Problems on Ages: Multiples, Families, Averages and Digit Reversal
Problems on Ages: Multiples, Families, Averages and Digit Reversal
The second family of age problems uses "k times as old", families of three or more people, average ages of groups that change over time, and two-digit ages with reversed digits. Recent NTPC papers (2025-26) favour the father-son and family-sum types.
1. Rules Box
"Father is k times as old as son": F = kS. "n years ago he was m times": F − n = m(S − n)
"When will F be twice S?": F + t = 2(S + t), so t = F − 2S (t negative means it happened in the past)
Average age of a group of k people rises by n after n years (if nobody joins or leaves)
A baby born now adds 0 to the total age but 1 to the count
Reversed digits: ages 10a + b and 10b + a differ by 9(a − b)
| Situation | Total age to use |
|---|---|
| Family of 5, average 28, six years ago | Now: 5 × 28 + 5 × 6 = 170 |
| Plus a baby born 4 years ago | 170 + 4 = 174 for 6 people |
| A member leaves | Subtract his present age and one from the count |
✗ Average of a family of 5 was 28 six years ago, so it is 28 now | ✓ Each member is 6 years older: the average is 34 if nobody joined
✗ A was 24 when B was born, so A is always twice B | ✓ Only the difference 24 is fixed; A = 2B only when B is 24
हिंदी नोट: "पिता पुत्र से k गुना" का अर्थ F = kS है। समूह की औसत आयु n वर्ष में n बढ़ जाती है, जब तक कोई सदस्य जुड़ता या जाता नहीं। अंक उलटने वाली आयु का अंतर हमेशा 9 का गुणज होता है।
Exam Pointer: Verified NTPC patterns: father k times the son's age in a family with a daughter and a past total (March 2026 Graduate CBT-1), "n years ago the father was k more than twice the son; when will he be twice" (August 2025 UG CBT-1) and a family's average age with a new birth (January 2021 CBT-1). Age-when-born and reversed-digit questions had no verified NTPC shift and are tagged syllabus-based.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: k times as old: a family with a past or future total
[PYQ: NTPC Graduate CBT-1 16-Mar-2026 Shift-1]
EXAM LEVEL
Q. A father is now 3 times as old as his son. Five years ago he was 4 times as old. Find their present ages.
F = 3S and 3S − 5 = 4(S − 5) give S = 15 and F = 45 years.
Answer: Son 15, father 45 years
EXAMATLAS LEVEL
Q. A father is 4 times as old as his daughter, and his son is 3 years older than the daughter. Four years from now their ages will add to 75. Find the present ages, and after how many years the father will be twice as old as the son.
Now the total is 75 − 12 = 63. With daughter d: 4d + d + (d + 3) = 63, so 6d = 60 and d = 10. Father 40, son 13. Twice: 40 + t = 2(13 + t), so t = 14 years.
Answer: Father 40, daughter 10, son 13; after 14 years
Pattern 2: When will one be twice (or k times) the other
[PYQ: NTPC UG CBT-1 28-Aug-2025 Shift-2]
EXAM LEVEL
Q. A father is 40 and his son 12. After how many years will the father be twice as old as the son?
40 + t = 2(12 + t), so t = 40 − 24 = 16 years.
Answer: 16 years
EXAMATLAS LEVEL
Q. Six years ago a father's age was 20 years more than twice his son's age. The father is now 50. After how many years will he be twice as old as his son?
Six years ago the father was 44, so 44 = 2S + 20 and the son was 12. Now the son is 18. 50 + t = 2(18 + t) gives t = 14 years.
Answer: 14 years
Pattern 3: Average age of a family or group over time
[PYQ: NTPC CBT-1 16-Jan-2021 Shift-2]
EXAM LEVEL
Q. The average age of a family of 5 is 24 years. A baby is born today. What is the new average?
The total stays 120 and the count becomes 6, so the average is 20 years.
Answer: 20 years
EXAMATLAS LEVEL
Q. Six years ago the average age of a family of 5 was 28 years. Two years later a baby was born. What is the family's average age now?
The five members now total 140 + 30 = 170 years. The baby is 4 years old. Total 174 over 6 people = 29 years.
Answer: 29 years
Pattern 4: Age when the other was born, and reversed digits
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. A was 24 years old when B was born. Now A is twice as old as B. Find their ages.
A − B = 24 always, and A = 2B, so B = 24 and A = 48 years.
Answer: A 48, B 24 years
EXAMATLAS LEVEL
Q. A father's age is his son's age with the digits reversed. One year ago the father was twice as old as the son. Find their present ages.
Let F = 10a + b and S = 10b + a. F − 1 = 2(S − 1) gives 10a + b − 1 = 20b + 2a − 2, so 8a = 19b − 1. Only b = 3 gives a digit: a = 7. Father 73, son 37. Check: 72 = 2 × 36.
Answer: 73 and 37 years
60-Second Revision
- F = kS now; shift both ages by the same years for past or future.
- Twice-as-old time: t = F − 2S.
- Group average rises by n in n years; a newborn adds 1 to the count only.
- Reversed digits: set up 10a + b and 10b + a and test digits.
Next Step: Ages done. Try the age questions in the ExamAtlas RRB NTPC 2026 mock tests; plug each option back into the statement and the right one fits in seconds.