SOLUTION: There were 34 nickels, dimes, and quarters whose total value came to $3.80. How many coins of each kind were there if there were 4 more nickels than quarters?

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Misc -> SOLUTION: There were 34 nickels, dimes, and quarters whose total value came to $3.80. How many coins of each kind were there if there were 4 more nickels than quarters?       Log On

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

   

Question 31828: There were 34 nickels, dimes, and quarters whose total value came to $3.80. How many coins of each kind were there if there were 4 more nickels than quarters?

Answer by stanbon(48558) About Me  (Show Source):
You can put this solution on YOUR website!
Let number of quarters be "x"
Then number of nickels is "x+4"
And number of dimes is "34-x-(x+4)"="30-x"
Value of quarters is 25x cents
Value of nickels is 5(x+4)= 5x+20 cents
Value of dimes is 10(30-x)=300-10x cents
EQUATION:
value + value + value = 380 cents
25x + 5x+20 + 300-10x = 380
20x + 320= 380
20x=60
x=3 (number of quarters)
x+4=7 (number of nickels)
30-x= 27 (number of dimes)
Cheers,
Stan H.