Age Aptitude Questions
Age-related problems are a common topic in almost every competitive exam, including SSC, RRB, IBPS, and other government job tests. These questions are designed to test a candidate’s logical thinking, algebraic skills, and ability to form and solve equations based on given conditions.
Typically, age problems involve the present, past, or future ages of individuals and their relationships—such as father-son, brother-sister, or friends. The key to solving these questions lies in understanding the scenario, forming the correct equations, and simplifying them efficiently.
Mastering this topic can give you a significant edge in the quantitative aptitude section of any exam. In this article, you’ll find a variety of solved age problems—from basic to advanced—complete with step-by-step explanations to help you grasp the concepts thoroughly.
Basic Age Problems with Solutions
1. The present age of a father is 3 times that of his son. Five years ago, the father was 4 times as old as his son. What is the present age of the son?
Solution:
Let son’s present age = x
Father’s present age = 3x
5 years ago:
Father = 3x – 5
Son = x – 5
=> 3x – 5 = 4(x – 5)
=> 3x – 5 = 4x – 20
=> x = 15
✅ Answer: 15 years
2. A is 2 years older than B, who is twice as old as C. If the sum of their ages is 58, find A’s age.
Let C’s age = x
Then B = 2x, and A = 2x + 2
=> x + 2x + (2x + 2) = 58
=> 5x + 2 = 58
=> x = 11.2
=> A = 2x + 2 = 24.4
✅ Answer: 24.4 years
3. The sum of ages of a father and son is 60 years. Five years ago, the father was four times as old as his son. Find the present age of the son.
Let son’s age = x
Then father = 60 – x
5 years ago:
(60 – x – 5) = 4(x – 5)
=> 55 – x = 4x – 20
=> 5x = 75
=> x = 15
✅ Answer: 15 years
4. The ratio of the present ages of a father and his son is 5:2. After 10 years, the ratio will be 3:2. Find the present age of the father.
Let ages be 5x and 2x
After 10 years:
(5x + 10)/(2x + 10) = 3/2
=> 2(5x + 10) = 3(2x + 10)
=> 10x + 20 = 6x + 30
=> 4x = 10
=> x = 2.5
=> Father’s age = 5x = 12.5
✅ Answer: 12.5 years
5. Ten years ago, the age of A was half the age of B. The sum of their present ages is 80. What is B’s present age?
Let B’s present age = x
Then A’s age 10 years ago = (x – 10)/2
=> Present A = (x – 10)/2 + 10
Total: [(x – 10)/2 + 10] + x = 80
=> (x – 10 + 20 + 2x)/2 = 80
=> (3x + 10)/2 = 80
=> 3x + 10 = 160
=> x = 50
✅ Answer: 50 years
🧠 Moderate to Difficult Age Questions
6. The present ages of three persons are in the ratio 4:7:9. Eight years ago, the sum of their ages was 56. What is their present age?
Let ages be 4x, 7x, 9x
Eight years ago:
(4x – 8) + (7x – 8) + (9x – 8) = 56
=> 20x – 24 = 56
=> 20x = 80
=> x = 4
=> Ages = 16, 28, 36
✅ Answer: 16, 28, 36 years
7. The difference between the present ages of a father and his son is 24 years. Four years ago, the ratio of their ages was 5:1. Find their present ages.
Let son = x
Father = x + 24
4 years ago:
(x + 24 – 4)/(x – 4) = 5/1
=> (x + 20)/(x – 4) = 5
=> x + 20 = 5x – 20
=> 4x = 40
=> x = 10
Father = 34
✅ Answer: Son – 10 years, Father – 34 years
8. A man is 24 years older than his son. In 2 years, his age will be twice that of his son. Find their present ages.
Let son = x
Father = x + 24
In 2 years:
x + 2 and x + 26
=> x + 26 = 2(x + 2)
=> x + 26 = 2x + 4
=> x = 22
Father = 46
✅ Answer: Son – 22 years, Father – 46 years
9. If 6 years are subtracted from the present age of A and the result is divided by 18, then you get B’s age. If B is 2 years younger than C whose age is 5, then find A’s age.
C = 5, B = 3
=> (A – 6)/18 = 3
=> A – 6 = 54
=> A = 60
✅ Answer: 60 years
10. The age of a father 5 years ago was 7 times the age of his son. Five years from now, the father’s age will be 3 times the age of his son. Find their current ages.
Let son = x
Father = y
5 years ago:
y – 5 = 7(x – 5)
=> y – 5 = 7x – 35 …(i)
After 5 years:
y + 5 = 3(x + 5)
=> y + 5 = 3x + 15 …(ii)
Solving:
From (i): y = 7x – 30
From (ii): y = 3x + 10
=> 7x – 30 = 3x + 10
=> 4x = 40
=> x = 10
=> y = 40
✅ Answer: Son – 10 years, Father – 40 years
