Question 937378
look at this problem which is similar but leaves out the  60%
http://www.algebra.com/algebra/homework/word/coins/Word_Problems_With_Coins.faq.question.348068.html

01p+10d+25q+50h=500
We have to use a trial and error
60 per cent are dimes
.6*100=60*10=600
makes $6 dollars too much

.6*100=60*05=300 but there are no nickels

.6*100=60*01=60 pennies
60 pennies

now we need $4.40 using dimes and quarters and halves 

we now need 40 coins which total $4.40
assuming we are only dealing with two types 
which of these has integer solutions
x+y=40,
25x+10y=440
quarters and dimes


x+y=40
25x+50y=440
quarters and halves
couldn't add up to 4.40

x+y=40
10x+50y=440
dimes and halves
this has a solution of y =1 and x=39

another way
now are we using dimes quarters or halves
lets try one half which would leave 390 which would work out with 39 dimes
so the answer is 
one half 
39 dimes
60 pennies