UPSC Prelims 2014 · CSAT
Question 25: Basic Numeracy
Six identical cards are placed on a table. Each card has number '1' marked on one side and number '2' marked on its other side. All the six cards are placed in such a manner that the number '1' is on the upper side. In one try, exactly four (neither more nor less) cards are turned upside down. In how many least number of tries can the cards be turned upside down such that all the six cards show number '2' on the upper side ?
AnswerUPSC’s answer is (a).
Each card must be turned an odd number of times. Two tries give 8 turns, but no card can be turned more than twice, so each would be turned exactly once: only 6 turns, not 8. So two tries fail. With three tries there are 12 turns: three cards turned in all three tries (3 turns each) and the other three cards turned once each. Turn cards 1, 2 and 3 every time, plus card 4, then 5, then 6.
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.
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) Right. Cards 1, 2, 3 in every try, plus one of cards 4, 5, 6 each time.
- (b) Five is more than the minimum; three tries are enough.
- (c) Seven is more than the minimum; three tries are enough.
- (d) It can be achieved, as the three-try plan shows.
Answer: UPSC’s key. Explanation: written by UPSC Grade, not by UPSC.
Source: UPSC, Civil Services (Preliminary) Examination 2014, General Studies Paper II (CSAT), Series A. The paper on upsc.gov.in.