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
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8th Grade Math Practice
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