25[a math cd]
DESCRIPTION
mathTRANSCRIPT
![Page 1: 25[A Math CD]](https://reader038.vdocument.in/reader038/viewer/2022102800/563db786550346aa9a8bd684/html5/thumbnails/1.jpg)
1 © Penerbitan Pelangi Sdn. Bhd.
Paper 1
1. y ∝ x y = kx 6 = k(2)
k = 6—2
= 3Hence, y = 3x
Answer: A
2. y ∝ x y = kx 10 = k(5)
k = 10–––5
= 2Hence, y = 2x
Answer: C
3. p ∝ t p = kt 4 = k(1) k = 4Hence, p = 4t
When t = 2, p = 4(2) p = 8Answer: D
4. y ∝ 1—x y = k—x
2 = k—3
k = 2 × 3 = 6
Hence, y = 6—x
Answer: B
5. Q ∝ 1—x Q = k—x
9 = k—4
k = 4 × 9 = 36
Hence, Q = 36–––x
Answer: D
6. w ∝ 1—v w = k—v
3 = k—4
k = 3 × 4 = 12
Hence, w = 12–––v
When v = 2, w = 12–––2
= 6
Answer: D
7. y ∝ x—t
y = kx–––t
15 = k × 6–––––4
k = 15 × 4––––––6
= 10
Hence, y = 10x––––t
Answer: B
CHAPTER
25 VariationsCHAPTER
![Page 2: 25[A Math CD]](https://reader038.vdocument.in/reader038/viewer/2022102800/563db786550346aa9a8bd684/html5/thumbnails/2.jpg)
2
Mathematics SPM Chapter 25
© Penerbitan Pelangi Sdn. Bhd.
8. P ∝ Q
–––R
P = kQ–––R
4 = k × 24––––––3
k = 3 × 4–––––24
= 1—2
Hence, P = Q
–––2R
Answer: C
9. T ∝ x—y
T = kx–––y
7 = k × 14––––––12
k = 7 × 12––––––14
= 6
Hence, T = 6x–––y
n = 6 × 16––––––8
= 12Answer: B
Paper 1
1. H ∝ 1x2
H = kx2
12
= k42
k = 12
× 42
= 8
Hence, H = 8x2
When H = 32, 32 = 8x2
x2 = 832
= 14
x = 14
= 12
Answer: C
2. Q ∝ H2
Q ∝ 1––––3 R
Hence, Q ∝ H2––––
3 RAnswer: B
3. H ∝ F2–––G
H = kF2–––G
12 = k × 62––––––
16
k = 12 × 16–––––––62
= 16–––3
Hence, H = 16F2––––3G
15 = 16F2––––––3 × 20
F2 = 15 × 3 × 20––––––––––16
= 225––––4
F = 225––––4
= 15–––2
Answer: D
4. T ∝ x3
T ∝ 1–––y
Hence, T ∝ x 3–––
y
x3 T ∝ ––– y
1—2
Compare with T ∝ x m
y n
Hence, m = 3, n = 12
Answer: C
5. E ∝ 1–––t
Answer: A
![Page 3: 25[A Math CD]](https://reader038.vdocument.in/reader038/viewer/2022102800/563db786550346aa9a8bd684/html5/thumbnails/3.jpg)
3
Mathematics SPM Chapter 25
© Penerbitan Pelangi Sdn. Bhd.
6. e ∝ 1––––f g
e = k––––f g
3 = k––––4 9
k = 3 × 4 9 = 36Hence, e = 36––––
f g
9—2 = 36––––
2 n
9—2 = 18–––
n
n = 2 × 18––––––9
= 4 n = 42
= 16
Answer: D
7. p ∝ q3
p = kq3
12 = k × 23
k = 12–––8
= 3––2
Hence, p = 3—2 q3
r = 3—2 (4)3
= 96
Answer: C
8. P ∝ 1x
P = kx
3 = k4
k = 12
Hence, P = 12x
Answer: A
9. P ∝ 1x2
P = k 1x2
6 = k 132
k = 54
Hence, P = 54 1x2
When x = 4, P = 54 142
= 278
Answer: A
10. y ∝ t
y = k t
2 = k 162 = 4k
k = 12
Hence, y = 12
t
When t = 25, y = 12
25
= 52
Answer: D
11. H ∝ xy
H = kxy
8 = k163
k = 3 ×84
= 6
Hence, H = 6xy
When x = 36 and y = 4
H = 6364
= 9Answer: B
12. MN 2 = k
M = k 1–––N 2
Hence, M ∝ 1–––N 2
Answer: B
![Page 4: 25[A Math CD]](https://reader038.vdocument.in/reader038/viewer/2022102800/563db786550346aa9a8bd684/html5/thumbnails/4.jpg)
4
Mathematics SPM Chapter 25
© Penerbitan Pelangi Sdn. Bhd.
13. R ∝ 1X3
R = k1 1X3 2
32
= k1 123 2
k = 12
Hence, R = 12X 3
m = 1243
= 316
Answer: A
14. y ∝ x3
y = kx3
6 = k × 23
k = 6–––23
= 3—4
Hence, y = 3—4
x3
When y = 48, 48 = 3—4
x3
x3 = 48 × 4––––––3
x = 3 64 = 4
Answer: C
15. w ∝ 1–––x2
w = k–––x2
6 = k–––32
k = 6 × 32
= 54
Hence, w = 54–––x2
When x = 6, w = 54–––62
= 3—2
Answer: A
16. y ∝ rt2
y = krt2
54 = k × 4 × 32
k = 54–––––4 × 9
= 3—2
Hence, y = 3—2
rt2
45 = 3—2
× h × 52
h = 2 × 45––––––3 × 52
= 1.2
Answer: A
Paper 1
1. y ∝ x y = kx 10 = k × 2
k = 10–––2
= 5Hence, y = 5x
When x = 4, y = 5 × 4 = 20
Answer: C
2. n ∝ p n = kp 12 = k × 3
k = 12–––3
= 4Hence, n = 4p
Answer: D
![Page 5: 25[A Math CD]](https://reader038.vdocument.in/reader038/viewer/2022102800/563db786550346aa9a8bd684/html5/thumbnails/5.jpg)
5
Mathematics SPM Chapter 25
© Penerbitan Pelangi Sdn. Bhd.
3. y ∝ t2
y = kt 2 36 = k × 32
k = 36–––9
= 4Hence, y = 4t2
When y = 100, 100 = 4t2
t2 = 100––––4
= 25 t = 25 = 5
Answer: B
4. T ∝ q2
T = kq2
24 = k × 42
k = 24–––16
= 3—2
Hence, T = 3—2
q2
Answer: A
5. H ∝ G3
H = kG3
216 = k × 33
k = 216––––33
= 8Hence, H = 8G3
Answer: B
6. e ∝ f 3 e = kf 3 36 = k × 33
k = 36–––27
= 4—3
Hence, e = 4—3
f 3
x = 4—3
× 63
= 288
Answer: C
7. L ∝ m L = k m 3 = k 144 3 = k(12)
k = 3–––12
= 1—4
Hence, L = 1—4
m
Answer: D
8. P ∝ h P = k h
15 = k 9–––16
15 = k × 3—4
k = 4 × 15––––––3
= 20
Hence, P = 20 h 60 = 20 w
w = 60–––20
= 3 w = 32
= 9
Answer: C
9. y ∝ m y = km y = k(5h − 3) 14 = k(5 × 2 − 3) 14 = 7k
k = 14–––7
= 2Hence, y = 2(5h − 3)
When y = 44, 44 = 2(5h − 3) = 10h − 6 44 + 6 = 10h
h = 50–––10
= 5
Answer: A
![Page 6: 25[A Math CD]](https://reader038.vdocument.in/reader038/viewer/2022102800/563db786550346aa9a8bd684/html5/thumbnails/6.jpg)
6
Mathematics SPM Chapter 25
© Penerbitan Pelangi Sdn. Bhd.
10. P—Q
= 2—1
= 3–––3—2
= 4—2
= 2
Hence, P—Q
= 2
P = 2Q
Answer: B
11. F–––e2 = 4–––
12 = 16–––22 = 36–––
32 = 4
Hence, F–––e2 = 4
F = 4e2
Answer: A
12. y ∝ n y = kn
y = k––––––8x – 5
5 = k–––––––8(1) – 5
5 = k—3
k = 3 × 5 = 15
Hence, y = 15––––––8x – 5
When x = 5, y = 15–––––––8(5) – 5
= 15–––35
= 3—7
Answer: C
13. J ∝ 1—e
J = k—e
8 = k—5
k = 5 × 8 = 40
Hence, J = 40–––e
When J = 10, 10 = 40–––e e = 40–––
10 = 4
Answer: D
14. m ∝ 1—t
m = k—t
8 = k—2
k = 2 × 8 = 16
Hence, m = 16–––t
Answer: A
15. Q ∝ 1–––x2
Q = k–––x2
10 = k––––– 1—
2
2
k = 1—2
2 × 10
= 5—2
Hence, Q = 5––––2x2
Answer: C
16. g ∝ 1–––r2
g = k–––r2
32 = k–––12
k = 32
Hence, g = 32–––r2
128 = 32–––h2
h2 = 32––––128
= 1—4
h = 1—4
= 1—2
Answer: D
![Page 7: 25[A Math CD]](https://reader038.vdocument.in/reader038/viewer/2022102800/563db786550346aa9a8bd684/html5/thumbnails/7.jpg)
7
Mathematics SPM Chapter 25
© Penerbitan Pelangi Sdn. Bhd.
17. G ∝ 1–––y3
G = k–––y3
2 = k–––43
k = 2 × 43 = 128
Hence, G = 128––––y3
Answer: D
18. h ∝ 1–––g3
h = k–––g3
3 = k–––23
k = 23 × 3 = 24
Hence, h = 24–––g3
1––9
= 24–––x3
x3 = 24 × 9 x = 3 216 = 6
Answer: D
19. y ∝ 1–––q
y = k–––q
1—6
= k––––81
k = 81
––––6
= 9—6
= 3—2
Hence, y = 3––––2 q
Answer: A
20. P ∝ 1–––y
P = k–––y
2 = k –––––121
k = 2 × 121 = 22
Hence, P = 22–––
y
When y = 16, P = 22––––16
= 22–––4
= 11–––2
Answer: B
21. R ∝ x–––p3
R = kx–––p3
3 = k(4)––––23
k = 23 × 3––––––4
= 6
Hence, R = 6x–––p3
2 = 6x–––33
x = 2 × 33––––––
6 = 9
Answer: C
22. D ∝ H 2–––E
D = kH 2––––E
10 = k × 52––––––
20
k = 20 × 10–––––––52
= 8
Hence, D = 8H 2––––E
18 = 8 × 62––––––
E E = 8 × 62
––––––18
= 16
Answer: C
![Page 8: 25[A Math CD]](https://reader038.vdocument.in/reader038/viewer/2022102800/563db786550346aa9a8bd684/html5/thumbnails/8.jpg)
8
Mathematics SPM Chapter 25
© Penerbitan Pelangi Sdn. Bhd.
23. F ∝ 1LM
F = kLM
45 = k12
× 6
45 = k3
k = 3 × 45 = 135
Hence, F = 135LM
27 = 13513
x
13
x = 13527
x = 135 × 327
= 15
Answer: C
24. f ∝ m2–––
n
f = km2
––––n
16 = k × 42––––––
9
k = 16 9–––––42
= 3
Hence, f = 3m2––––
n
Answer: B
25. p ∝ x y p = kx y 6 = k(6) 4 = 12k
k = 612
= 12
Hence, p = 12
x y
9 = 12
(2) a
9 = a a = 92
= 81
Answer: D
26. Q ∝ x3
Q ∝ 1y
Hence, Q ∝ x3
y
Q ∝ x3 y− 1—2
Compare with Q ∝ xmyn
Hence, m = 3 and n = − 12
.
Answer: C
27. Let d = density of the iron cube m = mass of the iron cube and x = lenght of the side of the iron cube
d ∝ mx3
d = k mx3
7.85 = k 7 850103 7.85 kg = 7 850 g
k = 1
Hence, d = mx3
When x = 6 cm,
7.85 = m63
m = 1695.6 g = 1.6956 kg
Answer: A
28. Let E = kinetic energy m = mass of the object and v = speed of the object
E ∝ mv2
E = kmv2
640 = k(5)(16)2 k = 0.5Hence, E = 0.5 mv2
When m = 5, E = 1000 1000 = 0.5(5)v 2 v2 = 400 v = 20 m s–1
Answer: B