Higher Tier - Grades 4-6

Higher Core A

Higher tier fundamentals, including crossover algebra and geometry.

16 questions - 60 marks - calculator allowed

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Questions

Question 1 [1 marks]

Indices and Standard Form

Write 9.1 x 10^4 as an ordinary number.

Question 2 [1 marks]

Fractions, Decimals and Percentages

Work out 3/8 + 2/8. Give your answer in its simplest form.

Question 3 [5 marks]

Statistics and Probability

Freya Sinclair organised a charity bake sale where 90 cakes were sold, either Sponge or Chocolate, in the morning or the afternoon. Some of the results are shown in the two-way table below.

Question 4 [2 marks]

Linear Algebra

Solve 2(x + 3) = 16 Show your working.

Question 5 [2 marks]

Graphs and Coordinates

The graphs of x = 3 and y = 2x - 4 are drawn on the grid. Write down the solution of the simultaneous equations x = 3 and y = 2x - 4.

Question 6 [2 marks]

Ratio and Proportion

A cyclist travels at a constant speed of 20 m/s. Work out this speed in km/h.

Question 7 [2 marks]

Area, Volume and Measures

Shape 1 has sides of length 4.5 cm and 6 cm. Shape 2 is mathematically similar to shape 1. The side on shape 2 corresponding to the 4.5 cm side has length 3 cm.

Question 8 [2 marks]

Linear Algebra

Show that the equation 4(2x - 3) = 2(x + 9) simplifies to x = 5

Question 9 [4 marks]

Quadratics

A student attempts to solve 3x^2 + 4x - 2 = 0 using the quadratic formula. Their working is shown below: x = (-4 +/- sqrt(4^2 - 4x3x(-2))) / (2x3) x = (-4 +/- sqrt(16 - 24)) / 6 Identify the error made by the student, and find the correct solutions to the equation, giving your answers correct to 2 decimal places.

Question 10 [5 marks]

Sequences

Here are the first four terms of a sequence. 50, 44, 38, 32

Write down the next term of the sequence.

Find an expression, in terms of n, for the nth term of the sequence.

Explain whether 0 is a term of the sequence.

Question 11 [6 marks]

Pythagoras and Trigonometry

A surveyor stands at point A and measures the angle of elevation to the top of a tower as 31 degrees. She then walks 48 m closer to the tower along the level ground to point B and measures the angle of elevation as 46 degrees. Work out the height of the tower to the nearest metre.

Find the height of the tower to the nearest metre.

Question 12 [6 marks]

Angles and Geometrical Reasoning

Two points A(-5, 0) and B(5, 0) are 10 cm apart on a coordinate grid (1 unit = 1 cm). (a) Describe the locus of points that are equidistant from A and B, and give its equation. (b) A circle has centre A and radius 13 cm. Find the coordinates of the points where this circle meets the locus from part (a). Show your working.

Describe the geometric locus in words and give its equation.

Find the coordinates of the points where the circle centre A radius 13 cm meets the locus x = 0. Show your working.

Question 13 [7 marks]

Graphs and Coordinates

Tom walks 3 km from home to the park in 40 minutes. He then jogs 5 km from the park to the leisure centre in 25 minutes.

Calculate Tom's average speed while walking, in km/h.

Calculate Tom's average speed while jogging, in km/h.

Calculate Tom's average speed for the whole journey from home to the leisure centre, in km/h. Give your answer correct to 1 decimal place.

Question 14 [3 marks]

Angles and Geometrical Reasoning

A search and rescue team searches a triangular area defined by three points X, Y and Z. XY = 12 km, XZ = 9 km, and the angle YXZ (the angle between the bearings from X to Y and from X to Z) is 105 degrees. Diagram: Point X is shown with a North arrow and two rays to Y (12 km) and Z (9 km), with angle YXZ marked 105 degrees.

Question 15 [4 marks]

Graphs and Coordinates

Each equation below is a transformation of y = sin x. Match each equation to the letter of the description that correctly describes its transformation. A) Translation by vector (-90, 0) B) Stretch, scale factor 4, parallel to the y-axis C) Translation by vector (0, -3) D) Stretch, scale factor 2, parallel to the x-axis

y = sin x - 3

y = sin(x + 90)

y = 4 sin x

y = sin(x/2)

Question 16 [8 marks]

Statistics and Probability

A bag contains 3 red counters and n blue counters only, where n > 0. Two counters are taken from the bag at random, one after another, without replacement.

Show that the probability that both counters are red is 6 / ((n+3)(n+2)).

Given that the probability that both counters are red is 1/5, show that n^2 + 5n - 24 = 0.

Hence find the number of blue counters in the bag.

Model solutions

Mark scheme for Question 1 [1 mark]
Question 1[1 mark]
Answer or workingMarks
91000B1cao
Mark scheme for Question 2 [1 mark]
Question 2[1 mark]
Answer or workingMarks
5/8B1cao
Mark scheme for Question 3 [5 marks]
Question 3[5 marks]
Answer or workingMarks
Morning Total = 42B1cao
Morning Chocolate = 20B1cao
Afternoon Sponge = 30B1cao
Afternoon Chocolate = 18B1cao
Chocolate column Total = 38B1cao
Final answer: Morning Total = 42, Morning Chocolate = 20, Afternoon Sponge = 30, Afternoon Chocolate = 18, Chocolate Total = 38
Mark scheme for Question 4 [2 marks]
Question 4[2 marks]
Answer or workingMarks
2x + 6 = 16 oe (expand correctly)M1
x = 5A1cao
Mark scheme for Question 5 [2 marks]
Question 5[2 marks]
Answer or workingMarks
x = 3B1
y = 2 (oe as a coordinate pair)B1
Final answer: x = 3, y = 2
Mark scheme for Question 6 [2 marks]
Question 6[2 marks]
Answer or workingMarks
20 x 3.6 oe, e.g. 20 x 3600 / 1000M1
72 (km/h)A1cao
Final answer: 72 km/h
Mark scheme for Question 7 [2 marks]
Question 7[2 marks]
Answer or workingMarks
scale factor = 3/4.5 oe (= 2/3)M1
4 (cm)A1cao
Final answer: 4 cm
Mark scheme for Question 8 [2 marks]
Question 8[2 marks]
Answer or workingMarks
expand both sides correctly, 8x - 12 = 2x + 18M1oe
cso, correct rearrangement shown to give x = 5 with no errorsA1
Final answer: x = 5 (shown)
Mark scheme for Question 9 [4 marks]
Question 9[4 marks]
Answer or workingMarks
identifies the sign error, e.g. 4 x 3 x (-2) = -24 so 4^2 - 4ac = 16 - (-24) = 16 + 24, not 16 - 24B1
correct discriminant of 40 used in the formula with a = 3, b = 4, c = -2M1
x = 0.39 (awrt 0.39)A1
x = -1.72 (awrt -1.72)A1
Final answer: x = 0.39 or x = -1.72
Mark scheme for Question 10 [5 marks]
Question 10[5 marks]
Answer or workingMarks
26B1cao
common difference of -6 used correctly, e.g. -6n + cM1oe
56 - 6n oeA1cao
56 - 6n = 0 oe, or 56 / 6 evaluatedM1
correct conclusion: no, since n = 9.33... (28/3) is not a positive integerA1oe
Final answer: 26 | 56 - 6n | No, 0 is not a term (n would be 9.33..., not a whole number)
Mark scheme for Question 11 [6 marks]
Question 11[6 marks]
Answer or workingMarks
use tan 31 = h / x and tan 46 = h / (x - 48) to form two equationsM1
eliminate x: h = x * tan31 and h = (x - 48) * tan46 then set x * tan31 = (x - 48) * tan46M1
rearrange to x (tan31 - tan46) = -48 * tan46 and solve for xM1
find h using h = x * tan31M1
height = 69 m awrt (1 m)A1
method and final value consistent with workingA1
Final answer: 69 m
Mark scheme for Question 12 [6 marks]
Question 12[6 marks]
Answer or workingMarks
identify locus as the perpendicular bisector of AB (the set of points equidistant from A and B)M1
state equation x = 0 (since AB is horizontal with midpoint (0,0))A1cao
set up equation of circle centre A(-5,0) radius 13: (x+5)^2 + y^2 = 169M1
substitute x = 0: 25 + y^2 = 169, so y^2 = 144M1
y = 12 or y = -12A1
final coordinates (0, 12) and (0, -12)A1cao
Final answer: The locus is the perpendicular bisector of AB: the straight line through the midpoint (0,0) at right angles to AB. Since AB lies along the x-axis, the perpendicular bisector is the y-axis, equation x = 0. | (0, 12) and (0, -12).
Mark scheme for Question 13 [7 marks]
Question 13[7 marks]
Answer or workingMarks
3 divided by (40/60)M1oe
4.5 (km/h)A1cao
5 divided by (25/60)M1oe
12 (km/h)A1cao
total distance = 8 (km)M1
total time = 65 minutes oe (13/12 hours, or 1.0833 hours)M1
awrt 7.4 (km/h)A1
Final answer: 4.5 km/h | 12 km/h | 7.4 km/h
Mark scheme for Question 14 [3 marks]
Question 14[3 marks]
Answer or workingMarks
area = 0.5 x 12 x 9 x sin(105)M1oe
correct processM1
awrt 52.2 (km^2)A1
Final answer: 52.2 km^2
Mark scheme for Question 15 [4 marks]
Question 15[4 marks]
Answer or workingMarks
CB1
AB1
BB1
DB1
Final answer: C | A | B | D
Mark scheme for Question 16 [8 marks]
Question 16[8 marks]
Answer or workingMarks
P(first red) = 3/(n+3)M1
P(second red | first red) = 2/(n+2)M1
cso: 3/(n+3) x 2/(n+2) = 6/((n+3)(n+2))A1
5 x 6 = (n+3)(n+2), i.e. (n+3)(n+2) = 30M1
cso: expands to n^2 + 5n + 6 = 30, giving n^2 + 5n - 24 = 0A1
factorises: (n+8)(n-3) = 0M1
n = 3 selected, rejecting n = -8 since n > 0M1
n = 3A1cao
Final answer: 6 / ((n+3)(n+2)) shown | n^2 + 5n - 24 = 0 shown | 3 blue counters