Question Detail

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.

1. 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 ?

30 hours
35 hours
40 hours
42 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}

2. A tank can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tank from empty state if B is used for the first half time and then A and B fill it together for the other half.

15 mins
20 mins
25 mins
30 mins

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}

3. 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}

4. 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

5. 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}