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 = \(\frac{-1}{m}\)

**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. **

**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.**

**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. **

**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. **

**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. **

● **Time and Work**

**Practice Test on Time and Work**

● **Time and Work - Worksheets**

**8th Grade Math Practice** **From Pipes and Cistern to HOME PAGE**

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