Chapter 4 Practice Paper: Algebraic Fractions and Formulae — mark scheme (print)

QuizQuipPractice Series

QQ/S2MATH/CH4

Secondary 2 · Lower Secondary Mathematics (G2 / G3)

Mark scheme

Algebraic Fractions and Formulae

Chapter 4 Practice Paper · Maximum mark: 30

Notes for markers

Award one mark for each point marked [1], up to the marks shown for the part.
Accept answers worded differently that show the same understanding.
Units are required where the answer line shows one. ecf: allow an error carried forward.

Section A: Multiple-choice questions[10 marks]

Question	Answer
1	BCancel the shared factor 2x, leaving 5x².	1
2	CThe numerators add over the shared denominator: 2x/4 = x/2.	1
3	AThe a cancels and 3/6 = 1/2.	1
4	DDividing by 2 multiplies the denominator: x over 14.	1
5	BMultiply both sides by 4: x = 20.	1
6	CThe top factorises as (x + 4)(x − 4), and x + 4 cancels.	1
7	AMultiply by 2 to remove the half, then divide by h.	1
8	DMultiply by x: 3 = 9x, so x = 1/3.	1
9	BOver 12: 4y/12 − 3y/12 = y/12.	1
10	CSubtract u, then divide by a, keeping (v − u) in brackets.	1

Section B: Structured questions[20 marks]

Question	Answer and marking points	Marks
1(a)	5x/15 + 3x/15 = 8x/15. Common denominator 15 used [1] 8x/15 [1]	2
1(b)	8x/12 − 3x/12 = 5x/12. Common denominator 12 used [1] 5x/12 [1]	2
1(c)	12 (the LCM of 3 and 4) [1]	1
	Total for Question 1	5
2(a)	(x + 5)(x − 5)/(x + 5) = x − 5. Factorises the numerator [1] x − 5 [1]	2
2(b)	x(x + 7)/(x + 7) = x. Factorises to x(x + 7) [1] x [1]	2
2(c)	The numerator x² + 9 is a sum of squares, not a product containing (x + 3), so nothing can cancel. Numerator has no (x + 3) factor [1]	1
	Total for Question 2	5
3(a)	3(x + 3) = 2(2x − 1). 3(x + 3) [1] 2(2x − 1) [1]	2
3(b)	3x + 9 = 4x − 2, so x = 11. Expands both brackets [1] x = 11 [1]	2
3(c)	Left: 14/4 = 3.5; right: 21/6 = 3.5 ✓. Both sides evaluated to 3.5 [1]	1
	Total for Question 3	5
4(a)	5y = 3x − 4, so 3x = 5y + 4 and x = (5y + 4)/3. Multiplies by 5 [1] Adds 4 to get 3x = 5y + 4 [1] x = (5y + 4)/3 [1]	3
4(b)	(35 + 4)/3 = 13 x = 13 [1]	1
4(c)	(9 − 4)/5 = 1 y = 1 [1]	1
	Total for Question 4	5

— End of mark scheme —