Latest TCS Aptitude Question SOLUTION: A person run from A to B. He took ¼ of the time less to reach B when compare to run at normal Speed.Then how many percentage he has increased his speed? a) 40 b) 44.4 c) 33.3 d) 2

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(#M40003071) TCS question Keep an EYE

A person run from A to B. He took ¼ of the time less to reach B when compare to run at normal Speed.Then how many percentage he has increased his speed?
a) 40 b) 44.4 c) 33.3 d) 22.2

Asked In TCS irfan (12 years ago)

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Read Solution (3) :

Rohit Khajuria  (12 years ago)

Let the distance be D
time for normal run = t
new time = t - t/4 = 3t/4
normal speed, s = D/t or D = s*t ...(1)
changed speed, s' = D/(3t/4) or D = s' *(3t/4) or D = 3t * s'/4 ...(2)
from 1 and 2 ,we get
s * t = 3t s'/4
or s' = 4s/3
or s' = 1.33333*s
hence he have increase his speed by
= [(changed speed - normal speed)/normal speed]*100
= [(1.3333s - s)/s]*100
= .3333 * 100
= 33.33
hence ANSWER is 33.3

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prashant  (12 years ago)

i have simple logic,1-(1/4)=(3/4),then reverse it so it is 4/3=1.333,
1.33-1=33.33

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sumat srivastav  (12 years ago)

initialy---
speed-s1 ,distance -d ,time -t1
s1=d1/t1
now speed-s2 ,distance - d, time- t2=(1-1/4)=3/4s1
so, s1/s2=3/4.
s2= 4/3s1 = 1.33s1
% increase = (1.33s1 - s1)/s1 *100 =33.3%

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