# mixture problems

• Mar 7th 2009, 10:28 AM
mandy123
mixture problems

how many ounces of a solution that is 90% alcohol need to be mixed with 5 ounces of a solution that is 50% alcohol in order to obtain a solution that is 80% alcohol?

I set it up this way, but don't think i did it right, can someone please help me???

.90x+.50(5)=.80y

but then how do I find the solution?
• Mar 7th 2009, 10:32 AM
JeWiSh
Quote:

Originally Posted by mandy123

how many ounces of a solution that is 90% alcohol need to be mixed with 5 ounces of a solution that is 50% alcohol in order to obtain a solution that is 80% alcohol?

I set it up this way, but don't think i did it right, can someone please help me???

.90x+.50(5)=.80y

but then how do I find the solution?

You're almost there, but that would be unsolvable with two unknowns. You do actually know the value of y, it's 5+x. So multiplying out gives you \$\displaystyle 0.9x+2.5=4+0.8x\$. Go from there...
• Mar 7th 2009, 10:38 AM
mandy123
oh, ok, I was missing that y=5-x, I got it, thank you so much!!!!
(Wink)