Question

Two taps t1 and t2 can fill in 5 hours and 20 respectively. If both taps are open then due to leakage it took 30 minutes more to fill the tank.If the tank is ful how long it take for the leakage alone to empty the tank.Easy methodk

Answer 1

Answer: 36 hours.

Step-by-step explanation:

Since, the tap $t_1$ alone fill the tank = 5 hours,

Work done by it in 1 hour = $\frac{1}{5}$

Similarly,  the tap $t_2$ alone fill the tank = 20 hours,

Work done by it in 1 hour = $\frac{1}{20}$

Thus, the total work done by both pipes when they Work simultaneously in 1 hour = $\frac{1}{5}+\frac{1}{20}=\frac{4+1}{20}=\frac{5}{20}=\frac{1}{4}$

Hence, the time taken by these pipe to fill the tank simultaneously = 4 hours.

According to the question,

Time taken by these pipe when the leakage is open = 4 - 1/2 = 9/2 hours

⇒ The total work done by both pipes and leakage in 1 hour = 2/9

⇒ $\frac{1}{5}+\frac{1}{20}-\frac{1}{x}=\frac{2}{9}$ (Let x be the time taken by leakage to empty the tank and negative sign shows the emptying the tank)

⇒  $\frac{1}{x}=\frac{1}{5}+\frac{1}{20}-\frac{2}{9}$

⇒ $\frac{1}{x}=\frac{36+9-40}{180}$

⇒ $\frac{1}{x}=\frac{5}{180}=\frac{1}{36}$

⇒ Time taken by leakage alone = 36 hours.