Question 937378: A student has 100 coins totaling 5.00 dollars. He had no nickels. What coins did he have if 60% were the same type? Found 2 solutions by josgarithmetic, richwmiller:Answer by josgarithmetic(39620) (Show Source):
If you used 100 pennies, this would only give $1.00.
Arranging for any higher denomination coin to replace some pennies, you would have FEWER than to needed 100 coins.
You can put this solution on YOUR website! 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
(