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! .
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): 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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