Question

how many liters of a 50% acid solution must be mixed with a 15% acid solution to get 315 L of a 40% acid solution?

Answer 1

First, define your variables:
A = liters of 50% acid solution
B = liters of 15% acid solution
A+B=320 liters
Now what do we do?

Answer 2

hmm

Answer 3

any ideas on how to represent 50% and 15% and 40% as a decimal?

Answer 4

and so

Answer 5

@Aveline its 315 L
you do .50+.015

Answer 6

Oops, 315 is correct.
Make sure that you have the correct decimal place for 15%
Here's the main equation:
0.50A+0.15B=(315)(0.40)
Can you find a way to substitute B for A using this equation?
A+B=315

Answer 7

@Directrix

Answer 8

@pphalke
Do you need me to walk you through it?

Answer 9

I'll write a detailed response. I can't reply after this because I'll be heading to dinner.
A = liters of 50% acid solution
B = liters of 15% acid solution
These are the variables
A+B=315
When you add the liters of A and B, they need to add up to 315 liters

B=315-A
Here, we are rearranging the previous equation. This way, we only have one unknown variable.

0.5A+0.15B=(315L)(0.40)

This is the equation you will start off with.
In words: one liter of a 50% acid solution plus one liter of a 15% acid solution equals 315 liters of a 50% acid solution

0.5A+0.15(315-A)=126

Substitute (315-A) for B.

0.5A+47.25-0.15A=126

Use the distributive property
...and simplify:
0.35A+47.25=126
0.35A=78.75
A=225 liters

Remember the equation A+B=315?
plug in A
225+B=315
B=90 liters

A is 225 liters and B is 90 liters