**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?

**Solution:**

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

$\Rightarrow \frac{4}{100}(640+x)<\frac{8}{100} \times 640+\frac{2}{100} \times x<\frac{6}{100}(640+x)$

Multiplying throughout by 100 :

$\Rightarrow 2560+4 x<5120+2 x<3840+6 x$

$\Rightarrow 2560+4 x<5120+2 x$ and $5120+2 x<3840+6 x$

$\Rightarrow 2 x<2560$ and $4 x>1280$

$\Rightarrow x<1280$ and $x>320$

$\Rightarrow 320

Thus, the amount of solution must be less than 1280 litres but more than 320 litres.