Problems on Ages: Aptitude Questions with Answers – Free Practice!
Problem on ages is a specialized case of Algebra Problems. You will need simple skills to convert basic word problem to algebraic equations to solve these problems.
Q.1 The ratio of the present ages of two brothers is 1 : 2. Five years ago, the ratio was 1 : 3. What will be the ratio of their ages after five years?
Check Solution
Ans: C
Let the present ages be x and 2x.
Five years ago, their ages were (x-5) and (2x-5). So, (x-5)/(2x-5) = 1/3 => 3x-15 = 2x-5 => x=10.
Their present ages are 10 and 20.
In five years, their ages will be 15 and 25, so the ratio is 15:25 or 3:5.
Q.2 A father is twice as old as his son. 20 years ago, the age of the father was 12 times the age of the son. The present age of the father is:
Check Solution
Ans: A
Let the son’s present age be x.
The father’s present age is 2x.
Twenty years ago, (2x-20) = 12(x-20) => 2x-20 = 12x-240 => 10x = 220 => x=22.
The father’s present age is 2x = 44 years.
Q.3 The sum of the ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
Check Solution
Ans: A
Let the age of the youngest child be x.
The ages are x, x+3, x+6, x+9, and x+12.
Their sum is 5x + 30 = 50 => 5x = 20 => x=4 years.
Q.4 The age of a man is four times that of his son. Five years ago, the man’s age was nine times that of his son’s age at that time. The present age of the man is:
Check Solution
Ans: B
Let the son’s age be x.
The man’s age is 4x.
Five years ago, (4x-5) = 9(x-5) => 4x-5 = 9x-45 => 5x = 40 => x=8.
The man’s present age is 4x = 32 years.
Q.5 If the total ages of Iqbal and Shikhar is 12 years more than the total age of Shikhar and Charu, then Charu is how many years younger than Iqbal?
Check Solution
Ans: B
Let the ages be I, S, and C.
Given: I + S = S + C + 12.
Subtracting S from both sides gives I = C + 12.
This means Charu is 12 years younger than Iqbal.
Q.6 The present age of a woman is 3 years more than three times the age of her daughter. Three years hence, the woman’s age will be 10 years more than twice the age of her daughter. Find the present age of the woman.
Check Solution
Ans: B
Let the daughter’s age be x.
The woman’s age is 3x+3.
In 3 years, (3x+3)+3 = 2(x+3) + 10 => 3x+6 = 2x+6+10 => x=10.
The woman’s present age is 3(10)+3 = 33 years.
Q.7 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 is 27, then how old is B?
Check Solution
Ans: D
Let C’s age be x. B’s age is 2x.
A’s age is 2x+2. The sum is x + 2x + (2x+2) = 27 => 5x+2=27 => 5x=25 => x=5.
B’s age is 2x = 10 years.
Q.8 The product of the ages of Ankit and Nikita is 240. If twice the age of Nikita is more than Ankit’s age by 4 years, what is Nikita’s age?
Check Solution
Ans: A
Let Ankit’s age be A and Nikita’s age be N.
We have A*N=240 and 2N = A+4. From the second equation, A = 2N-4.
Substitute into the first: (2N-4)*N=240 => 2N^2-4N-240=0 => N^2-2N-120=0.
Factoring gives (N-12)(N+10)=0.
Since age must be positive, N=12 years.
Q.9 Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar’s age at present?
Check Solution
Ans: A
Let their ages six years ago be 6x and 5x.
Their present ages are 6x+6 and 5x+6.
In four years, their ages will be 6x+10 and 5x+10. So, (6x+10)/(5x+10) = 11/10 => 60x+100 = 55x+110 => 5x=10 => x=2.
Sagar’s present age is 5x+6 = 5(2)+6 = 16 years.
Q.10 The average age of a husband and his wife was 23 years at the time of their marriage. After five years, they have a one-year-old child. The average age of the family now is:
Check Solution
Ans: A
At marriage, sum of ages = 23 * 2 = 46.
Five years later, the sum of the couple’s ages is 46 + (5*2) = 56.
With the one-year-old child, the total sum of ages is 56+1=57 for 3 people.
The average age is 57/3 = 19 years.
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