# Quantitative Aptitude: Problems on Ages Set 9

1. The ratio of ages of A and B is 3:8. After 10 years the ratio of B and C’s age is 13:10. If C’s eighth birthday was celebrated two years ago, then the present age of A must be
10yrs
8yrs
6yrs
5yrs
None
Option C
Solution:
C’s present age =8+2=10
After ten years , B:C=13:10
C after 10 years=20
10……… 20
13…. ? ==>26yrs(B’s age after 10 years)
Present A:B=3:8
B’s age before 10years=16yrs.
8……. 16
3… ? ==>6yrs (A’s age now).
2. 16 years ago, my grandfather was 8 times older than me. After 8 years from today, my grandfather will be thrice as old as I will be at that time. Eight years ago, what was the ratio of my age and my grandfather’s age?
53:11
11:53
15:22
22:15
None
Option B
Solution:
Let 16 years ago my age was x.
8 years of now, the ratio be
(x+24)/(8x = 24) = 1/3
x = 48/5
16 years ago the ages be 48/5, 8*48/5
After 8 years,the ratio of ages be
(48/5+8) / (8*48/5+8)=88/424=11/53.
3. A father said to his son, “I was as old as you are at the present at the time of your birth”. If the father’s age is 42 years now, the son’s age six years back was:
17yrs
15yrs
13yrs
20yrs
None
Option B
Solution:
Let the son’s present age be x years. Then, (42 – x) = x
2x = 42.
x = 21.
Son’s age 6 years back (21 – 6) = 15 years.
4. If twice the son’s age in years be added to the father’s age, the sum is 90 and if twice the father’s age is added to the son’s age, the sum is 120. Their age’s are
40yrs, 15yrs
45yrs, 20yrs
50yrs, 20yrs
50yrs, 15yrs
None
Option C
Solution:
2S+F=90
S+2F=120
Solving the above equation we get
F=50yrs.
Then S=120-100=20yrs
5. The average age of 11 students and their teacher is 20 yrs. The average age of the first six students is 22 yrs and that of the last five is 16 yrs. The teacher’s age is ?
28yrs
32yrs
30yrs
35yrs
None
Option A
Solution:
Total age of 11 students and a teacher = 12 x 20 = 240 yrs
Total age of first 6 students = 6 x 22 = 132 yrs
and total age of last 4 students = 5 x 16= 80 yrs
Age of teacher = 240 – 132 – 80 = 28 yrs.
6. The sum of ages of a father and son is 40 years. Four years ago, the product of their ages was four times the father’s age at that time. The present age of the father is :
32yrs
35yrs
28yrs
40yrs
None
Option A
Solution:
Let the fathers age be x, then sons age is 40-x
4 years ago,
(x-4)(36-x)=4(x-4)
X=32years.
7. Prem’s present age is four times his son’s and two times his mother’s present age. The sum of the present ages of all three of them is 78 yr. What is the difference between the present ages of Prem’s son and Prem’s mother?
38yrs
40yrs
36yrs
42yrs
None
Option D
Solution:
S:P=1:4, P:M=1:2
S:P:M=1:4:8
13 ……… 78
7(8-1).. ?==> 42yrs.
8. The ratio of the present ages of a son and his father is 1 : 4 and that of his mother and father is 8 : 7. After 4 years the ratio of the age of the son to that of his mother becomes 3 : 8. What is the present age of the father ?
35yrs
33yrs
32yrs
39yrs
None
Option C
Solution:
S:F=1:4, M:F=8:7
S:F:M=2:8:7
After 4 years S:M=3:8
(2x+4)/(7x+4)=3/8
X=4.
Mother’s age is 28
And father’s age is 32yrs
9. If the two digits of the age of Mr. X are reversed then the new age so obtained is the age of his wife. 1/11 of the sum of their ages is equal to the difference between their ages. If Mr. X is elder than his wife then find the difference their ages.
10yrs
13yrs
12yrs
9yrs
None
Option D
Solution:
Let the two digit no be 10x+y,and reverse is 10y+x
(10x + y + 10y + x) / 11 = 10x + y -10y-x
x + y = 9x – 9y
x/y = 5/4.
Then Diff of two number is (10x + y) – (10y – x)
= 9x – 9y
Substitute x and y here.
= 9(5-4)
= 9 yrs.
10. One year ago, Mosses was five times as old as his son kishore. Six years hence, Mosses’s age will exceed his son’s age by 20 years. The ratio of the present ages of Mosses and his son is
3:13
5:14
7:12
5:21
None
Option A
Solution:
One year ago, M:K=4:1
Then (5x+7)-(x+7)=20
X=5.
Ratio=6:26=3:13.

## 3 Thoughts to “Quantitative Aptitude: Problems on Ages Set 9”

1. jaga

thank u mam
q6..
(x-4)(35-x)=4(x-4)

1. jubilantjaggu24*7

2. jubilantjaggu24*7