Two taps X and Y can fill a water tank in 15 hours and 20 hours respectively. Both taps are opened together at 2 p.m. At what time should tap X be closed so that the tank is just full at exactly 2 a.m. the next day?

Introduction / Context:
This problem involves scheduling the operation of two taps so that a tank is full at a specified future time. Both taps start together, but one tap is turned off earlier. The question checks understanding of work rates over different time intervals, requiring careful tracking of how long each tap runs and how much work is done during those intervals.

Given Data / Assumptions:

Tap X alone fills the tank in 15 hours.

Tap Y alone fills the tank in 20 hours.

Both taps are opened at 2 p.m.

The tank must be exactly full at 2 a.m. the next day.

Flow rates are constant and there are no leaks.

Concept / Approach:
First, compute the total duration between 2 p.m. and 2 a.m., which is 12 hours. Let x be the number of hours that tap X remains open after 2 p.m. Tap Y runs the entire 12 hours. We then express the total work done as the sum of work contributed by each tap: tap X works for x hours at its rate, and tap Y works for 12 hours at its rate. This total must equal one full tank. Solving the resulting equation for x tells us when tap X should be closed.

Step-by-Step Solution:
Let the capacity of the tank be 1 unit.Rate of tap X = 1/15 tank per hour.Rate of tap Y = 1/20 tank per hour.Total available time from 2 p.m. to 2 a.m. is 12 hours.Let tap X be open for x hours (starting at 2 p.m.).Tap Y remains open for the entire 12 hours.Work done by X = x * (1/15) tank.Work done by Y = 12 * (1/20) tank.Total work must equal 1 tank, so x/15 + 12/20 = 1.Compute 12/20 = 3/5.The equation becomes x/15 + 3/5 = 1.Rearrange: x/15 = 1 - 3/5 = 2/5.Thus x = 15 * (2/5) = 30/5 = 6 hours.So tap X should be closed 6 hours after 2 p.m., that is at 8 p.m.
Verification / Alternative check:
From 2 p.m. to 8 p.m., both taps work for 6 hours.Work in first 6 hours = 6 * (1/15 + 1/20) = 6 * (7/60) = 42/60 = 7/10 of the tank.From 8 p.m. to 2 a.m., only tap Y works for another 6 hours.Work by Y in last 6 hours = 6 * (1/20) = 6/20 = 3/10 of the tank.Total work = 7/10 + 3/10 = 1 full tank, which confirms the timing.
Why Other Options Are Wrong:
Closing tap X at 7 p.m., 9 p.m., 10 p.m. or 11 p.m. would lead to total work different from exactly one tank. Closing too early would mean the tank does not reach full capacity by 2 a.m. Closing too late would cause the tank to overfill before 2 a.m. Only 8 p.m. balances the contributions correctly.

Common Pitfalls:
Some students miscalculate the total time from 2 p.m. to 2 a.m. or forget that tap Y runs throughout. Others attempt to average the times 15 hours and 20 hours instead of adding work contributions. Mismanaging fractions when solving x/15 + 3/5 = 1 is another common error. Writing out each step and converting to simple fractions helps to avoid mistakes.

Final Answer:
Tap X should be closed at 8 p.m. so that the tank is full at exactly 2 a.m.