Question 920612

Hello, i am having problems setting up this problem that consists of coins.

1. The amount of money in a bag is $7.70. There are 62 coins consisting of nickels, dimes, and quarters. If there are three times as many dimes as nickels, how many coins of each are there?
<pre>
Let number of nickels, and quarters, be N, and Q, respectively
Since there are 3 times as many dimes as nickels, then: D = 3N 
N + 3N + Q = 62 ------- Equation expressing the number of coins
    4N + Q = 62 ------- eq (i)

.05N + .1(3N) + .25Q = 7.7 ----- Equation expressing the value  of all the coins
.05N + .3N + .25Q = 7.7
      .35N + .25Q = 7.7 -------- eq (ii)

You now have 2 equations (i & ii) that you can solve. My choice would be either:
1) to solve equation (i) for Q, and substitute that value for Q in eq (ii), or
2) to apply the elimination method of solving simultaneous equations
Either way, the value derived for N: the number of nickels, MUST BE an INTEGER,
and so MUST the value for Q: the number of quarters.

Then do a check!! 
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