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. An electric pump can fill a tank in 3 hours. Because of a leak in the tank, it took 3 hours 30 min to fill the tank. In what time the leak can drain out all the water of the tank and will make tank empty ?

10 hours
13 hours
17 hours
21 hours

Answer: Option D

Explanation:

We can get the answer by subtrating work done by leak in one hour by subtraction of filling for 1 hour without leak and with leak, as

Work done for 1 hour without leak = 1/3
Work done with leak =
\begin{aligned}
3\frac{1}{2} = \frac{7}{2} \\
\text{Work done with leak in 1 hr= }\frac{2}{7} \\
\text{Work done by leak in 1 hr }\\
= \frac{1}{3} - \frac{2}{7} = \frac{1}{21}
\end{aligned}

So tank will be empty by the leak in 21 hours.

2. 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.

3. 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

4. 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}

5. A tap can fill a tank in 6 hours. After half the tank is filled then 3 more similar taps are opened. What will be total time taken to fill the tank completely.

2 hours 30 mins
2 hours 45 mins
3 hours 30 mins
3 hours 45 mins

Answer: Option D

Explanation:

Half tank will be filled in 3 hours
Lets calculate remaining half,

Part filled by the four taps in 1 hour = 4*(1/6) = 2/3

Remaining part after 1/2 filled = 1-1/2 = 1/2

\begin{aligned}
\frac{2}{3}:\frac{1}{2}::1:X \\

=> X = \left( \frac{1}{2}*1*{3}{2} \right) \\
=> X = \frac{3}{4} hrs = 45 \text{ mins} \\

\end{aligned}
Total time = 3 hours + 45 mins = 3 hours 45 mins