 0
Mark scheme for Question 6 [1 mark]
Question 6[1 mark]
Answer or workingMarks
4sqrt(2) oeB1cao
Final answer: 4sqrt(2)
Mark scheme for Question 7 [1 mark]
Question 7[1 mark]
Answer or workingMarks
750 ml bottle cheaper, 120p/0.75 = 160p per litreB1cao
Final answer: 750 ml bottle, 160p per litre (cheaper than 166.67p per litre for 1.5 L).
Mark scheme for Question 8 [1 mark]
Question 8[1 mark]
Answer or workingMarks
7/4 oe (e.g. 1 3/4)B1cao
Final answer: 7/4
Mark scheme for Question 9 [1 mark]
Question 9[1 mark]
Answer or workingMarks
8.3 x 10^-8B1oe
Mark scheme for Question 10 [2 marks]
Question 10[2 marks]
Answer or workingMarks
method: divide 160 by 2/5 or multiply by 5/2M1
400 gA1cao
Mark scheme for Question 11 [2 marks]
Question 11[2 marks]
Answer or workingMarks
correct substitution into V = pi x r^2 x h, e.g. pi x 3.5^2 x 9M1oe
awrt 346 cm^3A1cao
Final answer: 346 cm^3 (3 s.f.)
Mark scheme for Question 12 [2 marks]
Question 12[2 marks]
Answer or workingMarks
correct substitution into (y2 - y1)/(x2 - x1), e.g. (13-5)/(6-2)M1
2A1cao
Mark scheme for Question 13 [2 marks]
Question 13[2 marks]
Answer or workingMarks
identifies a pair of numbers with product -30 and sum -13 (-15 and 2), or one correct bracket seenM1
(5x + 2)(x - 3)A1cao
Mark scheme for Question 14 [3 marks]
Question 14[3 marks]
Answer or workingMarks
10 choices identified for each dialB1
(dep) 10 * 10 * 10M1oe
1000A1cao
Mark scheme for Question 15 [3 marks]
Question 15[3 marks]
Answer or workingMarks
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 markedA1cao
Final answer: Vertex (2, -1); x-intercepts (1, 0) and (3, 0); y-intercept (0, 3).
Mark scheme for Question 16 [3 marks]
Question 16[3 marks]
Answer or workingMarks
the perpendicular from the centre to a chord bisects the chord, so PM = 24 / 2 = 12 cmB1oe
sqrt(13^2 - 12^2)M1oe
5 cmA1cao
Final answer: The perpendicular from the centre to a chord bisects the chord, so PM = 24 / 2 = 12 cm. | 5 cm
Mark scheme for Question 17 [4 marks]
Question 17[4 marks]
Answer or workingMarks
4/3B1oe
27 / 3 (= 9), the value of one partM1
9 x (3 + 4), ft their value of one partM1
£63A1cao
Final answer: 4/3 | £63
Mark scheme for Question 18 [4 marks]
Question 18[4 marks]
Answer or workingMarks
5(x^2 + 4x) + 3 seen, factor of 5 taken out correctlyM1
(x + 2)^2 seen within the bracketM1
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 - 17A1cao
Final answer: 5(x + 2)^2 - 17 (a = 5, p = 2, q = -17)
Mark scheme for Question 19 [5 marks]
Question 19[5 marks]
Answer or workingMarks
multiplies one equation to match a coefficient, e.g. first equation x3M1
multiplies the other equation to match the same coefficient, e.g. second equation x2M1
subtracts or adds correctly to eliminate one variableM1
x = 2A1
y = 5A1cao
Final answer: x = 2, y = 5
Mark scheme for Question 20 [5 marks]
Question 20[5 marks]
Answer or workingMarks
half-chord AM = 14/2 = 7 cm and right triangle OMA usedM1
use Pythagoras: radius^2 = OM^2 + AM^2M1oe
compute radius^2 = 6^2 + 7^2 = 36 + 49 = 85M1
radius = sqrt(85)A1oe
give decimal to 3 s.f. 9.22 cm (sqrt(85)=9.2195...)A1awrt
Final answer: 9.22 cm
Mark scheme for Question 21 [5 marks]
Question 21[5 marks]
Answer or workingMarks
C = 15 + 0.05mB1oe
C = 9 + 0.08mB1oe
sets 15 + 0.05m = 9 + 0.08mM1
rearranges correctly to solve for mM1
m = 200 minutes and C = £25 (both requiredA1cao
Final answer: C = 15 + 0.05m and C = 9 + 0.08m | m = 200 minutes, common cost = £25
Mark scheme for Question 22 [6 marks]
Question 22[6 marks]
Answer or workingMarks
forms the equation x(x + 5) = 84M1
cso, rearranges correctly to x^2 + 5x - 84 = 0A1
correctly factorises to (x + 12)(x - 7) = 0M1
identifies both roots x = 7 and x = -12A1
(dep) selects x = 7, rejecting x = -12 since x > 0A1
12 cm, ft from x = 7 in part (b)B1
Final answer: x^2 + 5x - 84 = 0 (shown) | x = 7 | 12 cm
Mark scheme for Question 23 [1 mark]
Question 23[1 mark]
Answer or workingMarks
BB1oe
Final answer: B) angle A is acute and a >= b
Mark scheme for Question 24 [2 marks]
Question 24[2 marks]
Answer or workingMarks
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)).
Mark scheme for Question 25 [5 marks]
Question 25[5 marks]
Answer or workingMarks
area of triangle = (1/2) x 6 x 4 = 12 cm^2M1
12 x 11 = 132 cm^3 fully shown with no errorsA1cso
converts 3.168 m^3 to cm^3, e.g. 3.168 x 1,000,000 (oe), = 3,168,000 cm^3M1
3,168,000 / 132 (oe, ft their cm^3 volume and the value from part (a))M1
24000 doorstopsA1cao
Final answer: 132 cm^3 (shown) | 24000 doorstops
Mark scheme for Question 26 [6 marks]
Question 26[6 marks]
Answer or workingMarks
47 degreesB1
reason: alternate segment theoremB1
65 degreesB1
reason: alternate segment theoremB1
180 - 47 - 65M1oe
68 degreesA1cao
Final answer: 47 degrees | 65 degrees | 68 degrees
Mark scheme for Question 27 [7 marks]
Question 27[7 marks]
Answer or workingMarks
perimeter = 2 x 18 + arc length, i.e. 54 = 36 + arc lengthM1oe
arc length = 54 - 36 = 18 cm, correctly shownA1
sets up 18 = (theta/360) x 2 x pi x 18M1oe
theta = 57.3 degrees awrtA1cao
(their theta / 360) x pi x 18^2 oe, ft their part (b)M1
162 cm^2 awrtA1cao
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