Two taps can fill a cistern in 4 hours and 38 hours respectively. A third tap can empty it in 38 hours. How long (in hours) will it take to fill one-fourth of the empty cistern, if all of them are opened together?

The correct option is 1.

Let's solve the problem step-by-step to find the total time taken to fill one-fourth of the empty cistern when all three taps are opened together.

Step 1: Determine the work done by each tap in 1 hour.
Let Tap A be the first tap that fills the cistern in 4 hours.
Work done by Tap A in 1 hour = $\frac{1}{4}$ of the cistern.

Let Tap B be the second tap that fills the cistern in 38 hours.
Work done by Tap B in 1 hour = $\frac{1}{38}$ of the cistern.

Let Tap C be the third tap that empties the cistern in 38 hours.
Work done by Tap C in 1 hour (emptying) = -$\frac{1}{38}$ of the cistern.

Step 2: Calculate the net work done in 1 hour when all taps are open together.
Net work in 1 hour = Work done by Tap A + Work done by Tap B - Work done by Tap C

$\text{Net work in 1 hour}=\frac{1}{4}+\frac{1}{38}-\frac{1}{38}=\frac{1}{4}$

So, together all three taps fill $\frac{1}{4}$ of the cistern in 1 hour.

Step 3: Calculate the time taken to fill one-fourth ($\frac{1}{4}$) of the cistern.
Since $\frac{1}{4}$ of the cistern is filled in 1 hour, the time required to fill one-fourth of the cistern is 1 hour.

Therefore, it will take 1 hour to fill one-fourth of the empty cistern.