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. | 28 | |
B. | 32 | |
C. | 40 | |
D. | 45 | |
E. | None of these |
 - Let the mother’s present age be X years. Then, the person’s present age = 2X/5 years
2 X + 8 = 1 (X + 8) 2(2X + 40) = 5(X + 8) X = 40 5 2
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More Questions Related to Quantitative Aptitude > Problems On Ages :
Two years ago a man was 6 times as old as his son, After 18 years, he will be twice as old as his son. Their present ages (in years) are?
- 32 years and 7 years
- 45 years and 7 years
- 40 years and 8 years
- 32 years and 8 years
- None of these
 - Let son's age 2 years ago be X
Man's age 2 years ago = 6X
2 (X + 2 + 18) = (6X + 2 + 18)
4X = 20
X = 5
Their present ages are (6X + 2) and (X + 2),
i.e., 32 years and 7 years
A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of the son is:
- 22
- 24
- 25
- 30
- Noe of these
 - Let the son’s present age be X years. Then, man’s present age = (X +24) years
(X + 24) + 2 = 2 (X + 2) X + 26 = 2X + 4
X = 22
Mother's age today is thrice as her daughter's. After 10 years it would be just double. How much old is the daughter today?
- 5
- 8
- 10
- 15
- None of these
 - Let the age of the daughter today be ‘X’ years.
Mother’s age today = 3X years
After 10 years, we have
3X + 10 = 2(X + 10) ===> X = 10 years
If twice the son’s age in years be added to the father’s ages, the sum is 70 and if twice the father’s age is added to the son’s ages, the sum is 95. Father’s age is:
- 15
- 20
- 40
- 45
- None of these
 - Let son's age (in years) = X and father's age (in years) = Y
Given: 2X+Y = 70 and, X+2Y = 95
Solving for Y, we get Y = 40
The average age of class of 20 students is 20 years. If the teacher's age is included the average increases by 1. What is the age of the teacher?
- 27
- 32
- 41
- 48
- None of these
 - Avg x Number = Total
27 nos x 20 years = 400 ….(1)
21 nos x 21 years = 441 ….(2)
Teacher's age = (2) - (1) = 441 - 400 = 41 years.
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:
- 7 : 3
- 3 : 7
- 4 : 7
- 7 : 4
- None of these
 - 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