Latest TCS Aptitude Question SOLUTION: A and B are two places 210 km away. Ajay and Jay start simultaneously from these places. Ajay is faster than Jay. In 3 hours they meet 20 km from B. In how many hours Jay will retu

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A and B are two places 210 km away. Ajay and Jay start simultaneously from these places. Ajay is faster than Jay. In 3 hours they meet 20 km from B. In how many hours Jay will return to A?

Asked In TCS Mahesh Patil (9 years ago)

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Pallavi  (9 years ago)

Acc to ques, Jay starts from A and Ajay starts from B.
Since Ajay is faster than Jay, they will not meet in their first trip.
At the time of their meeting, Ajay would have been returning while Jay is still making his trip to B for the first tym.
By that tym, Ajay wud have covered (210+190)=400km and Jay covers 190 km.
speed of ajay=400/3 and speed of jay=190/3.
som required tym = 420/(190/3) [since total distance=420km].

Is this solution Helpfull? Yes (12) | No (2)  

Aditya Vyas  (9 years ago)

The que is in how many hours jay will return to A, that means jay has started from A and ajay from B. so jay's speed is 190/3 and ajay's speed is 20/3. so jay would take 210/(190/3) that is 3.31 hr to reach B and total 6.63 hr to return to A
plz correct me if i m wrong

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sanjana  (9 years ago)

20 km from B in 3 hrs means jay speed is 20/3 kmph and ajay speed is 190/3 .so to travel 210 km jay would take 210/(20/3) that is 31.5
please correct if im wrong

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Anurag Krishna  (9 years ago)

since we have to find time to return to A time should be 2*31.5=63

Is this solution Helpfull? Yes (3) | No (1)  

Lekha Tamvada  (9 years ago)

Jay's Speed = 20/3 kmph
distance to be covered by Jay = 210-20 = 190 km
therefore,
time = distance/speed => time = 190/(20/3) = 3*190/20 = 19*1.5 = 28.5 hrs

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manikandan  (9 years ago)

jay covers 20 km in 3hrs, so speed=20/3,
remaining distance to cover by jay=210-20=190km
remaining hours taken by jay=190*/20=19hrs
total time taken by jay=19+3=22hrs

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Shankar Jha  (9 years ago)

20:3=190:x
then
26.5

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