A solution of $8 %$ boric acid is to be diluted by adding a $2 %$ boric acid solution to it. Then 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

$640 ℓ o f 8 % s o l u t i o n$

$l e t x L o f 2 % s o l u t i o n a d d i d$

$N e t v o l u m e = (640 + x) L$

$A c i d i n 80 % \to (640 \times \frac{8}{100}) L$

$a c i d i n \to (x \times \frac{2}{100}) L$

$n e t a c i d v o l u m e \to (640 \times \frac{8}{100} + x \times \frac{2}{100}) L$

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

$\Rightarrow 4 \times 640 + 4 x < 640 \times 8 + 2 x < 6 \times 640 + 6 x$

$4 \times 640 - 8 \times 640 + 4 x - 2 x < 0 < 6 \times 640 - 8 \times 640 + 6 x - 2 x$

$- 4 \times 640 + 2 x < 0 < - 2 \times 640 + 4 x$

$2 x - 4 \times 640 < 0$

$x < 1280$

$0 < - 2 \times 640 + 4 x$

$4 x > 640 \times 2$

$x > 640 \times \frac{2}{4}$

$x > 320$

$320 < x < 1280$

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

A solution of $8 %$ boric acid is to be diluted by adding a $2 %$ boric acid solution to it. Then 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?