SOLUTION: Coin problem: a bank teller had 135 quarters, dimes and nickels in his till totaling $23.50. There were 3 times as many quarters as nickels. How many of each coin were there? I tr
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Question 110884: Coin problem: a bank teller had 135 quarters, dimes and nickels in his till totaling $23.50. There were 3 times as many quarters as nickels. How many of each coin were there? I tried N=3Q. N+D+Q = 135. 3Q+D+Q=135. 4Q+D=135. D=135-4Q. 5N+10D+25Q=2350. 5(3Q)+10(135-4Q)+25Q=2350. Once I simplify the expression I keep getting 0Q=1000. Please Help. Thanks
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
"----There were 3 times as many quarters as nickels----" I BELIEVE THIS WOULD BE WRITTEN AS: Q=3N
Let Q=number of quarters
D=number of dimes
N=number of nickels
N+D+Q=135----------------------------eq1
5N+10D+25Q=2350--------------------------eq2
Q=3N--------------------------------------eq3
Multiply eq1 by 10 and subtract it from eq2 and we get
5N+10D+25Q=2350
(-)10N+10D+10Q=1350
-5N+15Q=1000 Now substitute Q=3N from eq3 into this
-5N+15(3N)=1000
-5N+45N=1000
40N=1000
N=25----------------------------number of nickels
Q=3N
Q=3*25=75------------------------number of quarters
From eq1:
N+D+Q=135
25+D+75=135
D=35------------------------------------number of dimes
CK
10*35+75*25+25*5=2350
350+1875+125=2350
2350=2350
Hope this helps---ptaylor
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