Math, asked by seema23466, 10 months ago

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?

Answers

Answered by Anonymous
3

Answer:

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your answer is here !

Step-by-step explanation:

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 = −19 (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 36041 hours, when all the three are opened together.

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Answered by sonabrainly
3

Answer:

Step-by-step explanation:

 tap A to fill the tank = 8 hours.

B to fill the tank = 10 hours.

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 = −19 (cistern being emptied by C)

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

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