Question

HOTS
Pipe A takes 5 hours to fill a tank, and pipe B takes 15 hours to fill the same tank.
i. If the Pipes A and B are opened simultaneously how much time will take to
fill the tank
ii. If Pipe A is opened for one hour, and then Pipe B is also opened, in how
much time will the empty tank fill?​

Answer 1

Let the volume of tank = x

hence rate of filling by A = X/5

"". "" B = B/15

now I) when opened simultaneously both will be filling it that's why

x = (x/5 + x/15)T where T is time

of x = 4Tx/15

now solving it we get T = 15/4 hrs

A is opened for 1 hr hence X/5 amount of tank is filled

now volume left = 4X/5

now using above method

4X/5 = 4TX/15

hence T = 3 hrs

hence time taken will be 3hrs

Answer 2

Work done by Pipe A in 1 hour =1/5
Work done by Pipe B in 1 hour=1/15

i) Condition:They are opened simultaneously= Work done together

Therefore 1/5+1/15
LCM=15
3/15+1/15
4/15

They complete 4/15 of the work in one hour
Hence, time taken to complete the work=15/4= 3.75
Answer= Tank was filled in 3 hours 45 minutes.


ii)Condition:Pipe A is opened for 1 hour and then Pipe B is also opened.
Work done by Pipe A=20% (100%= 5 hours)
Work left=80%
100% is filled by both A and B in 15/4 hours
So by unitary method
15/4*100/80
=3hours (Time taken to complete remaining work)

Answer= Total time taken to fill the tank is 4 hours( A alone(1)+ A+B(3) )

Hope it helps