Problem On Ages - Examples (with Solutions), Quantitative Aptitude

Introduction 

These problems on ages are designed to help you develop a strong understanding of age-related word problems, which are commonly found in competitive exams like SBI PO, SBI Clerk, IBPS PO, IBPS Clerk, and IBPS RRB. These problems involve setting up and solving equations based on relationships between the ages of different individuals at various points in time. By practicing these types of problems, you will enhance your problem-solving skills, improve your ability to interpret information, and boost your chances of performing well in your upcoming exams. The problems cover a range of scenarios, including age differences, age ratios, and the relationship between present and past ages, offering you a comprehensive preparation experience.

Solved Examples

1. Rajeev's age after 15 years will be 5 times his age 5 years back. What is the present age of Rajeev?

Sol. Let Rajeev's present age be x years.
Rajeev's age after 15 years = x + 15.
Rajeev's age 5 years back = x - 5.
Given: x + 15 = 5(x - 5).
x + 15 = 5x - 25.
Bring terms together: 4x = 40.
x = 10.
Ans: Rajeev's present age = 10 years.

2. The ages of two persons differ by 16 years. If 6 years ago, the elder one is 3 times as old as the younger one, find their present ages.

Sol. Let the younger person's present age be x years.
Then the elder person's present age = x + 16 years.
Six years ago: younger = x - 6, elder = x + 16 - 6 = x + 10.
Given: elder (six years ago) = 3 × younger (six years ago): x + 10 = 3(x - 6).
x + 10 = 3x - 18.
Rearrange: 2x = 28 ⇒ x = 14.
Ans: Present ages are 14 years (younger) and 30 years (elder).

3. 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?

Sol. Let Ankit's age = x years. Then Nikita's age = 240 / x years.
Condition: 2 × (Nikita's age) - Ankit's age = 4.
So 2 × (240/x) - x = 4 ⇒ 480/x - x = 4.
Multiply both sides by x: 480 - x² = 4x.
Bring all terms to one side: x² + 4x - 480 = 0.
Factorise: (x + 24)(x - 20) = 0. Accept positive root x = 20.
Nikita's age = 240 / 20 = 12 years.
Ans: Nikita's age = 12 years.

4. The present age of a father is 3 years more than three times the age of his son. Three years hence, father's age will be 10 years more than twice the age of the son. Find the present age of the father.

Sol. Let the son's present age = x years.
Father's present age = 3x + 3 years.
After 3 years: son's age = x + 3; father's age = 3x + 3 + 3 = 3x + 6.
Given: father's age after 3 years = 2 × (son's age after 3 years) + 10:
3x + 6 = 2(x + 3) + 10.
3x + 6 = 2x + 6 + 10 = 2x + 16.
So x = 10.
Father's present age = 3x + 3 = 3×10 + 3 = 33 years.
Ans: Father's present age = 33 years.

5. Rohit was 4 times as old as his son 8 years ago. After 8 years, Rohit will be twice as old as his son. What are their present ages?

Sol. Let the son's age 8 years ago be x years.
Rohit's age 8 years ago = 4x years.
Son's present age = x + 8; son's age after 8 years = x + 16.
Rohit's present age = 4x + 8; Rohit's age after 8 years = 4x + 16.
Given: Rohit after 8 years = 2 × (son after 8 years): 4x + 16 = 2(x + 16).
4x + 16 = 2x + 32 ⇒ 2x = 16 ⇒ x = 8.
Son's present age = x + 8 = 16 years.
Rohit's present age = 4x + 8 = 4×8 + 8 = 40 years.
Ans: Son = 16 years; Rohit = 40 years.

6. One year ago, the ratio of Gaurav's and Sachin's age was 6: 7 respectively. Four years hence, this ratio would become 7: 8. How old is Sachin?

Sol. Let ages one year ago be Gaurav = 6x and Sachin = 7x.
Gaurav's age four years hence = (6x + 1) + 4 = 6x + 5.
Sachin's age four years hence = (7x + 1) + 4 = 7x + 5.
Given ratio: (6x + 5) : (7x + 5) = 7 : 8.
So 8(6x + 5) = 7(7x + 5) ⇒ 48x + 40 = 49x + 35 ⇒ x = 5.
Sachin's present age = 7x + 1 = 7×5 + 1 = 36 years.
Ans: Sachin is 36 years old.

7. Abhay's age after six years will be three-seventh of his father's age. Ten years ago the ratio of their ages was 1: 5. What is Abhay's father's age at present?

Sol. Let ages 10 years ago be Abhay = x and father = 5x.
Abhay's age after 6 years = (x + 10) + 6 = x + 16.
Father's age after 6 years = (5x + 10) + 6 = 5x + 16.
Given: (x + 16) : (5x + 16) = 3 : 7.
So 7(x + 16) = 3(5x + 16) ⇒ 7x + 112 = 15x + 48 ⇒ 8x = 64 ⇒ x = 8.
Father's present age = 5x + 10 = 5×8 + 10 = 50 years.
Ans: Abhay's father is 50 years old at present.

8. The Ratio of Ages of Mona and Sona is 4:5. Twelve Years hence, their ages will be in the ratio of 5:6. What will be Sona's age after 6 years?

Sol. Let their present ages be Mona = 4x and Sona = 5x.
After 12 years: Mona = 4x + 12, Sona = 5x + 12.
Given: (4x + 12) : (5x + 12) = 5 : 6.
So 6(4x + 12) = 5(5x + 12) ⇒ 24x + 72 = 25x + 60 ⇒ x = 12.
Sona's present age = 5x = 60 years.
Sona's age after 6 years = 60 + 6 = 66 years.
Ans: Sona will be 66 years old after 6 years.

9. Ramu was 4 times as old as his son 8 years ago. After 8 years, Ramu will be twice as old as his son. What their present ages?

Sol. Let the son's age 8 years ago be x years.
Ramu's age 8 years ago = 4x years.
Son's present age = x + 8; son's age after 8 years = x + 16.
Ramu's present age = 4x + 8; Ramu's age after 8 years = 4x + 16.
Given: Ramu after 8 years = 2 × (son after 8 years): 4x + 16 = 2(x + 16).
4x + 16 = 2x + 32 ⇒ 2x = 16 ⇒ x = 8.
Son's present age = x + 8 = 16 years.
Ramu's present age = 4x + 8 = 40 years.
Ans: Son = 16 years; Ramu = 40 years.

10. A man is four times as old as his son. Five years ago, the man was nine times as old his son was at that time. What is the present age of a man?

Sol. Let the son's present age = x years. Then the man's present age = 4x years.
Five years ago: son's age = x - 5; man's age = 4x - 5.
Given: man five years ago = 9 × (son five years ago): 4x - 5 = 9(x - 5).
4x - 5 = 9x - 45 ⇒ bring x terms together: 45 - 5 = 9x - 4x ⇒ 40 = 5x ⇒ x = 8.
Man's present age = 4x = 4 × 8 = 32 years.
Ans: The man's present age is 32 years.