Problems on Ages Notes for SSC Exams

Problems on Ages is a simple chapter of SSC Quantitative Aptitude, and one question is asked from it in some Tier 1 papers. The key is to form the right equation from the given statement. These problems on ages notes for SSC explain the standard methods with solved examples and practice questions.

Basic Rules

• If the present age is x, then the age after n years is x + n and the age n years ago was x - n.
• The difference between the ages of two persons always remains the same.
• The sum of the ages of two persons increases by 2n after n years.
• If the ratio of the ages is a : b, take the ages as ax and bx.
• Always define the present ages of the persons as the variable.

Steps to Solve Any Age Problem

1. Assume the present ages as x and y (or in the ratio ax and bx).
2. Write the age condition for the past or the future.
3. Form one or two equations.
4. Solve for x and put the value back into the ages.
5. Check the answer against the given conditions.

Common Types

Tricks for Speed

When the difference between two ages is known, use it as a constant. For example, if a father is 30 years older than his son, then in any year the difference is 30, so you only need to solve for one person. With ratios, you can use the ratio parts directly: if the ratio of ages is 5 : 7 and the difference is 8 years, then 2 parts = 8, so 1 part = 4, and the ages are 20 and 28.

Solved Examples

Example 1: The ages of a father and a son are in the ratio 4 : 1 and their total age is 50. Find their ages.
Solution: Total parts = 5. Son = 50 / 5 = 10 years. Father = 40 years.

Example 2: The present ages of A and B are in the ratio 3 : 5. After 6 years the ratio becomes 2 : 3. Find the present ages.
Solution: (3x + 6) / (5x + 6) = 2/3. So 9x + 18 = 10x + 12, giving x = 6. Ages are 18 and 30 years.

Example 3: Five years ago A was thrice as old as B. Five years from now A will be twice as old as B. Find their present ages.
Solution: a - 5 = 3(b - 5) gives a = 3b - 10. a + 5 = 2(b + 5) gives 3b - 5 = 2b + 10, so b = 15 and a = 35. Ages are 35 and 15 years.

Example 4: The sum of the ages of a mother and her daughter is 56 years. Four years ago the mother was three times as old as the daughter. Find their present ages.
Solution: m + d = 56 and m - 4 = 3(d - 4), so m = 3d - 8. Then 4d - 8 = 56 and d = 16. Daughter = 16 years, mother = 40 years.

Example 5: A is 6 years older than B. In 4 years, A will be twice as old as B. Find their present ages.
Solution: a = b + 6 and a + 4 = 2(b + 4). So b + 10 = 2b + 8 and b = 2. B = 2 years and A = 8 years.

Example 6: A father is three times as old as his son. After 10 years, the father will be twice as old as his son. Find the present ages.
Solution: f = 3s and f + 10 = 2(s + 10). So 3s + 10 = 2s + 20 and s = 10. Son = 10 years, father = 30 years.

Quick Tips to Solve Faster

• The difference between two ages always stays the same.
• For ratio questions, always assume the ages as ax and bx.
• Form the equation for the "after" or "before" condition separately.
• Use the options to test the answer when the equation looks long.
• Use the average age to find the total age and then solve the equation.

Common Mistakes to Avoid

• Adding the years to only one person.
• Forgetting that the age difference remains constant.
• Taking the ratio as the actual ages.
• Using "ago" as a plus instead of a minus.
• Not checking the answer against the original statement.

Quick Revision Points

• Age after n years = x + n; age before n years = x - n.
• Age difference is constant.
• Ratio a : b means ages ax and bx.
• Sum of ages after n years increases by 2n for two persons.
• Total age = average x number of persons.

Practice Questions With Answers

1. The age of A is twice that of B. The sum of their ages is 45. Find A's age.
   Answer: 30 years (B = 15).
2. The ages of two persons are in the ratio 4 : 5 and the sum is 36. Find the younger person's age.
   Answer: 16 years (36 x 4/9).
3. A father is 30 years older than his son. After 5 years, the father will be 4 times as old as the son. Find the son's present age.
   Answer: 5 years (s + 35 = 4(s + 5)).
4. The present ages of two persons are in the ratio 5 : 7 and their difference is 8 years. Find the age of the elder.
   Answer: 28 years (2 parts = 8).
5. The average age of a father and his son is 30 years. The father is 4 times as old as the son. Find the son's age.
   Answer: 12 years (Total = 60, parts = 5).
6. Ten years ago A was half as old as B. If A is 20 years old now, find B's present age.
   Answer: 30 years (A was 10, so B was 20).
7. After 8 years A will be twice as old as B. If B is 12 years old now, find A's present age.
   Answer: 32 years (2 x 20 - 8).
8. The ratio of the present ages of a father and his son is 7 : 2 and the sum is 63. Find the father's age.
   Answer: 49 years (63 x 7/9).
9. Four years ago A was 3 times as old as B. The sum of their present ages is 40. Find A's present age.
   Answer: 28 years (a - 4 = 3(b - 4)).
10. A is 3 years older than B and 2 years younger than C. If A is 20, find C's age.
    Answer: 22 years (C = A + 2).

How to Practise This Chapter

1. Learn the formulas in the table until you can write them without looking.
2. Solve the solved examples again on your own, without reading the solution.
3. Attempt the practice questions above with a timer of 10 minutes.
4. Solve the matching practice papers to test yourself in exam style.
5. Check the previous year papers to see how SSC frames this topic.

Related Notes and Practice Papers

• Ratio and Proportion Notes
• Average Notes
• Percentage Notes
• Partnership Notes
• SSC Percentage Practice Paper

Explore More SSC Subjects

SSC exams test many subjects, so connect this topic with the related subjects below:

• Reasoning Notes
• Statistics Notes
• Indian Economy Notes
• English Notes
• General Awareness Notes
• All SSC subjects and notes

Frequently Asked Questions

What are problems on ages?

These are questions in which the ages of persons are connected through ratios, sums or differences at different times, and we find their present or future ages.

How many questions come from ages in SSC exams?

Usually one question comes from ages in SSC Tier 1 in some papers.

What stays constant in age problems?

The difference between the ages of two persons always stays constant.

How do I solve a ratio-based age problem?

Take the ages as ax and bx, form an equation using the condition after or before some years and solve for x.

How do I find the age after n years?

Add n to the present age. For the age n years ago, subtract n.

What is the sum of ages after n years?

The sum of the ages of two persons increases by 2n after n years.

Can I use the options to solve ages questions?

Yes. Put the options in the conditions and check which one fits, especially in long equations.

How should I practise ages?

Solve the examples here and practise ratio and equation-based questions.