SOLUTION: Michelle has $2.45 in dimes, nickels, and quarters. If Michelle has 3 fewer nickels than quarters and one more dime than twice as many quarters then how many of each type of coin d

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: Michelle has $2.45 in dimes, nickels, and quarters. If Michelle has 3 fewer nickels than quarters and one more dime than twice as many quarters then how many of each type of coin d      Log On

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Question 1080044: Michelle has $2.45 in dimes, nickels, and quarters. If Michelle has 3 fewer nickels than quarters and one more dime than twice as many quarters then how many of each type of coin does she have?
Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let Q = the number of quarters.

Then the number of nickels is (Q-3), and the number of dimes is (2Q+1).


Therefore, your "value" equation is 

5*(Q-3) + 10*(2Q+1) + 25Q = 245   (cents).

5Q - 15 + 20Q + 10 + 25Q = 245,

50Q - 5 = 245  --->  50Q = 245 + 5 = 250  --->  Q = 250%2F50 = 5.


Answer.  5 quarters, 5-3 = 2 nickels, 2%+1 = 11 dimes.


There is entire bunch of lessons on coin problems
    - Coin problems
    - More Coin problems
    - Solving coin problems without using equations
    - Kevin and Randy Muise have a jar containing coins
    - Typical coin problems from the archive
    - More complicated coin problems
    - Solving coin problems mentally by grouping without using equations
    - Santa Claus helps solving coin problem
in this site.

You will find there the lessons for all levels - from introductory to advanced,
and for all methods used - from one equation to two equations and even without equations.

Read them and become an expert in solution of coin problems.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Coin problems".