Math, asked by DheerajMehlawat7906, 1 year ago

A cistern can be filled by two pipes a and b in 10 and 15 hours respectively and is then empited by a tap in 8 hours.If all the taps are opened,then cistern will be fill in

Answers

Answered by santy2
16

Answer:

A cistern can be filled by two pipes a and b in 10 and 15 hours respectively and is then empited by a tap in 8 hours.If all the taps are opened,then cistern will be filled in 24 hours

Step-by-step explanation:

Find the fraction that all the taps opened together will fill in an our:

Tap a fills the cistern in 10 hours:

If 10 hours = 1/1 of the cistern

Then 1 hour = 1 ×1/10

Tap a fills a tenth , 1/10 of the cistern in 1 hour

Tap b can fill the cistern is 15 hours:

If 15 hours = full cistern, 1

Then 1 hour = 1 × 1/15

Tap b in 1 hour fills 1/15 of the cistern

Tap C can empty the tank in 8 hours:

If 8 hours = 1

Then 1 hour = 1 × 1/8

Therefore tap c empties 1/8 of the cistern in 1 hour

⇒ Add tap a and b and minus c to find the fraction of the tap that will be filled in an hour:

a + b - c = 1/10 + 1/15 - 1/8

            =    (12 + 8 - 15) / 120

            =     5 / 120

            =    1 / 24

Therefore the fraction of the cistern filled when all 3 taps are opened in an hour is 1 / 24 .

Find the time it will take the whole cistern to be filled if in 1 hour the fraction filled is 1/24

     If 1/24 of the cistern = 1 hour

   Then 24/24 of the cistern = 1 hour × 24/24 ×24/1

                                            = 24 hours

Therefore if the three taps are opened the cistern will be full after 24 hours.

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