Let x litres of 2% boric and acid solution be added to 640 litres of 8% solution of boric acid.
%Strength = \(\frac{\frac{8}{100} \times 640 + \frac{2}{40} \times x}{640 +x}\)
= \(\frac{5120 + 2x}{100 (640 +x)}\)
It is given that boric acid content in the resulting mixture ranges from 4% to 6%
Therefore,
Multiplying throughout by 100 in the above equation
\(\frac{5120 + 2{\text{x}}}{640 +{\text{x}}} > 4\) and \(\frac{5120 + 2{\text{x}} } {640 + {\text{x}}}\) < 6
When,
\(\frac{5120 + 2{\text{x}} } {640 + {\text{x}}}\) > 4
Multiplying both the sides by (640 + x) in the above equation
\(\frac{5120 + 2{\text{x}} } {640 + {\text{x}}}\) (640 + x) > 4(640 + x)
5120 + 2x > 2560 + 4x
Subtracting 2x from both the sides in above equation
5120 + 2x – 2x > 2560 + 4x – 2x
5120 > 2560 + 2x
Subtracting 2560 from both the sides in above equation
5120 – 2560 > 2560 + 2x – 2560
2560 > 2x
Dividing both the sides by 2 in above equation
\(\frac{2560}{2} > \frac{2{\text{x}}}{2}\)
1280 > x
Now when
\(\frac{5120 + 2{\text{x}}}{640 + {\text{x}}}(640 + {\text{x}})\) < 6(640 + x)
5120 + 2x < 3840 + 6x
Subtracting 2x from both the sides in above equation
5120 + 2x – 2x < 3840 + 6x – 2x
5120 < 3840 + 4x
Subtracting 3840 from both the sides in above equation
5120 – 3840 < 3840 + 4x – 3840
1280 < 4x
Dividing both the sides by 4 in above equation
\(\frac{1280}{4} < \frac{4{\text{x}}}{4}\)
320 < x
Thus, the value of 2% boric acid solution to be added ranges from:
320 to 1280 litres
