A solution of 8%

Volume of the 8% solution = 640 litres

Boric acid present in the 8% solution = 8% of 640 …(i)

And the rest 92% of 640 litres is water in the 8% solution.

Let volume of 2% solution added to 640 liters be x.

Boric acid present in 2% solution = 2% of x …(ii)

New volume of 8% solution = 640 + x …(iii)

Boric acid present in the new solution (that is, after adding x litres of 2% solution to 8% solution) = Boric acid present in the 8% solution + Boric acid present in the 2% solution [from (i) & (ii)]

⇒ Boric acid present in the new solution = 8% of 640 + 2% of x

⇒Boric acid present in the new solution =

⇒Boric acid present in the new solution = …(iv)

According to the question,

The resulting mixture is to be more than 4% but less than 6% boric acid.

That is, the boric acid content in the resulting mixture must be more than 4% but less than 6% boric acid.

So, first let us take boric acid content in the resulting mixture to be more than 4%.

⇒ Boric acid present in the new solution > 4% of the new volume of 8% solution

[from (iii) & (iv)]

⇒ 2x + 5120 > 2560 + 4x

⇒ 5120 – 2560 > 4x – 2x

⇒ 2560 > 2x

⇒ 2x < 2560

⇒ x < 1280

Now, let us the take boric acid in the resulting mixture to be less than 6%.

⇒ Boric acid present in the new solution < 6% of the new volume of 8% solution

[from (iii) & (iv)]

⇒ 2x + 5120 < 3840 + 6x

⇒ 5120 – 3840 < 6x – 2x

⇒ 1280 < 4x

⇒ 4x > 1280

⇒ x > 320

We have

x < 1280 & x > 320

⇒ 320 < x < 1280

Hence, the required liters of 2% solution to be added to 8% of the solution is between 320 liters and 1280 liters.