Question

In a tank six taps of equal efficiency are fitted on equal intervals. The first pipe is at the base of the tank and the 6th pipe is at 5/6th of height of the tank. Then calculate in how many times the whole tank will empty. If the first pipe

Answer 1

Given:

In a tank six taps of equal efficiency are fitted on equal intervals. The first pipe is at the base of the tank and the 6th pipe is at 5/6th of height of the tank.  

To find:

Then calculate in how many times the whole tank will empty. If the first pipe can empty the tank in 12 hours.

Solution:

From given, we have,

The first pipe is at the base of the tank and the 6th pipe is at 5/6th of height of the tank.  

Therefore, the time taken to empty the tank is given by,

= 5/6 + 5/5 + 5/4 + 5/3 + 5/2 + 5/1

= 72/12 + 10/12 + 15/12 + 20/12 + 30/12

= 49/4

= 12.25

Therefore, the time required to empty the whole tank is 12 hours 15 minutes.

Answer 2

Answer:

as per time is not mentioned

Explanation:

let each pipe take 12 hours

and let each pipe have 6 unit eff

capacity =12*6 = 72

each = 12

time = 12/6+12/12+12/18+12/24+12/30+24/36

= 12(60+30+20+15+12+10)/360

= 12*147/360 = 147/30 *60 = 294 mint