I'm stuck! Suppose a lab needs to make 900 liters of a 39% acid solution, but the only solutions available to the lab are 15% acid and 70% acid. What system of equations can be used to find the number of liters of each solution that should be mixed to make the 39% solution? Let c represent the number of liters of 15% acid solution and let d represent the number of liters of 70% acid solution.
I can give you the equation in terms of one variable going from a table, but am having difficulty using two variables, for some odd reason.
oh wait...duh...this is it!
c + d = 900 (relating the amounts of liquid) .15c + .70d = .39(900)
Say that c = 900-d, fill in that amount for c in the second equation to get this:
.15(900-d)+.7d=351
Doing that math gives you
135-.15d+.7d=351 and .55d = 216 d = 392.7
That is how much of the 70% acid solution that needs to be mixed in with the 15% acid of which there is 900-392.7=507.3
these are the options but I say that it is B
507.3 liters of 15% acid + 392.7 liters of 70% acid = 900 liters of 39% acid
No its not B; it's the last choice; that one is exactly what I did to solve this. It's the very last choice. Check it against my work and then do the math yourself to see that it is correct.
okay Im going to try to do it and see but thank you so much you're a big help!
\(\large { \begin{array}{cccccllll} &liters\ needed&acidity&total\ liters \\\hline\\ 15\%&c&0.15&0.15c\\ 70\%&d&0.70&0.70d\\ mixture&c+d=900&0.39&0.39(900)=351 \end{array}\\ \quad \\ c+d=900\implies c=900-d\qquad thus \\ \quad \\ \begin{array}{cccccllll} &liters\ needed&acidity&total\ liters \\\hline\\ 15\%&900-d&0.15&0.15(900-d)\\ 70\%&d&0.70&0.70d\\ mixture&(900-d)+d=900&0.39&0.39(900)=351 \end{array} \\ \quad \\ \implies {\color{brown}{ 0.15(900-d)+0.70d=351}} }\)
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