Question 1177083
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(1) An algebraic setup using two variables....<br>
Let d = # of dimes
Let q = # of quarters<br>
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)<br>
Solve using either substitution or elimination....<br>
(2) An algebraic setup using a single variable....<br>
Let d = # of dimes
Then 70-d = # of quarters  (because the total number of coins is 70)<br>
The total value of the coins is $11.65:<br>
10(d)+25(70-d) = 1165<br>
Solve using basic algebra.<br>
(3) a quick mental solution, if formal algebra is not required, and/or if getting the answer quickly is advantageous....<br>
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.<br>
ANSWER: 31 quarters and 39 dimes.<br>
CHECK: 31(25)+39(10) = 775+390 = 1165<br>