Pipes and Cistern

In pipes and cistern suppose, a water tank or a cistern is connected with two types of pipes to fill and empty it.

(i) Inlet: The pipe which fills the tank up is called an inlet.

(ii) Outlet: The pipe which empties the tank is called an outlet.

If an inlet fills up the cistern in 5 hours, then in 1 hour it fills up 1/5₀ part of it. We say that the work done by inlet in 1 hour is ¹/₅.

Rule 1. Suppose a pipe fills a tank in n hours.

Then, part of the tank filled in 1 hour = $\frac{1}{n}$, i.e., work done by the inlet in 1 hour= $\frac{1}{n}$

Rule 2. Suppose an outlet empties a full tank in m hours.

Then, part of the tank emptied in 1 hour = $\frac{1}{m}$, i.e., work done by the outlet in 1. hour = $\frac{-1}{m}$

Solved Problems on Pipes and Cistern 1. A tap A can fill a cistern in 8 hours while tap B can fill it in 4 hours. In how much times will the cistern be filled if both A and B are opened together?

Solution:

Time taken by tap A to fill the cistern = 8 hours.

Work done by tap A in 1 hour = ¹/₈

Time taken by tap B to fill the cistern = 4 hours.

Work done by tap B in 1 hour = ¹/₄

Work done by (A + B) in 1 hour = (¹/₈ + ¹/₄) = ³/₈

Therefore, time taken by (A + B) to fill the cistern = ⁸/₃ hours = 2 hours 40 min.

2. A tap A can fill a cistern in 4 hours and the tap B can empty the full cistern in 6 hours. If both the taps are opened together in the empty cistern, in how much time will the cistern be filled up?

Solution:

Time taken by tap A to fill the cistern = 4 hours.

Work done by tap A in 1 hour = 1/4ᵗʰ

Time taken by tap B to empty the full cistern = 6 hours.

Work done by tap B in 1 hour = -1/6ᵗʰ (since, tap B empties the cistern).

Work done by (A + B) in 1 hour (¹/₄ - ¹/₆) = (3 - 2)/12 = 1/12th part of the tank is filled.

Therefore, the tank will fill the cistern = 12 hours.

3. A cistern can be filled by two taps A and B in 12 hours and 16 hours respectively. The full cistern can be emptied by a third tap C in 8 hours. If all the taps are turned on at the same time, in how much time will the empty cistern be filled completely?

Solution:

Time taken by tap A to fill the cistern = 12 hours.

Time taken by tap B to fill the cistern = 16 hours.

Time taken by tap C to empty the full cistern = 8 hours.

A’s 1 hour’s work = ¹/₁₂

B’s 1 hour’s work = ¹/₁₆

C’s 1 hour’s work = $\frac{-1}{8}$ (cistern being emptied by C)

Therefore, (A + B + C)’s 1 hours net work= (¹/₁₂ + ¹/₁₆ - ¹/₈) = ¹/₄₈

Therefore, time taken by (A + B + C) to fill the cistern = 48 hours.

4. A tank can be filled by two taps A and B in 8 hours and 10 hours respectively. The full tank can be emptied by a third tap C in 9 hours. If all the taps are turned on at the same time, in how much time will the empty tank be filled up completely?

Solution:

Time taken by tap A to fill the tank = 8 hours.

Time taken by tap B to fill the tank = 10 hours.

Time taken by tap C to empty the full tank = 9 hours.

A’s 1 hour’s work = ¹/₈

B’s 1 hour’s work = ¹/₁₀

C’s 1 hour’s work = $\frac{-1}{9}$ (cistern being emptied by C)

Therefore, (A + B + C)’s 1 hours net work= (¹/₈ + ¹/₁₀ - ¹/₉) = (45 + 36 – 40)/360 = ⁴¹/₃₆₀

Thus, tank will be filled completely in $\frac{360}{41}$ hours, when all the three are opened together.

5. A cistern can be filled by one tap in 5 hours and by another in 4 hours. How long will it take to fill if both the taps are opened simultaneously?

Solution:

Time taken by first tap to fill the cistern = 5 hours.

Work done by first tap in 1 hour = ¹/₅

Time taken by second tap to fill the cistern = 4 hours.

Work done by second tap in 1 hour = ¹/₄

Work done by (first tap + second tap) in 1 hour = (¹/₅ + ¹/₄) = ⁹/₂₀

Therefore, time taken by (first tap + second tap) to fill the cistern = $\frac{20}{9}$ hours.