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Previous Year Question (PYQs)

Using only 2, 5, 10, 25 and 50 paise coins, the smallest number of coins required to pay exactly 79 paise, 66 paise and Re 1.01 to three different persons is




Solution

Minimum Coins Required

Coins available: 2p, 5p, 10p, 25p, 50p

1) For 79 paise: 50p(1) + 25p(1) + 2p(2) = 4 coins

2) For 66 paise: 50p(1) + 10p(1) + 2p(3) = 5 coins

3) For Re 1.01 (101 paise): 50p(1) + 25p(2) + 10p(2) + 2p(3) = 8 coins
Total = 4+5+8=17 coins


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