Math, asked by kumarsarmendra42, 10 months ago

pipes A& B can fill a tank in 16hours and 24hours respectively and pipe C alone can empty the full tank in X hours all the pipes were opened together at 10:30am but C closed at 2:30 pm .if the tank was full at 8:30pm on the same day ,than what is the value of X?​

Answers

Answered by UmangThakar
5

Answer: 96 Hours.

Step-by-step explanation:

Part of the tank filled in one hour by pipe A = 1/16

Part of the tank filled in one hour by pipe B = 1/24

Pipe C, can empty the full tank in X hours,

Part of the tank emptied in one hour by pipe C = 1/X

When all the three pipes are opened in the beginning,

Net part filled in one hour = (1/15)+(1/20)-(1/X) = 1/16 + 1/24 - 1/X

In 4 hours (from 10.30 am to 2.30 pm) , the part filled, F1 = 4 x (1/16 + 1/24 - 1/X  )  

= 4/16 + 4/24 - 4/X

= 10/24 - 4/X

= 5/12 - 4/X

After 4 hours, pipes A and B together take 6 (from 2.30 pm TO 8.30 pm) hours to fill the remaining part,

   

Part filled by A and B in one hour = (1/16)+(1/24) = 5/48

Part filled by A and B in 6 hours F2 = 6 x (5/48) = 5/8

Total part filled = F1 + F2, which shall be equal to 1

(5/12 - 4/X) + 5/8  =1  

100/96 - 1 = 4/X

X = 96 Hours

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