Coins Collections

Question: How to solve these coin problems?

(Posted by: AnonymousHOE. on 2010-02-26 18:24:48)

A valuable collection of coins contained old nickels,dimes,quarters and pennies. The face value of the pennies was $8. There were seven more than three times as many quarters as dimes and sixteen less than twice as many nickels as quarters. If the face value of the entire collection was $38.40,how many of each kind of coin was there?

Posted by: hayharbr on 2010-02-26,18:54:31

You know $8 in pennies is 800 pennies. That leaves $30.40, or 3040¢ for the other coins. If d = number of dimes, worth 10d ¢, 3d + 7 = number of quarters worth 25(3d + 7) ¢ and 2(3d + 7) - 16 = number of nickels = 6d + 14 - 16 = 6d - 2 worth 5(6d - 2)¢ . So 10d + 25(3d + 7) + 5(6d - 2) = 3040 10d + 75d + 175 + 30d - 10 = 3040 115d + 165 = 3040 115d = 2875 so d = 2875 ÷ 115 See what that is then use it to answer the question.