A pot can be filled by pipe X in 2 hours and pipe Y in 6 hours. At 10

1.	A pot can be filled by pipe X in 2 hours and pipe Y in 6 hours. At 10 A.M. pipe X was opened. At what time will the pot be filled, if pipe Y is opened at 11 A.M.?1. 12.45 A.M.2. 5 P.M.3. 11.45 A.M.4. 12 P.M.
Answer» Correct Answer - Option 3 : 11.45 A.M. Given: A pot can be filled by pipe X in 2 hours and pipe Y in 6 hours. Pipe X was opened at 10 A.M. Pipe Y is opened at 11 A.M. Concept: If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank. Calculation: Part of the pot filled in 1 hours by pipe X = \(\frac{1}{2}\) Part of the pot filled by both pipes in 1 hour ⇒ \(\frac{1}{2}{\rm{}} + {\rm{}}\frac{1}{6}{\rm{}} = {\rm{}}\frac{{3{\rm{\;}} + {\rm{\;}}1}}{6}{\rm{}} = {\rm{}}\frac{2}{3}\) Time taken to fill \(\frac{2}{3}\) part = 60 minutes Time taken to fill \(\frac{1}{2}\) part ⇒ \(\frac{{60\; \times \;3}}{2} \times \frac{1}{2}{\rm{\;}}\)= 45 minutes ∴ The pot will be filled at 11:45 A.M.

A pot can be filled by pipe X in 2 hours and pipe Y in 6 hours. At 10 A.M. pipe X was opened. At what time will the pot be filled, if pipe Y is opened at 11 A.M.?1. 12.45 A.M.2. 5 P.M.3. 11.45 A.M.4. 12 P.M.

Answer» Correct Answer - Option 3 : 11.45 A.M.

Given:

A pot can be filled by pipe X in 2 hours and pipe Y in 6 hours.

Pipe X was opened at 10 A.M.

Pipe Y is opened at 11 A.M.

Concept:

If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank.

Calculation:

Part of the pot filled in 1 hours by pipe X = \(\frac{1}{2}\)

Part of the pot filled by both pipes in 1 hour

⇒ \(\frac{1}{2}{\rm{}} + {\rm{}}\frac{1}{6}{\rm{}} = {\rm{}}\frac{{3{\rm{\;}} + {\rm{\;}}1}}{6}{\rm{}} = {\rm{}}\frac{2}{3}\)

Time taken to fill \(\frac{2}{3}\) part = 60 minutes

Time taken to fill \(\frac{1}{2}\) part

⇒ \(\frac{{60\; \times \;3}}{2} \times \frac{1}{2}{\rm{\;}}\)= 45 minutes

∴ The pot will be filled at 11:45 A.M.