Question 120443: mr m has 3 times as many quarters as dimes, if he has a total of $6.80 in quarters and dimes, how many of each does he have.
i know this is an easy question i just dont know how to do it algebraically
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Let D be the number of dimes and Q be the number of quarters.
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Since there are 3 times as many quarters as dimes, if you multiply the number of dimes by 3
the answer will be the same as the number of quarters. In equation form this becomes:
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3D = Q
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Each dime is 0.1 of a dollar. So if you multiply the number of dimes D by 0.1 you get the
amount in dollars that come from dimes. The resulting term is 0.1D.
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Each quarter is 0.25 of a dollar. So if you multiply the number of quarters Q by 0.25 you get
the amount in dollars that come from quarters. The resulting term is 0.25Q.
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The problem tells you that adding these two terms will result in $6.80. In equation
form this is:
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0.1D + 0.25Q = 6.80
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From our first equation we know that Q = 3D. This tells us that we can go to our dollar equation
and substitute 3D for Q. This substitution leads to:
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0.1D + 0.25(3D) = 6.80
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Multiply out the second term on the left side and you get:
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0.1D + 0.75D = 6.80
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Add the two terms on the left side to get:
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0.85D = 6.80
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Solve for D by dividing both sides of this equation by 0.85:
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D = 6.80/0.85 = 8
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This tells you that the number of dimes equals 8
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Since the number of quarters is three times the number of dimes, then the number of quarters
is 8*3 = 24.
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Let's check ... each 4 quarters is a dollar ... and there are six groups of 4 quarters, then
the number of dollars from quarters is $6.00
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and the 8 dimes is 80 cents or $0.80.
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The sum of these two amounts is $6.00 + $0.80 = $6.80, just as the problem said it should.
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And with 24 quarters and 8 dimes, there are 3 times as many quarters as there are dimes.
This is also as the problem said it should be. Therefore, the answer of 24 quarters and
8 dimes fully checks out.
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Hope this helps you to understand the problem and how to solve it algebraically.
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