Higher Tier - Set B, Paper 3

Exam Structure Set B: Higher Paper 3

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 [2 marks]

Area, Volume and Measures

Rectangle A measures 5 cm by 8 cm.

Which of these rectangles is mathematically similar to rectangle A?

Write down the scale factor of enlargement from rectangle A to the rectangle you chose in part (a).

Question 2 [2 marks]

Fractions, Decimals and Percentages

Write 0.3333... (3 recurring) as a fraction in its simplest form.

Question 3 [5 marks]

Statistics and Probability

Priya Chandra manages Riverside Car Park. She recorded how many cars and vans were parked there in the morning and in the afternoon.

Question 4 [1 marks]

Angles and Geometrical Reasoning

A student is asked to enlarge a triangle by scale factor -2 from a given centre. The student draws the image on the same side of the centre as the original triangle, but twice the size. Explain the error in the student's diagram.

Question 5 [1 marks]

Graphs and Coordinates

Write down the equation of the graph obtained by reflecting y = f(x) in the y-axis.

Question 6 [1 marks]

Linear Algebra

Simplify 7y - 2y + 4y.

Question 7 [1 marks]

Indices and Standard Form

Work out the value of 16^(1/2).

Question 8 [2 marks]

Ratio and Proportion

A recipe for 6 portions uses 300 g of flour. Work out how much flour is needed for 15 portions, assuming the amount of flour is directly proportional to the number of portions.

Question 9 [2 marks]

Area, Volume and Measures

A sphere has radius 3 cm. Work out the volume of the sphere. Give your answer as a multiple of pi.

Question 10 [2 marks]

Linear Algebra

Make b the subject of the formula: a = (b + 11)/6

Question 11 [3 marks]

Number and Calculation

Find the highest common factor (HCF) of 24 and 36. You must show your method.

Question 12 [3 marks]

Quadratics

Consider the equation x^3 + 4x - 10 = 0 and the iteration x_{n+1} = (10 - x_n^3)/4. Starting with x_0 = 1.5, calculate x_1 and x_2. From these two iterations, state whether the sequence appears to be converging.

Question 13 [3 marks]

Pythagoras and Trigonometry

A cone has base radius 5 cm and perpendicular height 12 cm. Diagram: cone shown in cross-section, base radius marked 5 cm from the centre of the base to the edge, perpendicular height marked 12 cm from the centre of the base to the apex, and the slant height labelled l cm from the apex to the edge of the base. Calculate the slant height, l, of the cone.

Question 14 [4 marks]

Sequences

A sequence has nth term given by T_n = 5n - 3 for n odd, and T_n = 5n + 1 for n even. (a) Work out T_5 and T_6. (b) Explain briefly whether this sequence is arithmetic overall.

Question 15 [4 marks]

Indices and Standard Form

a = 3 x 10^5 b = 4 x 10^-2

Work out a x b. Give your answer in standard form.

Work out a / b. Give your answer in standard form.

Question 16 [4 marks]

Pythagoras and Trigonometry

An isosceles triangle has two equal sides of length 10 cm and a base of length 12 cm. A line is drawn from the apex of the triangle perpendicular to the base, meeting the base at its midpoint. Diagram: isosceles triangle with the two equal sides marked 10 cm, base marked 12 cm, and a dashed perpendicular line from the apex to the midpoint of the base labelled h cm.

Explain why the perpendicular line splits the base into two lengths of 6 cm each.

Calculate the perpendicular height, h, of the triangle.

Question 17 [4 marks]

Linear Algebra

Solve the simultaneous equations: y = 3x - 2 2x + y = 13

Question 18 [4 marks]

Number and Calculation

A rectangle has length 3 times its width. The perimeter of the rectangle is 96 cm. (a) Find the width and the length. (b) Work out the area of the rectangle.

Question 19 [4 marks]

Fractions, Decimals and Percentages

The table shows the number of apples picked at three orchards and the fraction of each orchard's apples that were sold at a farmers' market. Orchard | Apples picked | Fraction sold at market Ashcroft Farm | 96 | 3/8 Bramble Hill | 105 | 2/5 Cedar Grove | 90 | 7/9 Work out which orchard sold the most apples at the market. You must show your working to support your answer.

Question 20 [7 marks]

Statistics and Probability

Amelia recorded the number of text messages she sent on each of 13 days: 2, 5, 6, 6, 7, 8, 9, 9, 10, 12, 14, 15, 20

Find the median number of texts sent.

Find the lower quartile and the upper quartile of the number of texts sent.

Hence find the interquartile range of the number of texts sent.

Draw a box plot to represent Amelia's data on the grid provided.

Question 21 [1 marks]

Quadratics

An iteration formula is x_(n+1) = 3/(x_n - 2). Which of these values of x0 would make it impossible to calculate x1?

Question 22 [2 marks]

Statistics and Probability

In a capture-recapture study, n1 = 40 animals are marked and released. In the second sample, m = 5 of the recaptured animals are marked, and this gives a Lincoln-Petersen population estimate of N = 240. Find n2, the number of animals caught in the second sample.

Question 23 [3 marks]

Indices and Standard Form

Expand and simplify (3 + sqrt(5))(5 - sqrt(5)).

Question 24 [5 marks]

Area, Volume and Measures

A rectangular lawn ABCD has AB = 12 m and BC = 7 m. An isosceles triangular flower bed is attached to side AB, outside the rectangle, with the two equal slant sides of the triangle each measuring 7.5 m. Work out the total area of the lawn and the flower bed together.

Question 25 [3 marks]

Statistics and Probability

A histogram represents the finishing times, in minutes, of 200 athletes in a marathon, grouped into three classes: 180 <= t < 190 (width 10), 190 <= t < 210 (width 20), and 210 <= t < 240 (width 30). The heights of the three bars are in the ratio 3 : 2 : 1 respectively.

Question 26 [4 marks]

Graphs and Coordinates

A bus leaves a stop and accelerates uniformly to a speed of 12 m/s in 15 seconds. It then travels at 12 m/s for 40 seconds, before decelerating uniformly to rest in 10 seconds. The bus then waits at a second stop for 20 seconds, before accelerating uniformly back up to 12 m/s over the next 12 seconds.

Question 27 [3 marks]

Statistics and Probability

Using the cumulative frequency graph for the fun run times (60 runners, shown in Question 3), the organisers decide to award a medal to the fastest 20% of runners. Estimate the completion time below which a runner receives a medal.

Model solutions

Mark scheme for Question 1 [2 marks]
Question 1[2 marks]
Answer or workingMarks
B) 15 cm by 24 cmB1
3 cao ft their (a)B1
Final answer: a) B b) 3
Mark scheme for Question 2 [2 marks]
Question 2[2 marks]
Answer or workingMarks
3/9 seen or equivalent unsimplified fraction with denominator 9M1
1/3A1cao
Mark scheme for Question 3 [5 marks]
Question 3[5 marks]
Answer or workingMarks
Afternoon column Total = 60B1cao
Vans Morning = 15B1cao
Cars Afternoon = 35B1cao
Cars Total = 80B1cao
Vans Total = 40B1cao
Final answer: Afternoon Total = 60, Vans Morning = 15, Cars Afternoon = 35, Cars Total = 80, Vans Total = 40
Mark scheme for Question 4 [1 mark]
Question 4[1 mark]
Answer or workingMarks
the image should be drawn on the opposite side of the centre of enlargement (and inverted), because the scale factor is negativeB1oe
Final answer: The image should be on the opposite side of the centre of enlargement (and inverted), because the scale factor is negative.
Mark scheme for Question 5 [1 mark]
Question 5[1 mark]
Answer or workingMarks
y = f(-x)B1cao
Mark scheme for Question 6 [1 mark]
Question 6[1 mark]
Answer or workingMarks
9yB1cao
Final answer: 9y (cao)
Mark scheme for Question 7 [1 mark]
Question 7[1 mark]
Answer or workingMarks
4B1cao
Mark scheme for Question 8 [2 marks]
Question 8[2 marks]
Answer or workingMarks
use direct proportion: 300/6 * 15 or equivalent methodM1
750 gA1cao
Final answer: 750 g (grams). This is 300/6 * 15 = 50 * 15 = 750 g. Working check: 6 portions -> 300 g, so per portion 50 g; 15 portions -> 750 g.
Mark scheme for Question 9 [2 marks]
Question 9[2 marks]
Answer or workingMarks
correct substitution (4/3) x pi x 3^3M1oe
36 x pi cm^3 (oe, cao - exact value required, not a decimal)A1
Final answer: 36 x pi cm^3 (exact)
Mark scheme for Question 10 [2 marks]
Question 10[2 marks]
Answer or workingMarks
multiplies both sides by 6, e.g. 6a = b + 11M1
b = 6a - 11 oeA1cao
Final answer: b = 6a - 11
Mark scheme for Question 11 [3 marks]
Question 11[3 marks]
Answer or workingMarks
factors of 24 listed: 1, 2, 3, 4, 6, 8, 12, 24 (or use of prime factorisation)M1
factors of 36 listed: 1, 2, 3, 4, 6, 9, 12, 18, 36 and common factors identifiedM1
12A1cao
Mark scheme for Question 12 [3 marks]
Question 12[3 marks]
Answer or workingMarks
calculate x_1 = (10 - 1.5^3)/4 = 1.65625M1
calculate x_2 = (10 - x_1^3)/4 = 1.36415863M1awrt
conclude that the sequence does not appear to be converging (it oscillates)B1oe
Final answer: x_1 = (10 - 1.5^3)/4 = 6.625/4 = 1.65625; x_2 = (10 - 1.65625^3)/4 approx 1.36415863. The values jump (1.5 -> 1.65625 -> 1.36416), so the sequence appears to oscillate and not settle (not converging).
Mark scheme for Question 13 [3 marks]
Question 13[3 marks]
Answer or workingMarks
recognises right-angled triangle formed by the radius, height and slant heightM1oe
5^2 + 12^2 oe (= 169)M1
13 cmA1cao
Mark scheme for Question 14 [4 marks]
Question 14[4 marks]
Answer or workingMarks
substitute n = 5 and n = 6 into correct formula depending on parityM1
22 and 31A1cao
compare differences between consecutive terms and note they are not constantM1
statement: not arithmetic because differences are not constant (example given)A1
Final answer: T_5 = 22, T_6 = 31 | Not arithmetic, because consecutive differences are not constant (they alternate).
Mark scheme for Question 15 [4 marks]
Question 15[4 marks]
Answer or workingMarks
3 x 4 = 12 and 10^5 x 10^-2 = 10^3 seen, or 12 x 10^3 oe unsimplifiedM1
1.2 x 10^4A1cao
3 / 4 = 0.75 and 10^5 / 10^-2 = 10^7 seen, or 0.75 x 10^7 oe unsimplifiedM1
7.5 x 10^6A1cao
Final answer: 1.2 x 10^4 | 7.5 x 10^6
Mark scheme for Question 16 [4 marks]
Question 16[4 marks]
Answer or workingMarks
correct reason, e.g. the perpendicular from the apex of an isosceles triangle bisects the base (by symmetry), so each half is 12 / 2 = 6 cmB1oe
10^2 - 6^2 oe (= 64)M1
sqrt(their 64)M1
8 cmA1cao
Final answer: The perpendicular from the apex of an isosceles triangle bisects the base by symmetry, so each half is 12 divided by 2 = 6 cm. | 8 cm
Mark scheme for Question 17 [4 marks]
Question 17[4 marks]
Answer or workingMarks
substitutes y = 3x - 2 into the second equationM1
forms and simplifies a single linear equation in x, e.g. 5x - 2 = 13M1
x = 3A1
y = 7A1cao
Final answer: x = 3, y = 7
Mark scheme for Question 18 [4 marks]
Question 18[4 marks]
Answer or workingMarks
forms equation 2(3w + w) = 96 or 8w = 96M1
width = 12 cm and length = 36 cmA1cao
uses area = length × widthM1
432 cm^2A1cao
Final answer: width 12 cm, length 36 cm | 432 cm^2
Mark scheme for Question 19 [4 marks]
Question 19[4 marks]
Answer or workingMarks
96 / 8 = 12, 12 x 3 = 36 oe (Ashcroft Farm's apples sold)M1
105 / 5 = 21, 21 x 2 = 42 oe (Bramble Hill's apples sold)M1
90 / 9 = 10, 10 x 7 = 70 oe (Cedar Grove's apples sold)M1
cao, Cedar Grove sold the most, since 70 is greater than 42 and 36A1
Final answer: Cedar Grove sold the most, 70 apples (Ashcroft Farm = 36, Bramble Hill = 42, Cedar Grove = 70).
Mark scheme for Question 20 [7 marks]
Question 20[7 marks]
Answer or workingMarks
9B1cao
splitting the data into a lower half (2,5,6,6,7,8) and an upper half (9,10,12,14,15,20) and finding the median of each halfM1
LQ = 6 and UQ = 13, bothA1cao
ft: 7B1cao
ft: whiskers drawn correctly from the minimum (2) to the box, and from the box to the maximum (20)B1
ft: box drawn correctly from the lower quartile (6) to the upper quartile (13)B1
ft: median (9) correctly marked inside the boxB1
Final answer: 9 texts | LQ = 6 texts, UQ = 13 texts | 7 texts | Box plot with min 2, LQ 6, median 9, UQ 13, max 20
Mark scheme for Question 21 [1 mark]
Question 21[1 mark]
Answer or workingMarks
CB1cao
Mark scheme for Question 22 [2 marks]
Question 22[2 marks]
Answer or workingMarks
rearrange N = n1 * n2 / m to make n2 the subject: n2 = N * m / n1M1oe
30A1cao
Mark scheme for Question 23 [3 marks]
Question 23[3 marks]
Answer or workingMarks
at least 3 of the 4 terms 15, -3sqrt(5), 5sqrt(5), -5 correctM1
correct unsimplified expansion 15 - 3sqrt(5) + 5sqrt(5) - 5 (all 4 terms right)A1
10 + 2sqrt(5) oeA1cao
Final answer: 10 + 2sqrt(5)
Mark scheme for Question 24 [5 marks]
Question 24[5 marks]
Answer or workingMarks
half base = 6 m identifiedM1
height^2 = 7.5^2 - 6^2 (Pythagoras)M1
height = 4.5 mA1cao
triangle area = 1/2 x 12 x 4.5 (=27) and rectangle area = 12 x 7 (=84)M1
111 m^2A1cao
Mark scheme for Question 25 [3 marks]
Question 25[3 marks]
Answer or workingMarks
expressing frequencies in terms of k: 10 x 3k = 30k, 20 x 2k = 40k, 30 x k = 30kM1
30k + 40k + 30k = 200, so 100k = 200, k = 2M1
frequency for 190 <= t < 210 = 40k = 80A1cao
Final answer: 80 athletes
Mark scheme for Question 26 [4 marks]
Question 26[4 marks]
Answer or workingMarks
correct straight line from (0,0) rising to (15,12)B1
correct horizontal line from (15,12) to (55,12)B1
correct straight line falling from (55,12) to (65,0), then horizontal at 0 from (65,0) to (85,0)B1
correct straight line rising from (85,0) to (97,12), with key time and velocity values labelled on the axesB1
Final answer: Velocity-time graph with vertices at (0,0), (15,12), (55,12), (65,0), (85,0), (97,12).
Mark scheme for Question 27 [3 marks]
Question 27[3 marks]
Answer or workingMarks
identifying the required position as the 12th value (20% of 60 = 12)M1
correct interpolation method between (40,4) and (50,13)M1
awrt 49 minutes (accept 47-51)A1
Final answer: awrt 49 minutes