Problem on Ages

Maths-9 Problems on Ages: Questions and Step-by-Step Solutions for all Competitive Exams

by
Problems on Ages

Maths-9 Problems on Ages: Questions and Step-by-Step Solutions for all Competitive Exams

Age problems are one of the friendliest topics in maths. We’ll start with the basic idea, learn useful shortcuts, and solve several example problems step-by-step so nothing is left mysterious.

Problems on Ages

1. The basic idea

An age problem asks about how old someone is now, or will be, or was in the past — usually using relationships (like “A is 5 years older than B”, or “In 5 years, A will be twice B”).

Two things to remember every time:

  • Ages change by the same amount for everyone as time passes.
    • If you go forward 5 years, everyone’s age goes up by 5.
    • If you go back 3 years, everyone’s age goes down by 3.
  • The difference in ages between two people is always the same (it does not change with time).

2. Useful rules / formulas

  • If A’s present age is aand B’s present age is b
    • After x  years: A will be a +x , B will be b +x
    • x years ago: A was a- x B was b- x .
  • Age difference: A- B = a- b (constant for past or future).
  • When a problem gives relationships like “A is k times B”, convert to equations: e.g. a = k b
  • Often we set one unknown (say B’s age =x) and express others in terms of , then solve a linear equation.

3. Step-by-step examples

Example 1 — very simple (difference)

Q: Ravi is 5 years older than Sita. If Sita is 12 now, how old is Ravi?
Step-by-step:

  1. Sita’s age = 12. Ravi is 5 years older → Ravi = Sita + 5.
  2. Compute: Ravi = 12 + 5 = 17
    Answer: Ravi is 17 years old.

Example 2 — years ahead

Q: Meena is 10 now. How old will she be after 7 years?
Step-by-step:

  1. Present = 10. After 7 years → add 7.
  2. 10 + 7 =17
    Answer: Meena will be 17 years old.

Example 3 — years ago

Q: Arun is 15 now. What was his age 5 years ago?
Step-by-step:

  1. Present = 15. 5 years ago → subtract 5.
  2. 15-5 = 10
    Answer: Arun was 10 years old.

Example 4 — difference stays the same

Q: A is 8 years older than B. If A is 20 now, what was B’s age 3 years ago?
Step-by-step:

  1. A = 20. Difference A − B = 8 → so B = 20 − 8 = 12 (present).
  2. 3 years ago B = 12-3 = 9
    Answer: B was 9 years old 3 years ago.

Example 5 — set up equation (classic)

Q: Sum of ages of two children is 20. If one child is 6 years older than the other, find their ages.
Step-by-step:

  1. Let younger child = x , Then older = x + 6
  2. Sum given: x + (x +6 ) = 20
  3. Simplify: 2x + 6 = 20 Subtract 6: 2x = 14 , Divide by 2: x = 7
  4. Younger = 7. Older = 7+6 = 13
    Answer: Ages are 7 and 13.

Example 6 — future relation (twice, thrice)

Q: After 5 years, Rahul will be twice as old as his sister. Rahul is 11 now. How old is his sister now?
Step-by-step:

  1. Rahul after 5 years = 11+5 =16 . Let sister now be s . After 5 years sister = s +5
  2. Given: Rahul after 5 years = 2 × (sister after 5 years) → 16 = 2 x(s+5).
  3. Simplify: 16 = 2s + 10 Subtract 10: 6 = 2s So s = 3
    Answer: Sister is 3 years old now.

Example 7 — past relation (half, double)

Q: Ten years ago, John’s age was three times his son’s age. Now John is 40. Find the son’s present age.
Step-by-step:

  1. John now = 40. Ten years ago John = 40-10 =30 Let son now = x . Ten years ago son = x-10
  2. Given: 10 years ago John = 3 × (son 10 years ago) → 30= 3(x-10)
  3. Simplify: 30= 3x-30 Add 30: 60= 3x So x=20
    Answer: Son is 20 years old now.

Example 8 — tricky: use difference stays same

Q: A father is 36 years older than his son. After 6 years, father’s age will be 3 times the son’s age. Find their present ages.
Step-by-step:

  1. Let son now = x . Father now = x+ 36
  2. After 6 years: son = x+ 6, father = x + 36 +6 = x+42
  3. Given: father after 6 years = 3 × son after 6 years → x + 42 = 3(x + 6)
  4. Expand RHS: x + 42 = 3x + 18 . Move x terms: 42-18 = 3x-x → 24 =2x . So x=12
  5. Son now = 12. Father now = 12+36 =48
    Answer: Son = 12, Father = 48.

Example 9 — average & ages

Q: Average age of 4 friends is 10 years. If one friend is 14, what is the average age of the remaining three?
Step-by-step:

  1. Total age of 4 friends = average × number = 10 x 4= 40
  2. One friend = 14, so sum of other three = 40-14 =26.
  3. Average of 3 = 26 / 3 = years.
    Answer: Average = years (or 8.666… years).

4. Common types of exam problems & shortcuts

  • Sum of ages: convert average → total → subtract known ages.
  • Difference of ages: stays same always — useful when moving forward/back.
  • “After x years, A will be k times B”: compute A after x and B after x (add x to both), then set equation.
  • Set variable for the youngest often simplifies algebra.
  • Look for easy elimination: sometimes subtract two equations to eliminate variable.

5. Practice problems

Try these on your own, then check the answers below.

  1. Sita is 4 years older than Rani. If Rani is 9, how old is Sita?
  2. The sum of the ages of A and B is 30. If A is 6 years older than B, find their ages.
  3. After 3 years, Maya will be 15. How old is she now?
  4. Ten years ago, Raj was 3 times as old as Amit. Now Raj is 40. Find Amit’s present age.
  5. Father is 30 years older than his son. After 5 years, the father will be twice the son. Find present ages.
  6. The average age of 5 students is 12. If one student is 20, find the average age of the remaining four.
  7. The present ages of two people are in the ratio 3 : 5. If 5 years are added to each, the ratio becomes 2 : 3. Find their present ages.
  8. A is twice as old as B. In 4 years, A will be 6 years older than B. Find the present ages.
  9. The sum of the ages of 3 people is 60. If one is 10 years older than another and the third is 5 years younger than the second, find the individual ages.
  10. A father is three times as old as his son. After 15 years, he will be twice as old. Find present ages.

6. Answers (brief)

  1. Sita = 9+4 = 13 years.
  2. Let younger = x, x+(x+6) = 30→ 2x+6 = 30→ 2x= 24 → x= 12 . Ages: 12 and 18.
  3. Present = 15-3= 12 years.
  4. Same method as Example 7 → Amit now = 20 years.
    • (Raj 40 → 10 years ago 30 = 3 × (Amit 10 years ago) → Amit 10 years ago = 10 → Amit now 20).
  5. Son = 10, Father = 40. (Let son = x → father = x + 30; after 5 years: x+35 = 2(x+5) → solve → x=10.)
  6. Total = 5 x12 = 60. Remaining sum = 60-20 =40 . Average =40/4 =10
  7. Let ages = 3x , 5x . After 5 years: (3x+5)/(5x+5)=2/3. Solve: 3(3x+5)=2(5x +5)→ 9x + 15= 10x +10→ x=5. Ages: 15 and 25.
  8. Let B = x, A = 2x. In 4 years: 2x + 4 = x + 4 + 6 → 2x + 4 = x + 10 → x = 6. So B=6, A=12.
  9. Let ages be a,b,c with b= a + 10 , c= b-5 = a +5 Sum: a + (a +10) + ( a +5)= 60 → 3a + 15 = 60 → 3a = 45 → a = 15 . So ages: 15, 25, 20.
  10. Let son =x, father = 3x After 15 years: 3x + 15 = 2(x + 15)→ 3x +15 =2x + 30→ x = 15 . So son 15, father 45.

7. Final tips for students

  • Translate words into “present age” variables — write equations.
  • Always do a quick reality check: ages should be positive and sensible (no negative ages!).
  • Practice converting story → equation — that’s the key skill.
  • For ratio problems, let ages = ratio × some number and solve for that number.

Leave a Comment