SOLUTION: Jo has 37 coins (all nickels, dimes, and quarters) worth $5.50. She has four more quarters than nickels. How many dimes does Jo have? I am having problems with these mon

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Finance -> SOLUTION: Jo has 37 coins (all nickels, dimes, and quarters) worth $5.50. She has four more quarters than nickels. How many dimes does Jo have? I am having problems with these mon      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

Question 53602: Jo has 37 coins (all nickels, dimes, and quarters) worth $5.50. She has four more quarters than nickels. How many dimes does Jo have?

I am having problems with these money problems. I cannot seem to find a way to set this one up.

Answer by rchill(405) About Me  (Show Source):
You can put this solution on YOUR website!
We know that a nickel can be represented as .05, and similarly for a dime and a quarter: .10 and .25; so the equation becomes .05x%2B.10y%2B.25z=5.50, where x is the number of nickels, y is the number of dimes, and z is the number of quarters, and furthermore, x%2By%2Bz=37. But that has 3 unknowns, so let's go further: Jo has 4 more quarters than nickels. Because the number of nickels is x, that means she has (x+4) quarters; so our equation is now: .05x%2B.10y%2B.25%28x%2B4%29=5.50. Expanding that equation we get .05x%2B.10y%2B.25x%2B1=5.50. Combining like terms and simplifying we get .30x%2B.10y=4.50. Going back to the number of coins equation: x%2By%2Bz=37, we substitute in (x+4) for z (number of quarters) to get x%2By%2Bx%2B4=37 and combining like terms and simplifying to get 2x%2By=33, or y=33-2x. Now let's substitute that value of y into our previous equation to get .3x%2B.10%2833-2x%29=4.5. Expanding gives us .3x%2B3.3-.2x=4.5. Combining like terms and simplifying we get .1x=1.2, or x=12. That means we have 12 nickels, or $.60; because we have 4 more than 12 for the number of quarters, we have 16 quarters, or $4.00. So far we have 28 coins (12 nickels and 16 quarters); that means we must have 9 dimes (37-28=9), or $.90; adding up all the money, we get $.60+$4.00+$.90=$5.50. So that means are equations and answer of 9 dimes is correct.