# Problems on Ages Questions

Q:

A father said his son , " I was as old as you are at present at the time of your birth. " If the father age is 38 now, the son age 5 years back was :

 A) 14 B) 19 C) 33 D) 38

Explanation:

Let the son's present age be x years .Then, (38 - x) = x => x= 19.

Son's age 5 years back = (19 - 5) = 14 years

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Q:

In 10 years, A will be twice as old as B was 10 years ago. If A is now 9 years older than B, the present age of B is :

 A) 19 B) 29 C) 39 D) 49

Explanation:

Let B's present age = x years. Then, A's present age = (x + 9) years.
(x + 9) + 10 = 2(x - 10)

=> x + 19 = 2x - 20

=> x =39.

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Q:

The total age of A and B is 12 years more than the total age of B and C. C is how many years younger than A ?

 A) 12 B) 13 C) 14 D) 15

Explanation:

(A+B) - (B+C) = 12

$\inline \fn_jvn \Rightarrow$A - C = 12.

$\inline \fn_jvn \Rightarrow$C is younger than A by 12 years.

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Q:

The sum of the present ages of a father and his son is 60 years. five years ago, father's age was four times the age of the son. so now the son's age will be:

 A) 5 B) 10 C) 15 D) 20

Explanation:

Let the present ages of son and father be x and (60 -x) years respectively.

Then, (60 - x) - 5= 4(x - 5)

55 - x = 4x - 20

5x = 75  => x = 15

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Q:

The ratio of the present ages of P and Q is 3 : 4. Five years ago, the ratio of their ages was 5 : 7. Find their present ages.

As the ratio of their present ages is 3 : 4 , let their present ages be 3X and 4X.

So, 5 years ago, as the ratio of their ages was 5 : 7, we can write  (3x-5) : (4x-5) = 5 : 7. Solving, we get X = 10. Hence, their present ages are 3X = 30 and 4X = 40

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Q:

The age of a man is 4 times of his son. Five years ago, the man was nine times old as his son was at that time. The present age of man is?

Let the son's age be x years and the father's age be 4x years

$\inline \Rightarrow (4x-5)=9(x-5)$

5x = 40

x = 8

$\inline&space;\therefore$ present age of the father = $\inline&space;4x$ = $\inline&space;4\times&space;8$ = 32 years

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Q:

The average age of a group of 10 students is 15 years. When 5 more students join the group, the average age increase by 1 year. The average age of the new students is?

Total age of 10 students = 150 years

Total age of 15 students = 240 years

Total age of 5 new students = 240 - 150 = 90 years

$\inline&space;\therefore$ Average age of 5 new students = $\inline \frac{90}{5}$ = 18 years

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Q:

"I am five times as old as you were, when I was as old as you are", said a man to his son. Find out their present ages, if the sum of their ages is 64 years ?

 A) Father = 50; Son =14 B) Father = 40; Son =24 C) Father = 60; Son =4 D) Father = 48; Son =16

Explanation:

Let the present age of the man be 'P' and son be 'Q',
Given, P + Q = 64 or Q = (64 - P)
Now the man says "I am five times as old as you were, when I was as old as you are",
So, P = 5[B - (P - Q)]
We get 6P = 10Q,
Substitute value for Q,
6P = 10(64 - P),
Therefore P = 40, Q = 24.