Question 52766
A child has 40 coins worth more than $6.00.  If the coins are dimes and quarters only, what can you conclude about the number of quarters?
:
We can make two equations:
:
"child has 40 coins (dimes & quarters)
d  +  q = 40
and they are worth over $6
.10d + .25q > 6
:
Using the 1st equation, arrange so we are solving for q (quarters)
d  +  q = 40
      d = [40 - q]
:
Substitute [40-q] for d in the 2nd equation:
.10[40-q] + .25q  > 6
4 - .10q  + .25q  > 6
            .15q  > 6 - 4
               q  > 2/.15
               q  > 13.33 
               q  => 14  has to be an integer number

Check:  
14 quarters + 26 dimes
    3.50    + 2.60  = $6.10 [fulfills the requirement of > $6]  
:
Try 13 quarters + 27 dimes
3.25    +  2.70  = $5.95   less than the required $6