|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
01 Aug 2019, 00:02
Kudos
Bookmarks
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
100%
(03:06)
wrong
based on 4
sessions
History
Date
Time
Result
Not Attempted Yet
Official CAT 2018 Questions; Section: QA
A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?
A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?
10
03 Aug 2019, 03:43
Kudos
Bookmarks
Narenn
If x=efficiency of input pipe and y=efficiency of output pipe, then
36x-30y=1
300x-360y=1
t*2x-t*y=1
=> x= 1/12, and y=1/15
Thus, time taken when one draining and two filling pipes are on = t = 1/(2x-y) = 1/(1/6-1/15) = 10 hours.
07 Aug 2019, 02:33
Kudos
Bookmarks
Alternative approach:
If x = rate of input and y = rate of output [original].
Then:
(x-y) = 10*(5x/6 - 6y/5) [as the time taken is 10 times in the second case, so the rate in the first case must be 10 times the second]
=> y =(2/3)x
Put this in 6*(x-y)=1, to get x=1/2 and y=1/3
Then to find the required time t, we solve t*(x/3 - y/5) = 1, to get t=10 hours.
If x = rate of input and y = rate of output [original].
Then:
(x-y) = 10*(5x/6 - 6y/5) [as the time taken is 10 times in the second case, so the rate in the first case must be 10 times the second]
=> y =(2/3)x
Put this in 6*(x-y)=1, to get x=1/2 and y=1/3
Then to find the required time t, we solve t*(x/3 - y/5) = 1, to get t=10 hours.
