Higher Tier - Year 10

Year 10 Paper 6: Full Year 10 Review

Covers number and calculation, fractions, decimals and percentages, ratio and proportion, indices and standard form, linear algebra, graphs and coordinates, sequences, and angles and geometrical reasoning.

24 questions - 80 marks - calculator allowed

Year 10 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 10, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.

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Questions

Question 1 [1 marks]

Indices and Standard Form

Write the number 45000 in standard form.

Question 2 [2 marks]

Fractions, Decimals and Percentages

Work out 23% of £560.

Question 3 [2 marks]

Number and Calculation

A number, n, is rounded to the nearest whole number. The result is 15. Write down the error interval for n.

Question 4 [5 marks]

Linear Algebra

a = 4 and b = -3. Work out the value of each expression.

a + b

ab

a - 2b

Question 5 [1 marks]

Graphs and Coordinates

Find the gradient of the straight line joining the points (-3, -4) and (1, -4).

Question 6 [2 marks]

Number and Calculation

Find the LCM of 8 and 12.

Question 7 [2 marks]

Linear Algebra

Expand and simplify 2(x + 6) + 5(x - 2)

Question 8 [2 marks]

Number and Calculation

Work out sqrt(81) + 2^3.

Question 9 [3 marks]

Sequences

The nth term of a sequence is n^2 + 4n + 3. Find the value of n for which the term is 63.

Question 10 [3 marks]

Angles and Geometrical Reasoning

Column vectors are written in the form (x, y), where x is the top (horizontal) number and y is the bottom (vertical) number. a = (5, -2) and b = (-3, 4).

Write down 2a as a column vector.

Work out a + b as a column vector.

Work out a - b as a column vector.

Question 11 [3 marks]

Number and Calculation

Find the smallest positive integer n such that 360 x n is a perfect cube.

Question 12 [3 marks]

Fractions, Decimals and Percentages

After a pay rise of 3%, Kwame's monthly salary is £2,729.50. Work out his monthly salary before the rise.

Question 13 [3 marks]

Graphs and Coordinates

Use your table of values from Question 3 for y = x^3 - 4.

Draw the graph of y = x^3 - 4 for -2 <= x <= 2 on the grid provided.

Write down the coordinates of the point where the graph crosses the y-axis.

Question 14 [4 marks]

Ratio and Proportion

The table shows some values of x and y, where y is directly proportional to x^2. x: 2, 4, 6 y: 12, 48, ?

Find the value of k in y = kx^2.

Complete the table by finding the missing value of y when x = 6.

Question 15 [5 marks]

Graphs and Coordinates

The line L has equation y = ax + 4 and passes through the point (6, -2). Find the value of a. Then state the equation of the line perpendicular to L that passes through (6, -2).

State the equation of the line perpendicular to L through (6, -2).

Question 16 [6 marks]

Linear Algebra

These formulae involve a square or a square root.

The area of a circle of radius r is A = pi * r^2. Make r the subject of the formula.

v^2 = u^2 + 2as. Make u the subject of the formula.

Question 17 [6 marks]

Graphs and Coordinates

(a) Rearrange 3x - 2y = 6 to make y the subject. (b) Rearrange x + y = 7 to make y the subject. (c) By drawing suitable graphs on the grid, solve the simultaneous equations 3x - 2y = 6 and x + y = 7.

Question 18 [7 marks]

Angles and Geometrical Reasoning

The front elevation of a step-shaped prism is an L-shape, with these measurements: overall width 10 cm, overall height 8 cm, the height of the lower step (across the full width) is 3 cm, the width of the upper part is 4 cm, and the upper part rises a further 5 cm above the lower step. The prism is 6 cm deep (front to back).

Work out the area of the front elevation.

Work out the perimeter of the front elevation.

The prism is 6 cm deep. Work out the total surface area of the solid.

Question 19 [4 marks]

Indices and Standard Form

Simplify (5 + sqrt(3))/(sqrt(3) - 1) fully, writing your answer in the form a + bsqrt(3), where a and b are integers.

Question 20 [3 marks]

Number and Calculation

A padlock code has 3 digits. Each digit can be any number from 0 to 9 and digits may repeat, but the first digit cannot be 0 and the last digit cannot be 0. Work out the number of different codes possible.

Question 21 [3 marks]

Indices and Standard Form

Simplify sqrt(x^3) / x^(1/4). Give your answer as a single power of x.

Question 22 [4 marks]

Sequences

Here are the first five terms of a sequence. 3, 8, 15, 24, 35

Work out the first differences between consecutive terms.

Work out the second differences, and use this to explain why the sequence is quadratic.

Work out the next term in the sequence.

Question 23 [2 marks]

Number and Calculation

Explain why rounding 148 to 1 significant figure gives 100, and not 150.

Question 24 [4 marks]

Sequences

A quadratic sequence has nth term n^2 - 7n + 15.

Work out the 5th term of the sequence.

Find the smallest value in the sequence, and state the term number(s) at which it occurs.

Model solutions

Mark scheme for Question 1 [1 mark]
Question 1[1 mark]
Answer or workingMarks
4.5 x 10^4B1oe
Mark scheme for Question 2 [2 marks]
Question 2[2 marks]
Answer or workingMarks
0.23 x 560M1oe
128.80 (pounds)A1
Final answer: £128.80
Mark scheme for Question 3 [2 marks]
Question 3[2 marks]
Answer or workingMarks
one correct bound identified, 14.5 or 15.5M1oe
14.5 <= n < 15.5A1cao
Mark scheme for Question 4 [5 marks]
Question 4[5 marks]
Answer or workingMarks
1B1cao
4 x (-3) seenM1oe
-12A1cao
2 x (-3) (= -6) seenM1oe
10A1cao
Final answer: 1 | -12 | 10
Mark scheme for Question 5 [1 mark]
Question 5[1 mark]
Answer or workingMarks
0B1cao
Mark scheme for Question 6 [2 marks]
Question 6[2 marks]
Answer or workingMarks
8 = 2^3 and 12 = 2^2 x 3, or a correct list of multiples of each numberM1
24A1cao
Mark scheme for Question 7 [2 marks]
Question 7[2 marks]
Answer or workingMarks
2x + 12 + 5x - 10 (both brackets expanded correctly)M1oe
7x + 2A1cao
Mark scheme for Question 8 [2 marks]
Question 8[2 marks]
Answer or workingMarks
correct method e.g. sqrt(81)=9 and 2^3=8 shownM1
17A1cao
Mark scheme for Question 9 [3 marks]
Question 9[3 marks]
Answer or workingMarks
form equation n^2 + 4n + 3 = 63M1oe
rearrange and factorise to (n + 10)(n - 6) = 0 or equivalent methodM1
n = 6 cao (positive integer solution)A1
Final answer: 6
Mark scheme for Question 10 [3 marks]
Question 10[3 marks]
Answer or workingMarks
(10, -4)B1cao
(2, 2)B1cao
(8, -6)B1cao
Final answer: (10, -4) | (2, 2) | (8, -6)
Mark scheme for Question 11 [3 marks]
Question 11[3 marks]
Answer or workingMarks
prime factorise 360 = 2^3 x 3^2 x 5^1M1
choose multipliers to make exponents multiples of 3: need 3^1 and 5^2M1
n = 3 x 25 = 75A1cao
Final answer: 75
Mark scheme for Question 12 [3 marks]
Question 12[3 marks]
Answer or workingMarks
correct multiplier 1.03M1oe
complete method 2729.50 / 1.03M1
£2,650A1cao
Mark scheme for Question 13 [3 marks]
Question 13[3 marks]
Answer or workingMarks
all 5 points from the table plotted correctly, ft candidate's table from Q3B1
single smooth curve drawn through all the points, correct increasing s-shape with no turning pointsB1
(0, -4)B1cao
Final answer: Smooth curve through (-2,-12), (-1,-5), (0,-4), (1,-3), (2,4) | (0, -4)
Mark scheme for Question 14 [4 marks]
Question 14[4 marks]
Answer or workingMarks
uses a pair of values, e.g. 12 = k*2^2, or 48 = k*4^2M1
k = 3A1cao
substitutes x = 6 into y = 3x^2 (ft their k)M1
y = 108A1cao
Final answer: k = 3 | y = 108
Mark scheme for Question 15 [5 marks]
Question 15[5 marks]
Answer or workingMarks
substitute (6, -2) into y = ax + 4: -2 = 6a + 4 and form 6a = -6M1
a = -1A1cao
use perpendicular gradient product = -1: gradient of L is -1 so perpendicular gradient = 1M1
use point (6, -2) to form equation y = 1x + c and solve for c: -2 = 6 + cM1
y = x - 8A1cao
Final answer: a = -1 | y = x - 8
Mark scheme for Question 16 [6 marks]
Question 16[6 marks]
Answer or workingMarks
divides both sides by pi, e.g. r^2 = A/piM1
square roots both sides, dependent on the previous method markdM1
r = sqrt(A/pi) oe, cao (positive root only, since r is a length)A1
subtracts 2as from both sides, e.g. u^2 = v^2 - 2asM1
square roots both sides, dependent on the previous method markdM1
u = sqrt(v^2 - 2as) oeA1cao
Final answer: r = sqrt(A/pi) | u = sqrt(v^2 - 2as)
Mark scheme for Question 17 [6 marks]
Question 17[6 marks]
Answer or workingMarks
y = 1.5x - 3 (oe, e.g. y = (3x - 6) / 2)B1
y = 7 - xB1
draws y = 1.5x - 3 correctly, ft from (a), e.g. through (0,-3), (2,0), (4,3)M1
draws y = 7 - x correctly, ft from (b), e.g. through (0,7), (4,3), (7,0)M1
identifies intersection at (4,3)A1
states solution x = 4, y = 3A1ft
Final answer: y = 1.5x - 3 | y = 7 - x | x = 4, y = 3
Mark scheme for Question 18 [7 marks]
Question 18[7 marks]
Answer or workingMarks
splitting into two rectangles, e.g. 10 x 3 (= 30) and 4 x 5 (= 20)M1
50 cm^2A1cao
identifying the missing horizontal length, 10 - 4 = 6 cm, and attempting to sum all six sides (10 + 3 + 6 + 5 + 4 + 8)M1
36 cmA1cao
2 x 50 ft (their area from part a), the two L-shaped faces (= 100)M1
36 x 6 ft (their perimeter from part b, x depth) (= 216)M1
316 cm^2 caoA1ft
Final answer: 50 cm^2 | 36 cm | 316 cm^2
Mark scheme for Question 19 [4 marks]
Question 19[4 marks]
Answer or workingMarks
multiplies numerator and denominator by (sqrt(3) + 1)M1
denominator simplifies to 3 - 1 = 2M1
numerator simplifies to 8 + 6sqrt(3)A1
4 + 3sqrt(3) cao (a = 4, b = 3)A1
Final answer: 4 + 3sqrt(3) (a = 4, b = 3)
Mark scheme for Question 20 [3 marks]
Question 20[3 marks]
Answer or workingMarks
identifies 9 choices for the first digit and 9 choices for the last digit (excluding 0 in each case)M1
(dep) 9 * 10 * 9M1oe
810A1cao
Mark scheme for Question 21 [3 marks]
Question 21[3 marks]
Answer or workingMarks
sqrt(x^3) written as x^(3/2)M1
indices subtracted: 3/2 - 1/4M1
x^(5/4) oe (accept x^1.25)A1
Final answer: x^(5/4)
Mark scheme for Question 22 [4 marks]
Question 22[4 marks]
Answer or workingMarks
5, 7, 9, 11 all correctB1cao
second differences all equal to 2, ft from part (a)B1
correct explanation, e.g. a constant (non-zero) second difference means the sequence is quadraticB1
48 cao, ft from their pattern of differencesB1
Final answer: 5, 7, 9, 11 | Second differences are 2, 2, 2. Since the second differences are constant, the sequence is quadratic. | 48
Mark scheme for Question 23 [2 marks]
Question 23[2 marks]
Answer or workingMarks
identifying that the first significant figure is the hundreds digit, 1, and all following digits become 0 for 1 sfB1
correct reasoning that the next digit (4, the tens digit) is less than 5, so the 1 is not rounded up, giving 100; 150 is not a rounding to 1 significant figureB1
Final answer: 100, because the first significant figure (1) is followed by a 4, which is less than 5, so it rounds down; 150 has two significant figures, not one.
Mark scheme for Question 24 [4 marks]
Question 24[4 marks]
Answer or workingMarks
5B1cao
finding the vertex of the quadratic at n=3.5 using n=-b/(2a) (or an equivalent method)M1
evaluating the sequence at n=3 and n=4 (both give 3)M1
smallest value is 3, occurring at both n=3 and n=4A1
Final answer: 5 | Smallest value is 3, at n=3 and n=4