Understanding the Problem

This problem involves calculating the time required to fill a tank using a different number of pipes. The key concept is that the number of pipes and the time taken to fill the tank are inversely proportional. This means if you decrease the number of pipes, the time required will increase proportionally.

Calculations for Pipe Time

First, convert the given time into a single unit, minutes:

• Initial time: $1 \text{ hour } 20 \text{ minutes} = (1 \times 60) + 20 = 80 \text{ minutes}$.

Let $P_1$ be the initial number of pipes and $T_1$ be the initial time taken. Let $P_2$ be the new number of pipes and $T_2$ be the time we need to find.

• $P_1 = 6$ pipes
• $T_1 = 80$ minutes
• $P_2 = 5$ pipes
• $T_2 = ?$ minutes

Using the inverse proportion relationship ($P_1 \times T_1 = P_2 \times T_2$):

$ 6 \text{ pipes} \times 80 \text{ minutes} = 5 \text{ pipes} \times T_2 $

$ 480 \text{ pipe-minutes} = 5 \times T_2 $

Now, solve for $T_2$:

$ T_2 = \frac{480 \text{ pipe-minutes}}{5 \text{ pipes}} $

$ T_2 = 96 \text{ minutes} $

Converting Time Back to Hours and Minutes

Convert the calculated time ($96$ minutes) back into hours and minutes:

• $96 \text{ minutes} = 60 \text{ minutes} + 36 \text{ minutes}$
• $96 \text{ minutes} = 1 \text{ hour} + 36 \text{ minutes}$
• $T_2 = 1 \text{ hour } 36 \text{ minutes}$

Therefore, it will take 5 pipes 1 hour and 36 minutes to fill the tank.