There are two pipes which are functioning simultaneouly to fill a tank in 12 hours, if one pipe fills the tank 10 hours faster than other then how many hours second pipe will take to fill the tank ?

Answer: Option D

Explanation:

We can get the answer by subtrating work done by leak in one hour by subtraction of filling for 1 hour without leak and with leak, as

Work done for 1 hour without leak = 1/3
Work done with leak =
$$\begin{aligned} 3\frac{1}{2} = \frac{7}{2} \\ \text{Work done with leak in 1 hr= }\frac{2}{7} \\ \text{Work done by leak in 1 hr }\\ = \frac{1}{3} - \frac{2}{7} = \frac{1}{21} \end{aligned}$$

So tank will be empty by the leak in 21 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

Answer: Option C

Explanation:

When we have question like one is filling the tank and other is empting it, then we subtraction as,

Filled in 1 hour = 1/4
Empties in 1 hour = 1/9

Net filled in 1 hour = 1/4 - 1/9
= 5/36

So cistern will be filled in 36/5 hours i.e. 7.2 hours

Answer: Option D

Explanation:

Let the total time be x mins.
Part filled in first half means in x/2 = 1/40

Part filled in second half means in x/2 = $$\begin{aligned} \frac{1}{60}+\frac{1}{40} \\ = \frac{1}{24} \\ \text{ Total = } \\ \frac{x}{2}*\frac{1}{40} + \frac{x}{2}*\frac{1}{24} = 1 \\ => \frac{x}{2} \left(\frac{1}{40}+\frac{1}{24} \right) = 1 \\ => \frac{x}{2}*\frac{1}{15} = 1 \\ => x = 30 mins \end{aligned}$$