SOLUTION: There are 8 nickels and 8+9=17 dimes. A collection of dimes and quarters has a total value of $2.20. If there are three times as many quarters, how many of each coin is in the equ

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Question 1042100: There are 8 nickels and 8+9=17 dimes.
A collection of dimes and quarters has a total value of $2.20. If there are three times as many quarters, how many of each coin is in the equation?

Found 4 solutions by stanbon, solver91311, josgarithmetic, ikleyn:
Answer by stanbon(75887) About Me  (Show Source):
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A collection of dimes and quarters has a total value of $2.20. If there are three times as many quarters, how many of each coin is in the equation?
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10d + 25q = 220 cents
q = 3d
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Substitute for "q" and solve for "d"::
10d + 25(3d) = 220
10d + 75d = 220
85d = 220
d = is not a positive whole number
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Please check you post for accuracy.
Cheers,
Stan H.
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Answer by solver91311(24713) About Me  (Show Source):
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I'm assuming that the "There are 8 nickels and 8+9=17 dimes." is extraneous information that has nothing whatever to do with the problem you are asking about.

Next, I think you have left some words out of the problem or misstated it in some other way. If the total amount of money is $2.20 and there are three times as many quarters (as there are dimes one would assume from the way you wrote this) then you have an impossible situation. The number of quarters would have to be a multiple of 3 since the number of dimes must be an integer. That means that you either have 3 quarters (75 cents) or 6 quarters ($1.50). You could not have 9 quarters because that would be worth $2.25 which is more than the total of the dimes and quarters. If you have 3 quarters, then you would have one dime, 85 cents. If you have 6 quarters, then you have 2 dimes, $1.70. Neither adds up to $2.20.

What I think you meant to write was 'three times as many DIMES AS quarters". Then you can say let represent the number of quarters, then is the number of dimes. The value of the dimes is cents and the value of the quarters is cents. Since the total value of the money in the collection is 220 cents, you can write:







So 4 quarters is $1.00 and 3 times 4 equals 12 dimes is $1.20 and $1.00 plus $1.20 is $2.20. Checks.

John

My calculator said it, I believe it, that settles it


Answer by josgarithmetic(39620) About Me  (Show Source):
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There are 8 nickels and 8+9=17 dimes.
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What are you trying to answer in that?



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A collection of dimes and quarters has a total value of $2.20. If there are three times as many quarters, how many of each coin is in the equation?
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Count of dimes is d.
Count of quarterss is q.
System of equations for account of the money and of the coins is this:
system%280.1d%2B0.25q=2.20%2Cq=3d%29

Simplify the money account equation.
10d%2B25q=220
2d%2B5q=22%2A2%2A5%2F5
2d%2B5q=44
and substitute using the other equation, for q.
2d%2B5%2A3d=44
and solve first for d.
highlight_green%2817d=44%29-------this is bad because it will not give a whole numbered value for d, and the solution quantity of coins must be whole numbered values for each.

Something is wrong in your description.

Answer by ikleyn(52814) About Me  (Show Source):
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.
There are 8 nickels and 8+9=17 dimes.
A collection of dimes and quarters has a total value of $2.20. If there are three times as many quarters, how many of each coin is in the equation?
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For me, this formulation is a trash.

From the first phrase to the last word.