**Question:**

How many litres of water will have to be added to 600 litres of the 45% solution of acid so that the resulting mixture will contain more than 25%, but less than 30% acid content?

**Solution:**

Let x litres of water be added.

Then total mixture = x + 600

Amount of acid contained in the resulting mixture is 45% of 600 litres.

It is given that the resulting mixture contains more than 25% and less than 30% acid content.

Therefore,

$45 \%$ of $600>25 \%$ of $(x+600)$

And

30% of (x+600) > 45% of 600

When

45% of 600 > 25% of (x+600)

Multiplying both the sides by 100 in above equation

$\frac{45}{100} \times 600>\frac{25}{100} \times(\mathrm{x}+600)$

$45 \times 600>25(x+600)$

$27000>25 x+15500$

Subtracting 15500 from both the sides in above equation

$27000-15500>25 x+15500-15500$

$11500>25 x$

Dividing both the sides by 25 in above equation

$\frac{11500}{25}>\frac{25 x}{25}$

460 > x

Now when,

45% of 600 < 30% of (x+600)

Multiplying both the sides by 100 in the above equation

$\frac{45}{100} \times 600<\frac{30}{100} \times(x+600)$

$45 \times 600<30(x+600)$

$27000<30 x+18000$

Subtracting 18000 from both the sides in above equation

$27000-18000<30 x+18000-18000$

$9000<30 x$

Dividing both the sides by 30 in above equation

$\frac{9000}{30}>\frac{30 x}{30}$

300 < x

Thus, the amount of water required to be added ranges from 300 litres to 460 litres.