Solution of `9%` acid is to be diluted by adding `3%` acid solution to it. The resulting mixture is to be more than `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

Explanation

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.