Question

How many liters of a 30% acid solution should be mixed with an 80% acid solution to get 70 L of a solution that is 50% acid?

Answer 1

\(\large {\begin{array}{rccccclll}
&liters&acidity&total\ liters
\\\hline\\
30\% \ sol.&a&0.30&0.30\cdot a\\
60\%\ sol.&b&0.80&0.80\cdot b\\
\\\hline\\
mixture& a+b=70&0.50&70\cdot 0.50=35
\end{array}
\\ \quad \\
\qquad thus\qquad a+b=70\implies a=70-b\qquad then
\\ \quad \\
\begin{array}{rccccclll}
&liters&acidity&total\ liters
\\\hline\\
30\% \ sol.&{\color{brown}{ 70-b}}&0.30&0.30\cdot ({\color{brown}{ 70-b}})\\
60\%\ sol.&b&0.80&0.80\cdot b\\
\\\hline\\
mixture& a+b=70&0.50&70\cdot 0.50=35
\end{array}
\\ \quad \\
then\qquad 0.30( {\color{brown}{ 70-b}})+0.08b=35 }\)

solve for "b" to see how much you'd need of the 80% acid
and to get "a", well, a = 70-b

Answer 2

hmm kinda put 60%... should be 80% anyhow

Answer 3

\(\large {\begin{array}{rccccclll}
&liters&acidity&total\ liters
\\\hline\\
30\% \ sol.&{\color{brown}{ 70-b}}&0.30&0.30\cdot ({\color{brown}{ 70-b}})\\
80\%\ sol.&b&0.80&0.80\cdot b\\
\hline\\
mixture& a+b=70&0.50&70\cdot 0.50=35
\end{array}
\\ \quad \\
then\qquad 0.30( {\color{brown}{ 70-b}})+0.08b=35 }\)

Answer 4

Okay, I did .30(70-y) + .80y=35, but am still not coming up with the same answer. They show 28, 12, 58, and 42 as one of the answers and I can't seem to find any of them.

Answer 5

well, review your simplification, I can see the choices matching up

Answer 6

I see. After dividing I came up with 28. Would I then say 70-28 to get my final answer?

Answer 7

yeap