One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in

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 B

Explanation:

In this type of questions we first get the filling in 1 minute for both pipes then we will add them to get the result, as

Part filled by A in 1 min = 1/20
Part filled by B in 1 min = 1/30

Part filled by (A+B) in 1 min = 1/20 + 1/30
= 1/12

So both pipes can fill the tank in 12 mins.

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 C

Explanation:

Work done by the inlet in 1 hour =

$$\begin{aligned} \frac{1}{6}-\frac{1}{8} = \frac{1}{24} \\ \text{Work done by inlet in 1 min} \\ = \frac{1}{24}*\frac{1}{60}\\ = \frac{1}{1440} \\ => \text{Volume of 1/1440 part = 4 liters} \\ \end{aligned}$$

Volume of whole = (1440 * 4) litres = 5760 litres.

Answer: Option A

Explanation:

There are two important points to learn in this type of question,
First, if both will open then tank will be empty first.

Second most important thing is,
If we are calculating filling of tank then we will subtract as (filling-empting)

If we are calculating empting of thank then we will subtract as (empting-filling)

So lets come on the question now,

Part to emptied 2/5

Part emptied in 1 minute =

$$\begin{aligned} \frac{1}{6} - \frac{1}{10} \\ = \frac{1}{15} \\ => \frac{1}{15}:\frac{2}{5}::1:x \\ => \frac{2}{5}*15 = 6 mins \end{aligned}$$