Question 113454This question is from textbook algebra structure and method book 1
: leo has $4.00 IN DIMES AND QUARTERS. HE HAS 6 MORE DIMES THAN QUARTERS. HOW MANY QUARTERS DOES HE HAVE?
This question is from textbook algebra structure and method book 1
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! In doing money problems you need to keep track of the count of the coins and their value. Any easy way to think of this is that the number of dimes can be called d. The value of the times is 10*d. The number of quarters can be called q, and the value is 25*q. In both cases, the values are in cents.
So, we have dimes and quarters valued at $4.00. (given)
That can be depicted as 10d + 25q = 400 (Recall $4 = 400 cents.)
We also are told that the number of dimes is 6 more than the number of quarters.
So, we can say d = q+6.
Substituting, we have:
10(q+6) + 25q = 400
Multiply to remove the parenthesis:
10q + 60 + 25q = 400
Simplify by adding the q terms and subtracting 60 from both sides:
35q = 340
Unfortunately, at this point we do not have an integral solution because the number of quarters, q, is not a whole number. You cannot have a fraction of a quarter, so there is no solution.
However, if the total were to have been $4.10 (410 cents), then the problem works out fine.
35q = 350
q = 10
d = q+6 = 16
10(16) + 25(10) = 160 + 250 = 410.
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