SOLUTION: OK, I am banging my head on my desk over this one! I have 27 coins in nickles, dimes and quarters. There are five more dimes than nickles. The total value of the coins is $3.55.

Question 76021: OK, I am banging my head on my desk over this one!
I have 27 coins in nickles, dimes and quarters. There are five more dimes than nickles. The total value of the coins is $3.55. What is the number of quarters?
Ugg! Please help, thank you
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Write a simple equation for each sentence:
;
"I have 27 coins in nickles, dimes and quarters."
n + d + q = 27
:
"There are five more dimes than nickels."
d = (n + 5)
:
"The total value of the coins is $3.55."
.05n + .10n + .25q = 3.55
:
Let's try to get it down to two unknowns, substitute (n+5) for d in the 1st eq;
n + (n+5) + q = 27
2n + q = 27 - 5
2n + q = 22
:
Do the same with $$ equation, substitute (n+5) for d:
.05n + .10(n+5) + .25q = 3.55
.05n + .10n + .5 + .25q = 3.55
.15n + .25q = 3.55 - .5
.15n + .25q = 3.05
:
Mult the above equation by 4 (makes.25q = 1q) and subtract it from 2n + q + 22
2n + q = 22
.6n + q = 12.2
---------------- subtracting eliminates q
1.4n + 0 = 9.8
n = 9.8/1.4
n = 7 nickels
:
Remember we were given: d = n + 5:
d = 7 + 5
d = 12 nickels
:
" What is the number of quarters?"
Substitute for n & d in the 1st equation n + d + q = 27; find q
7 + 12 + q = 27
19 + q = 27
q = 27 - 19
q = 8 quarters
:
:
Check our solutions in the the $$ equation:
.05(7) + .10(12) + .25(8) =
.35 + 1.20 + 2.00 = 3.55 as given
:
A lot of steps, but each one is pretty simple, could you follow this OK?
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