Higher Tier - Grades 7-9

Higher Stretch A

Demanding Higher practice across algebra, geometry and probability.

32 questions - 100 marks - calculator allowed

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Questions

Question 1 [1 marks]

Indices and Standard Form

Work out the value of 3^4.

Question 2 [1 marks]

Graphs and Coordinates

A line has gradient -3/4. What is the gradient of a line perpendicular to it? A) -3/4 B) 3/4 C) -4/3 D) 4/3

Question 3 [2 marks]

Fractions, Decimals and Percentages

Work out 7/9 - 4/9. Give your answer in its simplest form.

Question 4 [3 marks]

Sequences

Three sequences are shown below. Sequence A: 7, 11, 15, 19 Sequence B: 4, 12, 36, 108 Sequence C: 3, 6, 11, 18

State whether Sequence A is arithmetic, geometric, or neither of these. Give a reason for your answer.

State whether Sequence B is arithmetic, geometric, or neither of these. Give a reason for your answer.

State whether Sequence C is arithmetic, geometric, or neither of these. Give a reason for your answer.

Question 5 [4 marks]

Ratio and Proportion

A recipe for lemonade that makes 6 glasses needs 3 lemons and 250 ml of water. Nadia has 8 lemons and 900 ml of water. Work out the greatest number of glasses of lemonade Nadia can make, assuming she can only make whole batches of the recipe.

Question 6 [1 marks]

Number and Calculation

Write down the lower bound and upper bound for a number which is 12, correct to the nearest integer, using inequality notation.

Question 7 [1 marks]

Linear Algebra

Solve t - 8 = 2.

Question 8 [1 marks]

Number and Calculation

Round 7,246 to the nearest 100.

Question 9 [2 marks]

Area, Volume and Measures

Two mathematically similar solids have a linear scale factor of 4 from the smaller solid to the larger solid. The volume of the smaller solid is 3 cm^3. Work out the volume of the larger solid.

Question 10 [2 marks]

Quadratics

Solve the inequality 2x^2 - 8x >= 0. Give your answer in inequality form.

Question 11 [2 marks]

Pythagoras and Trigonometry

From a point on level ground 18 m from the base of a lighthouse, the angle of elevation to the top of the lighthouse is 52 degrees. Calculate the height of the lighthouse. Give your answer correct to 3 significant figures.

Question 12 [2 marks]

Angles and Geometrical Reasoning

The exterior angle of a regular polygon is 15 degrees. Work out the number of sides of the polygon.

Question 13 [2 marks]

Indices and Standard Form

A storage box for a school science kit in Bristol is a cube with volume 125 cm^3. Work out the length of one edge of the box.

Question 14 [3 marks]

Quadratics

Solve the equation 2x^2 - x - 6 = 0.

Question 15 [3 marks]

Angles and Geometrical Reasoning

Triangles ABC and DEF satisfy angle A = 50 degrees, angle C = 70 degrees and AC = DF. In triangle DEF, angle D = 50 degrees and angle F = 70 degrees. Show that the triangles are congruent and state two pairs of corresponding equal sides.

Question 16 [3 marks]

Sequences

The table shows the number of matchsticks used to make patterns of joined squares in a row. Pattern number (n): 1, 2, 3 Number of matchsticks: 4, 7, 10

Write down the number of matchsticks needed for Pattern number 4.

Find an expression, in terms of n, for the number of matchsticks in Pattern number n.

Question 17 [2 marks]

Angles and Geometrical Reasoning

w is the column vector (5, 11). Work out |w|, giving your answer correct to 1 decimal place.

Question 18 [4 marks]

Graphs and Coordinates

Show that the cubic y = x^3 - 6x^2 + 11x - 6 factorises as (x - 1)(x - 2)(x - 3). Hence, sketch the cubic and state the x-intercepts.

Show that the cubic factorises as stated.

Hence sketch the cubic and state the x-intercepts.

Question 19 [4 marks]

Quadratics

Solve 3x^2 = 13x - 4 by factorising. Show your working.

Question 20 [4 marks]

Angles and Geometrical Reasoning

O is the origin. The position vectors of points P, Q and R are OP = 2a - b, OQ = 6a + kb and OR = a + 2b, where a and b are non-parallel vectors and k is a constant. Given that P, Q and R lie on a straight line, find the value of k.

Question 21 [4 marks]

Area, Volume and Measures

A recipe for flapjacks that serves 6 people uses 750 g of oats.

Write 750 g in kilograms.

Bilal wants to make enough flapjacks to serve 15 people, using the recipe in the same proportions. Work out how many kilograms of oats he needs.

Question 22 [4 marks]

Statistics and Probability

A deck contains only cards numbered 1 to n, one of each. A card is chosen at random. The probability it is a multiple of 3 is 1/4. Find n, given n is a positive integer and the deck starts at 1.

Question 23 [4 marks]

Angles and Geometrical Reasoning

TP is a tangent to a circle with centre O, touching the circle at T. OT = 5 cm and OP = 13 cm. Since OT is a radius and TP is a tangent at T, angle OTP = 90 degrees. Diagram: circle, centre O, radius OT drawn to the point of contact T, tangent TP drawn perpendicular to OT, OP drawn from O to the external point P, right angle marked at T, OT = 5 cm, OP = 13 cm.

Work out the length TP.

Work out the size of angle TPO. Give your answer correct to 1 decimal place.

Question 24 [5 marks]

Statistics and Probability

The frequency table shows the results for Class B in the same test as Question 11 (also marked out of 10, also 20 pupils).

Calculate the mean score for Class B. Give your answer correct to 2 decimal places.

Find the range of scores for Class A and the range of scores for Class B.

Question 25 [9 marks]

Angles and Geometrical Reasoning

A surveyor produces a scale drawing of a triangular field. On the drawing the sides are 7.2 cm, 5.4 cm and 4.2 cm. The scale is 1:500 for linear measurements. (a) Work out the actual perimeter of the field in metres to 3 significant figures. (b) Hence, or otherwise, find the actual area of the field in square metres. Give your answer to 3 significant figures. (You may use Heron's formula.)

Question 26 [2 marks]

Statistics and Probability

A scientist plots the reaction time (in seconds) of runners against their age (in years), for ages 10 to 80. The points curve upwards steeply for ages under 20 and over 60, but are roughly flat for ages 20 to 60. Explain why it would not be appropriate to draw a single straight line of best fit for this data.

Question 27 [3 marks]

Linear Algebra

The kinetic energy of a moving object is given by E = (1/2)mv^2, where E is the kinetic energy in joules, m is the mass in kilograms and v is the speed in metres per second. Make v the subject of the formula.

Question 28 [3 marks]

Sequences

The nth term of a sequence is given by An + B, where A and B are numbers to be found. The 2nd term of the sequence is 11. The 5th term of the sequence is 23. Work out the values of A and B.

Question 29 [4 marks]

Quadratics

The function f is defined by f(x) = x^2 - 3x + 7.

Express f(x) in the form (x - a)^2 + b, where a and b are exact values.

Hence state the minimum value of f(x).

Question 30 [4 marks]

Statistics and Probability

A wildlife survey estimates the population of red squirrels in a forest in two consecutive years using capture-recapture. Year 1: 60 squirrels tagged in the first sample; second sample of 50, of which 12 are tagged. Year 2: 80 squirrels tagged in the first sample; second sample of 70, of which 14 are tagged.

Calculate the population estimate for Year 1 and for Year 2.

Work out the percentage change in the population estimate from Year 1 to Year 2.

Question 31 [5 marks]

Angles and Geometrical Reasoning

In triangle ABC, D is a point on BC. E is the foot of the perpendicular from D to AB, and F is the foot of the perpendicular from D to AC, with DE = DF. Prove that AD bisects angle BAC.

Question 32 [6 marks]

Pythagoras and Trigonometry

A vertical tree TB stands on horizontal ground, with its base at B. From a point C on the ground, in line with B, the angle of elevation of the top of the tree, T, is 42 degrees. From a point D, further from the tree than C, with C between B and D and CD = 15 m, the angle of elevation of T is 26 degrees.

By forming two equations for the height, h, of the tree (using triangles TBC and TBD), calculate h. Give your answer correct to 3 significant figures.

Hence calculate the distance BC.

Model solutions

Mark scheme for Question 1 [1 mark]
Question 1[1 mark]
Answer or workingMarks
81B1cao
Mark scheme for Question 2 [1 mark]
Question 2[1 mark]
Answer or workingMarks
D selectedB1cao
Final answer: D (4/3)
Mark scheme for Question 3 [2 marks]
Question 3[2 marks]
Answer or workingMarks
3/9 oe (subtracting the numerators, same denominator)M1
1/3 cao (simplest form)A1
Final answer: 1/3
Mark scheme for Question 4 [3 marks]
Question 4[3 marks]
Answer or workingMarks
arithmetic, since the terms have a common difference of 4B1oe
geometric, since the terms have a common ratio of 3B1oe
neither, since the differences (3, 5, 7) are not constant and the ratios are not constantB1oe
Final answer: Arithmetic (common difference of 4) | Geometric (common ratio of 3) | Neither (it is a quadratic sequence)
Mark scheme for Question 5 [4 marks]
Question 5[4 marks]
Answer or workingMarks
8 / 3 (= 2.67...)M1oe
900 / 250 (= 3.6)M1oe
2 (recipes), identifying lemons as the limiting ingredientA1
12 (glasses) caoA1dep
Final answer: 12 glasses
Mark scheme for Question 6 [1 mark]
Question 6[1 mark]
Answer or workingMarks
11.5 <= x < 12.5B1cao
Mark scheme for Question 7 [1 mark]
Question 7[1 mark]
Answer or workingMarks
t = 10B1cao
Final answer: 10
Mark scheme for Question 8 [1 mark]
Question 8[1 mark]
Answer or workingMarks
7,200B1cao
Mark scheme for Question 9 [2 marks]
Question 9[2 marks]
Answer or workingMarks
volume scale factor = 4^3 (=64)M1
192 cao, units cm^3A1
Final answer: 192 cm^3
Mark scheme for Question 10 [2 marks]
Question 10[2 marks]
Answer or workingMarks
factorise to 2x(x-4) and identify roots x=0 and x=4M1
x <= 0 or x >= 4A1cao
Mark scheme for Question 11 [2 marks]
Question 11[2 marks]
Answer or workingMarks
height = 18 tan(52)M1oe
awrt 23 (m)A1
Final answer: height = 23 m (3 s.f.)
Mark scheme for Question 12 [2 marks]
Question 12[2 marks]
Answer or workingMarks
360 / 15M1
24A1cao
Final answer: 24 sides
Mark scheme for Question 13 [2 marks]
Question 13[2 marks]
Answer or workingMarks
take the cube root of 125 to find the edge lengthM1
5 cmA1cao
Mark scheme for Question 14 [3 marks]
Question 14[3 marks]
Answer or workingMarks
correct factorisation or rearrangement shown, e.g. 2x^2 +3x -4x -6 or (2x +3)(x -2)M1
obtain factors (2x +3)(x -2)M1oe
x = 2 or x = -3/2A1cao
Mark scheme for Question 15 [3 marks]
Question 15[3 marks]
Answer or workingMarks
find the remaining angle B = 60 degrees and E = 60 degrees or state two angles equal in both trianglesM1
identify two angles and included side AC = DF so ASA appliesM1
conclude triangles are congruent and so AB = DE and BC = EF (state two corresponding sides)A1
Final answer: Triangles are congruent by ASA. Corresponding equal sides: AB = DE and BC = EF.
Mark scheme for Question 16 [3 marks]
Question 16[3 marks]
Answer or workingMarks
13B1cao
common difference of 3 identified, e.g. 3n + ...M1oe
3n + 1A1cao
Final answer: 13 | 3n + 1
Mark scheme for Question 17 [2 marks]
Question 17[2 marks]
Answer or workingMarks
sqrt(5^2 + 11^2)M1oe
awrt 12.1A1
Final answer: 12.1
Mark scheme for Question 18 [4 marks]
Question 18[4 marks]
Answer or workingMarks
either show synthetic division by x=1 then x=2 or factor by grouping leading to factors (x-1),(x-2),(x-3)M1
(x - 1)(x - 2)(x - 3)A1cao
sketch shows cubic crossing x-axis at three points and general shape correctM1
x-intercepts (1, 0), (2, 0), (3, 0)A1cao
Final answer: (x - 1)(x - 2)(x - 3) | (1, 0), (2, 0), (3, 0)
Mark scheme for Question 19 [4 marks]
Question 19[4 marks]
Answer or workingMarks
rearranges to 3x^2 - 13x + 4 = 0M1oe
identifies a pair of numbers with product 12 and sum -13 (-12 and -1), leading to (3x - 1)(x - 4)M1
x = 4A1cao
x = 1/3 oe, ft from a correctly factorised bracketA1
Final answer: x = 4 or x = 1/3
Mark scheme for Question 20 [4 marks]
Question 20[4 marks]
Answer or workingMarks
finds PQ = OQ - OP = 4a + (k+1)bM1oe
finds PR = OR - OP = -a + 3bM1oe
sets PQ = m x PR and equates coefficients of a: 4 = -m, so m = -4M1
k = -13A1cao
Mark scheme for Question 21 [4 marks]
Question 21[4 marks]
Answer or workingMarks
0.75B1cao
750 / 6 (= 125), finding the oats needed per personM1
their 125 x 15M1dep
1.875 (kg) oe cao (accept 1875 g)A1
Final answer: 0.75 kg; 1.875 kg
Mark scheme for Question 22 [4 marks]
Question 22[4 marks]
Answer or workingMarks
identify count of multiples of 3 as floor(n/3) and form equation floor(n/3)/n = 1/4 or reason using integer k such that number of multiples = k and k/n = 1/4M1
use integer reasoning to set k = n/4 so n divisible by 4 and test small multiples of 4 against floor(n/3) = n/4M1
find smallest n satisfying integer constraints or demonstrate n = 4 is validM1
n = 4 cao (accept n = 8, the only other value satisfying floor(n/3)/n = 1/4)A1
Final answer: 4
Mark scheme for Question 23 [4 marks]
Question 23[4 marks]
Answer or workingMarks
sqrt(13^2 - 5^2)M1oe
12 cmA1cao
sin(TPO) = 5/13 oe, or use tan = 5/12 with their TPM1
awrt 22.6 degreesA1
Final answer: 12 cm | 22.6 degrees (1 d.p.)
Mark scheme for Question 24 [5 marks]
Question 24[5 marks]
Answer or workingMarks
at least 4 correct values of f x x, or sight of sum(fx) = 137M1
137 / 20 (dependent on previous M1)M1
6.85A1cao
Class A range = 4B1cao
Class B range = 3B1cao
Final answer: 6.85 | Class A range = 4; Class B range = 3
Mark scheme for Question 25 [9 marks]
Question 25[9 marks]
Answer or workingMarks
convert each side by factor 500: 7.2*500 = 3600 cm, 5.4*500 = 2700 cm, 4.2*500 = 2100 cmM1
sum perimeter in cm = 3600 + 2700 + 2100 = 8400 cm = 84.00 mM1
84.0 m cao to 3 s.f.A1
convert sides to metres: 36.00 m, 27.00 m, 21.00 mM1
use Heron: s = (36 + 27 + 21)/2 = 42.0 mM1
compute area = sqrt(s(s-a)(s-b)(s-c)) methodM1
numerical substitution: area = sqrt(42*(42-36)*(42-27)*(42-21)) = sqrt(42*6*15*21)M1
area = 281.7... m^2 awrt 282 m^2 to 3 s.f.A1
final area 282 m^2A1cao
Final answer: 84.0 m | 282 m^2
Mark scheme for Question 26 [2 marks]
Question 26[2 marks]
Answer or workingMarks
the relationship is not linear (it curves / changes shape across the range)B1
a single straight line would not model the curved trend well, giving poor estimates for the youngest and oldest runnersB1
Final answer: The data does not follow a linear pattern (it curves at the youngest and oldest ages), so a straight line of best fit would not fit the trend well and would give poor estimates at the extremes of age.
Mark scheme for Question 27 [3 marks]
Question 27[3 marks]
Answer or workingMarks
multiplies both sides by 2 and divides by m, e.g. v^2 = 2E/mM1
square roots both sides, dependent on the previous method markdM1
v = sqrt(2E/m) oe, cao (positive root only, since v is a speed)A1
Final answer: v = sqrt(2E/m)
Mark scheme for Question 28 [3 marks]
Question 28[3 marks]
Answer or workingMarks
forms two correct equations, 2A + B = 11 and 5A + B = 23M1
solves simultaneously, e.g. subtracts to get 3A = 12, dependent on the previous method markdM1
A = 4 and B = 3 both correctA1cao
Final answer: A = 4, B = 3
Mark scheme for Question 29 [4 marks]
Question 29[4 marks]
Answer or workingMarks
(x - 1.5)^2 oe seen (allow x - 3/2)M1
correct subtraction of (1.5)^2 = 2.25 from 7M1
(x - 1.5)^2 + 4.75 oe (accept (x - 3/2)^2 + 19/4)A1
ft 4.75 oe (accept 19/4)B1
Final answer: (x-1.5)^2 + 4.75; minimum value 4.75
Mark scheme for Question 30 [4 marks]
Question 30[4 marks]
Answer or workingMarks
250 cao (Year 1)B1
400 cao (Year 2)B1
(400 - 250) / 250 x 100 oe (ft their part (a) values)M1
60% increaseA1cao
Final answer: (a) 250, 400 (b) 60% increase
Mark scheme for Question 31 [5 marks]
Question 31[5 marks]
Answer or workingMarks
angle AED = angle AFD = 90 degrees (given)B1
DE = DF (given)B1
AD = AD (common side, the hypotenuse of each triangle); hence RHS, triangle ADE = triangle ADFB1
hence angle DAE = angle DAF (corresponding angles in congruent triangles)B1
therefore AD bisects angle BACB1cso
Final answer: Triangle ADE = Triangle ADF (RHS); AD bisects angle BAC
Mark scheme for Question 32 [6 marks]
Question 32[6 marks]
Answer or workingMarks
BC = h/tan(42) oe (correct equation from triangle TBC)M1
BD = h/tan(26) oe (correct equation from triangle TBD)M1
correct method to combine: BD - BC = 15, leading to h(1/tan(26) - 1/tan(42)) = 15 or equivalent rearrangementM1
awrt 16 (m)A1
BC = their h / tan(42)M1ft
awrt 17.7 (m)A1
Final answer: h = 16 m (3 s.f.) | BC = 17.7 m (3 s.f.)