Question
Download Solution PDFPipes A and B can fill a tank in 10 hours and x hours respectively. Pipe C alone can empty the full tank in 15 hours. When all the three pipes are opened simultaneously for 8 hours, 60% of the tank can be filled. In how many hours can pipes B and C together empty two-fifth of the tank?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
Pipes A and B can fill a tank in 10 hours and x hours respectively.
Pipe C alone can empty the full tank in 15 hours. When all the three pipes are opened simultaneously for 8 hours, 60% of the tank can be filled.
Concept Used:
By Fraction Method
Calculation:
Pipe A fill in the tank in 10 hours
In one hours = 1/10
Pipe B fill in the tank in x hours
In one hours = 1/x
Pipe C alone can empty the full tank in 15 hours.
In one hours = 1/15
When all the three pipes are opened simultaneously for 8 hours, 60% of the tank can be filled.
⇒ (1/10 + 1/x - 1/15)8 = 60/100
⇒ (15x + 150 - 10x)/150x = 60/800
⇒ 90x = 40x + 1200
⇒ 50x = 1200
⇒ x = 24
Total Work is 150x
⇒ 150 × 24 = 3600
B and C together empty two-fifth of the tank
⇒ {3600 × (2/5)}/(10x - 150)
⇒ 1440/90
⇒ 16 hours
∴ 16 hours can pipes B and C together empty two-fifth of the tank.
Last updated on May 9, 2025
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