Question 898072: you found $6.60 on the ground at school, all in nickels, dimes, and quarters. you have twice as many quarters as dimes and 42 coins in all. how many of each type of coin do you have?
Found 2 solutions by josgarithmetic, richwmiller:
Answer by josgarithmetic(39614)   (Show Source):
You can put this solution on YOUR website! Look at richwmiller's solution which is a correct solution. The solution I tried to present shows the kind of mess that can happen when someone (me,...) does not check the problem description carefully:
p,n,d,q for pennies nickels dimes quarters.
p+n+d+q=42;
d=2q;
0.01p+0.05n+0.10d+0.25q=6.60.
Simplify the money equation first. Multiply by 100.
 .
Use the d & q equation to eliminate d in the coin count and money count equations.
This system may have more than one solution. Notice from the initial set of equations, three equations but four unknown variables. The simplified system, two equations but three unknown variables.
Try assuming for constant value k, that k=q.
 and  .
Add  ;
You need WHOLE number values for k and n. Look for all the useful values for k, meaning q, that will work in this system and also in the original system. Realize you cannot have more than about k=20. Why?  , ignoring remainder, but you know this will not leave enough coins.
Work with it; and see what you find.
k=25?
k=24?
k=23?
k=22?
..
.
.
?
You should find that k must be less than 15.
.
.
?
Answer by richwmiller(17219)  (Show Source):
You can put this solution on YOUR website! There are no pennies so we have three equations in three unknowns.
The other tutor also had the number of quarters and dimes reversed
n+d+q=42,
2d=q,
5n+10d+25q=660
d = 10, n = 12, q = 20
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