Paradigm Specialising in O & A Level Mathematics
Answer Key
Estimation & Approximation
1 Solutions:
3
√999 √65
(c)(i) × 5.01 (ii) 3 × 19.9
9.8 √9
3 √64
√1000 ≈ × 20
≈ × 5 3
√8
10
8
10 = 2 × 20
= 10 × 5
= 80
=5
Ans: (ai) 3025.38(2d.p) (aii) 3000 (2s.f) (b) 0.0329 (2s.f) (ci) 5 (cii) 80
2
Ans: (a) 0.02087, (b) 0.021
3 Solution:
√101.3×64.231 10×60
(a) =
(1.98)3 8
= 75
Ans: (a) 75, (b) Underestimate. The numbers in the numerator have been rounded
down and the number in the denominator has been rounded up,
making the final answer smaller than the actual value.
Algebra
1 Solutions:
4(2𝑝−3𝑞)−5(3𝑝−5𝑞)
(a) 5𝑥 + 15𝑦 + 12𝑥 − 6𝑦 (b) 20
8𝑝−12𝑞−15𝑝+25𝑞
= 17𝑥 + 9𝑦 (b) = 20
−7𝑝+13𝑞
=
−7𝑝+13𝑞 20
Ans: (a) 17𝑥 + 9𝑦 (b)
20
2 Solutions:
2𝑥
(a) 3 = −54, (b) 6(𝑥 − 1) − 2(𝑥 + 2) = 12,
6𝑥 − 6 − 2𝑥 − 4 = 12
2𝑥 = −108 4𝑥 − 10 = 12
𝑥 = −5𝑥 4𝑥 = 22
1
𝑥 = 5 ∕ 5.5
2
1
Ans: (a) 𝑥 = −5𝑥 (b) 𝑥 = 5.5/5 2
Maths Secrets Page 5
Paradigm Specialising in O & A Level Mathematics
3 Solutions:
1
(a) 𝑤 = 3 [(−2)2 + 3)] (b) 32 = 8(𝑥 − 3)
32 = 8𝑥 − 24
7
=3 8𝑥 = 56
𝑥=7
7
Ans: (a) 𝑤 = 3, (b) 𝑥 = 7
4 Solution:
4 − 3[2 − 3 + 2𝑦]
= 4 − 3(2𝑦 − 1)
= 4 − 6𝑦 + 3
= 7 − 6𝑦
Ans: 7 − 6𝑦
5 Solutions:
(c) 5𝑢(2𝑣 − 3) − 6(2𝑣 − 3)
= (5𝑢 − 6)(2𝑣 − 3)
Ans: (a) 5𝑚(5𝑝 − 8𝑛) (b) 6𝑎(−3𝑥 + 4𝑦 − 1) (c) (5𝑢 − 6)(2𝑣 − 3)
6 Ans: (a) 𝑡 , (b) ℎ𝑚 + 15, (c) −𝑐 + 1
2𝑣
Algebra (Word Problem)
1 Solutions:
(b)(i) 2𝑝 + 4(𝑝 − 8) (b)(ii) Total cost
= 2𝑝 + 4𝑝 − 32 = 6(50) − 32
= $(6𝑝 − 32) = $268
Ans: (a) $(𝑝 − 8) (bi) $(6𝑝 − 32) (bii) $268
2 Solutions:
(a)(ii) 2(𝑥 + 8) = 2𝑥 + 16 years old
(b) 𝑥 + 𝑥 + 8 + 3𝑥 + 2𝑥 + 16 = 129
7𝑥 + 24 = 129
𝑥 = 15
Ans: (a) (𝑥 + 8) years old (aii) 2𝑥 + 16 years old (b) Mary’s age is 15 years old.
Maths Secrets Page 6
