SOLUTION: You have 55 coins totaling $10. There are more nickels than pennies, more dimes than nickles, and more quarters than dines. How many of each coin do you have?

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Question 486861: You have 55 coins totaling $10. There are more nickels than pennies, more dimes than nickles, and more quarters than dines. How many of each coin do you have?
Answer by Edwin McCravy(20059)   (Show Source): You can put this solution on YOUR website!
I will assume that you have some pennies.  Then
since $10.00 is 1000 cents, a multiple of 5, you
must have a multiple of 5 pennies.  So the smallest
number of pennies we can have is 5.  That makes that 
the smallest number of nickels we could have would
be 6.
 
We will have the greatest possible number of quarters 
if we have the least possible number of pennies, nickels and dimes.

That would be 5 pennies, 6 nickels and 7 dimes, which is 105 cents.
Since 1000-105 = 895.  When we divide that by 25, we get 35.8. So
that tells us we cannot have more than 35 quarters. 

Hmmm!  Let's see if that's possible.  5 pennies, 6 nickels, and 
35 quarters.

That's a total of: 

           coins   money
pennies      5    $ .05
nickels      6      .30
quarters    35    $8.75   
-----------------------
totals      46    $9.10

So we have 9 coins to go and 90 cents to go
in dimes, and so -- eureka!  That's easy. We 
can get that with 9 dimes.

So we end up with:
   
           coins   money
pennies      5    $ .05
nickels      6      .30
dimes        9      .90 
quarters    35    $8.75   
-----------------------
totals      55   $10.00

Edwin


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