Question 1150414: erin has a total of thirty coins worth $4.25. All of the coins are either nickels, dimes, or quarters. Of the number of quarters is the same as the number of dimes, how many of each kind of coin does she have?
Found 3 solutions by Alan3354, josmiceli, ikleyn:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! erin has a total of thirty coins worth $4.25. All of the coins are either nickels, dimes, or quarters. Of the number of quarters is the same as the number of dimes, how many of each kind of coin does she have?
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n + d + q = 30
q = d
---> n + 2d = 30
================================
5n + 10d + 25q = 425
q = d
---> 5n + 35d = 425
=============================
n + 2d = 30
n + 7d = 85
------------------------ Subtract
-5d = -55
d = q = 11
n = 8
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8*5 + 11*10 + 11*25 = 425
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It can be solved with a single unknown/variable, but so what?
There is no charge for variables. What is the advantage?
Answer by josmiceli(19441)  (Show Source):
Answer by ikleyn(52902)  (Show Source):
You can put this solution on YOUR website! .
It can be solved using one single unknown and one equation.
Let x be the number of dimes.
Then the number of quarters is the same, x.
The number of nickels is (30-x-x) = (30-2x).
The "money" equation is
5*(30-2x) + 10x + 25*x = 425 cents.
Simplify step by step
150 - 10x + 10x + 25x = 425
25x = 425 - 150 = 275.
x = 275/25 = 11.
ANSWER. 11 dimes, 11 quarters and 30-2*11 = 8 nickels.
CHECK. 11*10 + 11*25 + 8*5 = 425 cents. ! Precosely correct !
Solved.
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The lesson to learn:
This problem is to be solved using one single unknown and one equation.
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In his post, Alan notes
It can be solved with a single unknown/variable, but so what?
There is no charge for variables. What is the advantage?
Alan, it is true, variables are for free; but the knowledge on how to solve the problem correctly, think correctly
and in a right way   .
* * * And it is what I teach students for : to think CORRECTLY. * * *
The advantage is in that even the 6-th and 7-th grade students can solve such problems easily, although they do not know
yet about systems of equations.
Alan, when the person who positioned himself as a tutor, asks such questions as you do, he does not look good . . .
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