1. A person’s present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the
age of his mother. How old is the mother at present ?
A) 40 years
B) 41 yrs
C) 42 yrs
D) 43 yrs
Ans. A
Let the mother’s present age be x years.
Then, the person’s present age = 2 x years.
5
2 x + 8 = 1 (x + 8)
5 2
2(2x + 40) = 5(x + 8)
x = 40
2.The age of father 10 years ago was thrice the age of his son. Ten years hence, father’s age will be twice
that of his son. The ratio of their present ages is :
A) 7 : 3
B) 13 : 4
C) 9 : 2
D) 5 : 2
Ans. A
Let the ages of father and son 10 years ago be 3x and x years respectively.
Then, (3x + 10) + 10 = 2[(x + 10) + 10] 3x + 20 = 2x + 40
x = 20.
Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3
3. Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of
Ronit’s age. After further 8 years, how many times would he be of Ronit’s age ?
A) 2 times
B) 3 times
C) 4 times
D) 5 times
Ans. A
Let Ronit’s present age be x years. Then, father’s present age =(x + 3x) years = 4x years.
(4x + 8) = 5 (x + 8)
2
8x + 16 = 5x + 40
3x = 24
x = 8.
Hence, required ratio = (4x + 16) = 48 = 2.
(x + 16) 24
4. A person’s present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. How old is the mother at present
A) 38
B) 40
C) 42
D) 44
Ans. B
Let the mother’s present age be x years.
Then, the person’s present age = (2/5 x) years.
=> (2/5 x + 8/2) = 1 (x + 8)
=> 2(2x + 40) = 5(x + 8)
=> x = 40
5. The sum of the present ages of a father and his son is 60 years. Six years ago, father’s age was five times the age of the son. After 6 years, son’s age will be
A) 15 years
B) 18 years
C) 20 years
D) 22 years
Ans. C
HINT:
(60 – x) – 6 = 5(x – 6)
