Question 179941
Let number of dimes be d, and the number of nickels n
Then, d(.10)  +  n(.05)  =  6.05, and
2d(.10)  +  (n – 10).05  =  9.85, or .2d  +  .05n  -  .5  =  9.85


  .1d  +  .05n  =  6.05
  .2d  +  .05n  =  10.35


Subtract eq (ii) from eq (i)           - .1d    =    - 4.3
					       d    =   43    


Therefore, we have 43 dimes.


To find how many nickels there are, we just substitute 43 for d in eq (i).
This gives us:              4.3  +  .05n  =  6.05
			     	     .05n  =   1.75
				        n    =    35        


				        43 dimes  =  $4.30
				       35 nickels =  $1.75
     $4.30 + $1.75  =  $6.05


Doubling dimes    =    2(43)    =    86    =      $8.60
Decreasing nickels by 10 = (35 – 10) = 25 = $1.25
				$8.60  +  $1.25  =   $9.85