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?
According to the question,
Let x litres of 3% solution is to be added to 460 liters of the 9% of solution
Then, we get,
Total solution = (460 + x) litres
Total acid content in resulting solution
= (460 × 9/100 + x × 3/100)
= (41.4 + 0.03x)%
According to the question, we have,
Resulting mixture should be more than 5% acidic but less than 7% acidic
So we get,
⇒ 5 % of (460 + x) < 41.4 + 0.03x < 7% of (460 + x)
⇒ 5/100 × (460 + x) < 41.4 + 0.03x < 7/100 × (460 + x)
⇒ 23 + 0.05 x < 41.4 + 0.03x < 32.2 + 0.07x
Now, we have
⇒ 23 + 0.05x < 41.4 + 0.03x and 41.4 + 0.03x < 32.2 + 0.07x
i.e., 0.02x < 18.4 and 0.04x > 9.2
⇒ 2x < 1840 and 4x > 920
⇒ x < 920 and x > 230
⇒ 230 < x < 920
Hence, solution between 230 l and 920 l should be added.