How many gallons of a 8% acid solution should be mixed with

[patkwallace]

How many gallons of a 8% acid solution should be mixed with 40 gallsons of a 20% acid solution to obtain a mixture that is 10% acid?

What steps are needed to solve?

 

[soroban]

Re: Word problem

Hello, patkwallace!

How many gallons of a 8% acid solution should be mixed
with 40 gallons of a 20% acid solution to obtain a mixture that is 10% acid?

Consider the amount of acid at each phase.

We start with 40 gallons which is 20% acid.
. . It contains: \(\displaystyle \:20\%\,\times\,40\:=\0.20)(40) \:=\:8\) gallons of acid.

We add \(\displaystyle x\) gallons which is 8% acid.
. . It contains: \(\displaystyle \:8\%\,\times\,x\:=\:0.08x\) gallons of acid.

So the mixture contains: \(\displaystyle \:8\,+\,0.08x\) gallons of acid. .[1]


Let's look at the mixture again.
We started with 40 gallons of stuff; we added \(\displaystyle x\) gallons of stuff.
. . So we have: \(\displaystyle \,x\,+\,40\) gallons of stuff.

This mixture is supposed to be 10% acid.
. . So it contains: \(\displaystyle \:0.10(x\,+\,40)\) gallons of acid. .[2]


We just described the final amount of acid in two ways.

There is our equation! . . . \(\displaystyle \L8\,+\,0.08x\;=\;0.10(x\,+\,40)\)

Now solve for \(\displaystyle x\) . . .