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Practice Questions - Ratio Proportion

Q11) An alloy contains zinc and copper in the ratio 7:4. If the alloy contains 14 Kg copper, then the quantity of zinc in the alloy is
1) 24.5 Kg
2) 8 Kg
3) 28 Kg
4) 17.5 Kg

Solution:
Zinc = 7k, copper = 4k
4k = 14; k = 3.5
Quantity of Zinc in the alloy is 7k = 7(3.5) = 24.5 Kg
Hence, option 1.

Q12) If 3A = 4B = 5C, then A:B:C is equal to
1) 3:4:5
2) 9:8:7
3) 20:15:12
4) 1/3:1/4:1/5

Solution:
A = 20, B = 15, C = 12
Hence, option 3.

Q13) If , then the value of a:b is
1) 2:3
2) 3:1
3) 3:2
4) 1:3

Solution:

On solving, a = 3b
Hence, option 2.

Q14) A bag contains 1 rupee, 50 paise and 25 paise coins in the ratio 3:4:5. Total value of these coins is Rs 24. Find the number of 50 paise coins.
1) 12
2) 20
3) 16
4) 8

Solution:
Let 3k be 1 rupee coins, 4k be 50 paise coins, 5k be 25 paise coins.
A.T.Q. 3k(1) + 4k(0.5) + 5k(0.25) = 24
On solving, k = 4
Number of 50 paise coins = 4k = 16
Hence, option 3.

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Q15) If P:Q = 3:5 and Q:R = 7:8 then 5P:3Q:7R is equal to
1) 6:15:25
2) 1:1:6
3) 4:7:8
4) 3:3:8

Solution:
P/Q = 3/5 and Q/R = 7/8
P/Q = 21/35 and Q/R = 35/40
P = 21, Q = 35, R = 40
Hence, option 4.

Q16) When 50% of a number is added to a second number, the second number increases to its four-third. What is the ratio between the first number and the second number?
1) 3:2
2) 3:4
3) 2:3
4) Data Inadequate

Solution:
first number = f, second number = s
s + f/2 = 4s/3
f/2 = s/3
f/s = 2/3
Hence, option 3.

Q17) Divide 1162 into 3 parts such that 4 times the first is equal to 5 times the second and 7 times the third. Find the smallest part.
1) 280
2) 200
3) 420
4) 490

Solution:
A + B + C = 1162
4A = 5B = 7C = 4×5×7×k
A = 35k, B = 28k, C = 20k
35k + 28k + 20k = 1162
k = 14
Smallest part = 20k = 20×14 = 280
Hence, option 1.

Q18) Two candles of the same height are lighted at the same time. The first is consumed in 7 hours and the second in 6 hours. Assuming that each candle burns at a constant rate, in how many hours after being lighted, the ratio between the first and second candles becomes 3:1.
1) 5 hours 36 minutes
2) 5 hours 30 minutes
3) 5 hours
4) 6 hours

Solution:
Height = lcm(6,7) = 42 units.
As first candle is consumed in 7 hours so 6 units are consumed in an hour.
As second candle is consumed in 6 hours so 7 units are consumed in an hour.
't' hours after being lighted, the ratio between the first and second candles becomes 3:1 so,
(42 - 6t)/(42 - 7t) = 3/1
t = 28/5 hours = 5 hours 36 minutes
Hence, option 1.

Q19) From each of the two given numbers, half the smaller number is subtracted. After such subtraction, the larger number is 4 times as large as the smaller number. What is the ratio of the numbers?
1) 5:2
2) 1:4
3) 4:1
4) 4:5

Solution:
Let the smaller number be x and the larger number be y.
(y - x/2) = 4(x - x/2)
(y - x/2) = 4x/2
y = 5x/2
y:x = 5:2
Hence, option 1.
Q20) A sum of Rs 300 is divided among P, Q and R in such a way that Q gets Rs 30 more then P and R gets Rs 60 more than Q. Then, ratio of their shares is
1) 2:3:5
2) 3:2:5
3) 2:5:3
4) 5:3:2

Solution:
Share of P = x
Share of Q = x + 30
Share of R = (x + 30) + 60 = x + 90
Sum of money with P, Q, R = 300
x + x + 30 + x + 90 = 300
3x + 120 = 300
x = 60
Required ratio = 60:(60 + 30):(60 + 90) = 60:90:150 = 2:3:5
Hence, option 3.

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