Latest TCS Aptitude Question SOLUTION: one day eesha started 30min late from and reached her office 50min late while driving 25% slower than her usual speed.how much time in min does eesha usually take to reach her offi

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one day eesha started 30min late from and reached her office 50min late while driving 25% slower than her usual speed.how much time in min does eesha usually take to reach her office from home.

Asked In TCS vignesh (10 years ago)

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sruthi  (10 years ago)

actual late = 20 min. Distance = speed * time. d=3(s/4) * (t+20) . so, t*s=3(s/4)*(t+20) => t=60min.

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aditya  (10 years ago)

60 min z crrct..

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NARESH SOORA  (10 years ago)

she started 30 minutes late and reached the office 50 minute late
started late(30 min)+travelling time(20 min)=50 min late
so it takes 20 minutes reach the office when she is 25% slow
=>(20)*(25/100)=5 min
so it takes 5 min more when she is 25% slow
so
actual time-5=20-5=15
it takes 15 min to each the office

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lucky  (10 years ago)

60 min is correct

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babu rao ganpat rao apte  (10 years ago)

speed becomes 3/4 of original...thus time will become 4/3...
increase in time=1/3 of original...
thus given 1/3 of time=20 minutes
thus time = 60 minutes......this solution is given in arun sharma

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subash  (10 years ago)

is it this year question?

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shadankashif  (10 years ago)

I think 60min

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rahul thakur  (10 years ago)

here we see that it takes her total (50-30)min when she is 25% slow
i.e 20 min for 25% slow
total time= 20/25 *100 = 80min

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