A solution of 9% acid is to be diluted

Solution:

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.