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Home | Education Today | Ratio And Proportion Problems Made Easy

Ratio and proportion problems made easy

This article is in continuation to the last article focusing on the ratio and proportion topic. Here are some practice questions along with solutions on the ratio and proportion topic that will help you in your preparation for the State government recruitment jobs. 1. If (3a-5b)/(3a 5b) = (3c-5d)/(3c 5d), then which one of the […]

By Telangana Today
Published Date - 7 August 2022, 11:30 PM
Ratio and proportion problems made easy
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This article is in continuation to the last article focusing on the ratio and proportion topic. Here are some practice questions along with solutions on the ratio and proportion topic that will help you in your preparation for the State government recruitment jobs.

1. If (3a-5b)/(3a 5b) = (3c-5d)/(3c 5d), then which one of the following is true?


a) a/b = c/d b) a/b = d/c c) a/d = b/c d) a/d = c/b

Ans: a

Solution: (3a-5b)/(3a 5b) = (3c-5d)/(3c 5d)
Applying componendo and dividendo,
(3a – 5b 3a 5b)/ (3a – 5b – 3a -5b)
= (3c – 5d 3c 5d)/ (3c – 5d – 3c -5d)
= 6a/-10b = 6c/-10d
a/b = c/d

2. If (x^2 y^2)/(x^2 – y^2) = 17/8, then find the value of (x^3 y^3)/ (x^3 – y^3)

a) 76/49 b) 49/76 c) 289/64 d) 4913/512

Ans: a

Solution: (x^2 y^2)/(x^2 – y^2) = 17/8
Applying componendo and dividendo,
{(x^2 y^2) (x^2 – y^2)} / {(x^2 y^2) – (x^2 – y^2)} = 17 8 / 17 – 8
2x^2 / 2y^2 = 25/9
x^2 / y^2 = 25/9
x/y = 5/3
(x^3 y^3)/ (x^3 – y^3) = 125 27 / 125 – 27
= 152/98
= 76/49

3. A hostel has 500 students and their monthly consumption of cereals is 6000 kg. If 300 students leave the hostel, find how much cereals will be required per month for the students?

a) 2400 kg b) 2500 kg c) 2600 kg d) 2700 kg

Ans: a

Solution: Remaining number of students = 500 – 300 = 200
500 students –>; 6000 kg
200 students –>; ?
500/200 = 6000/?
? = 6000 × 200 / 500
= 2400 kg

4. 6 is the mean proportion between two numbers x and y and 48 is the third proportional of x and y. The values of x and y are

a) x = 3, y = 12 b) x = 4, y = 12 c) x = 10, y = 4 d) x = 5, y = 10

Ans: a

Solution:
case (i) : x/6 = 6/y
xy = 36
case (ii) : x/y = y/48
y² = 48x
(36/x²) = 48x
x³ = 27 –>; x = 3
3 × y = 36 –>; y = 12
Therefore, x = 3, y = 12

5. If xy = 27, then which of the following is correct?

a) x : 9 = 3 : y b) 9 : x = 3 : y c) x : 3 = y : 9 d) None of these

Ans: a

Solution: xy = 27
xy = 3 × 9
x/9 = 3/y
x : 9 = 3 : y

6. The price of branded PC (personal computer) and assembled PC in the ratio of 64: 27. If from now on, every year price of branded PC goes on decreasing by 10 percent and price of assembled PC goes on increasing by 20 percent, after how many years the price of both will be equal?

a) 1 year b) 2 years 3) 3 years 4) 3 1/2 years

Ans: c

Solution: Let the price of branded PC be = 64x and
the price of assembled PC be = 27x
Let after n years the prices of both will become equal
(90/100)^n × 64x = (1200/100)^n × 27x
64/27 = (4/3)^n
(4/3)^3 = (4/3)^n
n = 3 years

7. When a bus started from the first stop, the number of male passengers to the number of female passengers was 3 : 1. At the first stop, 16 passengers got down and 6 more female passengers got in. The ratio of the male to female passengers now become 2 : 1. What was the total number of passengers in the bus when it started from the first stop?

a) 48 b) 54 c) 64 d) 72

Ans: c

Solution: Let, initially, the number of males = 3x and
The number of females = x
At the first stop, suppose m males and f females left the bus
m f = 16 ……….. (i)
3x – m/(x – f) 6 = 2/1
3x – m = 2x – 2f 12
x = m – 2f 12
x = 16 – f – 3f 12 ( from (i))
x = 28 – 3f
f = 4, x = 16
Therefore, the total number of passengers in the beginning
4x = 64.
For no other value of f any of the other alternatives holds good.

To be continued…

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