Which of the following is the fully simplified form of (21x4y3)/(28x6y1)?
A) 3y2/(4x2)
B) 3x2y2/4
C) 3/(4x2y2)
D) 21y2/(28x2)
(Total for Question 1 is 1 mark)
2
Simplify fully (16a5b2)/(24a3b6)
(Total for Question 2 is 1 mark)
3
Simplify fully (12x2y4)/(18x5y)
(Total for Question 3 is 1 mark)
4
Simplify fully (35p3q5)/(20p6q2)
(Total for Question 4 is 1 mark)
5
Simplify fully (7x+21)/(x2+3x)
(Total for Question 5 is 1 mark)
6
Simplify fully (9x-27)/(x2-3x)
(Total for Question 6 is 1 mark)
7
Simplify fully (5x+30)/(x2+6x)
(Total for Question 7 is 1 mark)
8
Simplify fully (x2+8x+15)/(x2+2x-15)
(Total for Question 8 is 2 marks)
9
Simplify fully (x2-64)/(x2+3x-40)
(Total for Question 9 is 2 marks)
10
Simplify fully (x2-36)/(x2+x-30)
(Total for Question 10 is 2 marks)
11
Simplify fully (5x/4y) x (8y3/15x2)
(Total for Question 11 is 2 marks)
12
Simplify fully (9a2/5b3) / (3a/10b)
(Total for Question 12 is 2 marks)
13
Simplify fully (x2-49)/(x2-4x-21)
(Total for Question 13 is 2 marks)
14
Simplify fully (x+5)/(x2-16) x (x-4)/(2x+10)
(Total for Question 14 is 3 marks)
15
Simplify fully (x2-9)/(x+2) / (x-3)/(3x+6)
(Total for Question 15 is 3 marks)
16
Ben is asked to simplify (x2-25)/(x-5) and writes the answer as x-25.
(a)Explain the mistake Ben has made.(1)
(b)Work out the correct fully simplified answer.(2)
(Total for Question 16 is 3 marks)
17
Show that (3x2+11x+6)/(x2-3x-18) simplifies to (3x+2)/(x-6)
(Total for Question 17 is 3 marks)
18
A rectangular banner has area (x2-9)/(x+4) m2 and length (x-3)/(x2-16) m. Find an expression for the width of the banner in its simplest form.
(Total for Question 18 is 4 marks)
19
(x2-4x-21)/(x2-49) x (x2+7x)/(x2+10x+21)
(a)Simplify the expression above fully.(3)
(b)Hence find the value of the expression when x=7, giving your answer as a fraction in its simplest form.(2)
(Total for Question 19 is 5 marks)
Mark scheme · A1D Algebraic Fractions: Simplifying, Multiplying and Dividing: Fluency and Exam Drill
Question 1
B1 correct option identified (A) oe
Answer: A) 3y2/(4x2)
Question 2
B1 2a2/(3b4) cao
Answer: 2a2/(3b4)
Question 3
B1 2y3/(3x3) cao
Answer: 2y3/(3x3)
Question 4
B1 7q3/(4p3) cao
Answer: 7q3/(4p3)
Question 5
B1 7/x cao, factorising 7(x+3) and x(x+3) then cancelling (x+3)
Answer: 7/x
Question 6
B1 9/x cao, factorising 9(x-3) and x(x-3) then cancelling (x-3)
Answer: 9/x
Question 7
B1 5/x cao, factorising 5(x+6) and x(x+6) then cancelling (x+6)
Answer: 5/x
Question 8
M1 numerator factorised as (x+3)(x+5) and denominator as (x-3)(x+5)
A1 (x+3)/(x-3) cao
Answer: (x+3)/(x-3)
Question 9
M1 numerator factorised as (x-8)(x+8) and denominator as (x-5)(x+8)
A1 (x-8)/(x-5) cao
Answer: (x-8)/(x-5)
Question 10
M1 numerator factorised as (x-6)(x+6) and denominator as (x-5)(x+6)
A1 (x-6)/(x-5) cao
Answer: (x-6)/(x-5)
Question 11
M1 multiplies numerators and denominators: 40xy3/(60x2y) oe
A1 2y2/(3x) cao
Answer: 2y2/(3x)
Question 12
M1 inverts second fraction and multiplies: (9a2/5b3)(10b/3a) oe
A1 6a/b2 cao
Answer: 6a/b2
Question 13
M1 numerator factorised as (x-7)(x+7) and denominator as (x-7)(x+3)
A1 (x+7)/(x+3) cao
Answer: (x+7)/(x+3)
Question 14
M1 x2-16 factorised as (x-4)(x+4)
M1 2x+10 factorised as 2(x+5); cancels (x+5) and (x-4) correctly
A1 1/(2x+8) oe cao
Answer: 1/(2x+8)
Question 15
M1 inverts second fraction and factorises x2-9 as (x-3)(x+3)
M1 3x+6 factorised as 3(x+2); cancels (x-3) and (x+2) correctly
A1 3x+9 oe cao
Answer: 3x+9
Question 16
(a) B1 identifies that x2-25 must first be factorised as (x-5)(x+5) as a difference of two squares before any cancelling, rather than cancelling the 25 and the 5 as separate terms oe
(a) Answer: Ben cancelled individual terms (25 and 5) instead of factorising the numerator first.
(b) M1 factorises x2-25 as (x-5)(x+5)
(b) A1 x+5 cao
(b) Answer: x+5
Question 17
M1 numerator factorised as (3x+2)(x+3)
M1 denominator factorised as (x+3)(x-6)
A1 cancels (x+3) correctly to reach (3x+2)/(x-6) cso
Answer: (3x+2)/(x-6)
Question 18
M1 sets up width = area / length
M1 factorises x2-9 as (x-3)(x+3) and x2-16 as (x-4)(x+4)
M1 inverts the length fraction and multiplies, cancelling (x-3) and (x+4) correctly
A1 (x+3)(x-4) oe, or x2-x-12, cao
Answer: (x+3)(x-4), which expands to x2-x-12
Question 19
(a) M1 factorises x2-4x-21 as (x-7)(x+3) and x2-49 as (x-7)(x+7)
(a) M1 factorises x2+7x as x(x+7) and x2+10x+21 as (x+3)(x+7), cancelling (x-7) and (x+3)
(a) A1 cancels the remaining (x+7) to reach x/(x+7) oe cao
(a) Answer: x/(x+7)
(b) M1 substitutes x=7 into x/(x+7) ft from part (a)