DSSSB TGT CS 2023 Shift 2 – Question 055

Pipes A and B can fill a tank in 20 hours and 30 hours, respectively. Pipe C is an emptying pipe. When all the three pipes are opened together, \(\frac{2}{9}\text{ th}\) part of the tank is filled in 8 hours. Pipe C alone can empty \(\frac{1}{3}\text{ rd}\) part of the tank in:

• 4 hours
• 3 hours
• 5 hours
• 6 hours

Step 1: Calculate the individual rates of pipes A and B:
\[ \text{Rate of A} = \frac{1}{20}\text{ tank/hour} \]
\[ \text{Rate of B} = \frac{1}{30}\text{ tank/hour} \]
Combined filling rate of A and B:
\[ \text{Rate of }(A + B) = \frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{5}{60} = \frac{1}{12}\text{ tank/hour} \]

Step 2: Determine the net combined rate of all three pipes:
In 8 hours, \(\frac{2}{9}\) of the tank is filled.
\[ \text{Net rate of }(A + B – C) = \frac{2/9}{8} = \frac{2}{72} = \frac{1}{36}\text{ tank/hour} \]

Step 3: Calculate the emptying rate of pipe C:
\[ \frac{1}{12} – \text{Rate of C} = \frac{1}{36} \]
\[ \text{Rate of C} = \frac{1}{12} – \frac{1}{36} = \frac{3 – 1}{36} = \frac{2}{36} = \frac{1}{18}\text{ tank/hour} \]
This means pipe C alone can empty the entire tank in 18 hours.

Step 4: Find time taken by C to empty \(\frac{1}{3}\) of the tank:
\[ \text{Time} = \frac{1}{3} \times 18 = 6\text{ hours} \]
Hence, the correct answer is option d).