Question

A cistern is filled in 30 minutes by the first tube and emptied in 25 minutes by the second tube. The cistern The second tube was played for 10 minutes in a half-filled state. Then turn it off and the first tube is 6 The minutes were played. Now if the two pipes are opened together, how long will the tank be waterless?​

Answer 1

Answer:

45min

Step-by-step explanation:

work done by the first tap in one minute =1/30

work done by second in one minute =1/25

if second tube played for 10min then work done =10/25=2/5

this happened when tank if half so that means at the beginning the tank = 1/2

so after 10 min of second tank =1/2-2/5(since second tap emptying)

= (5-4)/10=1/10 of tank is there

so first tube played for 6 min that means work done = 6/30=1/5

so we have to add that to above condition

so at present the water should be =1/10 +1/5= 3/10 is there

so if two tubes are started at a time then it can empty in one minute = 1/25-1/30 =(6-5)/150=1/150

that means it can empty 1/150 of the tank water in one minute if both tubes are working at the same time

and out tank was at a level=3/10

so it will take a time of (3/10)/(1/150)=3*150/10=3*15=45min

so it will take 45 min to waterless

Answer 2

Answer:

tysm dear........ ♥♥♥