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?

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

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