Taps A and B can fill a bucket in 12 minutes and 15 minutes respectively. If both are opened and A is closed after 3 minutes, how much further time would it take for B to fill the bucket?

Answer: Option C

Explanation:

Part filled without leak in 1 hour = 1/9
Part filled with leak in 1 hour = 1/10

Work done by leak in 1 hour

$$\begin{aligned} = \frac{1}{9} - \frac{1}{10} \\ = \frac{1}{90} \end{aligned}$$

We used subtraction as it is getting empty.

So total time to empty the cistern is 90 hours

Answer: Option B

Explanation:

Net part filled in 1 hour =

$$\begin{aligned} \left(\frac{1}{5}+\frac{1}{6}-\frac{1}{12}\right) \\ = \frac{17}{60} hrs \\ = 3\frac{9}{17} \end{aligned}$$

Answer: Option A

Explanation:

Lets suppose tank got filled by first pipe in X hours,
So, second pipe will fill the tank in X + 10 hours.

$$\begin{aligned} => \frac{1}{X} + \frac{1}{X} + 10 = \frac{1}{12} \\ => X = 20 \end{aligned}$$

Answer: Option C

Explanation:

How we can solve this question ?
First we will calculate the work done for 10 mins, then we will get the remaining work, then we will find answer with one tap work, As

Part filled by Tap A in 1 min = 1/20
Part filled by Tap B in 1 min = 1/60

(A+B)'s 10 mins work =

$$\begin{aligned} 10*\left(\frac{1}{20}+\frac{1}{60}\right) \\ = 10*\frac{4}{60} = \frac{2}{3} \\ \text{Remaining work = } 1-\frac{2}{3} \\ = \frac{1}{3} \\ \text{METHOD 1} \\ => \frac{1}{60}:\frac{1}{3}=1:X \\ => X = 20 \\ \text{METHOD 2} \\ 1/60 \text{ part filled by B in} = 1 min \\ 1/3 \text{ part will be filled in} \\ = \frac{\frac{1}{3}}{\frac{1}{60}} \\ = \frac{60}{3} = 20 \end{aligned}$$

Answer: Option A

Explanation:

Let the slower pipe alone fill the tank in x minutes
then faster will fill in x/3 minutes.

Part filled by slower pipe in 1 minute = 1/x
Part filled by faster pipe in 1 minute = 3/x

Part filled by both in 1 minute =

$$\begin{aligned} \frac{1}{x} + \frac{3}{x}= \frac{1}{36} \\ => \frac{4}{x} = \frac{1}{36} \\ x = 36*4 = 144 mins \end{aligned}$$