A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely ?

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 =

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

Answer: Option B

Explanation:

(A+B)'s 2 hour's work when opened =

$$\begin{aligned} \frac{1}{6}+\frac{1}{4} = \frac{5}{12} \\ (A+B)'s \text{ 4 hour's work} = \frac{5}{12}*2 \\ = \frac{5}{6} \text{Remaining work = } 1-\frac{5}{6} \\ = \frac{1}{6} \\ \text{Now, its A turn in 5 th hour} \\ \frac{1}{6} \text{ work will be done by A in 1 hour}\\ \text{Total time = }4+1 = 5 hours \end{aligned}$$

Answer: Option D

Explanation:

Part filled by A in 1 hour = 1/5
Part filled by B in 1 hour = 1/10
Part filled by C in 1 hour = 1/30

Part filled by (A+B+C) in 1 hour =

$$\begin{aligned} \frac{1}{5}+\frac{1}{10}+\frac{1}{30} \\ = \frac{1}{3} \\ \end{aligned}$$

So all pipes will fill the tank in 3 hours.

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:

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