acid mixture

[cheergirl01]

how much pure acid should be added to 18 L of 30% acid to increase the concentration to 50%.


i have guesses to how to go about the problem but i honestly dont know.
i did

.30 + x = .50
which is .20
and then started the problem as:
.30(18) + .20(x) = .50(18+x)

 

[galactus]

You're on the right track, except you're adding x amount of 20% instead of x amount of pure.

Pure is 100%, so the decimal in front of the x would be a 1. See?.

\(\displaystyle .3(18)+x=.5(18+x)\)

 

[soroban]

Hello, cheergirl01!

Galactus is absolutely correct.

Here's the way I baby-talk my way through "mixture problems".

How much pure acid should be added to 18 L of 30% acid to increase the concentration to 50%?

\(\displaystyle \text{We start with 18 liters which is 30\% acid.}\)
. . \(\displaystyle \text{This contains: }\:0.30 \times 18 \:=\:5.4\) liters of acid.

\(\displaystyle \text{We add }x\text{ liters which is 100\% acid.}\)
. . \(\displaystyle \text{This contains: }\:1 \times x \:=\:x\text{ liters of acid.}\)

\(\displaystyle \text{Hence, the mixture contains: }\:5.4 + x\text{ liters of acid.}\) .[1]


\(\displaystyle \text{But we know that the mixture will be }18 + x\text{ liters which is 50\% acid.}\)

. . \(\displaystyle \text{That is, it contains: }\:0.50(18+x)\text{ liters of acid.}\) .[2]


\(\displaystyle \text{We just described the final amount of acid in }two\;ways.\)

. . \(\displaystyle There\text{ is our equation! } \;\hdots \quad 5.4 + x \:=\:0.50(18+x)\)

 

[lookagain]

how much pure acid should be added to 18 L of 30% acid to increase the concentration to 50%.


i have guesses to how to go about the problem but i honestly dont know.
i did

\(\displaystyle Regardless \ of \ the \ incorrectness \ of \ the \ following:\)

.30 + x = .50
which is .20

cheergirl,

define your variables(s).

State this or its equivalent:

Let x = the number of liters of pure acid to be added.