UPSC Prelims 2009 · General Studies
Question 99: Basic Numeracy
There are four persons A, B, C, D; and A has some coins. A gave half of the coins to B and 4 more besides. B gave half of the coins to C and 4 more besides. C gave half of the coins to D and 4 more besides. Both B and D end up with same number of coins. How many coins did A have originally?
AnswerThe answer is (c).
Let A have x coins. B gets x/2 + 4. B keeps half of that minus 4: x/4 - 2, and passes on x/4 + 6 to C. C passes half of that plus 4 to D, which is x/8 + 7. B and D end equal: x/4 - 2 = x/8 + 7, so x/8 = 9 and x = 72. Check: B gets 40, keeps 16; C gets 24, gives 16 to D.
How you could have got this
No shortcut here: in a “how many” question, ruling out options doesn’t help. Every statement has to be judged.
This follows an unofficial answer: UPSC has not published its key for this year, and if that answer is wrong, so is this path.
Which options go is worked out from UPSC’s key and the options, not by a model.
Why each option is right or wrong
- (a) 96 leaves B with 22 coins and D with 19, not equal.
- (b) 84 gives C 27 coins, an odd number that cannot be halved.
- (c) Right: 72 leaves B and D with 16 each.
- (d) 64 leaves B with 14 coins and D with 15, not equal.
Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2009, General Studies. UPSC's own paper isn't online; the words are from a published compilation. The paper on upsc.gov.in.