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CSAT 2019 paper

UPSC CSE CSAT 2019 · Paper II (CSAT) · Question 18 · Arithmetic: percentage, ratio, averages, time & work

In 2002, Meenu’s age was one-third of the age of Meera, whereas in 2010, Meenu’s age was half the…

UPSC Civil Services (Preliminary) Examination 2019, General Studies Paper II (CSAT), Question 18 — Arithmetic: percentage, ratio, averages, time & work.

In 2002, Meenu’s age was one-third of the age of Meera, whereas in 2010, Meenu’s age was half the age of Meera. What is Meenu’s year of birth?

Answer & explanation

Answer: (b) 1994

Writing Meera's 2002 age as three times Meenu's and applying the 'half' relation eight years later gives Meenu's 2002 age as 8, so she was born in 1994.

  1. Let Meenu be m years old in 2002; Meera is then 3m.
  2. In 2010 both are 8 years older: m + 8 = (3m + 8)/2.
  3. 2m + 16 = 3m + 8, so m = 8.
  4. Meenu was 8 in 2002, so she was born in 2002 − 8 = 1994.
  5. Check: 2002 — 8 and 24 (one-third); 2010 — 16 and 32 (half).

Remember · Both ages move by the same number of years; write them with one variable and shift both together.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·

Cite this question: UPSC CSE CSAT 2019, Q18 — minimalistias.com/pyqs/csat/2019/q18

In the UPSC syllabus: Arithmetic: percentage, ratio, averages, time & work › Averages & ages See it and tick it covered →

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