Problems on ages for RRB NTPC: methods, patterns and practice

Age problems turn a short story into an equation. The maths inside is almost always a one-step linear equation. What trips candidates up is the translation: adding years to one person but not the other, or answering with x when the question wanted an age.

So this guide spends most of its time on setting the question up correctly. Once the equation is right, the answer is two lines away. If you are shaky with ratios, revise ratio and proportion first; half of all age questions are ratio questions.

Three facts that do most of the work

1. Time moves for everyone equally. After n years every person is n years older; n years ago every person was n years younger.
2. The difference between two ages never changes. If one person is 8 years older today, they were 8 years older 10 years ago and will be 8 years older in 30 years.
3. A sum of ages moves with the number of people. For a group of k people, the sum of ages rises by k × n after n years.

Why fact 2 is so useful: ratios change over time, but differences do not. If a ratio question also gives you the difference, you can find the multiplier directly without forming any equation.

Setting up: one variable, not two

Choose the variable for the present age of the person the question mentions most, and write everyone else in terms of it.

The ratio multiplier

When ages are in a ratio a : b, write them as ax and bx. Then:

• If a future or past ratio is given, set up (ax ± n) : (bx ± n) and cross-multiply.
• If the difference is given, x = difference ÷ (b − a). This is the fastest case of all.
• If the sum is given, x = sum ÷ (a + b).

Always check your answer against the second condition before marking it. It takes five seconds and catches most slips.

Testing the options

Age questions are ideal for testing options, because ages are whole numbers and the conditions are easy to check.

1. Take an option and check the first condition.
2. If it passes, check the second.
3. The option that fits both is the answer.

With four options, you will often find the answer faster than by solving, especially when the question has two past-and-future conditions.

Worked questions at NTPC level

Q1. The ages of A and B are in the ratio 4 : 5. After 6 years, the ratio will be 5 : 6. Find their present ages.

Present ages: 4x and 5x. So (4x + 6) : (5x + 6) = 5 : 6.
6(4x + 6) = 5(5x + 6), so 24x + 36 = 25x + 30, and x = 6.
Ages: 24 and 30. Check: 30 : 36 = 5 : 6.

Q2. A parent is three times as old as their child. After 12 years, the parent will be twice as old as the child. Find their present ages.

Child = c, parent = 3c. Then 3c + 12 = 2(c + 12), so c = 12.
Child 12, parent 36. Check: after 12 years, 48 and 24.

Q3. The sum of the ages of A and B is 50. Five years ago, their ages were in the ratio 3 : 5. Find their present ages.

Five years ago both were 5 years younger, so the sum was 50 − 10 = 40.
3k + 5k = 40, so k = 5. Ages then: 15 and 25.
Present ages: 20 and 30.

Q4. The present ages of A and B are in the ratio 3 : 4. Ten years ago, A was half as old as B. Find the sum of their present ages.

Present: 3x and 4x. Ten years ago: 3x − 10 = ½(4x − 10).
6x − 20 = 4x − 10, so x = 5. Ages 15 and 20; sum = 35.
Check: ten years ago they were 5 and 10.

Q5. A is 2 years older than B, and B is twice as old as C. The total of their ages is 27. How old is B?

C = x, B = 2x, A = 2x + 2. So 5x + 2 = 27, and x = 5.
B = 10 (A = 12, C = 5).

Q6. Six years ago, A was four times as old as B. Four years from now, A will be twice as old as B. Find their present ages.

A − 6 = 4(B − 6), so A = 4B − 18.
A + 4 = 2(B + 4). Substituting: 4B − 14 = 2B + 8, so B = 11 and A = 26.
Ages: 26 and 11. Check: six years ago 20 and 5; in four years 30 and 15.

Common mistakes

• Adding years to one person only. Time passes for everyone in the question.
• Changing the sum by n instead of k × n. Two people five years ago means subtracting 10 from the sum.
• Answering with x. In Q1, x = 6 but the ages are 24 and 30. Options often include x as a trap.
• Reading "years hence" as "years ago". "Hence" means from now into the future.
• Using two variables when one is enough. It doubles the working and the chance of error.

Practice set

1. The ages of two people are in the ratio 3 : 4 and their sum is 56. Find their ages.
2. A is 5 years older than B. The sum of their ages is 45. Find their ages.
3. A parent is four times as old as their child. After 5 years, the parent will be three times as old. Find their present ages.
4. Two ages are in the ratio 5 : 7. After 4 years, the ratio will be 3 : 4. Find the present ages.
5. The average age of three siblings is 12. The youngest is 8 and the eldest is 16. How old is the middle one?
6. Two ages are in the ratio 5 : 3 and differ by 12 years. What will the ratio be after 6 years?
7. The sum of the present ages of four family members is 100. What was the sum five years ago?
8. The average age of 30 students is 14. When the teacher's age is included, the average rises by 1. Find the teacher's age.

Answers:

1. 24 and 32. x = 56 ÷ 7 = 8.
2. 25 and 20. B + (B + 5) = 45.
3. Child 10, parent 40. 4c + 5 = 3(c + 5); after 5 years, 15 and 45.
4. 20 and 28. 4(5x + 4) = 3(7x + 4) gives x = 4; after 4 years, 24 and 32.
5. 12. Total 36; 36 − 8 − 16.
6. 3 : 2. x = 12 ÷ 2 = 6, so ages 30 and 18; after 6 years, 36 : 24.
7. 80. Four people, each 5 years younger: 100 − 20.
8. 45. 31 × 15 − 30 × 14 = 465 − 420.

What to do next

• For your next 15 age questions, write the "now" and "then" ages in two rows before forming any equation.
• Solve five questions twice: once by equation, once by testing options. Note which was faster.
• Revise averages, because family and class average-age questions sit between the two chapters.