SOLUTION: A bag has 274 coins of nickels, dimes and quarters. There are twice as many nickels as quarters. If the total value of the coins was $30.3, how many nickels, dimes and quarters wer

Question 1196588: A bag has 274 coins of nickels, dimes and quarters. There are twice as many nickels as quarters. If the total value of the coins was $30.3, how many nickels, dimes and quarters were there?

Found 3 solutions by ikleyn, josgarithmetic, MathTherapy:
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
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x quarters; 2x nickels, and (274-x-2x) = (274-3x) dimes. Total money equation is 25x + 5*(2x) + 10*(274-3x) = 3030 cents. Simplify and solve 25x + 10x + 2740 - 30x = 3030 5x = 3030 - 2740 = 290 x = 290/5 = 58. ANSWER. 58 quarters; 2*58 = 116 nickels and the rest coins, 274-58-116 = 100, are dimes. CHECK. 25*58 + 5*116 + 10*100 = 3030 cents, in total. ! Correct !

Solved.

The lesson to learn
        this problem is to solve using one equation in one single unknown.

Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!

COIN COIN COUNTS MONEY(dollars) Nickels 2Q 0.05*2Q Dimes 274-2Q-Q=274-3Q 0.1*(274-3Q) Quarters Q 0.25Q TOTAL 30.30

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and counts for nickels and dimes are easily found.
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

A bag has 274 coins of nickels, dimes and quarters. There are twice as many nickels as quarters. If the total value of the coins was $30.3, how many nickels, dimes and quarters were there?

Let number of quarters be Q Then number of nickels = 2Q Also, number of dimes = 274 - Q - 2Q = 274 - 3Q We then get the following VALUE equation: .25Q + .05(2Q) + .1(274 - 3Q) = 30.3 .25Q + .1Q + 27.4 - .3Q = 30.3 .25Q + .1Q - .3Q = 30.3 - 27.4 .05Q = 2.9 Number of quarters or Number of nickels: 2(58) = 116 Number of dimes: 274 - 3(58) = 274 - 174 = 100

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