Question 1177083: “A purse contains $11.65 in quarters (25 cents) and dimes(10cents). If the total number of
coins is 70, find how many quarters and dimes are there.”
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
(1) An algebraic setup using two variables....
Let d = # of dimes
Let q = # of quarters
d+q=70 (the total number of coins is 70)
10d+25q = 1165 (the total value of the coins -- 10 cents for each dime and 25 cents for each quarter -- is $11.65, or 1165 cents)
Solve using either substitution or elimination....
(2) An algebraic setup using a single variable....
Let d = # of dimes
Then 70-d = # of quarters (because the total number of coins is 70)
The total value of the coins is $11.65:
10(d)+25(70-d) = 1165
Solve using basic algebra.
(3) a quick mental solution, if formal algebra is not required, and/or if getting the answer quickly is advantageous....
If all 70 coins were dimes, the value would be 700 cents; the actual value is 465 cents more than that.
Exchanging a dime for a quarter keeps the number of coins at 70 but increases the total value by 25-10 = 15 cents.
The number of times a dime needs to be exchanged for a quarter to make up the addition 465 cents is 465/15 = 31.
So there must be 31 quarters, which means 70-31 = 39 dimes.
ANSWER: 31 quarters and 39 dimes.
CHECK: 31(25)+39(10) = 775+390 = 1165
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