Question

Two pipes can fill a tank in 3 and 4
hrs resp. Third pipe can empty it in 1
hrs. If the pipes are open in the order
of 3, 4 and 5 pm resp. How soon the
tank will be empty?​

Answer 1

Step-by-step explanation:

Given Two pipes can fill a tank in 3 and 4  hrs resp. Third pipe can empty it in 1  hrs. If the pipes are open in the order  of 3, 4 and 5 pm resp. How soon the  tank will be empty?

• Let the tank be emptied in x hrs after 3 P.M
• Work done by A in x hrs, by B in x – 1 hrs and x – 2 hrs = 0
• So x / 3 + (x – 1) / 4 - (x – 2)  = 0
•  So 4x + 3x – 3 / 12 = x – 2
• So 7x – 3 = 12 x – 24
• 5x = 21
• Or x = 21/5

Therefore x = 4 hrs 12 min after 3 P.M. so it will be 7:12 P.M

Reference link will be

https://brainly.in/question/12245583

Answer 2

Efficiency of first pipe = $\frac{1}{3}$

Efficiency of second pipe  = $\frac{1}{4}$

Efficiency of third pipe = -1

Let the total time from 3P.M. be x hrs for tank to get emptied then

First pipe works for full x hours

second pipe works for x-1 hours

third pipe works for x-2 hours

Total work done = $\frac{1}{3}$.x + $\frac{1}{4}$.(x-1) + (-1)(x-2) = 0

$\frac{4x + 3x-3-12x+24}{12}$ = 0

5x = 21

x = $\frac{21}{5}$

x = $4\frac{1}{5}$

i.e 4 hours and one fifth of hour i.e. 12 mins

so after 4 hrs and 12 mins from 3P.M.

i.e. at 7:12PM