Math, asked by manjotsandhu9494, 4 months ago

A cistern can be filled by two taps A and B in 12 hours and 16 hours , respectively , The full cistern can be empited by a third tap C in 8 hours . If all the taps be turned on at the same time , in how much time will the empty cistern be filled up completely ?​

Answers

Answered by Salmonpanna2022
6

Answer:

Therefore, the Cistern will be filled up completely in 48 hours, if all the three taps are opened at the same time.

Step-by-step explanation:

Given that:- A cistern can be filled by two taps A and B in 12 hours and 16 hours respectively The full cistern can be emptied by a third tap C in 8 hours.

To find out:- If all the taps areturned on at the same time, in how much time will the empty cistern be filled up completely?

Solution:-

The time taken by A to fill the cistern = 12 hours

The time taken by B to fill the Cistern = 16 hours

The time taken by C to fill the Cistern = 8 hours.

Now, we have to find their first hour work,

∴ A's 1 hour's work = 1/12

B's 1 hour's work = 1/16

C's 1 hour's work = -1/8

∴ (ABC)'s 1 hour's work = (1/12)+(1/16)-(1/8)

LCM of 12,16 and 8 is 48.

= (3+4-6)/84

= (7-6)/48

= 1/48

Therefore, the Cistern will be filled up completely in 48 hours, if all the three taps are opened at the same time.

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