|
|
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.
21 Jun 2021, 06:15
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
64% (02:58) correct
36%
(03:25)
wrong
based on 175
sessions
History
Date
Time
Result
Not Attempted Yet
Two taps can separately fill a tank in 4 minutes and 5 minutes respectively. Due to a small hole at the bottom of the tank, the two taps together take 30 seconds more time to fill the tank. The hole can empty the completely filled tank in
(A) 980/71 minutes
(B) 980/81 minutes
(C) 980/91 minutes
(D) 980/61 minutes
(E) 981/51 minutes
(A) 980/71 minutes
(B) 980/81 minutes
(C) 980/91 minutes
(D) 980/61 minutes
(E) 981/51 minutes
21 Jun 2021, 09:01
Kudos
Bookmarks
Bunuel
Let the capacity of the tank be 20 units (LCM of 4 & 5)
Combined efficiency of both the taps is 5units/hr + 4units/hr = 9units/hr
Time required to fill the tank is 20/9 , however due to a leak the time taken is 1/2 minute more..
Thus, \(\frac{20}{(9 - x)} - \frac{20}{9} = \frac{1}{2}\)
Or, \(\frac{180 - 180 + 20x}{(81 - 9x)} = \frac{1}{2}\)
Or, \(40x = 81 - 9x\)
Or,\( x = \frac{81}{49}\)
So, Time required to empty the whole tank is 20*49/81 = 980/81, Answer must be (B)
21 Jun 2021, 09:38
Kudos
Bookmarks
Solution:
Let the capacity of tank = \(x\) lts
Tap A can fill it in 4 minutes.
Rate of work on tap A = \(\frac{x}{4}\) lts per minute.
Tap B can fill it in 5 minutes.
Rate of work on tap B = \(\frac{x}{5}\) lts per minute.
Together both the taps can fill = \(\frac{x}{4}+\frac{x}{5}=\) lts per minute
Due to the leakage, taps are taking 0.5 extra minute.
In this 0.5 minute taps will fill extra \(= 0.5\times \frac{9x}{20} =\frac{9x}{40}\) lts
The leak has been emptying the tank in = \(\frac{20}{9}+0.5=\frac{49}{18} \)minutes.
The leak will drain \(\frac{9x}{20}\) lts in \(\frac{49}{18}\) minutes
The leak will drain x lts in \(\frac{49}{18}\times \frac{20}{9} = \frac{980}{81}\) minutes
Hence the right answer is Option B.
Let the capacity of tank = \(x\) lts
Tap A can fill it in 4 minutes.
Rate of work on tap A = \(\frac{x}{4}\) lts per minute.
Tap B can fill it in 5 minutes.
Rate of work on tap B = \(\frac{x}{5}\) lts per minute.
Together both the taps can fill = \(\frac{x}{4}+\frac{x}{5}=\) lts per minute
Due to the leakage, taps are taking 0.5 extra minute.
In this 0.5 minute taps will fill extra \(= 0.5\times \frac{9x}{20} =\frac{9x}{40}\) lts
The leak has been emptying the tank in = \(\frac{20}{9}+0.5=\frac{49}{18} \)minutes.
The leak will drain \(\frac{9x}{20}\) lts in \(\frac{49}{18}\) minutes
The leak will drain x lts in \(\frac{49}{18}\times \frac{20}{9} = \frac{980}{81}\) minutes
Hence the right answer is Option B.
