Problems on Ages

Present, past and future age puzzles

102 practice questionsPage updated 27 Mar 2026Review metadata pendingPractice questions

Problems on ages is a classic quantitative aptitude topic that involves calculating ages of individuals based on relationships given at different points in time (present, past, or future). These problems require setting up equations based on age ratios and solving for unknown ages using algebraic methods.

Key formulas and rules

If the present age of a person is x years, then: n years ago their age was (x - n) years, and n years hence their age will be (x + n) years.

Age ratio A : B = m : n means A's age / B's age = m / n, or A's age = m x k and B's age = n x k for some positive constant k.

If the sum of ages of two people is S and their ratio is m : n, then their ages are (m x S)/(m + n) and (n x S)/(m + n).

The difference between ages remains constant over time: if A is 5 years older than B today, A will always be 5 years older than B.

If A's age is n times B's age today, after/fefore x years: (A +/- x) = n x (B +/- x) — adjust signs based on past/future.

Average age of a group = (Sum of all ages) / (Number of people in the group).

Key concepts

Present, Past, and Future Ages

Always define variables for present ages. If a problem mentions '5 years ago', subtract 5 from the present age. If it mentions 'after 8 years' or '8 years hence', add 8 to the present age. Be careful with double negatives: '5 years before 3 years ago' means 8 years ago from present.

Ratio-Based Age Problems

When ages are given in ratio m:n, represent the ages as mx and nx where x is a common multiplier. Set up equations using other given information (sum of ages, difference, or relationship after certain years) to solve for x, then calculate actual ages.

Constant Age Difference

The age gap between two people never changes. If Person A is 10 years older than Person B, this remains true 5 years ago, today, or 20 years in the future. Use this property to simplify problems involving multiple time periods.

Worked examples

Example 1

The present ages of A and B are in the ratio 5:7. Eight years ago, the ratio of their ages was 2:3. Find their present ages.

Solution

Let the present ages of A and B be 5x and 7x years respectively.
Eight years ago, A's age was (5x - 8) and B's age was (7x - 8).
Given that eight years ago, the ratio was 2:3.
So: (5x - 8) / (7x - 8) = 2 / 3
Cross-multiply: 3(5x - 8) = 2(7x - 8)
15x - 24 = 14x - 16
15x - 14x = 24 - 16
x = 8
Therefore, A's present age = 5 x 8 = 40 years
B's present age = 7 x 8 = 56 years
Verification: Eight years ago, A was 32 and B was 48. Ratio 32:48 = 2:3

Example 2

A father is 4 times as old as his son. In 20 years, he will be only twice as old as his son. Find their present ages.

Solution

Let the son's present age be x years.
Then the father's present age is 4x years.
In 20 years: Son's age = x + 20, Father's age = 4x + 20
Given: In 20 years, father will be twice as old as son.
So: 4x + 20 = 2(x + 20)
4x + 20 = 2x + 40
4x - 2x = 40 - 20
2x = 20
x = 10
Therefore, Son's present age = 10 years
Father's present age = 4 x 10 = 40 years
Verification: In 20 years, son will be 30 and father will be 60. 60 = 2 x 30

Example 3

The sum of the ages of a father and his son is 50 years. Five years ago, the father was 7 times as old as his son. Find the present age of the son.

Solution

Let the son's present age be x years.
Then the father's present age = (50 - x) years.
Five years ago: Son's age = x - 5, Father's age = 50 - x - 5 = 45 - x
Given: Five years ago, father was 7 times as old as son.
So: 45 - x = 7(x - 5)
45 - x = 7x - 35
45 + 35 = 7x + x
80 = 8x
x = 10
Therefore, the son's present age is 10 years.
Father's present age = 50 - 10 = 40 years.
Verification: Five years ago, son was 5 and father was 35. 35 = 7 x 5

Example 4

The average age of a husband and wife was 25 years when they married 6 years ago. Now the average age of the husband, wife, and their child is 22 years. Find the present age of the child.

Solution

Six years ago, average age of husband and wife was 25 years.
Sum of their ages 6 years ago = 25 x 2 = 50 years.
Present sum of husband and wife ages = 50 + 6 + 6 = 62 years.
Let child's present age be x years.
Now, average of all three = 22 years.
So: (62 + x) / 3 = 22
62 + x = 66
x = 4
Therefore, the child's present age is 4 years.

Representative solved questions

See the kind of question in this topic before opening the full practice set.

Question 1

The present ages of A and B are in the ratio 4:5. Five years ago, their ages were in the ratio 3:4. What will be the ratio of their ages after 10 years?

Answer: B. 6:7

Explanation

Step 1: Let present ages be 4x and 5x.
Step 2: Five years ago: (4x - 5)/(5x - 5) = 3/4
Step 3: 4(4x - 5) = 3(5x - 5)
Step 4: 16x - 20 = 15x - 15
Step 5: x = 5
Step 6: Present ages: 20 and 25 years.
Step 7: After 10 years: 30 and 35 years.
Step 8: Ratio = 30:35 = 6:7
Answer: 6:7

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 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. What is the present age of the father?

44 years
46 years
48 years
50 years

Answer: A. 44 years

Explanation

Step 1: Let son's present age = x years.
Step 2: Father's present age = 2x years.
Step 3: 20 years ago: Father's age = 2x - 20, Son's age = x - 20
Step 4: 2x - 20 = 12(x - 20)
Step 5: 2x - 20 = 12x - 240
Step 6: 10x = 220
Step 7: x = 22
Step 8: Father's age = 2 x 22 = 44 years.
Answer: 44 years

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Question 3

The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages was four times the father's age at that time. What is the present age of the father?

36 years
42 years
44 years
45 years

Answer: A. 36 years

Explanation

Let son's age = s, father's age = f.\nf + s = 45\n5 years ago: (f-5)(s-5) = 4(f-5)\nDividing both sides by (f-5): s - 5 = 4\ns = 9, f = 45 - 9 = 36 years.\nAnswer: 36 years

Sources and review notes

This is an AISEA-authored practice question.

Review status: accepted · Reviewed 2026-08-13 · structure and answer-key checks, editorial quality checks, duplicate screening

Problems on Ages

Key formulas and rules

Key concepts

Present, Past, and Future Ages

Ratio-Based Age Problems

Constant Age Difference

Multiple Time Period Relationships

Group Age Problems

Worked examples

Example 1

Example 2

Example 3

Example 4

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

Key formulas and rules

Key concepts

Present, Past, and Future Ages

Ratio-Based Age Problems

Constant Age Difference

Multiple Time Period Relationships

Group Age Problems

Worked examples

Example 1

Example 2

Example 3

Example 4

Representative solved questions

Question 1

Question 2

Question 3

Common mistakes and useful habits

Ready to test your understanding?

Keep building the same skill family.