SOLUTION: Sue has two more nickels than dimes and three more quarters than nickels. She has $3.35 in all. How many coins of each kind does she have?

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: Sue has two more nickels than dimes and three more quarters than nickels. She has $3.35 in all. How many coins of each kind does she have?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 131957: Sue has two more nickels than dimes and three more quarters than nickels. She has $3.35 in all. How many coins of each kind does she have?
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
Number of nickels: n
Number of dimes: d
Number of quarters: q

Given:
n=d%2B2, two more nickels than dimes,
q=n%2B3, three more quarters than nickels

Value of n nickels is 5n cents
Value of d dimes is 10d cents
Value of q quarters is 25q cents

Total money is $3.35 which can also be expressed as 335 cents.

5n+%2B+10d+%2B+25q=335

Since n=d%2B2 and q=n%2B3, by substitution we can say that q+=+%28d%2B2%29%2B3=d%2B5

Now take the value equation and substitute the two expressions for nickels and quarters in terms of dimes in place of the n and q variables:

5%28d%2B2%29+%2B+10d+%2B+25%28d%2B5%29=335

Distribute and collect terms:
5d%2B10%2B10d%2B25d%2B125
40d%2B135=335

Add -135 to both sides:
40d=200

Divide by 40:
d=5

So there are 5 dimes. That means there are 5+%2B+2=7 nickels and 5%2B5=10 quarters.

Check the answer:
10 quarters is $2.50,
5 dimes is $.50,
7 nickels is $.35, and finally,
$2.50 plus $.50 plus $.35 = $3.35.