63. A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution toit. 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?

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63. Let x litre of 20 % boric acid solution is required to added. Since, we have 640 litres of 8 % boric acid solution.

Total mixture will have (x + 640) litre.

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

$\begin{array}{l} \Rightarrow \frac{2 x}{100} + \frac{8 * 640}{100} > \frac{4 * (x + 640)}{100} \end{array}$

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

$\Rightarrow x + 2560 > 2 (x + 640)$

$\Rightarrow 2 x + 2560 > 2 x + 1280$

$\Rightarrow 2560 - 1280 > 2 x - x$

$\Rightarrow 1280 > x$ (1)

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

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

$\Rightarrow 2 x + 8 * 640 < 6 x + 6 * 64 0(multiplyingby 10 0throughout)$

$\Rightarrow 8 * 640 - 6 * 640 < 6 x - 2 x$

$\Rightarrow 640 (8 - 6) < 4 x$

$\begin{array}{l} \Rightarrow \frac{640 * 2}{4} < \frac{4 x}{4} 4 (dividing by 4 throughout) \\ \Rightarrow 320 < x(2) \end{array}$

So, from (1) and (2) we get,

320

Thus, the number of litre of 2 % boric acid solut

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