Thread: mixture problems

I am asked to find

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?

Originally Posted by mandy123

I am asked to find

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 . Go from there...

oh, ok, I was missing that y=5-x, I got it, thank you so much!!!!