18
Q:

# Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

 A) 10 min. 20 sec. B) 11 min. 45 sec. C) 12 min. 30 sec. D) 14 min. 40 sec.

Answer:   D) 14 min. 40 sec.

Explanation:

Part filled in 4 minutes =${\color{Black}&space;4\left&space;(&space;\frac{1}{15}+\frac{1}{20}&space;\right&space;)=\frac{7}{15}}$

Remaining part =${\color{Black}&space;\left&space;(&space;1-\frac{7}{15}&space;\right&space;)=\frac{8}{15}}$

Part filled by B in 1 minute =${\color{Black}&space;\frac{1}{20}}$

${\color{Black}\therefore&space;\frac{1}{20}:\frac{8}{15}::1:x}$

${\color{Black}x=\left&space;(&space;\frac{8}{15}&space;\times&space;1\times&space;20\right&space;)=10\frac{2}{3}min=10min.40sec.}$

${\color{Black}&space;\therefore&space;}$The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec

Q:

There is an empty reservoir whose capacity is 30 litres. There is an inlet pipe which fills at 5 L/min and there is an outlet pipe which empties at 4 L/min. Both the pipes function alternately for 1 minute. Assuming that the inlet pipe is the first one to function, how much time will it take for the reservoir to be filled up to its capacity?

 A) 49.5 min B) 50 min C) 51 min D) 52 min

Explanation:

The work to be done = Capacity of reservoir  = 30 litres.

1st Minute $\inline \fn_jvn \small \Rightarrow$ inlet pipe opened $\inline \fn_jvn \small \Rightarrow$ 5 lit filled

2nd minute $\inline \fn_jvn \small \Rightarrow$ inlet pipe closed; outlet pipe opened $\inline \fn_jvn \small \Rightarrow$ 4 lit emptied

In 2 minutes (5 litres - 4 litres = 1lit) is filled into the reservoir.

It takes 2 minutes to fill 1lit $\inline \fn_jvn \small \Rightarrow$ it takes 50 minutes to fill 25 litres into the reservoir.

In the 51st minute inlet pipe is opened and the reservoir is filled.

6 30
Q:

A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 ${\color{Blue} m^{3}}$. The emptying capacity of the tank is 10 $\inline {\color{Blue} m^{3}}$ per minute heigher than its filling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump?

 A) 50 m^3/min B) 60 m^3/min C) 72 m^3/min D) None of these

Explanation:

Let the filling capacity of the pump be x $\inline {\color{Black} m^{3}}$/min.

Then, emptying capacity of the pump=(x+10) $\inline {\color{Black} m^{3}}$/min.

so,$\inline {\color{Black} \frac{2400}{x}-\frac{2400}{x+10}=8\; \; \Leftrightarrow x^{2}+10x-3000=0}$

$\inline {\color{Black} \Leftrightarrow \left ( x-50 \right )+\left ( x+60 \right )=0\; \; \Leftrightarrow x=50}$

11 2966
Q:

Three taps A,B and C can fill a tank in 12,15 and 20 hours respectively. If A is open all the time and B ,C are open for one hour each alternatively, the tank will be full in:

 A) 6 hrs B) 20/3 hrs C) 7 hrs D) 15/2 hrs

Explanation:

$\inline {\color{Black} (A+B)'s\: 1\: hour\: work=\left ( \frac{1}{12}+\frac{1}{15} \right )=\frac{9}{60}=\frac{3}{20}}$$\inline {\color{Black} (A+C)'s\: 1\: hour\: work=\left ( \frac{1}{12}+\frac{1}{20} \right )=\frac{8}{60}=\frac{2}{15}}$  $\inline {\color{Black} Part \: \: filled \: in \: 2 \: hrs=\left ( \frac{3}{20} +\frac{2}{15}\right )=\frac{17}{60}}$

$\inline {\color{Black} Part \: \: filled \: in \: 6 \: hrs=(3\times \frac{17}{60})=\frac{17}{20}}$

$\inline {\color{Black} Remaining\; Part =\left ( 1-\frac{17}{20} \right )=\frac{3}{20}}$

Now, it is  the turn of A and B (3/20) part is filled by A and B in 1 hour.

therefore, Total  time taken to fill the tank =(6+1)hrs= 7 hrs

10 844
Q:

12 buckets of water fill a tank when the capacity of each tank is 13.5 liters. How many buckets will be needed to fill the same tank,if the capacity of each bucket is 9 liters?

 A) 8 B) 15 C) 16 D) 18

Explanation:

Capacity of the tank =(12 x 13.5) liters =162 liters.

Capacity of each bucket =9 liters

Number of buckets needed = 162/9 =18.

4 1268
Q:

One pipe can fill a tank  three times as fast as another pipe. If together the two pipes can fill the tank in 36 min, then the slower alone will be able to fill the tank in:

 A) 81 min B) 108 min C) 144 min D) 192 min

$\inline \fn_jvn {\color{Black}\therefore\; \; \frac{1}{x}+\frac{3}{x}=\frac{1}{36}\: \: \: \: \Leftrightarrow \: \: \: \frac{4}{x}=\frac{1}{36}\: \: \: \Leftrightarrow \: \: \: x=144\: min}$