12
Q:

Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:

 A) 10 B) 12 C) 14 D) 16

Explanation:

Part filled in 2 hours =${\color{Black}&space;\frac{2}{6}=\frac{1}{3}}$

Remaining part =${\color{Black}&space;\left&space;(&space;1-\frac{1}{3}&space;\right&space;)=\frac{2}{3}}$

${\color{Black}&space;\therefore&space;}$ (A + B)'s 7 hour's work =${\color{Black}&space;\frac{2}{3}}$

(A + B)'s 1 hour's work =${\color{Black}&space;\frac{2}{21}}$

${\color{Black}&space;}$${\color{Black}\therefore&space;}$C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

=${\color{Black}&space;\left&space;(&space;\frac{1}{6}-\frac{2}{21}&space;\right&space;)=\frac{1}{14}}$

${\color{Black}&space;\therefore&space;}$C alone can fill the tank in 14 hours.

Q:

Pipe K fills a tank in 30 minutes. Pipe L can fill the same tank 5 times as fast as pipe K. If both the pipes were kept open when the tank is empty, how much time will it take for the tank to overflow ?

 A) 3 minutes B) 2 minutes C) 4 minutes D) 5 minutes

Explanation:

Let the total capacity of tank be 90 liters.
Capacity of tank filled in 1 minute by K = 3 liters.
Capacity of tank filled in 1 minute by L = 15 liters.
Therefore, capacity of the tank filled by both K and L in 1 minute = 18 liters.
Hence, time taken by both the pipes to overflow the tank = 90/18 = 5 minutes.

4 26
Q:

Taps X and Y can fill a tank in 30 and 40 minutes respectively.Tap Z can empty the filled tank in 60 minutes.If all the three taps are kept open for one minute each,how much time will the taps take to fill the tank?

 A) 48min B) 72min C) 24min D) None of these

Explanation:

Given taps X and Y can fill the tank in 30 and 40 minutes respectively. Therefore,

part filled by tap X in 1 minute = 1/30

part filled by tap Y in 1 minute = 1/40

Tap Z can empty the tank in 60 minutes. Therefore,

part emptied by tap Z in 1 minute = 1/60

Net part filled by Pipes X,Y,Z together in 1 minute =

$\inline \fn_jvn \small \frac{1}{30}+\frac{1}{40}-\frac{1}{60}$

= 5/120 = 1/24

i.e., the tank can be filled in 24 minutes.

1 18
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 67
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 3233
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