# SOLUTION: 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

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 Click here to see ALL problems on Money Word Problems 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(15659)   (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? : Originally x=no.of nickels, y=dimes, z=quarters : Three equations: x + y + z = 25 : 5x + 10y + 25z = 275; using cents : 25x + 5y + 10z = 375 : 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 : 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 : We have two, 2 unknown equations y + 4z = 30 4y + 3z = 50 : 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