Question 775794: Erica has a collection of nickels, dimes, and quarters worth a total value of $11.50. If she has twice as many dimes as nickels, and seven more quarters than dimes, how many of each type of coin does she have?
So far I have(Not sure if I did this right):
X= Nickels
2x= Dimes
2x +7= Quarters
So x +2x +7= 11.50
I got x=1.50
Not sure where to go from there, that is if I even did it right. Please help.
Answer by MathTherapy(10552)   (Show Source):
You can put this solution on YOUR website! Erica has a collection of nickels, dimes, and quarters worth a total value of $11.50. If she has twice as many dimes as nickels, and seven more quarters than dimes, how many of each type of coin does she have?
So far I have(Not sure if I did this right):
X= Nickels
2x= Dimes
2x +7= Quarters
So x +2x +7= 11.50
I got x=1.50
Not sure where to go from there, that is if I even did it right. Please help.
That obviously could not be the answer to the amount of nickels, as an INTEGER is required. There is no such thing as 1.5 nickels.
You were able to find the amount of nickels: x, amount of dimes: 2x, and quarters: 2x + 7
Now, your next step is to determine the VALUE to EACH denomination.
In other words, with the amount of nickels being x, the value of the nickles would be .05x
Likewise, the value of the dimes would be: .1(2x), or .2x
Additionally, the value of the quarters would be: .25(2x + 7), or .5x + 1.75
Now, since the value of all the coins is $11.50, then we can say that:  .
You now need to solve this for x to determine the amount of nickels, represented by an INTEGER value. Then go on to find the amount of dimes and quarters.
You then do a check!!!
Further help is available, online or in-person, for a fee, obviously. Send comments, “thank-yous,” and inquiries to “D” at MathMadEzy@aol.com
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