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There is a escalator and 2 persons move down it.A takes 50 steps and B takes 75 steps
while the escalator is moving down. Given that the time taken by A to take 1 step is equal to
time taken by B to take 3 steps. Find the no. of steps in the escalator while it is staionary.
Read Solution (Total 5)
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- Let A take 1 step/min.
Hence B takes 3 steps/min. Let the escalator take n steps/min
Given that A takes 50 steps and hence in the same time escalator will
take 50n steps. So total no. of steps must be 50 + 50n
It is given that B takes 75 steps (which means he takes 25 min)
So in the same time escalator will cover 25n steps.
Hence no. of steps must be 75 +25n
Equating 50+50n = 75+25n we get n = 1.
Hence total no. of steps must be 100. - 13 years agoHelpfull: Yes(18) No(0)
- x-50/x-75=3/1
x=88 - 14 years agoHelpfull: Yes(2) No(12)
- suppose steps taken by B=75 at the same time by A=25 :( B=3A)
that means a still have to cover 25 more steps
(remove 25 from each which is due to escalator)
now
now the relativity of speed due to escalator will be B+E=2(A+E)
or,3A+E=2A+2E
, A=E
so step covered by A and E was equal :50+50=100
thank you hope i helped:) - 11 years agoHelpfull: Yes(2) No(0)
- If A takes 1 step in one second, then B takes 3 steps in one second.
If A takes t1 seconds to take 50 steps, then B takes 150 steps in t1 seconds.
For B, to take 150 steps he requires t1 seconds,
then to take 75 steps he requires t1/2 seconds.
So now, s1=50, t1 = t1 & s2=75, t2=t1/2
ans= (s1*t2 ~ s2*t1) / (t1 ~ t2) which gives 100.
so 100 steps is the answer - 7 years agoHelpfull: Yes(0) No(0)
- I didnt get it. can someone please explain it correctly.
Thanku - 6 years agoHelpfull: Yes(0) No(1)
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