8
Q:

A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?

 A) 3 hrs 15 min B) 3 hrs 45 min C) 4 hrs 15 min D) 4 hrs 1

Answer:   B) 3 hrs 45 min

Explanation:

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the four taps in 1 hour =${\color{Black}&space;\left&space;(&space;4\times&space;\frac{1}{6}&space;\right&space;)=\frac{2}{3}}$

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

${\color{Black}&space;\therefore&space;\frac{2}{3}:\frac{1}{2}&space;::1:x}$

${\color{Black}&space;\Rightarrow&space;x=\left&space;(&space;\frac{1}{2}&space;\times&space;1\times&space;\frac{3}{2}\right&space;)=\frac{3}{4}}$

So, total time taken = 3 hrs. 45 mins.

Q:

Three taps I, J and K can fill a tank in 20,30and 40 minutes respectively. All the taps are opened simultaneously and after 5 minutes tap A was closed and then after 6 minutes tab B was closed .At the moment a leak developed which can empty the full tank in 70 minutes. What is the total time taken for the completely full ?

 A) 24.315 minutes B) 26.166 minutes C) 22.154 minutes D) 24 minutes

Explanation:

Upto first 5 minutes I, J and K will fill => 5[(1/20)+(1/30)+(1/40)] = 65/120
For next 6 minutes, J and K will fill => 6[(1/30)+(1/40)] = 42/120
So tank filled upto first 11 minutes = (65/120) + (42/120) = 107/120
So remaining tank = 13/120
Now at the moment filling with C and leakage @ 1/60 per minute= (1/40) - (1/70) = 3/280
So time taken to fill remaining 13/120 tank =(13/120) /(3/280) = 91/6 minutes

Hence total time taken to completely fill the tank = 5 + 6 + 91/6 = 26.16 minutes.

3 17
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 32
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 22
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 72
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,$\inline {\color{Black} \frac{2400}{x}-\frac{2400}{x+10}=8\; \; \Leftrightarrow x^{2}+10x-3000=0}$

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