Book 4B Chapter 9 Variations Exam Focus Plus (Full Solutions) 9 Variations Exam-type Questions Suggested Solutions Paper I k 1. (a) Let y 3 , where k is a non-zero constant. x When x 6 and y 90 , k 90 3 6 k 19 440 19 440 ∴ y x3 Marks 1A 1M 1A 19 440 (by (a)) 33 720 The increase in the value of y 720 90 (b) The final value of y 1M 630 2. Remarks 1A (a) Let f ( x) k1 x 2 k2 x , where k1 and k2 are non-zero constants. f (4) 80 1M k1 (4)2 k2 (4) 80 16k1 4k2 80 f (3) 3 ......(1) 1M for either one* 3k1 k2 1 ......(2) (1) + (2) 4: 28k1 84 for either one* k1 (3) 2 k2 (3) 3 9k1 3k2 3 k1 3 By substituting k1 3 into (2), we have 3(3) k2 1 k2 8 ∴ (b) f ( x) 3 x 2 8 x 1A f ( x) 11 3x 2 8 x 11 0 (3x 11)( x 1) 0 11 x or x 1 3 NSS Mathematics in Action (2nd Edition) 1M 1A 1 © Pearson Education Asia Limited 2020 Book 4B Chapter 9 Variations 3. Exam Focus Plus (Full Solutions) Suggested Solutions Marks (a) Let g ( x) k1 x k2 x 2 , where k1 and k2 are non-zero constants. g (1) 7 1M Remarks k1 (1) k2 (1) 2 7 k1 k2 7 g (1) 11 ......(1) 1M for either one* ......(2) for either one* k1 (1) k2 (1) 2 11 k1 k2 11 (1) + (2): 2k2 18 k2 9 By substituting k2 9 into (2), we have k1 9 11 k1 2 ∴ g ( x) 9 x 2 2 x 1A g ( x) 5 2 x (b) 9x 2x 5 2x 2 1M 9 x2 4 x 5 0 ( x 1)(9 x 5) 0 x 1 or x 4. 5 9 1A (a) Let f ( x) k1 k2 x 2 , where k1 and k2 are non-zero constants. f (4) 69 1M k1 k2 (4)2 69 k1 16k2 69 f (2) 21 ......(1) 1M for either one* k1 4k2 21 ......(2) (1) (2): 12k2 48 for either one* k1 k2 (2) 2 21 k2 4 By substituting k2 4 into (2), we have k1 4(4) 21 k1 5 ∴ f ( x) 5 4 x 2 1A f (3) 5 4(3)2 41 NSS Mathematics in Action (2nd Edition) 1A 2 © Pearson Education Asia Limited 2020 Book 4B Chapter 9 Variations Exam Focus Plus (Full Solutions) Suggested Solutions Marks (b) By (a), we have a 41. ∵ 1M for either one* f ( x) 5 4 x 2 ∴ b f (3) f (3) 41 AB 3 (3) 6 for either one* 1M 1 Area of ABC (6)(41) 2 123 5. Remarks 1M 1A (a) Let g ( x) k1 x k2 x3 , where k1 and k2 are non-zero constants. g (3) 72 1M k1 (3) k2 (3)3 72 3k1 27k2 72 1M for either one* k1 9k2 24 ......(1) g (2) 8 k1 (2) k2 (2)3 8 2k1 8k2 8 k1 4k2 4 (1) (2): 5k2 20 for either one* ......(2) k2 4 By substituting k2 4 into (2), we have k1 4(4) 4 k1 12 ∴ g ( x) 12 x 4 x3 1A g (2) 12(2) 4(2) 8 3 1A (b) By (a), we have p 8 . 1M for either one* ∵ g ( x) 12 x 4 x3 ∴ q g (1) 12(1) 4(1)3 8 PQ 2 (1) 3 1M 1 Area of PQR (3)(72 8) 2 96 NSS Mathematics in Action (2nd Edition) for either one* 1M 1A 3 © Pearson Education Asia Limited 2020 Book 4B Chapter 9 Variations Exam Focus Plus (Full Solutions) Suggested Solutions 6. Marks (a) Let C k1 x k2 x 2 , where k1 and k2 are non-zero constants. 2.28 k1 (15) k2 (15) Remarks 1M 2 15k1 225k2 2.28 ......(1) 9.06 k1 (30) k2 (30) 1M for either one* 2 30k1 900k2 9.06 ......(2) for either one* (2) (1) 2: 450k2 4.5 k2 0.01 By substituting k2 0.01 into (1), we have 15k1 225(0.01) 2.28 k1 0.002 C 0.002 x 0.01x 2 1A (b) When C 25.1, 25.1 0.002 x 0.01x 2 1M ∴ 5 x 2 x 12 550 0 ( x 50)(5 x 251) 0 251 (rejected) 5 ∴ The length of the ruler is 50 cm. x 50 or x 7. 1A (a) Let C k1 k2 w , where k1 and k2 are non-zero constants. 1M 1480 k1 k2 16 k1 4k2 1480 ......(1) 1M for either one* 1600 k1 k2 25 k1 5k2 1600 ......(2) (2) (1): k2 120 By substituting k2 120 into (1), we have k1 4(120) 1480 for either one* k1 1000 C 1000 120 w The required cost $(1000 120 36) $1720 ∴ 1A 1A (b) The total cost of three wooden wardrobes of weight 10.24 kg 3 $(1000 120 10.24) $4152 2 $1720 ∴ The claim is incorrect. NSS Mathematics in Action (2nd Edition) 1M 1A 4 © Pearson Education Asia Limited 2020 Book 4B Chapter 9 Variations Exam Focus Plus (Full Solutions) Suggested Solutions 8. Marks (a) Let E k1n k2 , where k1 and k2 are non-zero constants. Remarks 1M 52 500 k1 (5) k2 5k1 k2 52 500 ......(1) 78 500 k1 (9) k2 1M for either one* 9k1 k2 78 500 ......(2) (2) (1): 4k1 26 000 for either one* k1 6500 By substituting k1 6500 into (1), we have 5(6500) k2 52 500 k2 20 000 ∴ E 6500n 20 000 The required expenditure $[6500(16) 20 000] $124 000 1A 1A (b) Suppose the expenditure of the company in that month is $200 000. 200 000 6500n 20 000 1M 180 000 6500n 360 n , which is not an integer. 13 ∴ It is not possible that when the company employs a certain number of workers in a month, its expenditure in that month is $200 000. 1A 9. (a) f ( x) 4 x 3 79 x 105 (2 x 3)(2 x 2 3 x 35) (2 x 3)(2 x 7)( x 5) 1M 1A (b) (i) Let P k1a 3 k2 a , where k1 and k2 are non-zero constants. 1M 78 k1 (0.5)3 k2 (0.5) 0.125k1 0.5k2 78 ......(1) 1M for either one* 150 k1 (1) k2 (1) 3 k1 k2 150 ......(2) for either one* (2) (1) 2: 0.75k1 6 k1 8 By substituting k1 8 into (2), we have 8 k2 150 k2 158 ∴ P 8a 3 158a NSS Mathematics in Action (2nd Edition) 1A 5 © Pearson Education Asia Limited 2020 Book 4B Chapter 9 Variations Exam Focus Plus (Full Solutions) Suggested Solutions Marks Remarks (ii) When P 210 , 210 8a3 158a 4a 79a 105 0 (2a 3)(2a 7)(a 5) 0 a 1.5 or a 3.5 (rejected) or a 5 (rejected) 1M 3 ∴ The area of the carpet is 1.5 m2. 1M 1A Paper II 10. C Let t kr 2 s , where k is a non-zero constant. 108 k (3) 2 9 k 4 ∴ t 4r 2 s When r 6 and s 25 , t 4(6) 2 25 720 11. C ky 2 Let x , where k is a non-zero constant. z 1 k (2) 2 2 6 3 k 4 3y2 ∴ x 4z When x 6 and y 12 , 3(12) 2 6 4z z 18 12. C Let z kx2 3 y , where k is a non-zero constant. 72 k (4) 2 3 27 k 1.5 ∴ z 1.5x2 3 y When x 9 and y 64 , z 1.5(9) 2 3 64 486 NSS Mathematics in Action (2nd Edition) 6 © Pearson Education Asia Limited 2020 Book 4B Chapter 9 Variations Exam Focus Plus (Full Solutions) Suggested Solutions 13. C Let d k1 k2c 3 , where k1 and k2 are non-zero constants. 42 k1 k2 (2)3 k1 8k2 42 ......(1) 357 k1 k2 (5)3 k1 125k2 357 ......(2) (2) (1): 133k2 399 k2 3 By substituting k2 3 into (1), we have k1 8(3) 42 k1 18 ∴ d 18 3c3 When c 4 , d 18 3(4)3 210 14. A k2 , where k1 and k2 are non-zero constants. q k 12 k1 2 2 2k1 k2 24 ......(1) k 4 k1 2 10 10k1 k2 40 ......(2) (2) (1): 8k1 64 Let p k1 k1 8 By substituting k1 8 into (1), we have 2(8) k2 24 k2 40 40 ∴ p 8 q When p 10 , 40 10 8 q q 20 NSS Mathematics in Action (2nd Edition) 7 © Pearson Education Asia Limited 2020 Book 4B Chapter 9 Variations Exam Focus Plus (Full Solutions) Suggested Solutions 15. C k , where k is a non-zero constant. v2 Let v0 be the original value of v. Let u New value of v (1 40%)v0 0.6v0 k k New value of u 2 (0.6v0 ) 0.36v0 2 k k 2 2 0.36v0 v0 100% Increase % of u k v0 2 7 177 % 9 16. C k a , where k is a non-zero constant. b2 Let a0 and b0 be the original values of a and b respectively. Let c New value of a (1 84%)a0 0.16a0 New value of b (1 50%)b0 0.5b0 New value of c k 0.16a0 0.4k a0 1.6k a0 (0.5b0 )2 0.25b0 2 b0 2 1.6k a0 k a0 b0 2 b0 2 % change of u 100% k a0 b0 2 ∴ 60% c is increased by 60%. NSS Mathematics in Action (2nd Edition) 8 © Pearson Education Asia Limited 2020 Book 4B Chapter 9 Variations Exam Focus Plus (Full Solutions) Suggested Solutions 17. D z . kx 2 be the original values of x and z respectively. Let z kx 2 y , where k is a non-zero constant. Then y Let x0 and z0 New value of x (1 20%) x0 0.8x0 New value of z (1 20%) z0 1.2 z0 1.2 z0 1.875 z0 New value of y 2 k (0.8 x0 ) kx0 2 1.875 z0 z 02 2 kx0 kx0 % change of y 100% z0 kx0 2 ∴ 18. 87.5% y is increased by 87.5%. A kq3 , where k is a non-zero constant. Then r p r k 3 q Let p k2 p2r q6 1 q6 k 2 p2r ∴ 19. q6 must be constant. p2r D Let s k t 2 u3 , where k is a non-zero constant. Then k st 2 u 3 k 2 s 2t 4u 3 ∴ s 2t 4u 3 must be constant. 20. C ∵ The equation of the straight line in the figure is y mx c , where m and c are non-zero constants. ∴ y is partly constant and partly varies directly as x. NSS Mathematics in Action (2nd Edition) 9 © Pearson Education Asia Limited 2020