A municipality made a tank for water preservation and added pumps with it. The pumps separately can fill the empty tank in 16, 20 & 30 hours respectively. Today when the 3 pumps started together at 7 am, 1 part of the tank was filled with water. After 1 hour 36 minutes the first pump and 3 after 2 more hours the third pump was stopped. Calculate when the tank was totally filled up.
Answers
Given : A municipality made a tank for water preservation and added pumps with it.
The pumps separately can fill the empty tank in 16, 20 & 30 hours respectively. Today when the 3 pumps started together at 7 am, 1/3 part of the tank was filled with water.
After 1 hour 36 minutes the first pump and after 2 more hours the third pump was stopped.
To Find : when the tank was totally filled up.
Solution:
The pumps separately can fill the empty tank in 16, 20 & 30 hours
=> Can fill in 1 hr together = 1/16 + 1/20 + 1/30
= ( 15 + 12 + 8)/240
= 35/240
= 7/48
filled in 1hr 36 mins = 1 + 36/60 hrs = 1 + 3/5 = 8/5 hrs
= (8/5) (7/48)
= 7/30 filled in 1 hr 36 mins
1/3 was already filled
Hence filled = 7/30 + 1/3 = 17/30
in next 2 hrs = 2 ( 1/20 + 1/30) = 2( 3 + 2) /60 = 1/6
17/30 + 1/6 = 22/30 = 11/15
Left = 1 - 11/15 = 4/15
Last pump n hrs = n * 1/20 = 4/15
=> n = 16/3 hrs = 5 hrs 20 mins
7 AM + 1 h 36 mins + 2 hrs + 5 hr 20 mins
= 3 : 56 PM
tank was totally filled up at 3 : 56 PM
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