Valuable Nickels

by admin on March 4, 2010

valuable nickels valuable nickels
How to solve these coin problems?

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?

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.

Magic and Innovation: Transforming Nickel Ideas to Dollar Innovations

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