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

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QQ/S1MATH/CH4

Secondary 1 · Lower Secondary Mathematics (G2 / G3)

Mark scheme

Algebraic Expressions 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	AAll three are like terms with coefficient 8 - 3 + 1 = 6, giving 6m. The last m counts as 1m.	1
2	BMultiply both terms inside the bracket: 5 × 2x = 10x and 5 × (-4) = -20.	1
3	AThe coefficient is the number multiplied by the letter. Since the y-term is -3y, the coefficient is -3.	1
4	CThe HCF of 12a and 18 is 6, so 12a + 18 = 6(2a + 3). The other factorisations leave a common factor inside (or use a fraction).	1
5	BSubstitute: 3 × 5 + 2 × 4 = 15 + 8 = 23.	1
6	AThe terms rise by 3, so the rule starts with 3n; testing n = 1 gives 5 = 3 + 2, so the rule is 3n + 2.	1
7	DCollect the k-terms: 6k - 2k = 4k. Collect the numbers: 4 - 9 = -5. So the result is 4k - 5.	1
8	DAli is y and his sister is y - 3, so the total is y + (y - 3) = 2y - 3.	1
9	CThe terms fall by 5, so the rule contains -5n. The zeroth term would be 101, so the rule is 101 - 5n.	1
10	BSubstitute: P = 2 × (9 + 4) = 2 × 13 = 26. The brackets must be worked out first.	1

Section B: Structured questions[20 marks]

Question	Answer and marking points	Marks
1(a)	7x + 4x - 2x = 9x 9x (coefficients 7 + 4 - 2) [1]	1
1(b)	5a - 3a = 2a and 2b + b = 3b, so the expression simplifies to 2a + 3b. a-terms collected to 2a [1] b-terms collected to 3b [1]	2
1(c)	4 × 3x - 4 × 5 = 12x - 20 Both terms multiplied by 4 [1] 12x - 20 [1]	2
	Total for Question 1	5
2(a)	3(3x + 4) [1]	1
2(b)	The HCF of 15y and 25 is 5, so 15y - 25 = 5(3y - 5). Identifies the HCF as 5 [1] 5(3y - 5) [1] Factorisations that leave a common factor inside, such as 15(y - 5/3) or 5(3y - 5) written as 1(15y - 25), do not earn the second mark.	2
2(c)	The student multiplied only the first term by 4 and copied the 12. Expanding 4(2x + 12) gives 8x + 48, not 8x + 12. Both terms must be divided by the common factor: 8x + 12 = 4(2x + 3). Identifies that only one term was divided (or expands to show 8x + 48) [1] Correct factorisation 4(2x + 3) [1]	2
	Total for Question 2	5
3(a)	F = 4 + 0.6 × 12 = 4 + 7.2 = 11.2, so the fare is $11.20. Substitutes d = 12 into the formula [1] 11.2 with dollars [1]	2
3(b)	13 = 4 + 0.6d, so 0.6d = 9, and d = 9 ÷ 0.6 = 15 km. Forms the equation 4 + 0.6d = 13 [1] d = 15 km [1]	2
3(c)	The fixed starting charge (flag-down fare) paid before any distance is travelled. Fixed/basic charge paid at the start [1]	1
	Total for Question 3	5
4(a)	24 and 29 (the terms rise by 5 each time). [1]	1
4(b)	The terms rise by 5, so the rule starts 5n. Since 5 × 1 = 5 and the first term is 4, the nth term is 5n - 1. Recognises the constant difference 5 [1] 5n - 1 [1]	2
4(c)	If 5n - 1 = 100, then 5n = 101 and n = 20.2, which is not a whole number. Term numbers must be positive integers, so 100 is not a term. Forms and solves 5n - 1 = 100 [1] Concludes n is not a whole number, so 100 is not a term [1]	2
	Total for Question 4	5

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