SOLUTION: Joan has some dimes and quarters. If she has 30 coins worth a total of $4.95, how many of each type of coin does she have?

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Question 1050355: Joan has some dimes and quarters. If she has 30 coins worth a total of $4.95, how many of each type of coin does she have?
Found 2 solutions by josmiceli, advanced_Learner:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
20 quarters = $5, so too many quarters
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19 quarters = $4.75
2 dimes = $.20
only 21 coins
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18 quarters = $4.50
Can't make $4.95 with dimes
The number of quarters has to be odd
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17 quarters = $4.25
7 dimes = $.70
only 24 coins
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15 quarters = $3.75
12 dimes = $1.20
only 27 coins
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13 quarters = $3.25
17 dimes = $1.70
This is the answer

Answer by advanced_Learner(501) About Me  (Show Source):
You can put this solution on YOUR website!
D+Q=30
10D+25=495
Solved by pluggable solver: SOLVE linear system by SUBSTITUTION
Solve:
+system%28+%0D%0A++++1%5CD+%2B+1%5CQ+=+30%2C%0D%0A++++10%5CD+%2B+25%5CQ+=+495+%29%0D%0A++We'll use substitution. After moving 1*Q to the right, we get:
1%2AD+=+30+-+1%2AQ, or D+=+30%2F1+-+1%2AQ%2F1. Substitute that
into another equation:
10%2A%2830%2F1+-+1%2AQ%2F1%29+%2B+25%5CQ+=+495 and simplify: So, we know that Q=13. Since D+=+30%2F1+-+1%2AQ%2F1, D=17.

Answer: system%28+D=17%2C+Q=13+%29.