There are 3 taps attached to a tank. Tap A takes 24 hours to fill the tank. Tap B takes 12 hours to fill the tank. Tap C takes 6 hours to empty the tank. Now, it is given that Tap A was opened at 1 am, Tap B was opened at 2 am, Tap C was opened at 4 am of same day. In how many hours the tank will be emptied, after Tap C is opened?
Answers
Answer:
Let us name the 3 Taps be T
A
,T
B
,T
C
where T
B
,T
C
are emptying pipes and T
A
is filling Pipe.
Water Emptied by T
B
in 6 hours=Volume of Tank or V
Water Emptied by T
B
in 1 Hour=
6
V
Similarly, Water Emptied by T
C
in 8 Hour=V
Water Emptied by T
C
in 1 Hour=
8
V
Similarly, Water Supplied by T
A
in 2 Hour=V
Water Supplied by T
A
in 1 Hour=
2
V
After 2 Hours,
⇒2
2
V
−2
8
V
−2
6
V
=
12
5V
After 3 Hours,
⇒
12
5V
+
2
V
−
6
V
=
4
3V
Now, only T
A
is opened
Remaining Volume=V−
4
3V
=
4
V
Now Time taken to Fill Remaining Tank
⇒
2
V
4
V
=
2
1
hrs=30min
Answer:
3 hours 26 minutes.
Step-by-step explanation:
Given:- Tap A takes 24 hours to fill the tank.
Tap B takes 12 hours to fill the tank.
Tap C takes 6 hours to empty the tank.
To Find:- Amount of time in which the tank will be emptied after Tap C is opened.
Solution:-
In 1 hour, Tap A can fill 1/24th of the tank.
Tap B can fill 1/12th of the tank.
Tap C can empty 1/6th of the tank.
Total amount of work by Tap A, Tap B and Tap C in 1 hour
= 1/24 + 1/12 - 1/6
= (1 + 2 - 4)/24
= -1/24
So, when all the three taps are opened together, in an hour the whole tank + 1/24th amount of tank can be emptied.
Tap A was opened at 1 am, Tap B was opened at 2 am and Tap C was opened at 4 am.
Therefore, amount of tank that was filled before Tap C was opened
= 1/24 + 1/24 + 1/24 + 1/12 + 1/12
= (1 + 1 + 1 + 2 + 2)/24 = 7/24
Therefore, number of hours in which the tank will be emptied
= 24/7 hours ≈ 3 hours 26 minutes
#SPJ3
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