Question Detail

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}

1. 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}

2. A tank can be filled by a tap in 20 minutes and by another tap in 60 minutes. Both the taps are kept open for 10 minutes and then the first tap is shut off. After this, the tank will be completely filled in what time ?

10 mins
15 mins
20 mins
25 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}

3. 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}

4. 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}

5. Taps A and B can fill a bucket in 12 minutes and 15 minutes respectively. If both are opened and A is closed after 3 minutes, how much further time would it take for B to fill the bucket?

8 min 15 sec
7 min 15 sec
6 min 15 sec
5 min 15 sec

Answer: Option A

Explanation:

Part filled in 3 minutes =
\begin{aligned}
3*\left(\frac{1}{12} + \frac{1}{15}\right) \\
= 3*\frac{9}{60} = \frac{9}{20}\\
\text{Remaining part }= 1-\frac{9}{20} \\
= \frac{11}{20} \\
=> \frac{1}{15}:\frac{11}{20}=1:X \\
=> X = \frac{11}{20}*\frac{15}{1} \\
=> X = 8.25 mins
\end{aligned}

So it will take further 8 mins 15 seconds to fill the bucket.