Problems with Solutions: Set 01
Question
No. 01
A
father was 4 times as old as his son 8 years ago. Eight years hence, father
will be twice as old as his son. Find their present ages.
Solution:
Let son's age 8 years ago be x years.
Thus, father's age at that time = 4x years
After 8 years, son's age = (x + 8) + 8 = (x+16) years
After 8 years, father's age = (4x + 8) + 8 = (4x+16) years
So, According to Question, 2(x + 16) = 4x + 16 or x = 8
Therefore, The present age of the son = x + 8 = 16 years
The present age of the
father = 4x + 8 = 32 + 8= 40
years
Question
No. 02
Father
is aged three times more than his son. After 8 years, he would be two and a
half times of his son's age. After further 8 years, how many times would he be
of his son's age?
Solution:
Let Son's present age be ‘x’ years.
Then, father's present age =(x + 3x)
years = 4x years.
Therefore, (4x + 8) = (2/5) (x +
8)
=> 8x + 16 = 5x + 40
=> 3x = 24 => x = 8.
After further 8 years, Son's age will be
(x + 16) = 24 years.
And father's age will be (4x +16) = 48 years.
Hence, the required ratio is (4x
+16)/(x+16) = 48/24 = 2.
Question
No. 03
The
sum of ages of 5 children born at the intervals of 3 years each is 50 years.
What is the age of the youngest child?
Solution:
Let the ages of children be x, (x
+ 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x +
6) + (x + 9) + (x + 12) = 50
=> 5x = 20
=> x = 4.
Age of the youngest child = x = 4
years.
Question
No. 04
‘A’
is twice as old as ‘B’ was two years ago. If the difference in their ages be 2
years, find A's age.
Solution:
Let B's age 2 years ago be x years
Therefore, A's present age = 2x years
Also 2x - (x + 2) = 2 or x=4
So, A's age = 2x = 2 × 4 = 8 years
Question
No. 05
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 38 years now, the son's age
five years back was:
Solution:
Let the son's present age be x years.
Then, (38 - x) = x
=> 2x = 38
=> x = 19.
Son's age 5 years back (19 - 5) = 14
years.
Question
No. 06
A
is two years older than B who is twice as old as C. If the total of the ages of
A, B and C be 27, the how old is B?
Solution:
Let C's age be x years.
Then, B's age = 2x years. A's age
= (2x + 2) years.
Therefore, (2x + 2) + 2x + x
= 27
=> 5x = 25
=> x = 5.
Hence, B's age = 2x = 10 years.
Question
No. 07
Present
ages of ‘A’ and ‘B’ are in the ratio of 5 : 4 respectively. Three years hence,
the ratio of their ages will become 11 : 9 respectively. What is B's present
age in years?
Solution:
Let the present ages of ‘A’ and ‘B’ be 5x
years and 4x years respectively.
Then, (5x + 3)/ (4x + 3) = 11/9
=> 9(5x + 3) = 11(4x +
3)
=> 45x + 27 = 44x + 33
=> 45x - 44x = 33 - 27
=> x = 6.
B's present age = 4x = 24 years.
Question
No. 08
The
age of a father 10 years ago was thrice the age of his son. Ten years hence,
the father's age will be twice that of his son. What is the ratio of their
present ages?
Solution:
Let the present ages of father and son be
‘x’ and ‘y’ years respectively.
Then (x - 10) = 3 (y - 10) or 3y - x = 20 ------ (1)
And (x+10) = 2 (y + 10) or x -
2y = 10 ------- (2)
Adding eq. (1) + (2) => y
= 30
Substituting the value of y = 30 in
equation (1) we get x = 70
Ratio of their ages = 70: 30 or 7:3
Question
No. 09
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 his son is:
Solution:
Let the son's present age be x years.
Then, man's present age = (x + 24)
years.
Therefore, (x + 24) + 2 = 2(x +
2)
=> x + 26 = 2x + 4
=> x = 22.
Hence, the present age of his son is 22
years.
Question
No. 10
Six
years ago, the ratio of the ages of ‘A’ and ‘B’ was 6 : 5. Four years hence,
the ratio of their ages will be 11 : 10. What is B's age at present?
Solution:
Let the ages of ‘A’ and ‘B’ 6 years ago
be 6x and 5x years respectively.
Then, [(6x + 6) + 4[/ [(5x +
6) + 4] =11/10
=> 10(6x + 10) = 11(5x +
10)
=> 5x = 10
=> x = 2.
Therefore, B's present age = (5x +
6) = 16 years.
Problems on Ages: Next Tests: