UPSC Civil Services · 2016 · Prelims CSAT (Paper II) · Q27 · Past-year questions with answers

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Arrangements, Relations & Puzzles · 2016 · Prelims CSAT (Paper II) · Q27

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PYQ · UPSC Civil Services 2016 · Prelims CSAT (Paper II) · Q27

Arrangements, Relations & Puzzles› Arithmetical Reasoning & Puzzles

Four friends A, B, C and D need to cross a bridge. A maximum of two persons can cross it at a time. It is night and they just have one lamp. Persons that cross the bridge must carry the lamp to find the way. A pair must walk together at the speed of slower person. After crossing the bridge, the person having faster speed in the pair will return with the lamp each time to accompany another person in the group. Finally, the lamp has to be returned at the original place and the person who returns the lamp has to cross the bridge again without lamp. To cross the bridge, the time taken by them is as follows : A: 1 minute, B: 2 minutes, C: 7 minutes and D: 10 minutes. What is the total minimum time required by all the friends to cross the bridge ?

Answer a.

Explanation Answer · a

(a) 23 minutes. With A shuttling every time, all four are across in 21 minutes, and the extra rule — the lamp goes back and its carrier crosses again — adds two more 1-minute trips by A. First read the added rule carefully: the person who returns must be the faster member of the pair that has just crossed. That switches off the memorised shortcut of sending the two slowest together, because if C and D crossed as a pair, C would have to walk back and cost 7 minutes. So A, the fastest, becomes the permanent ferry. Step 1: A and B cross together — a pair moves at the slower person's speed, so 2 minutes. A returns with the lamp, 1 minute. Running total 3; B is across. Step 2: A and C cross, 7 minutes; A returns, 1 minute. Running total 11; B and C are across. Step 3: A and D cross, 10 minutes. Running total 21; all four are now across. Step 4: the lamp is still on the far side, so A carries it back, 1 minute (total 22), leaves it at the original place, and crosses again without it, 1 minute (total 23). The order of the middle pairings does not matter — A with D first and B last also gives 10 + 1 + 7 + 1 + 2 = 21. Sending C and D together instead pushes the crossings to 27 minutes, so 21 + 2 = 23 is the minimum.

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