Question 223797: A mixture of dimes and quarters has a total value of $10.20. There are 57 coins in all. How many of each type are present? Answer by drj(1380) (Show Source):
You can put this solution on YOUR website! A mixture of dimes and quarters has a total value of $10.20. There are 57 coins in all. How many of each type are present?
Step 1. Let x be the number of quarters.
Step 2. Let 0.25x be the dollar value of quarters.
Step 3. Let y be the number of dimes.
Step 4. Let 0.10y be the dollar value of dimes.
Step 5. Then 0.25x+0.10y=10.20 be the total dollar value.
Step 6. Also, x+y=57 since there are 57 coins
Step 7. Our linear system of equation is given in Steps 4 and 5 is shown below:
Solve: We'll use substitution. After moving 0.1*y to the right, we get: , or . Substitute that
into another equation: and simplify: So, we know that y=27. Since , x=30.
Answer: .
So x=30 and y=27. The difference is 3 and the total dollar value is 0.25*30+0.10*27=7.50+2.70=10.20 which is a true statement. And the total number of coins is 57.
Step 8. ANSWER: The number of quarters is 30 and the number of dimes is 27.
I hope the above steps were helpful.
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