UPSC Prelims 2022·CSAT·Quantitative Aptitude·Algebra

A has some coins. He gives half of the coins and 2 more to B. B gives half of the coins and 2 more to C. C gives half of the coins and 2 more to D. The number of coins D has now, is the smallest two-digit number. How many coins does A have in the beginning?

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Last updated 23 May 2026, 3:31 pm IST
  1. A76
  2. B68
  3. C60
  4. D52Correct

Explanation

Here's a step-by-step explanation working backward: 1. Determine D's coins: The smallest two-digit number is 10. So, D has 10 coins. 2. Calculate C's coins before giving to D: C gave half of his coins and 2 more to D, resulting in D having 10 coins. Let C's coins be 'x'. C gave (x/2) + 2 to D. So, (x/2) + 2 = 10 x/2 = 10 - 2 x/2 = 8 x = 8 * 2 = 16. C had 16 coins before giving to D. 3. Calculate B's coins before giving to C: B gave half of his coins and 2 more to C, resulting in C having 16 coins. Let B's coins be 'y'. B gave (y/2) + 2 to C. So, (y/2) + 2 = 16 y/2 = 16 - 2 y/2 = 14 y = 14 * 2 = 28. B had 28 coins before giving to C. 4. Calculate A's initial coins: A gave half of his coins and 2 more to B, resulting in B having 28 coins. Let A's initial coins be 'z'. A gave (z/2) + 2 to B. So, (z/2) + 2 = 28 z/2 = 28 - 2 z/2 = 26 z = 26 * 2 = 52. A had 52 coins in the beginning. Therefore, A had 52 coins in the beginning. The final answer is D) 52.
Quantitative Aptitude: A has some coins. He gives half of the coins and 2 more to B. B gives half of the coins and 2 more to C. C gives half of
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