Latest Elitmus Aptitude Question SOLUTION: In a mall Rohit goes to 2nd floor from 1st floor by walking on a escalator.While going up time taken by him is 30 sec and while coming down he takes 120sec.Speed of Rohit is same i

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(#M40093834) ELITMUS Numerical Ability question Keep an EYE

In a mall Rohit goes to 2nd floor from 1st floor by walking on a escalator.While going up time taken by him is 30 sec and while coming down he takes 120sec.Speed of Rohit is same in both cases. If the escalator would stop then how much time Rohit will take to go up ???

Asked In Elitmus Paresh Kumar Nayak (9 years ago)

Unsolved Read Solution (7)

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Read Solution (7) :

shailendra krishna mishra  (9 years ago)

suppose length of escalator =x m
Let speed of rohit, sr ( m/s)
Let speed of escalator , se ( m/s)
* Net spped while going up while going up , sr + se = x/30 ----------eq (a) ( law of relative velocity)
* Net spped while coming down , se -sr = x/120 ----------eq (b) ( law of relative velocity)

Adding eq(a) and eq(b) we get 2sr= x/24 => sr=x/48 ---------------eq(c)

when escalator is running time take to go up , t= x/(sr+ se)
when escalator is stopped time take to go up , t= x/(sr+ 0) as se=0 m/s

therefore , t= x/sr ------------eq(d)
substitute value of sr from eq(c) in eq(d), we have

t=x/(x/48) => t=48 sec

hence rohit takes 48 sec to go up when escalator is stopped.

Is this solution Helpfull? Yes (31) | No (4)  

siddharth prakash singh  (9 years ago)

Let the distance between 1st floor and 2nd floor be D.
let the speed of Rohit be x m/s
and speed of escalator be y m/s
so the speed in upwards motion ( consider upstream movement as in case of boat nd stream) =(x-y) m/s
and the speed in downwards motion ( consider downstream movement as in case of boat nd stream) =(x+y) m/s
so a/c to given data in ques.
D/(x+y)=30------------(i)
and D/(x-y)=120------------(ii)
on dividing (i) by (ii) we get, y=3x/5
using eq. (i) we get,
D/[x+(3x/5)]=30
so, D/8x=6
=> D/x= 48
So the time taken by Rohit to go when escalator would stop =D/x =48 sec

Is this solution Helpfull? Yes (23) | No   

Amit kumar rana  (9 years ago)

Shailendra... How did u get... 2sr=x/48.... Ii might be 2se=x/48 and sr= x/80...
so.. ans wiil be 80 sec..

Is this solution Helpfull? Yes (7) | No (3)  

Ramdas Singathiya  (9 years ago)

Vr= speed of rohit
Va=speed of accelerator
D= distance covered from 1st floor to 2nd floor

Va+Vr= D/30
Vr-Va= D/120

add both equations
2Vr=D/24 => Vr=D/48.......(1)
now time taken by rohit to go up side is given by
t=D/(Vr+Va)...
now if accelerator stopped t=D/(Vr+0) => t=D / Vr
now using equation 1st
t=D/(D/48) = 48 sec.!!!!!!!!!!!!!!!!!!!!!!!! :)

Is this solution Helpfull? Yes (4) | No   

PRATIM JANA  (8 years ago)

Let the distance between 1st floor and 2nd floor be D.
let the speed of Rohit be x m/s
and speed of escalator be y m/s
so the speed in upwards motion ( consider upstream movement as in case of boat nd stream) =(x-y) m/s
and the speed in downwards motion ( consider downstream movement as in case of boat nd stream) =(x+y) m/s
so a/c to given data in ques.
D/(x-y)=30------------(i)
and D/(x+y)=120------------(ii)
on dividing (i) by (ii) we get, y= - 3x/5
using eq. (i) we get,
D/[x+(3x/5)]=30
so, D/8x=6
=> D/x= 48
So the time taken by Rohit to go when escalator would stop =D/x =48 sec

Is this solution Helpfull? Yes (4) | No   

Shock  (9 years ago)

The options were (a)45, (b)60, (c)75 and (d)cannot be determined.

Is this solution Helpfull? Yes | No   

soni kumari  (9 years ago)

let speed of escalator be x and that of rohit y
distance is same in both cases=d
for going up d/(x+v)=30 and for coming down d/(x-v)=120
solving we get x=(5/3)v
now d=30*(8v/3)=80 v
time taken=d/v=80 v/v=80 second when escalator is still

Is this solution Helpfull? Yes | No (2)  

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