Aspire Faculty ID #10245 · Topic: NIMCET 2015 · Just now
NIMCET 2015

Using only 2, 5,10, 25 and 50 paise coins, what is the smallest number of coins required to pay exactly 78 paise, 69 paise and Rs. 1.01 to three different persons?

Solution

1) 78 paise

Take highest coin:
50 → remaining = 28
28 = 10 + 10 + 2 + 2 + 2 + 2

Coins used:
50, 10, 10, 2, 2, 2, 2
Number of coins = 7

2) 69 paise

50 → remaining = 19
19 = 10 + 5 + 2 + 2

Coins used:
50, 10, 5, 2, 2
Number of coins = 5

3) Rs. 1.01 = 101 paise

50 + 25 = 75 → remaining = 26
26 = 10 + 10 + 2 + 2 + 2

Coins used:
50, 25, 10, 10, 2, 2, 2
Number of coins = 7

Total minimum coins:

78 paise → 7 coins
69 paise → 5 coins
101 paise → 7 coins

Total = 7 + 5 + 7 = 19 coins

Answer: 19 coins

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