Latest TCS Aptitude Question SOLUTION: A bus started from bustand at 8.00am, and after staying for 30 minutes at a destination, it returned back to the busstand. The destination is 27 miles from the busstand. The speed

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A bus started from bustand at 8.00am, and after staying for 30 minutes at a destination, it returned back to the busstand. The destination is 27 miles from the busstand. The speed of the bus is 18mph. During the return journey bus travels with 50% faster speed.At what time does it return to the busstand?

Asked In TCS Divya (10 years ago)

Unsolved Read Solution (6)

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Read Solution (6) :

SIVA KUMAR  (10 years ago)

Time = Distance/Speed
Reaching Time(destination) = 27/18 = 1.5hrs
50% faster the speed = (18*50)/100=9mph
Return speed = 18+9 = 27mph
Return time(Bus stand) = Distance/new speed = 27/27 = 1hr
Total = Reaching time + Staying time + return time
= 1.5hrs + 30min + 1hr = 3hrs
Therefore answer is 11am

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Rakesh Kumar Kohli  (10 years ago)

Time taken to Reach destination is 27/18 hrs. = 3/2 hrs
50% faster speed means speed is 3/2 times the original speed, so it will take 2/3 times time to cover the same distance.
So time taken during return journey = (3/2)*(2/3) = 1hr
Stay time = 30 minutes = 1/2 hrs
Total time = Time for onward journey + Staying time + Time for return journey
= 3/2 hrs + 1/2 hrs. + 1 hr = 3 hrs
It will return to the bus stand after 3 hrs.
Hence it will return to the bus stand at 11am

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aditi gakhar  (10 years ago)

can any of u explain speed unit ???
is it meters/hour or miles/hour???

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Pooja Chourey  (10 years ago)

bus will return in 3 hours i.e. at 11 am

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Sangeethab  (10 years ago)

t=d/s
(d1/s1)+(d2/s2)=time taken +stoppage time
(27/18)+(27/27)=x+(30/60)
x=3hrs
hence 8am +3hrs=11am

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sapan kumar paul  (10 years ago)

2hrs 42 min

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