Problems on ages for RRB Group D

Age problems turn a short story into an equation. "Five years ago, the parent was four times as old as the child" sounds like a puzzle, but it is really one line of algebra. The candidates who find these hard are usually trying to do them in their heads. Write the ages down in a small table and the equation almost writes itself.

This topic also borrows from two others. Ratio questions ("ages are in the ratio 2 : 3") need the parts method from the ratio guide, and some questions hide an average inside the story.

Three facts that solve most questions

• After n years, everyone is n years older. n years ago, everyone was n years younger. The change applies to every person.
• The difference between two people's ages never changes. If a parent is 28 years older than their child today, they were 28 years older ten years ago and will be 28 years older in ten years.
• If ages are in the ratio 2 : 3, write them as 2x and 3x. Never as 2 and 3.

The table method

Make three columns: "years ago", "now" and "years later". Fill in the "now" row with letters, then add or subtract for the other columns. Example 1 below uses this layout:

Then turn the sentence in the question into an equation, using the right column.

Worked examples

Example 1: The ages of a parent and a child add up to 50. Five years ago, the parent was four times as old as the child. Find their ages.

• Child = C, parent = 50 − C.
• Five years ago: 50 − C − 5 = 4(C − 5).
• 45 − C = 4C − 20, so 5C = 65 and C = 13.
• Child 13, parent 37. Check: five years ago they were 8 and 32, and 32 = 4 × 8.

Example 2: Two ages are in the ratio 2 : 3. After 8 years, the ratio will be 3 : 4. Find the ages.

• Ages 2x and 3x. (2x + 8) ÷ (3x + 8) = 3 ÷ 4.
• 4(2x + 8) = 3(3x + 8), so 8x + 32 = 9x + 24, and x = 8.
• Ages 16 and 24. Check: after 8 years, 24 and 32, which is 3 : 4.

Example 3: The present ages of two friends are in the ratio 5 : 7. Six years ago, the ratio was 3 : 5. Find their present ages.

• Ages 5x and 7x. (5x − 6) ÷ (7x − 6) = 3 ÷ 5.
• 5(5x − 6) = 3(7x − 6), so 25x − 30 = 21x − 18, and 4x = 12, x = 3.
• Ages 15 and 21. Check: six years ago, 9 and 15, which is 3 : 5.

Example 4: A parent is 30 years older than their child. In 5 years, the parent will be three times as old as the child. Find their present ages.

• Child = c, parent = c + 30.
• In 5 years: c + 35 = 3(c + 5), so c + 35 = 3c + 15, and c = 10.
• Child 10, parent 40. Check: in 5 years, 15 and 45, and 45 = 3 × 15.

Example 5: One sibling is 6 years older than the other. In 4 years, their ages will add up to 40. Find their present ages.

• In 4 years, both are 4 years older, so the sum grows by 8. Present sum = 40 − 8 = 32.
• Sum 32, difference 6: older = (32 + 6) ÷ 2 = 19, younger = 13.
• Ages 19 and 13.

Example 6: The average age of three siblings is 12. When their parent is included, the average becomes 18. How old is the parent?

• Totals: 4 × 18 = 72 and 3 × 12 = 36.
• Parent = 72 − 36 = 36 years.

Testing the options

In the exam, it is often faster to try each option than to set up an equation:

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

In Example 2, an option of "16 and 24" passes the 2 : 3 test at once, and adding 8 to both gives 24 : 32 = 3 : 4. Done in ten seconds. This works best when the options are actual ages rather than a single value.

Common mistakes

• Adding the years to only one person.
• For "years ago", subtracting from one person but not the other.
• Writing a ratio 2 : 3 as ages 2 and 3 instead of 2x and 3x.
• Giving x instead of the actual ages.
• Forgetting that a sum of two ages grows by 2n after n years, not by n.

Practice set

1. Two ages are in the ratio 5 : 6 and add up to 44. Find them.
2. One person is 8 years older than another. Their ages add up to 36. Find them.
3. A parent is three times as old as their child. In 10 years, the parent will be twice as old. Find their ages now.
4. Two ages are in the ratio 3 : 5. After 6 years, the ratio will be 2 : 3. Find the ages.
5. The average age of 4 people is 20. When a fifth person joins, the average becomes 22. How old is the fifth person?
6. Ten years ago, a parent was five times as old as their child. Now the parent is three times as old. Find their present ages.
7. The ages of two siblings add up to 25. In five years, their ages will be in the ratio 3 : 4. Find their present ages.
8. In 12 years, a person will be twice as old as they were 4 years ago. Find their present age.

Answers:

1. 20 and 24. 11 parts; one part = 4.
2. 22 and 14. (36 + 8) ÷ 2 = 22; 36 − 22 = 14.
3. Child 10, parent 30. 3c + 10 = 2(c + 10), so c = 10.
4. 18 and 30. (3x + 6) ÷ (5x + 6) = 2/3 gives 9x + 18 = 10x + 12, x = 6. Check: 24 : 36 = 2 : 3.
5. 30. 5 × 22 − 4 × 20 = 110 − 80.
6. Child 20, parent 60. Parent = 3c; 3c − 10 = 5(c − 10), so 2c = 40. Check: ten years ago, 10 and 50.
7. 10 and 15. In five years the sum is 35, split 3 : 4 as 15 and 20; subtract 5 from each.
8. 20. a + 12 = 2(a − 4), so a = 20. Check: 32 = 2 × 16.

What to do next

• Draw the "ago, now, later" table for the next ten age questions you solve, even the easy ones.
• Practise five ratio-type age questions and check each answer against both conditions.
• Time yourself using the option-testing method on five questions.
• Revise averages for the mixed questions, and see the maths plan for what to study next.