Question 17129: Find 5 coins with a value of 57 cents. You must use quarters, dimes, nickles, and pennies.
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! I'm thinking through this problem with you. Bear with me, and see how the reasoning sounds:
First of all, notice that you have quarters, dimes, nickels and pennies, and the 57 cents is two more than a multiple of 5. Therefore, there must be 2 pennies, with 55 cents in quarters, dimes, and nickels. (Note: you cannot have more than 2 pennies, since 7 pennies would be more than 5 coins!!)
There can only be 1 quarter since two quarters would not leave enough to have at least one dime and nickel. Subtract out 25 cents from 55 cents, and you have 30 cents left over in dimes and nickels.
So far you have 2 pennies and a quarter, which accounts for 3 of the 5 coins. However, only 2 coins cannot give you 30 cents. So this is NOT possible. Perhaps my assumption that there is at LEAST one nickel and dime is in error.
So, let's try 2 quarters, 2 pennies, which accounts for 4 of the 5 coins and a value of 52 cents out of the 55 cents. Only one coin remains, which would have to be a nickel.
Final answer: 2 pennies, 2 quarters, 1 nickel, and no dimes.
Check: 5 coins and 57 cents.
R^2 at SCC
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