Question

@Jonask and @kc_kennylau
can any of you guys explain?

Answer 1

A chemist needs to mix a 12% acid solution with a 32% acid solution to obtain an 8-liters mixture consisting of 20% acid. How many liters of each of the acid solutions must be used?

Answer 2

WHAT?!

Answer 3

Yes?

Answer 4

WHAT?!

Answer 5

Why are you typing my username over and over again? can you just help me instead?

Answer 6

Let the amount of acid A needed be a L. The amount of acid B needed is 8-a L

Answer 7

Oh, okay, @kc_kennylau.

Answer 8

\[\frac{12\%\times a+32\%\times(8-a)}8=20\%\]

Answer 9

Alright, let me try to do this one...

Answer 10

\[\frac{12 \text{%} \times a + 32 \text {%} \times (8-a) }{8} =20\text{%}\]

no idk, can you continue? I see how you got the formula, but idk....

Answer 11

what's so hard solving an equation? :)

Answer 12

12% and 32% of what? of 8 and of 8-a ?!

Answer 13

NVM about this q....

Answer 14

Here is how I think of these problems

type 1 solution is 12% acid
type 2 sol'n is 32% acid
type 3 sol'n is 20% acid. It is 8 liters, so it has 0.20 * 8 liters= 1.6 liters of acid

say we take A liters from type 1 and B liters from type 2.
we want A + B to add up to how much ? 8 liters of the final solution

First equation: A+B=8

How much acid is in A liters of type 1? 0.12*A
How much acid in B liters of type 2? 0.32*B

How much acid should this add up to? the amount of acid in type 3, 1.6 liters

Second equation: 0.12A + 0.32B = 1.6
if we multiply by 100 (both sides of the equation, and all terms) we could write this as
12 A + 32 B = 160

we could simplify this a bit by dividing all the terms by 4:
3A + 8B= 40


now solve
A+B=8
3A+8B=40

Answer 15

TY, that's just what I wanted!