SOLUTION: Tommy change the five dollar bill for dimes and quarters. He got eight more dimes and quarters. How many of each coin did he get?

Question 1124525: Tommy change the five dollar bill for dimes and quarters. He got eight more dimes and quarters. How many of each coin did he get?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Solution #1, using logical reasoning....

(1) Count the "extra" 8 dimes. Their total value is 80 cents; there is $4.20 (420 cents) remaining; there are now equal numbers of dimes and quarters.
(2) The value of one quarter and one dime is 35 cents.
(3) 420/35 = 12; that means the remaining coins are 12 groups each consisting of one quarter and one dime.
(4) So the number of quarters is 12; and the number of dimes, including the 8 we counted to start, is 12+8 = 20.

CHECK: 12(25)+20(10) = 300+200 = 500

Solution #2, using formal algebra....

I show both solutions so that you can see the formal algebra does virtually the same thing the logical reasoning does. Certainly you want to learn the methods of algebra so you can use them to solve more difficult problems. But solving problems like this using informal logical reasoning is good brain exercise.

Let x = number of quarters
then x+8 = number of dimes

25(x)+10(x+8) = 500
25x+10x+80 = 500
35x = 420
x = 420/35 = 12

The number of quarters is x=12; the number of dimes is x+8 = 20.
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