NIMCET 2019 Question Paper with solutions

Question : 78 of 120

Marks: +1, -0

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

Validate

Solution:

For 78 paise.

1×50+2×10+4×2=7

coins
For 69 paise,

1×50+1×10+1×5+2×2=5

coins
For 101 paise,

1×50+1×25+2×10+3×2=7

coins
Hence, required total number of coins

=7+5+7=19

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