SOLUTION: a pay phone is holding its maximum number of 500 coins consisting of nickels, dimes and quarters. the number of quarters is twice the number of dimes. if the value of all the coins

Question 190515: a pay phone is holding its maximum number of 500 coins consisting of nickels, dimes and quarters. the number of quarters is twice the number of dimes. if the value of all the coins is $88 how many nickels were in the pay phone?
Answer by jonvaliente(64) (Show Source): You can put this solution on YOUR website!
Let d=no of dimes in the pay phone
. 2*d=no of quarters ("the number of quarters is twice the number of dimes")
. n=no of nickels
.
We know that all coins total 500, so:
or..

.
We also know that the value of the all the coins is $88, so:

.
Simplifying, we get:

.
Let's try to express n as a function of d:
Subtract 0.6d from both sides:
.

.
Divide both sides by 0.05:
.

.
We now have n expressed as a function of d.
Substitute this value of n, in the first equation:
.

.
Subtract 1760 from both sides:
.

.
Divide both sides by -9:
.

.
We have 140 dimes, 140*2=280 quarters, 500-140-280=80 nickels.
.
To check:
140*(0.1)+280*(0.25)+80*(0.05)=?88
14+70+4=?88
88=88
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