A solution of ` 9%` acid is to be diluted by adding `3%` acid solution to it. The resulting mixture is to be more then ` 5%` but less than ` 7%` acid. If there is 460 L of the ` 9%` solution, how many litres of ` 3%` solution will have to be added?

A solution of ` 9%` acid is to be diluted by adding `3%` acid solution

1.	A solution of ` 9%` acid is to be diluted by adding `3%` acid solution to it. The resulting mixture is to be more then ` 5%` but less than ` 7%` acid. If there is 460 L of the ` 9%` solution, how many litres of ` 3%` solution will have to be added?
Answer» Let x L of ` 3%` solution be added to 460 L of ` 9%` solution of acid. Then, total quantity of mixture = (460+x)L Total acid content in the (460 + x) L of mixture ` (460 xx 9/100+x xx 3/100)` It is given that acid content in the resulting mixture must be more than ` 5%` but less than `7%` acid. Therefore, ` 5% "of "(460+x) lt 460 xx 9/100+(3x)/100 lt 7%" of "(460+x)` `rArr 5/100 xx(460+x) lt 460 xx 9/100 x lt 7/100 xx (460+x)` `rArr 5 xx (460+x) lt 460 xx 9+3x lt 7 xx (460+x)` [ multiplying by 100] ` rArr 2300+5x lt 4140 +3x lt 3220+7x` Taking first two inequalities, `2300+ 5x lt 4140+3x` `rArr 5x-3x lt 4140-2300` `rArr 2x lt 1840` `rArr x lt (1840)/2` `rArr x lt 920` ....(i) Taking last two inequalities, ` 4140+3x lt 3220+7xx` `rArr 3x-7x lt 3220-4140` `rArr -4x lt - 920` `rArr 4x gt 920` `rArr x gt (0=920)/4` `rArr x gt 230` ...(ii) Hence, the number of litres of the ` 3%` solution of acid must be more than 230 L and less than 920 L.