HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) Solution Guide

9 Rate and Ratio Instant Practice 2

1. Let x and y be the original numbers of boys and girls in
the classroom respectively.
Conventional Questions x:y=6:7 [1A]
Instant Practice 1 x 6
=
y 7
1. 6 30 + k 45 – 30 = 4 [1M] + [1A] 7x = 6y
60 60
3+k =4
7x – 6y = 0 ...... (1)
4 (x + 9) : (y + 2) = 11 : 10 [1A]
k=4 [1A] x + 9 11
=
y + 2 10
2 15 + k 20 = 1.5
10x + 90 = 11y + 22
2. [1M] + [1A]
60 60 10x – 11y = –68 ...... (2)
0.5 + k = 1.5 On solving (1) and (2), we have y = 28. [1M]
3
k=3 [1A] ` Original number of girls = 28 [1A]
Alternative method:
3. Let y hours be the time taken for the car to travel from Let 6k be the original numbers of boys, where k is a
city A to city B. positive constant. Then the original number of girls is 7k.
[1A]
Then the time taken for the car to travel from city B to
city C is 84 – y hours
(6k + 9) : (7k + 2) = 11 : 10 [1M] + [1A]
[1M] + [1A]
60 11(7k + 2) = 10(6k + 9)
75y + 90 84 – y = 114 [1M] + [1A] 77k + 22 = 60k + 90
60
k=4
126 – 15y = 114
y = 0.8 [1A] ` Original number of girls = 7(4) = 28 [1A]

` Time taken = 0.8 hours (Or 48 minutes)

2. Let x and y be the original numbers of male and female
staff respectively.
4. Let y hours be the time taken for the man to walk from
x : y = 13 : 12 [1A]
city B to city C.
12x = 13y
Then the time taken for him to walk from city A to city
B is 80 – y hours.
12x – 13y = 0 ...... (1)
[1M] + [1A]
60 (x + 8) : (y + 7) = 12 : 11 [1A]
3 80 – y + 4y = 5 [1M] + [1A] 12(y + 7) = 11(x + 8)
60
4+y=5 11x – 12y = –4 ...... (2)
y=1 [1A] On solving (1) and (2), we have x = 52. [1M]
` Time taken = 1 hour ` Original number of male staff = 52 [1A]

5. Let y hours be the time taken for the car to travel from 3. Let x and y be the original numbers of boys and girls
city X to city Y. respectively.
Then the time taken for the car to travel from city Y to x:y=7:6 [1A]

city Z is 64 – y hours.
6x = 7y
[1M] + [1A]
60 6x – 7y = 0 ...... (1)
80y + 72 64 – y = 80 [1M] + [1A] (x – 15) : (y – 10) = 12 : 11 [1A]
60
76.8 + 8y = 80 11(x – 15) = 12(y – 10)
y = 0.4 [1A] 11x – 12y = 45 ...... (2)
` Time taken = 0.4 hours (Or 24 minutes) On solving (1) and (2), we have x = 63 and y = 54. [1M]
` The original numbers of boys and girls are 63 and 54
respectively.
Original number of students = 63 + 54
= 117 [1A]

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 9 Rate and Ratio

4. Let x and y be the original numbers of adults and 2. (a) Distance between car A and town X
= 90 (75)
children in the playground respectively.
[1M]
x:y=3:2 [1A] 180
75
2x = 3y = km (Or 37.5 km) [1A]
2
2x – 3y = 0 ...... (1)
(b) Suppose that two cars meet at the time t minutes
(x – 10) : (y – 8) = 8 : 5 [1A] after 6:00 in the morning.
5(x – 10) = 8(y – 8) 90 t = 42 [1M]
5x – 8y = –14 ...... (2) 180
On solving (1) and (2), we have x = 42 and y = 28. [1M] t = 84
` The original numbers of adults and children are 42 ` Two cars meet at 7:24 in the morning. [1A]
and 28 respectively. (c) Speed of car B during 6:00 to 7:15 in the morning
New number of children = 28 – 8 42
=
= 20 [1A] 75
14
= km/minute
25
5. Let x and y be the original numbers of male and female Time interval that car B come to rest
members respectively.
= [1M]
x : y = 13 : 7 [1A]
7x – 13y = 0 ...... (1)
(x – 24) : (y + 14) = 8 : 7 [1A] 135
= minutes [1A]
7(x – 24) = 8(y + 14) 7
7x – 8y = 280 ...... (2)
On solving (1) and (2), we have x = 104 and y = 56. [1M] 3. (a) Speed of Andy
` Original numbers of male and female members are 24
= [1M]
104 and 56 respectively. 2
= 12 km/h [1A]
New number of members = 104 + 56 – 24 + 14
(b) Suppose that Andy and Billy meet at the time t
= 150 [1A] minutes after 8:00 in the morning.
12 t = 15 [1M]
Instant Practice 3 60
1. (a) Distance between car A and town X t = 75

= 90 (80)
` They meet at 9:15 in the morning. [1A]
[1M]
120 (c) Speed of Andy = 12 km/h
= 60 km [1A] Speed of Billy during the period 9:20 to 10:00
(b) Distance between car B and town X 15
= [1M]
= 30 80 [1M] 40
60
= 40 km 60
Distance between the two cars = 22.5 km/h
= 60 – 40  2(12) km/h
= 20 km [1A] ` The claim is disagreed. [1A]
(c) Speed of car B
90 − 40 4. (a) Speed of Amy
= [1M]
40 21
= [1M]
60 105
= 75 km/h [1A]
60
= 12 km/h [1A]

43
HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) Solution Guide

(b) Speed of Betty 2. 2d = 3e

21 − 12 `d:e=3:2
= [1M]
75 d : e =3 : 2 =3 : 2
d : f =1 : 5=3 : 15
60
= 7.2 km/h [1A] `d : e : f = 3 : 2 : 15
(c) Suppose that Amy and Betty meet at the time t Let d = 3k, where k is a non-zero constant. Then e = 2k
minutes after 8:00 in the morning. and f = 15k.
(d + 2e) : (2e + f ) = [3k + 2(2k)] : [2(2k) + 15k]
12 t = 12 + 7.2 t – 45 [1M]
60 60 = 7k : 19k
12t = 720 + 7.2t – 324 = 7 : 19
t = 82.5 ` The answer is D.
` They meet at 9:22:30 in the morning.
The claim is disagreed. [1A] 3. p = 3r
`p:r=3:1
5. (a) Speed of car A p : r=3 : 1=9 : 3
72 q : r= 4 : 3= 4 : 3
=
3
`p : q : r= 9 : 4 : 3
= 24 km/h
Distance of car A from town P Let p = 9k, where k is a non-zero constant. Then q = 4k
and r = 3k.
= 24 90 [1M]
60 (p + q + r) : (2p + q) = (9k + 4k + 3k) : [2(9k) + 4k]
= 36 km [1A] = 16k : 22k
(b) Speed of car B = 8 : 11
42 − 6 ` The answer is C.
= [1M]
90
60 4. 2t = 5u
= 24 km/h [1A] `t:u=5:2
(c) Suppose that two cars meet at the time t minutes v = 4t
after 9:30 in the morning. `t:v=1:4
36 + 24 t = 42 – 24 t [1M] t : u =5 : 2 =5 : 2
60 60 t : v=1 : 4=5 : 20
2 2
36 + t = 42 – t ` t : u : v= 5 : 2 : 20
5 5
t = 7.5 Let t = 5k, where k is a non-zero constant. Then u = 2k
` Two cars meet at 9:37:30 in the morning. and v = 20k.
The claim is agreed. [1A] (3v – 2t) : (5u + t) = [3(20k) – 2(5k)] : [5(2k) + 5k]
= 50k : 15k
Instant Practice 4 = 10 : 3
1. a = 2b ` The answer is A.
`a:b=2:1
a : b =2 : 1 =6 : 3 5. 4x = z
b : c= 3 : 4= 3 : 4 `x:z=1:4
`a : b : c= 6 : 3 : 4 3y = 5z
Let a = 6k, where k is a non-zero constant. Then b = 3k `y:z=5:3
and c = 4k. x : z=1 : 4=3 : 12
(a + b) : (b + c) = (6k + 3k) : (3k + 4k) y : z= 5 : 3= 20 : 12
= 9k : 7k `x : y : z= 3 : 20 : 12
=9:7
` The answer is C.

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 9 Rate and Ratio

Let x = 3k, where k is a non-zero constant. Then y = 20k Instant Practice 6

and z = 12k. 1 2
1. = and 2 = 3
(3y – z) : (z – 2x) = [3(20k) – 12k] : [12k – 2(3k)] a 3b 3b 2c
3b
= 48k : 6k `a= and 4c = 9b
2
=8:1 3 9
a = b and c = b
` The answer is B. 2 4
Since a, b and c are positive numbers,
a  b, c  b and c  a
Instant Practice 5
` ac
b
1. (a + b) : (a – b) = 5 : 1
The answer is C.
a + b = 5(a – b)
a + b = 5a – 5b
2 3
4a = 6b 2. = and 3 = 4
a 2b 2b 3c
`a:b=6:4=3:2 ` 4b = 3a and 8b = 9c
The answer is D. 4 8
a = b and c = b
3 9
Since a, b and c are positive numbers,
2. (4d + 2e) : (2d – e) = 4 : 1
a  b and c  b
4d + 2e = 4(2d – e)
`cba
4d + 2e = 8d – 4e
The answer is D.
6e = 4d
`d:e=6:4=3:2
2 4 4 6
3. = and =
The answer is D. 3p 5q 5q 7r
` 12p = 10q and 30q = 28r
3. (2m + n) : (m + 3n) = 2 : 3 5 15
p = q and r = q
3(2m + n) = 2(m + 3n) 6 14
Since p, q and r are positive numbers,
6m + 3n = 2m + 6n
p  q and r  q
4m = 3n
`pqr
`m:n=3:4
The answer is A.
The answer is B.

3 2 2 1
(p + 4q) : (2p + q) = 2 : 1 4. = and =
4. p 3q 3q 5r
p + 4q = 2(2p + q) ` 2p = 9q and 3q = 10r
p + 4q = 4p + 2q 9 3
p = q and r = q
3p = 2q 2 10
Since p, q and r are positive numbers,
`p:q=2:3
p  q and r  q
The answer is A.
`rqp
The answer is D.
5. (r + s) : (2r + 3s) = 3 : 8
8(r + s) = 3(2r + 3s) 5 1 1 3
5. = and =
8r + 8s = 6r + 9s x 3y 3y 5z
2r = s ` x = 15y and 5z = 9y
` (2r + s) : (4r + s) = (2r + 2r) : (4r + 2r) 9
x = 15y and z = y
5
= 4r : 6r
Since x, y and z are positive numbers.
=2:3
x  y, z  y and x  z
The answer is B.
` zx
y
The answer is B.

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HKDSE Exam Series — Master 1, Tackle 5 Hot Question Types for Mathematics (Compulsory Part) Solution Guide

Instant Practice 7 5. Let 1 : n be the scale of the map.

1. Scale of the map = 1 cm : 500 cm 1 : n2 = 3 cm2 : (2.7 × 105) m2
= 1 cm : 5 m 1 : n2 = 3 cm2 : (2.7 × 105)(1002) cm2
` Area on the map : Actual area 3n2 = 2.7 × 109
= 1 cm2 : (5 m)2 ` n = 9 × 10
2 8

= 1 cm2 : 25 m2 n = 3 × 104
Actual area of the park = 30 000
= 4(25) ` Scale of the map = 1 : 30 000
= 100 m2 The answer is A.
The answer is B.
Instant Practice 8
2. Scale of the map = 1 cm : 20 000 cm 1. Let $y/kg be the cost of tea Y.
= 1 cm : 200 m 5(63) + 7y
= 84
` Area on the map : Actual area 5+7
315 + 7y = 1008
= 1 cm2 : (200 m)2
y = 99
= 1 cm2 : 40 000 m2
` The cost of tea Y is $99/kg.
Actual area of the forest
The answer is B.
= 5(40 000)
= 200 000 m2
2. Let $y/kg be the cost of rice B.
= 2 × 105 m2
7(51) + 2y
The answer is B. = 49
7+2
357 + 2y = 441
3. Scale of the map = 1 cm : 50 000 cm y = 42
= 1 cm : 500 m ` The cost of rice B is $42/kg.
` Area on the map : Actual area The answer is A.
= 1 cm2 : (500 m)2
= 1 cm2 : 250 000 m2 3. The cost of the mixed flour
Actual area of the zoo 3(42) + 9(30)
=
= 6(250 000) 3+9
= $33/kg
= 1 500 000 m2
The answer is A.
= 1.5 × 106 m2
The answer is C.
10(64) + 92y
4. = 74.5
10 + y
4. Let 1 : n be the scale of the map. 640 + 92y = 745 + 74.5y
1 : n2 = 6 cm2 : 24 m2 y=6
1 : n2 = 6 cm2 : 24(1002) cm2 The answer is B.
6n2 = 240 000
n2 = 40 000 50a + 64b
5. = 56
n = 200 a+b
50a + 64b = 56a + 56b
` Scale of the map = 1 : 200
6a = 8b
The answer is A.
a:b=8:6=4:3
The answer is D.

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