Two pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full ?

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:

Part filled in 3 minutes =
$$\begin{aligned} 3*\left(\frac{1}{12} + \frac{1}{15}\right) \\ = 3*\frac{9}{60} = \frac{9}{20}\\ \text{Remaining part }= 1-\frac{9}{20} \\ = \frac{11}{20} \\ => \frac{1}{15}:\frac{11}{20}=1:X \\ => X = \frac{11}{20}*\frac{15}{1} \\ => X = 8.25 mins \end{aligned}$$

So it will take further 8 mins 15 seconds 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 C

Explanation:

Capacity of the tank = (12*13.5) litres
= 162 litres
Capacity of each bucket = 9 litres.
So we can get answer by dividing total capacity of tank by total capacity of bucket.
Number of buckets needed = (162/9) = 18 buckets

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