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04 Nov 2021, 04:45
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
71% (03:15) correct
29%
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wrong
based on 279
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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
Updated on: 16 Nov 2021, 04:13
Originally posted by sanjitscorps18 on 04 Nov 2021, 07:31.
Last edited by sanjitscorps18 on 16 Nov 2021, 04:13, edited 1 time in total.
Last edited by sanjitscorps18 on 16 Nov 2021, 04:13, edited 1 time in total.
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Two taps can separately fill a tank in 4 minutes and 5 minutes respectively.
Tank filled by both taps simultaneously (in 1 min) = 1/4 + 1/5 = 9/20 mins
Total time taken by both to fill the tank = 20/9 mins
The hole in the bottom makes them work for 1/2 min more hence total time including the hole to fill the tank
= 20/9 + 1/2 = 49/18
If the hole takes x minutes to empty the tank then in 1 minute 18/49 part of the tank gets filled
=> 1/4 + 1/5 - 1/x = 18/49
=> x = 980/81 mins
Option B
Tank filled by both taps simultaneously (in 1 min) = 1/4 + 1/5 = 9/20 mins
Total time taken by both to fill the tank = 20/9 mins
The hole in the bottom makes them work for 1/2 min more hence total time including the hole to fill the tank
= 20/9 + 1/2 = 49/18
If the hole takes x minutes to empty the tank then in 1 minute 18/49 part of the tank gets filled
=> 1/4 + 1/5 - 1/x = 18/49
=> x = 980/81 mins
Option B
General Discussion
16 Nov 2021, 04:04
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t1=4,t2=5 LCM(4,5)=20units=W(Capacity)
R1=5,R2=4,Rfill(1+2)=9,
Time to fill should be=20/9, but due to leak, time=(20/9+1/2)=49/18
Rfill-Rleak*49/18=20
(9-x)*49/18=20
x=81/49
time take to empty the tank=Capacity/x=980/81
Option B
R1=5,R2=4,Rfill(1+2)=9,
Time to fill should be=20/9, but due to leak, time=(20/9+1/2)=49/18
Rfill-Rleak*49/18=20
(9-x)*49/18=20
x=81/49
time take to empty the tank=Capacity/x=980/81
Option B
03 Nov 2025, 03:16
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I did with and assumption of the tank's capacity:
If tank = 20 liters, tap A fills in 4 min ⇒ rate = 20/4 = 5 liters/min.
Tap B fills in 5 min ⇒ rate = 20/5 = 4 liters/min.
Combined rate (no hole) = 5 + 4 = 9
5+4=9 L/min.
Time to fill without hole = 20/9 min.
With the hole, the two taps together take 30 seconds more or 0.5 min more, so total time = 20/9 + 0.5 = 49/18 mins
If the hole’s outflow rate is h liter/min. Then the filling rate with the hole = (9 − h)
We can write 20 / (9 - h) = 49/18
Solving this we get h = 81/49 Liters/min
So the hole alone will drain out the water in = 20/h
=980/ 81 minutes
And: B.
If tank = 20 liters, tap A fills in 4 min ⇒ rate = 20/4 = 5 liters/min.
Tap B fills in 5 min ⇒ rate = 20/5 = 4 liters/min.
Combined rate (no hole) = 5 + 4 = 9
5+4=9 L/min.
Time to fill without hole = 20/9 min.
With the hole, the two taps together take 30 seconds more or 0.5 min more, so total time = 20/9 + 0.5 = 49/18 mins
If the hole’s outflow rate is h liter/min. Then the filling rate with the hole = (9 − h)
We can write 20 / (9 - h) = 49/18
Solving this we get h = 81/49 Liters/min
So the hole alone will drain out the water in = 20/h
=980/ 81 minutes
And: B.
