SOLUTION: Jim has a total of 50 coins. Some are nickels and some are quarters. Altogether he has $4.10. How many quarters does he have? How many nickels?

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Question 312492: Jim has a total of 50 coins. Some are nickels and some are quarters. Altogether he has $4.10. How many quarters does he have? How many nickels?
Found 2 solutions by checkley77, ankor@dixie-net.com:
Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website!
N+Q=50 OR N=50-Q
.05N+.25Q=4.10
.05(50-Q)+.25Q=4.10
2.50-.05Q+.25Q=4.10
.20Q=4.10-2.50
.20Q=1.60
Q=1.60/.20
Q=8 QUARTERS.
50-8=42 NICKELS.
PROOF:
.05*42+8*.25=4.10
2.10+2=4.10
4.10=4.10

Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
Write an equation for each statement; n = no.of nickels; q = no. of quarters
:
"Jim has a total of 50 coins."
n + q = 50
n = (50-q); use this form for substitution
:
"Some are nickels and some are quarters. Altogether he has $4.10."
.05n + .25q = 4.10
Substitute (50-q) for n
.05(50-q) + .25q = 4.10
:
2.5 - .05q + .25q = 4.10
:
-.05q + .25q = 4.10 - 2.50
:
.20q = 1.60
q =
q = 8 quarters
then
50 - 8 = 42 nickels
:
:
Check solution:
42(.05) + 8(.25)
2.10 + 2.00 = 4.10