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UPSC CSE CSAT 2019 · Paper II (CSAT) · Question 40 · Arithmetic: percentage, ratio, averages, time & work

दिया गया है कि 100 विद्यार्थियों का औसत अंक 40 है । बाद में यह पाया गया कि एक विद्यार्थी का अंक 53…

UPSC Civil Services (Preliminary) Examination 2019, General Studies Paper II (CSAT), Question 40 — Arithmetic: percentage, ratio, averages, time & work.

दिया गया है कि 100 विद्यार्थियों का औसत अंक 40 है । बाद में यह पाया गया कि एक विद्यार्थी का अंक 53 था जिसे भूल से 83 पढ़ा गया । संशोधित औसत अंक कितना है ?

Read this question in English

The average marks of 100 students are given to be 40. It was found later that marks of one student were 53 which were misread as 83. The corrected mean marks are

(a) 39 · (b) 39.7 · (c) 40 · (d) 40.3

उत्तर और व्याख्या · explanation in English

Answer: (b) 39·7

The total was overstated by 83 − 53 = 30 marks, so the correct total is 3970 and the mean is 39.7.

  1. Recorded total = 100 × 40 = 4000.
  2. Excess from the misreading = 83 − 53 = 30.
  3. Correct total = 4000 − 30 = 3970; correct mean = 3970/100 = 39.7.
  4. Shortcut: the mean falls by 30/100 = 0.3.

Remember · One wrong entry shifts the mean by (wrong − right) ÷ n.

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, Q40 — minimalistias.com/pyqs/csat/2019/q40

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