Question

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?

Answer 1

$640\ell of8\%solution$

$letxLof2\%solutionaddid$

$Netvolume=(640+x)L$

$Acidin80\%\to \left(640\times \frac{8}{100}\right)L$

$acidin\to \left(x\times \frac{2}{100}\right)L$

$netacidvolume\to \left(640\times \frac{8}{100}+x\times \frac{2}{100}\right)L$

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

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

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

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

$2x-4\times 640<0$

$x<1280$

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

$4x>640\times 2$

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

$x>320$

$320<x<1280$