SOLUTION: RAY HAS $7.60 IN QUARTERS AND DIMES. IN ALL, HE HAS 40 COINS. HOW MANY COINS OF EACH KIND DOES HE HAVE?
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Question 108430: RAY HAS $7.60 IN QUARTERS AND DIMES. IN ALL, HE HAS 40 COINS. HOW MANY COINS OF EACH KIND DOES HE HAVE?
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
D+Q=40 OR D=40-Q
.10D+.25Q=7.60
.10(40-Q)+.25Q=7.60
4-.10Q+.25Q=7.60
.15Q=7.60-4
.15Q=3.60
Q=3.60/.15
Q=24 NUMBER OF QUARTERS
40-24=16 NUMBER OF DIMES.
PROOF
.10*16+24*.25=7.60
1.60+6=7.60
7.60=7.60
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