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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
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
Coins For 79 Paise :25 paise × 3 coins + 2 paise × 2 coins
Total number of coins for 79 paise = 5
Coins
For 66 Paise :
50 paise × 1 coin + 10 paise × 1 coin + 2 paise × 3 coin
Total number of coins for 66 paise = 5 coins
For 1.01 Paise :
50 paise × 1 coin + 25 paise × 1 coin + 10 paise × 2 coin
+ 2 paise × 3 coin
Total number of coins for Rs. 1.01 paise = 7 coins
So, total coins used = (5 + 5 + 7) coins
= 17 coins
Ans.(A) (because given that smallest no. of coin)
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