SOLUTION: A chemical manufacturer wishes to fill an order for 700 gallons of a 24% acid solution. Solutions of 20% and 30% are in stock. How many gallons of 30% acid solution will be used in

Question 1097569: A chemical manufacturer wishes to fill an order for 700 gallons of a 24% acid solution. Solutions of 20% and 30% are in stock. How many gallons of 30% acid solution will be used in the desired mixture?
(I understand there may not be such a thing as "24% acid solution", et cetera. I'm just focusing on what solution needs to be used to get the number being asked for.
Honestly, I've seen problems like these before but just have trouble when it's worded differently. Thank you!)
Found 2 solutions by ikleyn, jim_thompson5910:
Answer by ikleyn(52795)   (Show Source): You can put this solution on YOUR website!
.
One of basic principles of philosophy and education (from the ancient Greece times) is:

Identify identical and distinct different.

Your English is quite clear and logical.
I can not believe that you have difficulties reading different wording.

Make an effort.

Otherwise, what is the reason for us to solve it for you ??????????

----------------
Your sample is Problem 1 of the lesson
    - Solving typical word problems on mixtures for solutions
in this site.

Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!

"A chemical manufacturer wishes to fill an order for 700 gallons of a 24% acid solution" means they want 700*0.24 = 168 gallons of pure acid

Let
x = amount of the 20% acid solution
y = amount of the 30% acid solution
both amounts are in gallons

Since we can only use the 20% or the 30% solution, this means that x and y must add to the total of 700
So we have the equation
x+y = 700

Now solve for y. Subtract x from both sides
x+y = 700
x+y-x = 700-x
y = 700-x
Keep this equation in mind. We'll use this equation later

If we use x gallons of the 20% solution, then we have 0.20*x gallons of pure acid. If we use y gallons of the 30% solution, then we have 0.30*y gallons of pure acid. Combined we have 0.20*x+0.30*y gallons of pure acid. Recall that above we want the target amouunt of total pure acid to be 168 gallons. That is how we're able to get the equation below

0.20*x + 0.30*y = 168

Now use the equation y = 700-x to perform substitution
0.20*x + 0.30*y = 168
0.20*x + 0.30*( y ) = 168
0.20*x + 0.30*( 700 - x ) = 168 ... y has been replaced with 700-x

With y gone from the equation, we can now solve for x
0.20*x + 0.30*( 700 - x ) = 168
0.20*x + 0.30*( 700 ) + 0.30*( - x ) = 168
0.20*x + 210 - 0.30x = 168
0.20x - 0.30x + 210 = 168
-0.10x + 210 = 168
-0.10x + 210-210 = 168-210
-0.10x = -42
-0.10x/(-0.10) = -42/(-0.10)
x = 420

Use this x value to find y (again use y = 700-x)
y = 700-x
y = 700-420 ... replace x with 420
y = 280

----------------------------------------------------

In summary we found that
x = 420
y = 280

So we should use 420 gallons of the 20% solution, and 280 gallons of the 30% solution.

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