1115.Kavita is twice as old as Sunita. Three years ago she was three times as old as Sunita. How old is Kavita now?#
- A6 years
- B12 years
- C18 years
- D24 years
All 26 Ages questions the Punjab PSC asked between 2015 and 2025, each with a worked answer and the exam it came from. First 10 free to read.
4 of these 26 were set in more than one exam — the Commission reuses this topic, which is the argument for working through it rather than skimming it.
Free to read, with the worked answer and the exam each one came from. The book prints this topic in the order the Commission favours, so these are the ones to know first.
Let Sunita's present age be s, so Kavita is 2s. > 2s − 3 = 3(s − 3) > 2s − 3 = 3s − 9 → s = 6 > Kavita = 2 × 6 = 12 Check: three years ago they were 9 and 3, and 9 = 3 × 3.
Let the son's present age be x. The father was x years old when the son was born, so the father is always x years older: 36 − x = x, giving x = 18. Five years back the boy was 18 − 5 = 13. > 36 − x = x → x = 18 > 18 − 5 = 13.
Let the daughter be D years. Father = 3D, mother = 3D − 9. The son is half the mother: (3D − 9)/2. He is also D + 7. > (3D − 9)/2 = D + 7 → 3D − 9 = 2D + 14 → D = 23 > mother = 3 × 23 − 9 = 60 The mother is 60 years old (father 69, son 30, daughter 23).
Let the elder son be x years, so the father is 2x and the younger son is x − 15. > 2x + 10 = 3[(x − 15) + 10] = 3x − 15 > x = 25 → father = 50 years Check: in ten years the father is 60 and the younger son (now 10) is 20, and 60 = 3 × 20.
Let Sudha's age 16 years ago be s, so her grandfather was 8s. > Eight years from now is 24 years later: 8s + 24 = 3(s + 24) > 5s = 48, so s = 9.6 > Eight years ago Sudha was 9.6 + 8 = 17.6 and he was 76.8 + 8 = 84.8 > 17.6 : 84.8 = 11 : 53
Let the ages be 5x and 7x. > 5x + 7x = 12x = 60 → x = 5 > younger = 5x = 25 years So the younger brother is 25 years old. The elder is 35.
When the daughter was born Maria's age equalled the daughter's present age, so Maria is now exactly twice her daughter's age. > daughter now = 58 ÷ 2 = 29 > 10 years back = 29 − 10 = 19 So 19 years.
> Let the ages be 3x and 4x > 3x + 4x = 7x = 49, so x = 7 > A = 3 × 7 = 21 A is 21 years old and B is 28.
Let the common interval be d years. The children's ages are 4, 4 + d, 4 + 2d, 4 + 3d and 4 + 4d. > Sum of children = 20 + 10d; with father: 60 + 20 + 10d = 160 → d = 8 > eldest = 4 + 4 × 8 = 36 So the eldest child is 36 years old.
Let Meenu's age in 2002 be x, so Meera's was 3x. Eight years later, in 2010, Meenu was half of Meera. > x + 8 = (3x + 8)/2 → 2x + 16 = 3x + 8 → x = 8 > Meenu was 8 in 2002, so she was born in 2002 − 8 = 1994
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Yes. 4 of these 26 were set in more than one exam, which is why they are worth working rather than reading. Each question here carries a badge showing how many sittings asked it.
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The exams behind these questions, most first. A paper here is a sitting: a question set in two papers of the same sitting counts once.
| Exam | Year | Questions |
|---|---|---|
| Assistant Director, Industrial Training Principal Junior Scale Industrial Traini | 2018 | 6 |
| Enquiry Officer | 2018 | 6 |
| Drugs Control Officer | 2018 | 6 |
| Senior Scientific Officer (Drugs) | 2018 | 6 |
| Senior Scientific Officer (Food Safety) Microbiology | 2018 | 6 |
| Analysts | 2019 | 3 |
| Extra Assistant Conservator of Forests in the the Department of Forest and Wild | 2019 | 3 |
| Works Manager | 2018 | 2 |
| Scientific Assistant Physics | 2018 | 2 |
| District Sainik Welfare Officer | 2016 | 2 |
| Punjab Civil Services (Prelims) — CSAT | 2015 | 2 |
| Punjab Civil Services (Prelims) — CSAT | 2025 | 1 |