Page 39 - ACE YR IGCSE A TOP APPR' TO MATHS

(c) Time car C arrives at City Q x = 2.54 or x = –0.29 [3]
420 (x + 4)(x – 4)
= 21.30 + (c)
67.5 3(x + 4)
= 21.30 + 56 = x – 4 [2]
9 3
= 21.30 + 6 h 13 min (d) 1 – 4x + 4x – xy + 4x
2
= 3.43 a.m. [3] = 4x – xy + 1 [2]
2
(d) Time car A arrives at City Q 2 (a) 3g(1 – 4g + 2g ) [2]
2
= 21.30 + 470 (b) a(c + d) + b(c + d)
67.2 = (a + b)(c + d) [3]
= 21.30 + 6.25 (c) 5(x – 2) + (x – 2)
2
= 21.30 + 6 h 15 min = (x – 2)[5(x – 2) + 1]
= 3.45 a.m. = (x – 2)(5x – 9) [2]
4.04 a.m. – 3.45 a.m. 3 (a) x = 3 – 6y
2
= 19 min [3] x = ±√3 – 6y [2]
15 1 cm : 280000 cm
1 cm : 2800 m (b) 3y – 2 = √x – 1
1 cm : 2.8 km x – 1 = (3y –2)
2
2
(a) 17 = 6.071 cm x = 9y – 12y + 4 + 1
2
2.8 = 9y – 12y + 5 [3]
∼ 6.1 cm [2] (c) −2xy – 6y = 1 + x
(b) 1 cm : 7.84 km x + 2xy = –6y – 1
2
2
3.2 = 0.4082 cm x(1 + 2y) = –6y – 1
2
7.84 x = – 6y + 1 [3]
∼ 0.41 cm 2 [2] 1 + 2y
(c) 6.25 × 7.84 = 49 km ( ) 2
1
2
= 49 × 1000 × 100 cm 4 (a) –3 = 4 2 – p
2
2
2
= 4.9 × 10 cm 2 [2] p = 1+ 3
11
(d) Distance between Cyberjaya and Putrajaya = 4 [2]
= 12 × 0.75 (b) p = ± – r
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= 9 km

q
Speed on the return journey 1
9 = ± – 2
=
0.75 – 0.15 –3
= 15 km/h 1
Percentage increased = ± 6 [2]
15 – 12
= × 100% p + 2qr
12 (c) p =
= 25% [4] 2q
2pq = p + 2qr
p(2q – 1) = 2qr
2 2 Algebra and Graphs p = 2qr
2q – 1
1
Algebra 2(–3) ( )
2
1 (a) x + 4x + 4 – 15 + 6x = 0 = 2(–3) – 1
2
2
7x + 4x – 11 = 0
2
(7x + 11)(x – 1) = 0 = 3 [2]
7
x = – 11 or x = 1 [2]
7 x × 1
81
3x – 15 – 8x – 12 3
(b) = 2 5 (a)
2x – 7x – 15 1 1
2
9
3
–5x – 27 = 4x – 14x – 30 3 × 3 x × 3
2
4x – 9x – 3 = 0 = x 27
2
2
–(–9) + √(–9) – 4(4)(–3) 3x 3
x = x 24
2(4) = [3]
3
Answers 159

Answers.indd 159 15/03/2022 11:08 AM