getting half of right answer

[relapse]

I'm getting 45 for the following problem, am i forgetting to double it somewhere?

the problem is:

How many liters of a 10% alcohol solution must be mixed with 90 liters of a 90% solution to get a 50% solution.

I'm setting the equation up like this:

.10x+.90y=90(.50)

x=90-y

i use subsitution.

the solution should be 90.

 

[Unco]

Not defining variables has its consequences.

Let W be the volume of the 10% (weak) alcohol.
Let N be the volume of the 50% (new) alcohol.

amount = concentration * volume

amount of 10% alc + amount of 90% alc = amount of 50% alc:
0.10 * W + 0.90 * 90 = 0.50 * N

And volume of 10% alc + volume of 90% alc = volume of 50% alc:
W + 90 = N

 

[soroban]

Hello, relapse!

How many liters of a 10% alcohol solution must be mixed with 90 liters of a 90% solution to get a 50% solution?

You're confusing the facts. \(\displaystyle \;\) 90 is not the final amount.

First of all, we need only one variable.

Let \(\displaystyle x\) = number of liters of 10% solution.
\(\displaystyle \;\;\)This will contain: \(\displaystyle 0.10x\) liters of alcohol.

There is 90 liters of a 90% solution.
\(\displaystyle \;\;\)This contains: \(\displaystyle 0.90\,\times\,90\:=\:81\) liters of alcohol.

The mixture will contain: \(\displaystyle 0.10x\,+\,81\) liters of alcohol.


The mixture is \(\displaystyle x + 90\) liters of solution which is 50% alcohol.
\(\displaystyle \;\;\)That is, it will have: \(\displaystyle 0.50(x\,+\,90)\) liters of alcohol.


And there is our equation . . . \(\displaystyle \;0.10x\,+\,81\:=\:0.50(x\,+\,90)\)

 

[Unco]

Nonsense. When you dilute a solution with alcohol present, the volume of alcohol does not stay constant. The amount of alcohol is constant, its volume increases --> its concentration decreases.