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24.
(b)
2a = b
a/b = 1/2
a : b = 1 : 2 ...... [1]

3b = 2c
b/c = 2/3
b : c = 2 : 3 ...... [2]

From [1] and [2] :
a : b : c = 1 : 2 : 3

(c)
2a = 5b
a/b = 5/2
a : b = 5 : 2
a : b = 10 : 4 ...... [1]

5b = 4c
b/c = 4/5
b : c = 4 : 5 ...... [2]

From [1] and [2] :
a : b : c = 10 : 4 : 5

(d)
1/a : 1/b : 1/c = 2 : 3 : 4
a : b : c = 1/2 : 1/3 : 1/4
a : b : c = (1/2)×12 : (1/3)×12 : (1/4)×12
a : b : c = 6 : 4 : 3

Alternative method :
1/a : 1/b = 2 : 3
2/b = 3/a
a/b = 3/2
a : b = 3 : 2
a : b = 6 : 4 ...... [1]

1/b : 1/c = 3 : 4
3/c = 4/b
b/c = 4/3
b : c = 4 : 3 ...... [2]

From [1] and [2] :
a : b : c = 6 : 4 : 3

====
32.
Some words have been missed in the question :
"...... If B obtains $YYY than A ......"

====
34.
(a)
Let t kg be the weight of the tortoise, c kg be that of the cat, and d kg bethat of the dog.

t = (1/2)c
t/c = 1 : 2 ...... [1]

c = (1/4)d
c/d = 1/4
c : d = 1 : 4
c : d = 2 : 8 ...... [2]

From [1] and [2] :
(Weight of the tortoise) : (Weight of the cat) : (Weight of the dog)
= t : c : d
= 1 : 2 : 8

(b)(i)
Let k kg, 2k kg and 8k kg be the weights of the tortoise, cat and dogrespectively.

8k = 20
k = 2.5

The weight of the tortoise = k kg = 2.5kg

(b)(ii)
The total weight of the three pets
= (k + 2k + 8k) kg
= 11k kg
= 11 × 2.5 kg
= 27.5 kg

====
35.
Let n be the number of books that Angie borrowed from Leslie.

(48 + n) : (72 - n) = 13 : 17
17 × (48 + n) = 13 × (72 - n)
816 + 17n = 936 - 13n
30n = 120
n = 4

The number of books that Angie borrowed from Leslie = 4

====
37.
(a)
Let a, b and c be the number of $5-coins, $2-coins and 50¢-coins respectively.

5a : 2b : 0.5c = 35 : 24 : 11
5a/5 : 2b/2 : 0.5c/0.5 = 35/5 : 24/2 : 11/0.5
a : b : c = 7 : 12 : 22

Let 7k, 12k and 22k be the number of $5-coins, $2-coins and 50¢-coinsrespectively.

7k = 28
k = 4

The number of $2-coins
= 12 × 4
= 48

(b)
From (a), the numbers of $5-coins to $2-coins to 50￠-coins
= 7 : 12 : 22

參考資料 土扁

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