A 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 litres of the 9% solution, how many litres of 3% solution will have to be added?

#### Solution

Let x litres of 3% solution be added to 460 litres of 9%.

∴ Total amount of mixture = (460 + x) litres

Given that the acid contents in the resulting mixture is more than 5% but less than 7% acid.

∴ 5% of (460 + x) < `9/100 + 3/100 xx x < 7%` of (460 + x)

⇒ `5/100 (460 + x) < (4140 + 3x)/100 < 7/100 (460 + x)`

⇒ 5(460 + x) < 4140 + 3x < 3220 + 7x

⇒ 2300 + 5x < 4140 + 3x < 3220 + 7x

⇒ 2300 + 5x < 4140 + 3x and 4140 + 3x < 3220 + 7x

⇒ 5x – 3x < 4140 – 2300 and 3x – 7x < 3220 – 4140

⇒ 2x < 1840 and –4x < –920

⇒ `x < 1840/2` and 4x > 920

⇒ x < 920

∴ `x > 920/4` and x > 230

Hence the required amount of acid solution is more than 230 litres and less than 920 litres.