Test: Pipes And Cistern- 1 - CAT MCQ

Suppose the slower pipe can fill the tank in x minutes.
Then the faster pipe can fill in x/4 minutes.
Part filled by the slower pipe in 1 min =1/x
Part filled by the faster pipe in 1 min =4/x
Part filled by the both the pipes in 1 min =1/x +4/x
 
Therefore, both the pipes can fill together in 36 minutes.
Part filled by both in 1 minute = 1/36.
1/x + 4/x = 1/36
5/x = 1/36
 x = 5 * 36 = 180

Time taken by one tap to fill half of the tank = 3 hrs.
Part filled by the four taps in 1 hour = 
Remaining part = 

 hours i.e, 45 mins.
So, total time taken = 3 hrs. 45 mins.

A tap can fill a tank in 4 hours.
Therefore the tap can fill half the tank in 2 hours.
Remaining part = 1/2
After half the tank is filled, two more similar taps are opened.
Hence, total number of taps becomes 3.
Part filled by one tap in 1 hour = 1/4
Part filled by three taps in 1 hour 
= 3 × 1/4 = 3/4
i.e., 4 taps can fill remaining half in 40 minutes.
Total time taken = 2 hour + 40 minute 
= 2 hour 40 minutes

part filled in 2 hours = 
Remaining  part =  
∴ (A + B)'s 7 hour's work = 2/3
(A + B)'s 1 hour's work = 2/21
∴ C's 1 hour's work = {(A + B + C)'s 1 hour's  work} - {(A + B)'s 1 hour's work}

∴ C alone can fill the tank in 14 hours.

Part filled by (A + B) in 1 minute = 
Suppose the tank is filled in x minutes.
Then, 
⇒ x = 30 min.

Part emptied by the leak in 1 hour = 1/6
Net part emptied by the leak and the inlet pipe in 1 hour = 1/24
Part filled by the inlet pipe in 1 hour 
i.e., inlet pipe fills the tank in 8 hours = (8 * 60) minutes = 480 minutes
Given that the inlet fills water at the rate of 4 liters a minute
Hence, water filled in 480 minutes = 480 * 4 = 1920 litre
i.e., The cistern can hold 1920 litre

Part filled by first tap in 1 hour = 1/3
Part emptied by second tap 1 hour = 1/8
Net part filled by both these taps in 
1 hour
= 1/3 − 1/8 = 5/24
i.e, the cistern gets filled in 24/5 hours 
= 4.8 hours.

Part filled by pipe A in 1 hour = 1/5
Part filled by pipe B in 1 hour = 1/20
Part filled by pipe A and B in 1 hour = 
i.e., A and b together can fill the tank in 4 hours
Given that due to the leakage, it took 40 minutes more to fill the tank.
i.e., due to the leakage, the tank got filled in 
⇒ Net part filled by pipe A and B and the leak in 1 hour = 3/14
⇒ Part emptied by the leak in 1 hour = 
i.e., the leak can empty the tank in 28 hours.

Let capacity of bucket P = x
Then capacity of bucket Q =  x/3
Given that it takes 80 turns for bucket P  to fill the empty drum
⇒ capacity of the drum = 80x
Number of turns required if both P and Q are used = 

(A + B)'s 1 hour's work =  = 9/60 = 3/20.
(A+C)'s hour's work =   = 8/60 = 2/15.
Part filled in 2 hrs = 

Part filled in 6 hrs = 
Remaining part = 
Now, it is the turn of A and B and 3/20 part is filled by A and B in 1 hour.
∴ Total time taken to fill the tank = (6+1) hrs = 7 hrs