Elitmus
Exam
Numerical Ability
Pipes and Cistern
If a cistern has 3 pipes of 1 cm , 4/3 cm and 2 cm thickness ...where the pipe with 2 cm thickness can fill the cistern in 2 hrs and 2 min..then in how much time they can fill the tank?
Read Solution (Total 12)
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- 2cm can fill in 122min
thickness inversly propotional to time
total thickness of the pipe=1+4/3+2=4.3
we know that 2cm can fill in 122min
for 4.3cm time =56.7min - 8 years agoHelpfull: Yes(13) No(2)
- total efficiency of all the 3 pipes in 1 min when opened together = 1/122+1/244+4/(244*3) =13/732 part of the cistern =732/13 mins = 56.3 mins
- 8 years agoHelpfull: Yes(6) No(1)
- its innerverse of its area .So
(4+16/9+1)T=4*122.
T =72 minutes. - 8 years agoHelpfull: Yes(1) No(5)
- pipe a= 3/3 cms
pipe b=4/3 cms
pipe c=6/3 cms
given that pipe c takes 122 minutes
so for 122 parts 122*6=732
at a time three pipes means 6+4+3=13 parts
so 732/13 = 56.3 minutes - 8 years agoHelpfull: Yes(1) No(0)
- 1 hr ..lcm
- 8 years agoHelpfull: Yes(0) No(8)
- 2 cm can fill in 122 min
it means, 1 cm can fill in 61 min and 4/3 cm can fill in 81.33 min.
find part in 1 min for each like this, for 2 cm fill in 1 min = 1/122 = 0.0081
now add each part and you got 0.0368, so together 3 pipes can fill 0.0368 part in 1 min
divide it by 1, and you got 27.17 min. that should be an answer. - 8 years agoHelpfull: Yes(0) No(14)
- for 2 cm pipe surface area @(pai)=122 min
for 1 cm pipe S A=0.25@
for 4/3 pipe S.A=1.77@
so 3@=122/3=40.66-~40min(apx) - 8 years agoHelpfull: Yes(0) No(2)
- 2cm thick pipe can fill in 2hrs 2 min(61/30hr) ==> 1cm fills in 1hr 1 min(61/60hr) and 4/3cm fills in 4/3hr 4/3min (61/45hr)
So, (i) 61/30 hr --> 1 tank ==> 1hr --> 30/61 part ; (ii) 1hr --> 60/61 part (iii) 1hr --> 45/61 part
Therefore, in 1hr 3 pipes can fill ---> 135/61 part
So, time taken = 61/135 hrs = 26.8 min
- 8 years agoHelpfull: Yes(0) No(1)
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- 8 years agoHelpfull: Yes(0) No(1)
- let us assume,x cm height of liquid can pass in all the pipes in one minute.
so,volume of the tank is=3.14*2*2*x*122(as 2 hours 22 mins=122 mins)
let say,y mins will be needed to fill the tank using all the pipes..
so,volume of the tank would be,3.14*x*(2*2+1*1+4/3*4/3)*y
so, both volumes are equal.
after solving,y=72 mins or 1hr 12 mins - 8 years agoHelpfull: Yes(0) No(1)
- capacity of pipes is not given........ we cannot say ,it depends upon force of water
- 8 years agoHelpfull: Yes(0) No(1)
- we can do approximation to reduce calculation of fractions
2cm -> 2hr
4/3cm -> 3hr
1cm -> 4hr
took L.C.M of 2,3,4 = 12
divide 12/2+12/3+12/4 =13
12/13hr or 55.58 min. - 5 years agoHelpfull: Yes(0) No(0)
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