**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 we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?

**Solution:**

Let *x* litres of 2% boric acid solution is required to be added.

Then, total mixture = (*x* + 640) litres

This resulting mixture is to be more than 4% but less than 6% boric acid.

$\therefore 2 \% x+8 \%$ of $640>4 \%$ of $(x+640)$

And, $2 \% x+8 \%$ of $640<6 \%$ of $(x+640)$

$2 \% x+8 \%$ of $640>4 \%$ of $(x+640)$

$\Rightarrow \frac{2}{100} x+\frac{8}{100}(640)>\frac{4}{100}(x+640)$

$\Rightarrow 2 x+5120>4 x+2560$

$\Rightarrow 5120-2560>4 x-2 x$

$\Rightarrow 5120-2560>2 x$

$\Rightarrow 2560>2 x$

$\Rightarrow 1280>x$

$2 \% x+8 \%$ of $640<6 \%$ of $(x+640)$

$\frac{2}{100} x+\frac{8}{100}(640)<\frac{6}{100}(x+640)$

$\Rightarrow 2 x+5120<6 x+3840$

$\Rightarrow 5120-3840<6 x-2 x$

$\Rightarrow 1280<4 x$

$\Rightarrow 320

$\therefore 320

Thus, the number of litres of 2% of boric acid solution that is to be added will have to be more than 320 litres but less than 1280 litres.