Question 69526: (This is on page 105.) I'm going to type the question and then type how I tried to solve it.
"Marcela has a collection of nickels, dimes, and quarters worth $24.75. The number of dimes is 10 more than twice the number of nickels, and the number of quarters is 25 more than the number of dimes. How many coins of each kind does she have?"
I tried to work it like this:
Let nickels= x @ .05
Let dimes= 2x+10 @ .10
Let quarters= 2x+10+25 @ .25
x+2x+10+2x+10+25=?
Everytime I try this...I can not seem to get it right. I have to test on this on Thurs. and I just can not get it.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! "Marcela has a collection of nickels, dimes, and quarters worth $24.75. The number of dimes is 10 more than twice the number of nickels, and the number of quarters is 25 more than the number of dimes. How many coins of each kind does she have?"
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Let number of nickels be "x"; Value of these is 5x cents
Number of dimes is "2x+10 ; Value of these is 10(2x+10)=20x+100 cents
Number of quarters is "2x+10+25"=2x+35: Value is 25(2x+35)= 50x+875 cents
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EQUATION:
value + value + value = 2475 cents
5x + 20x+100 + 50x+875 = 2475
75x + 975 = 2475
75x=1500
x=20 (number of nickels)
2x+10= 2*20+10 = 50 dimes
2x+35=2*20+35=75 quarters
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Cheers,
Stan H.
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