Question

A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If there are 640 litres of the 8% solution, how many litres of 2% solution will have to be added?

Answer 1

Suppose x litres of 2% solution is added in the existing solution of 8% of boric acid.
Resulting mixture = (640 + x) L
Therefore, as per given conditions: 

`4% "of" (640+x)< 8% "of" 640 +2% of x< (640+x)` 

\[ \Rightarrow \frac{4}{100}(640 + x) < \frac{8}{100} \times 640 + \frac{2}{100} \times x < \frac{6}{100}(640 + x)\]
\[\text{ Multiplying throughout by } 100: \]
\[ \Rightarrow 2560 + 4x < 5120 + 2x < 3840 + 6x\]
\[ \Rightarrow 2560 + 4x < 5120 + 2x \text{ and } 5120 + 2x < 3840 + 6x\]
\[ \Rightarrow 2x < 2560 \text{ and } 4x > 1280\]
\[ \Rightarrow x < 1280 \text{ and } x > 320\]
\[ \Rightarrow 320 < x < 1280\]
\[\text{ Thus, the amount of solution must be less than 1280 litres but more than 320 litres } .\]