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
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Jun 21, 24 02:19 AM
Jun 21, 24 01:59 AM
Jun 21, 24 01:30 AM
Jun 21, 24 01:00 AM
Jun 19, 24 09:49 AM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.