Answer :

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.


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