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| 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(52879)
      (Show Source): 
You can put this solution on YOUR website! . 
 
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 learnthis problem is to solve using one equation in one single unknown.
 
 
Answer by josgarithmetic(39630)
      (Show Source): Answer by MathTherapy(10556)
      (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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