## 7 Nov 2016

### IBPS RRB/Clerk 2016 - Practice Quantitative Aptitude Questions

IBPS RRB/Clerk 2016 - Practice Quantitative Aptitude Questions (Ages) Set-33:
Dear Readers, Important Practice Aptitude Questions for IBPS Clerk and Upcoming Exams was given here with Solutions. Aspirants those who are preparing for the examination can use this.

1). Sum of present ages of A and B is 60 years. 5 years hence, their ages will be in the ratio 3:4. Find A’s present age.
a)    25 years
b)    30 years
c)    35 years
d)    40 years
e)    None of these

2). The sum of ages of Sunil and his father is 46 years. 3 years back, father’s age was 4 times sunil’s age. find present age of Sunil.
a)    7 years
b)    8 years
c)    11 years
d)    12 years
e)    None of these

3). Sum of present age of A, B and C is 72 years. If 4 years ago, their ages were in the ratio 1 : 2 : 3, find A’s present age.
a)    7 years
b)    10 years
c)    12 years
d)    14 years
e)    None of these

4). Present ages of A and B are 50 years and 18 years respectively. In how many years will A be twice as old as B?
a)    14 years
b)    15 years
c)    16 years
d)    18 years
e)    None of these

5). The sum of ages of Mohan and his father is 35 years. When Mohan’s age will be equal to present age of his father, then sum of their ages will be 85 years. Find present age of the father.
a)    20 years
b)    25 years
c)    30 years
d)    32 years
e)    None of these

6). The sum of present ages of a father and his son is 36 years. When the son reaches father’s present age, the sum of their ages will be 80 years. What is the present age of the son?
a)    4 years
b)    7 years
c)    9 years
d)    12 years
e)    None of these

7). The ages of A and B are in the ratio 8 : 5. If the sum of their ages is 39 years, what will be the ratio of their ages after 9 years?
a)    3 : 2
b)    8 : 7
c)    10: 7
d)    11 : 8
e)    None of these

8). A is 2 years older than B who is twice as old as C. If sum of ages of A, B and C is 37 years, find the age of A.
a)    7 years
b)    9 years
c)    14 years
d)    16 years
e)    None of these

9). At the time of marriage, a man was 6 years older to his wife. 12 years after their marriage, his age is 6/5 times the age of his wife. What was wife’s age at the time of marriage?
a)    18 years
b)    24 years
c)    30 years
d)    36 years
e)    None of these

10). A and B are 3 years and 2 years old respectively. Their father is 40 years. After how many years, father’s age would be twice of combined age of A and B?
a)    5 years
b)    10 years
c)    15 years
d)    20 years
e)    None of these

1).a)   2).c)  3).d)  4).a)  5).c)  6).b)  7).d)  8).d)  9).a)  10).b)

Solution:
1). Sum of their present ages = 60 years
sum of their ages, 5 years hence = 60 + (5 × 2) = 70 years.
after 5 years, ratio of ages of A and B will be 3 : 4.
A’s age after 5 years = (3/7) × 70 = 30 years
A’s present age = 30 – 5 = 25 years

2). Sum of their ages, 3 years ago = 46 – (2  × 3) = 40 years
sunil’s age, 3 years ago = 1/5 × 40 = 8 years
sunil’s present age = 8 + 3 = 11 years

3). Sum of present ages of A, B and C is 72 years.
sum of their ages (4 years ago) = 72 – (3  × 4) = 60 years
4 years ago, ratio of ages of A, B and C was 1 :  2 :3.
A’s age, 4 years ago = (1/6) ×60 = 10 years
A’s present age = 10 + 4 = 14 years

4). Let the required time be x years.
then (50+x) = 2 (18 + x) = 50 + x = 36 + 2x à x = 14 years.
Alternative method : Double of B’s age = 2  × 18 = 36 years
difference between A’s present age and double of B’s age = 50 – 36= 14 years.
A will be twice of B’s age after 14 years.

5). Sum of present ages = 35 years
sum of their ages will be 85 years after (85-35) / 2 = 25 years from now.
the father is 25 years older than his son.
Present age of father = (35 + 25)/2 = 30 years.

6). Difference in two sums = 80- 36 = 44
Age of both are increased by 44/2 = 22 years
Difference in ages of father and son = 22 years
sum of present ages of father and son = 36 years
Son’s present age = 1/2  × (36 - 22) = 7 years

7). Sum of ratios = 8 +5 = 13.
Sum of ages = 39 years
13 ratio = 39 years à 1 ratio = 3 years
9 years = 3 ratio
new ratio = (8+3) : (5+3) = 11 :  8

8). Let the age of C be x years. Then age of B = 2x and age of A = 2x + 2
sum of ages of A, B and C = (2x + 2) + (2x) + x = 37
(5x + 2) = 37 à x = 7
Age of A = 2x + 2 = 16 years

9). ratio of their ages, 12 years hence = 6 : 5 à Difference in ages = 1 ratio = 6 years
wife’s age (12 years hence) = 5 ratio = 5 × 6 years = 30 years
wife’s age at the time of marriage = 30 – 12 = 18 years

10). Father’s age = 40 years. Children’s combined age = 3 + 2= 5 years
Double of their ages = 2 × 5 = 10 years
Difference between father’s age and combined age = 40 – 10 = 30 years
required time = 1/ (2 + 1) × 30 = 10 years