Question 225843: ECE Board November 1999/2000 In a box, there are 25 coins consisting of quarter, nickels and dimes with a total amount of $2.75. if the nickel were dimes ad the dimes were quarters and the quarters were nickles, the total amount would be $3.75. how many quarters are there? note:Nickel = 5cents, dime = 10 cents and quarter = 25 cents.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! In a box, there are 25 coins consisting of quarter, nickels and dimes with a total amount of $2.75.
if the nickels were dimes, and the dimes were quarters, and the quarters were nickels, the total amount would be $3.75.
how many quarters are there?
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Originally x=no.of nickels, y=dimes, z=quarters
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Three equations:
x + y + z = 25
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5x + 10y + 25z = 275; using cents
:
25x + 5y + 10z = 375
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Multiply the 1st equation by 5, subtract from the 2nd equation
5x + 10y + 25z = 275
5x + 5y + 5z = 125
-----------------------eliminate x
5y + 20z = 150
simplify, divide by 5
y + 4z = 30
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Multiply 1st equation by 25, subtract the 3rd equation
25x + 25y + 25z = 625
25x + 5y + 10z = 375
------------------------eliminate x again
20y + 15z = 250
Simplify, divide by 5
4y + 3z = 50
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We have two, 2 unknown equations
y + 4z = 30
4y + 3z = 50
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Multiply the 1st equation by 4, subtract the 2nd
4y + 16z = 120
4y + 3z = 50
:
There is something wrong with this problem, it does not have integer solutions
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