Latest TCS Aptitude Question SOLUTION: If i decrease my speed by 20% of my original speed. I reach my office 7 min. late. What is my usual time taken to reach the office?

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If i decrease my speed by 20% of my original speed. I reach my office 7 min. late. What is my usual time taken to reach the office?

Asked In TCS Arpita Arora (11 years ago)

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ramya  (11 years ago)

Its 28seconds.
we know that distance=speed*time.
here distance to office is constant, then speed will be inversely proportional to time.let actual speed be s, then speed at now will be (4s)/5 i.e., s-(20% of s).and if t is the actual time taken, then time taken now will be (t+7).
now product of speed and time is constant, that implies
s*t=(4s/5)*(t+7)
5t=4t+28
t=28

Is this solution Helpfull? Yes (21) | No (1)  

Akshay Mundargi  (11 years ago)

Here let original speed be x. (1)
speed decreased by 20% mean 0.8x. (2)
time taken b t (3)
and after speed decreased, time= t-7min. (4)
Using 1-2 = 3-4, will get 0.2x= 7min.
so x=(7min/0.2)
=35min

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

saurabh agarwal  (11 years ago)

vt=.8v(t-7)
t=28

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Subhashini VKR  (11 years ago)

d=s*t1
d=(s-(20/100)s)*t2=0.8t2
s*t1=0.8s*t2
t1=0.8t2
since t2-t1=7min
sub t1, t2-0.8t2=7min
t2=7min/0.2=70/2min=35min
t1=35min-7min=28min

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Neha  (11 years ago)

if a man walks with (x/y) of his usual speed and he takes 't' hours to cover a certain distance then the time to cover the same distance with usual speed

=xt/(y-x)here x=1,y=5 and t=7mins
=7/4
=1.75 mins
hence usual time he takes to reach office is 1.75 mins.. pls confirm if this was an option...!!!

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Neha  (11 years ago)

sorry.. there's a correction in the explanation that-
he takes 't' hours more time to cover a distance.....

answer is otherwise correct to my expectations..!!!

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