Question 1097741: Kobe went to local bank to get $20 in quarters, dimes and nickels for his restaurant. He got twice as many quarters as dimes. Also, he got 10 more nickels than dimes. How many of each type of coin did he get?
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
First an informal solution, using logical analysis and a little arithmetic....
He has some number of dimes, that same number plus 10 more of nickels, and twice that number of quarters.
So "take away" those 10 "extra" nickels, with a value fo 50 cents. The remaining coins then have a value of $19.50.
The remaining coins are equal numbers of dimes and nickels, and twice that many quarters. So make "units" each consisting of 2 quarters, 1 dime, and 1 nickel. The value of each of those units is 65 cents.
The number of those units required to make $19.50 is 19.50/.65 = 30.
Those 30 "units" give you 60 quarters, 30 dimes, and 30 nickels.
Now add back in those "extra" 10 nickels.
You end up with 60 quarters, 30 dimes, and 40 nickels.
Now a formal solution using algebra....
let x = number of dimes
then 2x = number of quarters
and x+10 = number of nickels
Since the total value is $20, or 2000 cents,
# of dimes: x = 30
# of quarters: 2x = 60
# of nickels: x+10 = 40
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