A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. Find the time required by the first pipe to fill the tank ?

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 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 B

Explanation:

Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.

As per question, we get

$$\begin{aligned} \frac{1}{x} + \frac{1}{x-5} = \frac{1}{x-9} \\ => \frac{x-5+x}{x(x-5)} = \frac{1}{x-9}\\ => (2x - 5)(x - 9) = x(x - 5)\\ => x^2 - 18x + 45 = 0 \end{aligned}$$

After solving this euation, we get
(x-15)(x+3) = 0,
As value can not be negative, so x = 15

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 =

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

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.