1. ## Mixture word problem

How many liters of water must be added to 35 liters of 80% hydrochloric acid solution to reduce its strength to 75%?
Ans: 6.53 liters

solution:
x (0%water) +35 (80% HCL) = (35+x)(75% strength HCL)
35*(0.8) = (35+x)0.75
x = 2.3333 liters
incorrect

2. ## Re: Mixture word problem

Hello, MathMinors!

How many liters of water must be added to 35 liters of 80% HCl solution
to reduce its strength to 75%?

Ans: 6.53 liters . . . . Wrong!

We concerned with the amount of water to be added.

The amount of water is our concern.

. . It contains:. $0.20 \times 35 \:=\:7$ liters of water.

We add $x$ liters of water.

The mixture contains:. $x + 7$ liters of water. .[1]

We know that the mixture will contain $35 + x$ liters of which 25% is water.
. . The mixture contains:. $0.25(35+x)$ liters of water. .[2]

We just described the final amount of water in two ways.

There is our equation . . . $\boxed{x + 7 \:=\:0.25(35 + x)}$

Solve for $x\!:\;\;x + 7 \:=\:8.75 + 0.25x \quad\Rightarrow\quad 0.75x \:=\:1.75$

. . . . . . . . . . $x \:=\:\frac{1.75}{0.75} \:=\:\frac{175}{75} \:=\:\frac{7}{3}$

$\text{Therefore, we must add }2\tfrac{1}{3}\text{ liters of water.}$

3. ## Re: Mixture word problem

Using a concentration line of 0%, 75%, 80%
80 -75 = parts H2O =5 , 75-0 = parts of 80% to make 80 parts of 75%
parts of water required for 35 parts 80% = 35/75 *5 = 2.333 parts water.

Water solutions of HCL exceeding 40% do not exist at normal temp and pressure