A cistern is provided with two pipes A and B. A can fill it in 20 minutes and B can empty it in 30 minutes. If A and B be kept open alternately for one minute each, how soon will the cistern be filled ? (1) 121 minutes (2) 110 minutes (3) 115 minutes (4) 120 minutes
Answers
Answer:
(3) 115 minutes
Step-by-step explanation:
Given,
A cistern is provided with two pipes A and B
A can fill the tank in 20min
B can empty the tank in 30min
The part of tank filled by pipe A in 1 min =
The part of tank filled by pipe B in 1 min =
The part of tank filled by two pipes alternatively in 2 mins
= -
=
=
Therefore,
It takes 2 min to fill 1 litre
If both pipes works continuosly..Then,
The the time taken to fill 57 litres is
= 57×2
= 114 mins
Next, at the 115th minute, remaining three litres are filled by the pipe A
So, it takes 115 mins to fill the tank!
Answer: 115 Minutes
Step-by-step explanation:
Let us suppose that capacity of tank is 60 ltr. (LCM of 20 and 30)
Now,
A can fill the tank in 20 minutes,
It means rate of A = 60/20 = 3 ltr./minutes.
B can empty the tank in 30 minutes,
It means working rate of B = 60/30 = 2 ltr./Minutes
Here given that, both pipes be kept open alternately for one minute each.
It means that,
First we open A for 1 minute and close.
Then, open B for 1 minute and close
Again open A for 1 minute and close,
open B for 1 ...
And so on
A - B + A - B + A - B ......................
Let us Discuss first two minutes:-
Open A for 1 minutes;- Tank filled 3 ltrs.
Open B for next 1 minutes:- Tank got empty = 2 ltrs.
Total water in 2 minutes = 3(Filled) - 2(Got empty)
= 1 ltr.
Here we can conclude that, We are filling 1 ltr. of tank in 2 minutes.
So, 57 ltr. of tank can be fillted 57 × 2 = 114.
Just after this 144 minutes, A will be opened and it will fill the tank with 3ltrs. in next one minutes. So after 57 ltrs. tank will be filled upto 57+3=60
ltrs.
So total time taken = 114 + 1
= 115 Minutes.