SOLUTION: A parking meter contains quarters and dimes worth ​$8.50. There are 46 coins in all. Find how many of each there are.

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Question 1162726:
A parking meter contains quarters and dimes worth ​$8.50. There are 46 coins in all. Find how many of each there are.
Found 3 solutions by ikleyn, math_helper, greenestamps:
Answer by ikleyn(52793) About Me  (Show Source):
You can put this solution on YOUR website!
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https://www.algebra.com/algebra/homework/word/age/Age_Word_Problems.faq.question.1162718.html



Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!



Q + D = 46   (46 coins in all)

25Q + 10D = 850   (total worth is $8.50 or 850 cents)



One approach:

Solve the top equation for Q or D then substitute into the bottom equation

then solve that for the unknown variable.  Once you do that, plug into top

equation and solve for the other variable.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


An quick and easy informal solution, if formal algebra is not required....

(1) Imagine 46 dimes; the value of the 46 coins would be $4.60. The actual value is $8.50, which is $3.90 more than $4.60.
(2) Now exchange a dime for a quarter. There are still 46 coins, but the total value has increased by (25-10) = 15 cents.
(3) The number of times you need to do that to make up the additional $3.90 (390 cents) is 390/15 = 26.

ANSWER: 26 quarters and 46-26=20 dimes.

CHECK: 26($0.25)+20($0.10) = $6.50+$2.00 = $8.50