Latest TCS Aptitude Question SOLUTION: john walks on an escalator at a rate of 5 steps per second and reaches the other end in 10 secs. while coming back , walking at the same speed he reaches the starting point in 40se

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john walks on an escalator at a rate of 5 steps per second and reaches the other end in 10 secs. while coming back , walking at the same speed he reaches the starting point in 40seconds . what is the number of steps on the escalator?

Asked In TCS ratna (11 years ago)

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Swatantra Jain  (11 years ago)

ans must be 80
speed of john 5 steps per second
let speed of escalator be x steps per second
his speed while going up (5+x) per second
while coming down speed will be (5-x) per second
so (5+x)*10=(5-x)*40 by using distance = speed*time
giving x= 3 steps per second
so distance = (5*3)*10=(5-3)*40 = 80 steps

Is this solution Helpfull? Yes (25) | No   

soumik banerjee  (11 years ago)

starting tym step is total 50
and when conig back it is 200
so tatal step is 250

Is this solution Helpfull? Yes (8) | No (6)  

namita   (11 years ago)

ans 80 steps

Is this solution Helpfull? Yes (2) | No   

ratna  (11 years ago)

the answer given is 40

Is this solution Helpfull? Yes (1) | No (2)  

nancyjain  (11 years ago)

its a que of to and fro so 2xy/x+y,
2*10*40/10+40=16 ans

Is this solution Helpfull? Yes | No (7)  

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