# Quantitative Aptitude: Problems on Ages Set 2

Problems on Ages for SBI PO/IBPS PO, NIACL Assisatant, NICL AO and other upcoming exams.

1. Four times the difference in ages of C and A is one more than the age if B. Percentage of A’s age to C’s age is 75%. If ratio of B’s age 5 years hence to C’s age 1 year ago is 4 : 3. Find the average of ages A and C.
A) 20
B) 19
C) 12
D) 14
E) 8
Option D
Solution:

4 (C-A) = B + 1
A/C * 100 = 75
(B+5)/(C-1) = 4/3
Solve
A = 12, C = 16
2. 10 years ago daughter’s age was two-fifth of her mother’s age that time. While 10 years hence her age will be three-fifth of her mother’s age then. Find the difference in the ages of the two.
A) 24
B) 19
C) 26
D) 38
E) 16
Option A
Solution:

(x-10) = 2/5 (y-10)
(x+10) = 3/5 (y+10)
Solve, x = 26 and y = 50
3. B is as more younger than C as he is elder than A. Ratio of ages of A to C is 5 : 9. If B’s age after 10 years will be 24, find the average of all of their present ages.
A) 15
B) 16
C) 14
D) 22
E) 19
Option C
Solution:

B’s present age = 24-10 = 14
So C’ age = 14+x
And A’ age = 14-x
(14+x)/(14-x) = 9/5
Solve, x = 4
So average age = (10+14+18)/3 = 14 years
4. Kaira is 4 years younger to his brother. Her father was 30 years old when her sister was born while her mother was was 30 years old when she was born. If her sister was 4 years old when their brother was born, find the age of her father when her mother was born.
A) 11
B) 12
C) 4
D) 10
E) 8
Option E
Solution:

When Kaira was born:
Mother was 30.
She is 4 years younger to her brother, so brother was 4 years old.
Sister was 4 years old when brother was born, so sister is 4 years elder to brother, so sister was 8 years old.
Father was 30 when sister was born, so father is 30 years elder to sister, so father was 30+8 = 38 years old.
Now when Kaira was born, mother was 30 and father was 38
So difference in their ages is 8 years. So when mother was born, father was 8.
5. 6 years ago, three times the age of B was 2 more than the age if A that time. 6 years hence, twice age of B will be equal to A’s age that time. Find the total of their ages.
A) 48
B) 66
C) 56
D) 65
E) 60
Option B
Solution:

3 * (B-6) = 2 + (A-6)
2 * (B+6) = A + 6
Solve, A = 46, B = 20
6. If 6 years are subtracted from the present age of Babita and the remainder is divided by 18, then the present age of her granddaughter Geeta is obtained. If Geeta is 2 years younger to Sita whose age is 5 years, then what is Babita’s present age?
A) 77
B) 65
C) 84
D) 43
E) 79
Option A
Solution:

Geeta’s age = (5-2) = 3 years
Let age of Babita = x years
So (x-6)/18 = 3
Solve, x = 60
7. A’s age is twice C’ age. Ratio of age of B 2 years hence to age of C 2 years ago is 5 : 2. C is 14 years younger than D. Difference in ages of D and A is 4 years. Find the average of their ages.
A) 36
B) 25
C) 27
D) 13
E) 18
Option E
Solution:

A = 2C
(B+2)/(C-2) = 5/2
C = D – 14
D – A = 4
Solve, A = 20, B = 18, C = 10, D = 24
8. When the couple was married the average of their ages was 25 years. When their first child was born, the average age of family became 18 years. When their second child was born, the average age of the family became 15 years. Find the average age of the couple now.
A) 31
B) 27
C) 28
D) 29
E) 30
Option D
Solution:

Sum of ages of couple = 25*2 = 50
When 1st child born, total age of 3 = 18*3 = 54 years
At this time the child’s age was 0, so age of father and mother would have increased by same. So increased by 2 years each. So 50 +2 + 2 = 54
Now when 2nd child born, total age of 4 = 15*4 = 60
So this time second child’s age = 0 and age of father, mother and first child would have increased by same. So increased by 2 each such that 54 + 2+2+2 = 60
So now this time (after 4 years from age 50), total age of couple is 50+4+4 = 58
So average = 29 years
9. Ratio of age of A to B is 3 : 2 and that of A to C is 1 : 2. Difference in ages of B and C is 24 years. Find the average of their present ages.
A) 24
B) 22
C) 14
D) 26
E) 31
Option B
Solution:

B/A = 2/3 and A/C = 1/2
So B : A : C = 2*1 : 3*1 : 3*2 = 2 : 3 : 6
So 6x – 2x = 24, 4x = 24, x = 6
So total of their present ages = (2+3+6)*6. So average = (2+3+6)*6 / 3 = 22 years
10. Ratio of ages of A 5 years hence to B’s age 3 years ago is 5 : 3. Also ratio of ages of A 4 years ago to B’s age 2 years hence is 4 : 5. Find the age of the elder.
A) 15
B) 18
C) 21
D) 20
E) 24
Option D
Solution:

(A+5)/(B-3) = 5/3
(A-4)/(B+2) = 4/5
Solve A = 20, B = 18

## 12 Thoughts to “Quantitative Aptitude: Problems on Ages Set 2”

1. Deepak sahu

Thank you mam

2. Deepak sahu

Ap apne studendts ka bahut khyaal rakhe h uske liye dhanyawaad
Apka matterial acha rahta h

3. Yogi ji

Option se ho sakte ye ques

4. Nishi

@suraj sir plz check ques 3

1. Shubhra

Ratio is C to A

1. Nishi

ok mam

2. jazzzz

in ques 3 value of X is coming in -ve

5. Deepak sahu

mam/sir first question ki equetion nahi solve kar pa raha hu plz ap detail me bata dijiye

1. Shubhra

4 (C-A) = B + 1
4C – 4A = B + 1 ……….(1)
A/C = 3/4
4A = 3C …………….(2)
(B+5)/(C-1) = 4/3
3B + 15 = 4C – 4
So 3B – 4C = -19 ………….(3)

Now hume B ki value nhi chaiye, so (1) me se B ki value (3) me pot karo
Ab 2 equations bach jaengi – this new obtained and (2)
ye dono A and C me hai
solve now

6. Ati

So average age = (10+14+18)/3 = 12 years
it will be 14 yrs not 12 yrs correct it in 3rd ques

1. Shubhra

Yes 14

1. Ati

mam 6th ques ka ans 60 hi aa rha … u have given a option