Directions: Read the following information to answer the question.
Solution:
Let the number of fill pipes be ‘n'. Therefore, there will be 8−n, waste pipes.
Each of the fill pipes can fill the tank in 8 hours. Therefore, each of the fill pipes will fill (1/8)th of the tank in an hour.
Hence, n fill pipes will fill (n/8)th of the tank in an hour.
Similarly, each of the waste pipes will drain the full tank in 6 hours. That is, each of the waste pipes will drain (1/6)th of the tank in an hour.
Therefore, (8−n) waste pipes will drain (8−n/6)th of the tank in an hour.
Between the fill pipes and the waste pipes, they drain the tank in 6 hours. That is, when all 8 of them are opened, (1/6)th of the tank gets drained in an hour.
(Amount of water filled by fill pipes in 1 hour - Amount of water drained by waste pipes 1 hour)
= (1/6)th capacity of the tank drained in 1 hour.
n/8−8−n/6=−1/6
⇒ 6n−64+8n/48
=−1/6
⇒ 14n−64=−8 or 14n=56 or n=4
Note: In problems pertaining to Pipes and Cisterns, as a general rule find out the amount of the tank that gets filled or drained by each of the pipes in unit time (say in 1 minute or 1 hour).
Directions: Read the following information to answer the question.
There are four varieties of pipes Pipe A, Pipe B, Pipe C and Pipe D. Each pipe can be either an inlet pipe or an outlet pipe but cannot be both. there are 5 tanks of equal volume.
Tank P is filled by Pipe A and Pipe B
Tank Q is filled by Pipe A and Pipe C
Tank R is filled by Pipe A and Pipe D
Tank S is filled by Pipe B and Pipe C
Tank T is filled by Pipe C and Pipe D
Time taken for the first 3 tanks (P, Q and R) to get filled are in the ratio 1:2:4 and the time taken for the S and T tanks to be filled are in the ratio 7:10.
Find the outlet pipes among the 4 varieties.
Solution:
Let A,B,C and D do a,b,c and d with of work in an hour.
Let A and B fill the tank in 1 hour.
Then A and C would fill the tank in 2 hours while A and D in 4 hours.
a+b=1 ---------- (i)
a+c=1/2 ----------- (ii)
a+d=1/4 ------------ (iii)
Let B and C take 7k hours while c and D take 10k hours to fill the tank
⇒ b+c=1/7k --------------- (iv)
c+d=1/10k -------------- (v)
a=[(i)+(ii)−(iv)]/2 = [(ii)+(iii)−(v)]/2
⇒ a={1+1/2−1/7k}/2={1/2+1/4−1/10k}/2 ---------- (vi)
⇒ k=4/70
On substituting value of k in equation (vi) we get a<0
⇒ a<0,b>0,c>0 and d>0
Hence only A is the outlet pipe.
Directions: Read the following information to answer the question.
Solution:
P1 and P2 can fill the tank in
24/5 hr.[In one hour these fill (1/8+1/12) part of tank].
It takes 12/5 hr in filling half the tank.
For remaining half of the tank P3 will open and this will takes 6 hour.
Supervisor has gone out for (12/5+6) hr.
Now, (1/3)rd tank will fill in 8/5 hr.
In remaining 42/5hr only (33/60)th part of the tank will fill.
Empties part of tank =1−(1/3+33/60)=1/10
Which is 10% of tank.
Directions: Read the following information to answer the question.
Solution:
When A and B work together, they will take √5×15=√225= 15 days.
Where 5 and 45 are the extra time that A and B take to complete the job if they work alone compared to the time that they will take if they worked together.
Directions: Read the following information to answer the question.
Solution:
As per the information given, the ratio of efficiencies of Dr. Gupta, Dr. Sharma and Dr. Singh are 42:30:25
Hence, the ratio of time taken by Dr. Gupta and Dr. Sharma =5:7
Dr. Gupta takes 10 days less than Dr. Sharma, time taken by them will be 25 days and 35 days respectively.
Hence, time taken by Dr. Singh will be 42 days.<br>Part of work completed by Dr. Gupta =2/5
Part of work completed by Dr. Sharma =3/7
Remaining work =1−(2/5+3/7)=6/35
will be completed by Dr. Singh
He can complete it in 42×6/35
= 7.2 days.