## Fraction Solved Questions For IBPS, SBI and Other Bank Exams - Page 4

You will find 18 problems in 5 pages..

Fraction Problems Solved Questions (Page 4 of 5)

Question 1

There are two fractions such that fraction X is thrice the fraction Y and their product is 25/12. Which of the following is equal to Y?

a) 5/6 b) 4/5 c) 3/4 d) 7/9

Answer : a) 5/6

Solution :

Let X and Y are given fractions.

Given that, fraction X is thrice the fraction Y;
i.e., X = 3Y ...(1)

And their product = X x Y = 25/12
Sub. eqn (1) in above, (3Y)x(Y) = 25/12
3(Y2) = 25/12
(Y2) = 25/36
Y = 5/6

Hence the required fraction is 5/6.

Question 2

The sum of the numerator and denominator of a fraction is 13. If we add 3 and 9 to numerator and denominator respectively, it becomes 2/3. The fraction is:

a) 5/8 b) 6/7 c) 7/6 d) 8/5

Answer : c) 7/6

Solution :

Let the required fraction be p/q.

Given that, sum of the numerator and denominator of a fraction is 13;
i.e., p + q = 13 ...(1)

Adding 3 to numerator and 9 to denominator, we have,

(P+3)/(q+9) = 2/3
3(P+3) = 2(q+9)
3p+9 = 2q+18
3p-2q = 9 ...(2)

Multiply eqn (1)by 2, 2p + 2q = 26...(3)

Adding (2) and (3),

5p = 35
p = 7.

Putting p value in (1), we get q = 13 - p = 13 - 7 = 6.

Hence the required fraction is p/q = 7/6

Question 3

The denominator of a fraction exceeds its numerator by 7. If the numerator as well as the denominator is increased by 9, the fraction becomes 2/3. Which of the
following is the original fraction?

a) 5/12 b) 4/11 c) 6/13 d) none of these

Answer : a) 5/12

Solution :

Let the required fraction be p/q.

Given that, q exceeds p by 7; i.e., q - 7 = p
q - p = 7 ...(1)

If we increase the numerator as well as the denominator by 9 it becomes (p + 9)/(q + 9)

Given that, (p + 9)/(q + 9) = 2/3
3(p+9) = 2(q+9)
3p + 27 = 2q + 18
2q-3p = 9 ...(2)

Multiplying eqn(1) by 2, we get 2q - 2p = 14......(3)

Now, subtracting eqn(2) from eqn (3), we get p = 14 - 9 = 5
Putting p value in (1), we get q = 12.

Hence, the required fraction is p/q = 5/12

Question 4

The numerator of a fraction is 3 less than its denominator. If we add 10 to the numerator, the fraction is increased by 1 3/7. What was the original fraction?

a) 2/5 b) 11/14 c) 4/7 d) 16/19

Answer : c) 4/7

Solution :

Let the denominator of the required fraction be X.
Given that, the numerator of a fraction is 3 less than its denominator; then its numerator becomes X-3.

Then the required fraction (X-3)/X ...(1)

If we add 10 to numerator(X-3), original fraction (X-3)/X is increased by 1 3/7 (i.e., 10/7).

i.e., (X-3)/X + 10/7 = (X-3+10)/X
(X-3)/X + 10/7 = (X+7)/X
(X+7)/X - (X-3)/X = 10/7
10/X = 10/7
X = 7.

Therefore, X-3 / X = 4/7.
Hence the required fraction is 4/7.

Fraction Problems Solved Questions (Page 4 of 5)

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