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Numerical Ability
Pipes and Cistern
A cistern has three pipes A,B and C the pipes A and B can fill it in 4 and 5 hours respectively and C can empty it in 2 hours. If the pipes are opened in order at 1, 2 and 3 a.m. Respectively, when will the cistern be empty?
[A] 5 a.m
[C] 5 p.m
[B] 6 p.m
[D] 6 a.m
Read Solution (Total 9)
-
- 5 PM
tank filled up to 3 am = 2*1/4 + 1/5 = 1/2 +1/5 = 7/10
tank drained in 1 hr when 3 pipes are opened = 1/2 -1/4 -1/5 = 1/20
so already filled 7/10 tank is drained in 14 hrs i.e. at 3+14 = 1700 hrs or 5PM. - 13 years agoHelpfull: Yes(26) No(10)
- A can fill it in 4hrs i.e in 1hr it can fill 25% of tank
now in the 2nd hr both opened ie 25% a and 20% b contributes to fill the tank i.e the tank is fill upto 25+25+20 ie 70% of tank is full in 2 hrs now 30 % is empty .
what we hav to see here if all the pipes are opened what is the net filling i.e.
25+20-50= -5 hence we can say that frm 3rd hr 5% of reduction happens only .
hence we have 70% full tank so 70/5 =14 hrs so frm 3am to 3pm it is 12hrs and 2 more hrs ie 5 pm is our ans. - 10 years agoHelpfull: Yes(9) No(1)
- In 1 hour A can fill the cistern in = 1/4hr.
In 1 hour B can fill the cistern in = 1/5hr.
In 1 hour C can empty the cistern in =1/2hr.
Let the tank be empty after 1a.m. then:
x/4 + (x-1)/5 -(x-2)/2=0
x=16:9.
so after 1 a.m. cistern will be empty in 1+16:9=1709hrs.
Ans=5 p.m.
- 10 years agoHelpfull: Yes(6) No(4)
- Work done by a in 1 hour = 1/4
Work done by (a+b) in 1 hour=9/20,
work done from 1 A.M to 3 A.M is 14/20
drained in 1 hour = -1/20,
So, to drain 14/20 it requires 14 hours.
i.e, 5 p.m - 10 years agoHelpfull: Yes(4) No(2)
- 5 pm by hit and trial
- 9 years agoHelpfull: Yes(3) No(3)
- 1 hour A can fill the cistern in = 1/4hr.
In 1 hour B can fill the cistern in = 1/5hr.
In 1 hour C can empty the cistern in =1/2hr.
Let the tank be empty after 1a.m. then:
x/4 + (x-1)/5 -(x-2)/2=0
x=16:9.
so after 1 a.m. cistern will be empty in 1+16:9=1709hrs.
Ans=5 p.m. - 9 years agoHelpfull: Yes(3) No(0)
- its 5 am
14/20-1/20 - 9 years agoHelpfull: Yes(1) No(1)
- In 1 hour A can fill the cistern in = 1/4hr.
In 1 hour B can fill the cistern in = 1/5hr.
In 1 hour C can empty the cistern in =1/2hr.
Now consider in X hrs cistern will be empty after opening C
then C is working X hrs
B is working (X+1) hrs
A is working (X+2) hrs
So,
(X+2)/4 + (X+1)/5 - x/2=0
X=14hrs
So, Ans is 5 p.m - 8 years agoHelpfull: Yes(0) No(0)
- Suppose work is 20
Now
A--> 5 Work/Hour
B--> 4 Work/Hour
C--> (-10) Work/Hour
Work done till 3 pm is = (2*5) + (1*4) = 14
After that, every hour work done will be = (5+4-10) = -1
So It will take another 14 hours to completely empty the cistern. - 4 years agoHelpfull: Yes(0) No(0)
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