SOLUTION: A collection of 18 coins, each of which is a nickel,dime or quarter, is worth $2.6. If there are 2 more quarters than nickels, how many of each are there?

Question 1036071: A collection of 18 coins, each of which is a nickel,dime or quarter, is worth $2.6. If there are 2 more quarters than nickels, how many of each are there?
Found 2 solutions by stanbon, fractalier:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A collection of 18 coins, each of which is a nickel,dime or quarter, is worth $2.6. If there are 2 more quarters than nickels, how many of each are there?
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n + d + q = 18 coins
5n + 10d + 25q = 260 cents
q = n + 2
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Substitute for "q" and modify to get:
n + d + n+2 = 18
n + 2d + 5(n+2) = 52
=======
2n + d = 16
6n + 2d = 42
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2n + d = 16
3n + d = 21
Subtract and solve for "n"::
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n = 5 (# of nickels)
q = n+2 = 7 (# of nickels)
d = 18 -n-q = 18-12 = 6 (# of dimes)
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Cheers,
Stan H.
Answer by fractalier(6550) (Show Source): You can put this solution on YOUR website!
If we call their numbers n, d, and q, we can write
n + d + q = 18
and the value equation is
5n + 10d + 25q = 260 (we work in cents) and we also have
q = n + 2
let us substitute this third fact into the other two and get
n + d + n+2 = 18 and
5n + 10d + 25(n+2) = 260
which gives us
2n + d = 16 and
30n + 10d = 210
If we divide this second one by ten we can then subtract the first one from it...like this...
3n + d = 21
-(2n + d = 16)
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n = 5 so that
q = 7 and then
d = 6
5 nickels = 25 cents
7 quarters = 175 cents
6 dimes = 60 cents
Total = 260 cents
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