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 liters of the 8% solution,how many liters of 2% solution will have to be added ?

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Solution

Let x liters of 2% solution will have to be added to 640 liters of the 8% solution of acid.

Total quantity of mixture=(640+x)

Total acid in the (640+x) litres of mixture $\frac{2}{100} x + \frac{8}{100} 640$

It is given that acid content in the resulting mixture must be more than 4% but less than 6%.

$\frac{4}{100} [640 + x] < (\frac{2}{100} x + \frac{8}{100} 640) < \frac{6}{100} [640 + x] \Rightarrow 4 [640 + x] < (2 x + 8640) < 6 [640 + x] \Rightarrow 2560 + 4 x < 2 x + 8640 a n d 2 x + 8640 < 3840 + 6 x \Rightarrow 2560 - 8640 < 2 x < 4 x a n d 2 x - 6 x < 3840 - 8640 \Rightarrow x < 1280 a n d x > 320$

More than 320 liters but less than 1280 liters of 2% is to be added.

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