Question Detail

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 ?

10 hours
15 hours
17 hours
18 hours

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

1. 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 ?

6 min to empty
7 min to full
6 min to full
7 min to empty

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}

2. A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time cistern will get filled ?

7 hours
7.1 hours
7.2 hours
7.3 hours

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

3. 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 ?

3 hours
5 hours
7 hours
10 hours

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}

4. Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long it will take to fill the tank ?

10 mins
12 mins
15 mins
20 mins

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.

5. Pipes A and B can fill a tank in 5 hours and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in.

\begin{aligned} 3\frac{9}{5} \end{aligned}
\begin{aligned} 3\frac{9}{17} \end{aligned}
\begin{aligned} 3\frac{7}{5} \end{aligned}
\begin{aligned} 3\frac{7}{17} \end{aligned}

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}