Revision Library

revisionlibrary.co.uk

13+ · Mathematics

CE.M4 Algebra

ISEB CE CE.M4 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Simplify each expression by collecting like terms.

(a)Simplify 5x + 3y - 2x + 7y.(1)

(b)Simplify 4a² + 3a - a² + 5a.(1)

(Total for Question 1 is 2 marks)

p = 4, q = -3 and r = 5.

(a)Work out the value of 2p + q.(1)

(b)Work out the value of r² - q.(2)

(Total for Question 2 is 3 marks)

Expand the brackets in each expression, simplifying where possible.

(a)Expand 3(2x + 5).(1)

(b)Expand and simplify 4(x - 2) + 3(x + 6).(3)

(Total for Question 3 is 4 marks)

Solve each equation.

(a)Solve 3x + 7 = 22.(2)

(b)Solve 5(x - 3) = 20.(2)

(Total for Question 4 is 4 marks)

Expand and simplify each expression.

(a)Expand and simplify (x + 3)(x + 5).(2)

(b)Expand and simplify (x - 4)(x + 4).(2)

(Total for Question 5 is 4 marks)

Factorise each expression fully.

(a)Factorise fully 6x + 9.(1)

(b)Factorise fully 8x^2y - 12xy.(2)

(Total for Question 6 is 3 marks)

Solve each equation, showing your working.

(a)Solve 4(x + 2) = 2(x + 9).(3)

(b)Solve 3x - 5 = x + 11.(2)

(Total for Question 7 is 5 marks)

Here are the first four terms of a sequence: 5, 8, 11, 14, ...

(a)Write down the next two terms of the sequence.(1)

(b)Find an expression for the nth term of the sequence.(2)

(c)Is 101 a term in this sequence? Show working to justify your answer.(2)

(Total for Question 8 is 5 marks)

Use the laws of indices to simplify each expression.

(a)Simplify x⁵ * x^3.(1)

(b)Simplify (y⁴)^3.(1)

(c)Simplify 12x⁷ / 4x^2.(2)

(Total for Question 9 is 4 marks)

Solve the equation (x + 4)/2 - (x - 1)/3 = 3. Show each step of your working clearly.

(Total for Question 10 is 4 marks)

A rectangle has length (x + 5) cm and width (x - 2) cm. The perimeter of the rectangle is 46 cm.

(a)Write an equation in x for the perimeter of the rectangle.(1)

(b)Solve your equation to find the value of x.(2)

(c)Hence find the width of the rectangle.(1)

(Total for Question 11 is 4 marks)

Rearrange each formula to make the given letter the subject.

(a)Make x the subject of y = 3x + 7.(2)

(b)Make r the subject of A = π*r^2.(2)

(Total for Question 12 is 4 marks)

Answer the following, showing your working.

(a)Solve the inequality 3x - 4 < 11.(2)

(b)n is an integer such that -2 ≤ n < 3. List all the possible values of n.(2)

(Total for Question 13 is 4 marks)

Solve the simultaneous equations:
3x + 2y = 16
x - y = 2

(Total for Question 14 is 4 marks)

Factorise each quadratic expression.

(a)Factorise x² + 7x + 12.(2)

(b)Factorise x² - 5x - 24.(2)

(Total for Question 15 is 4 marks)

Solve the equation x² + 2x - 15 = 0. Give both solutions.

(Total for Question 16 is 4 marks)

Simplify each algebraic fraction fully.

(a)Simplify (x² - 9)/(x + 3).(2)

(b)Simplify (3x² + 6x)/(3x).(2)

(Total for Question 17 is 4 marks)

Freya is x years old. Her brother Tom is 4 years older than Freya. Their mother's age is three times Freya's age. The sum of all three ages is 64 years.

(a)Write an expression for Tom's age in terms of x.(1)

(b)Write an expression for the mother's age in terms of x.(1)

(c)Form an equation in x using the information about the sum of the three ages, and solve it to find x.(3)

(d)State the mother's age.(1)

(Total for Question 18 is 6 marks)

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

(a)Find the next term of the sequence.(1)

(b)Show that the nth term of the sequence is n² + 2n.(3)

(c)Use your formula to find the 10th term of the sequence.(1)

(Total for Question 19 is 5 marks)

A repair shop, Sanjay's Repairs, charges a call-out fee of 15 pounds plus 8 pounds per hour of labour. Let C be the total cost in pounds for h hours of labour.

(a)Write a formula for C in terms of h.(1)

(b)A repair for Priya costs 59 pounds in total. Form and solve an equation to find the number of hours of labour, h.(3)

(c)A rival company, TechFix, charges a flat fee of 27 pounds plus 2 pounds per hour of labour. Find the number of hours, h, for which both companies would charge the same total amount, showing your working.(3)

(d)A repair takes exactly 1 hour of labour. State which company charges less, and justify your answer with a calculation.(2)

(Total for Question 20 is 9 marks)

Mark scheme · CE.M4 Algebra

Question 1

(a) B1 3x + 10y oe cao
(a) Answer: 3x + 10y
(b) B1 3a² + 8a oe cao
(b) Answer: 3a² + 8a

Question 2

(a) B1 5 cao
(a) Answer: 5
(b) M1 r² = 25 seen
(b) A1 28 cao
(b) Answer: 28

Question 3

(a) B1 6x + 15 oe cao
(a) Answer: 6x + 15
(b) M1 4(x - 2) = 4x - 8 seen
(b) M1 3(x + 6) = 3x + 18 seen
(b) A1 7x + 10 oe cao
(b) Answer: 7x + 10

Question 4

(a) M1 3x = 15 seen
(a) A1 x = 5 cao
(a) Answer: x = 5
(b) M1 5x - 15 = 20 (expand) or x - 3 = 4 (divide by 5) seen
(b) A1 x = 7 cao
(b) Answer: x = 7

Question 5

(a) M1 x² + 5x + 3x + 15 seen (or three of the four terms correct)
(a) A1 x² + 8x + 15 oe cao
(a) Answer: x² + 8x + 15
(b) M1 x² + 4x - 4x - 16 seen
(b) A1 x² - 16 cao
(b) Answer: x² - 16

Question 6

(a) B1 3(2x + 3) cao, fully factorised
(a) Answer: 3(2x + 3)
(b) M1 4xy taken out as a common factor, e.g. 4xy(2x - 3) attempted
(b) A1 4xy(2x - 3) cao, fully factorised
(b) Answer: 4xy(2x - 3)

Question 7

(a) M1 4x + 8 = 2x + 18 (both brackets expanded)
(a) M1 2x = 10 (unknowns and numbers correctly collected)
(a) A1 x = 5 cao
(a) Answer: x = 5
(b) M1 2x = 16 seen
(b) A1 x = 8 cao
(b) Answer: x = 8

Question 8

(a) B1 17 and 20 both correct cao
(a) Answer: 17, 20
(b) M1 3n seen (common difference identified)
(b) A1 3n + 2 oe cao
(b) Answer: 3n + 2
(c) M1 3n + 2 = 101 set up, leading to 3n = 99
(c) A1 n = 33, a whole number, so yes 101 is a term (cao, ft from part b)
(c) Answer: Yes, 101 is a term (n = 33)

Question 9

(a) B1 x⁸ cao
(a) Answer: x⁸
(b) B1 y¹² cao
(b) Answer: y¹²
(c) M1 12 / 4 = 3 seen, or x⁷ / x² = x⁵ seen
(c) A1 3x⁵ oe cao
(c) Answer: 3x⁵

Question 10

M1 multiply every term by 6 (the lowest common multiple of 2 and 3): 3(x + 4) - 2(x - 1) = 18
M1 expand both brackets: 3x + 12 - 2x + 2 = 18
dM1 simplify to x + 14 = 18 (dependent on previous method)
A1 x = 4 cao
Answer: x = 4

Question 11

(a) B1 2(x + 5) + 2(x - 2) = 46 oe (accept simplified form 4x + 6 = 46)
(a) Answer: 2(x + 5) + 2(x - 2) = 46
(b) M1 4x + 6 = 46 leading to 4x = 40
(b) A1 x = 10 cao
(b) Answer: x = 10
(c) B1 8 cm ft from their value of x in part b
(c) Answer: 8 cm

Question 12

(a) M1 y - 7 = 3x seen
(a) A1 x = (y - 7)/3 oe cao
(a) Answer: x = (y - 7)/3
(b) M1 r² = A/π seen
(b) A1 r = √A/&π; oe cao
(b) Answer: r = √A/&π;

Question 13

(a) M1 3x < 15 seen
(a) A1 x < 5 cao
(a) Answer: x < 5
(b) B1 at least 3 correct values with no more than one error, or one end value missing
(b) B1 -2, -1, 0, 1, 2 all listed and no extra values cao
(b) Answer: -2, -1, 0, 1, 2

Question 14

M1 rearrange or substitute to eliminate one variable, e.g. x = y + 2 substituted into the first equation
M1 correct equation in one variable formed, e.g. 5y = 10
A1 y = 2 cao
A1 x = 4 cao (ft from their y)
Answer: x = 4, y = 2

Question 15

(a) M1 two numbers that multiply to give 12 and add to give 7 identified (3 and 4)
(a) A1 (x + 3)(x + 4) cao
(a) Answer: (x + 3)(x + 4)
(b) M1 two numbers that multiply to give -24 and add to give -5 identified (-8 and 3)
(b) A1 (x - 8)(x + 3) cao
(b) Answer: (x - 8)(x + 3)

Question 16

M1 factorise as (x + 5)(x - 3) = 0
M1 set each factor equal to zero: x + 5 = 0 or x - 3 = 0
A1 x = -5 cao
A1 x = 3 cao
Answer: x = -5 or x = 3

Question 17

(a) M1 numerator factorised as (x - 3)(x + 3) using the difference of two squares
(a) A1 x - 3 cao
(a) Answer: x - 3
(b) M1 numerator factorised as 3x(x + 2)
(b) A1 x + 2 cao
(b) Answer: x + 2

Question 18

(a) B1 x + 4 oe cao
(a) Answer: x + 4
(b) B1 3x oe cao
(b) Answer: 3x
(c) M1 x + (x + 4) + 3x = 64 formed (oe, ft from parts a and b)
(c) M1 5x + 4 = 64 leading to 5x = 60
(c) A1 x = 12 cao
(c) Answer: x = 12
(d) B1 36 years ft from their value of x in part c
(d) Answer: 36 years

Question 19

(a) B1 48 cao
(a) Answer: 48
(b) M1 second differences found to be constant and equal to 2, giving coefficient of n² equal to 1 (i.e. n²)
(b) M1 sequence minus n² calculated for at least two terms, giving 2, 4, 6, ... i.e. 2n
(b) A1 nth term = n² + 2n stated, confirmed against given terms (cso, answer given)
(b) Answer: n² + 2n
(c) B1 120 ft from their formula in part b
(c) Answer: 120

Question 20

(a) B1 C = 15 + 8h oe cao
(a) Answer: C = 15 + 8h
(b) M1 15 + 8h = 59 formed
(b) M1 8h = 44 seen
(b) A1 h = 5.5 cao (accept 5 1/2)
(b) Answer: h = 5.5
(c) M1 15 + 8h = 27 + 2h formed
(c) M1 6h = 12 seen
(c) A1 h = 2 cao
(c) Answer: h = 2
(d) B1 both costs correctly calculated: Sanjay's = 23 pounds, TechFix = 29 pounds
(d) B1 correct conclusion with reason: Sanjay's Repairs is cheaper, by 6 pounds, ft from their two costs
(d) Answer: Sanjay's Repairs is cheaper (23 pounds vs 29 pounds)

Algebra - Worksheets, Questions and Revision