Question

Two solutions of salt water contain 0.07% and 0.23% salt respectively. A lab technician wants to make 1 liter of solution which contains 0.11% salt. How much of each solution should she use?

Answer 1

0.75 liter of 0.07 % is used and 0.25 liter of 0.23 % salt is used

Solution:

We have to 1 liter of final solution

Let "x" be the salt solution of 0.07 %

Then, (1 - x) is the salt solution of 0.23 %

Then final solution should be 1 liter of solution which contains 0.11 % salt

Thus, 0.07 % of "x" is mixed with 0.23 % of (1 - x) to make a solution 0.11 % of 1 liter

Thus a equation is framed as:

0.07 % of x + 0.23 % of (1 - x) = 0.11 % of 1

$0.07 \% \times x + 0.23 \% \times (1-x) = 0.11 \% \times 1\\\\\frac{0.07}{100} \times x + \frac{0.23}{100} \times (1-x) = \frac{0.11}{100} \times 1\\\\0.0007x + 0.0023(1-x) = 0.0011\\\\0.0007x + 0.0023 - 0.0023x = 0.0011\\\\-0.0016x = -0.0012\\\\x = 0.75$

Thus 0.75 liter of 0.07 % is used

And,

1 - x = 1 - 0.75 = 0.25

Thus 0.25 liter of 0.23 % salt is used