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)22.2

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(#M40153420) TCS Numerical Ability 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 divyasree (8 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

diksha  (8 years ago)

33.3 as distance is constant
s1*t=s2*3/4t
s2=s1*4/3
increase=change*100/original
=(s2-s1)*100/s1
=33.3%

Is this solution Helpfull? Yes (7) | No   

charu singhania  (8 years ago)

time becomes 3/4 th time of the original.
speed will rise reciprocally and the final speed will be 4/3 times of the original.. i.e a 1/3 time increase or 33.33% increase.

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shakeel ahamad  (8 years ago)

let normal time=x
run time=x-0.25x=(3x/4)
normal speed u= d/x
running speed v=d/0.75x
v/u%=133.33
increament=33.33.......

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MATLOOB HUSAIN  (8 years ago)

for example if any body can go 400 km in 4 hours that means his speed will become 100 km/h.so this is normal speed of that man.if he take 1/4 of the time less taht means 3 hours for 400 km so
distance is equal then
v1xt1=v2xt2
100x4=v2x3
v2=400/3=133.33
that means his speed is increase 33.33 this is my answer.

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priyadharshini  (8 years ago)

ratio b/w time is: 1: 3/4
speed: 3/4: 1
therefore, ((3/4-1)/ 3/4) *100 =33.3 %

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KARTIK SHARMA  (8 years ago)

let T = normal time
old speed/new speed = T-(1/4)T/T
old speed/new speed = 3/4
new speed=(4/3)old speed
% increase = 33.3%

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kunwar utkarsh singh  (8 years ago)

(v+(v*x/100))/v=100/75 => x=33.33%

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Parameswari  (8 years ago)

decrease is 1/4 that is 1/x
so the increase will be 1/x-1 that is 1/4-1=1/3
(i.e)100/3=33.33

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pradeep patidar  (7 years ago)

speed of a is 1/4 less than b
than 1/(4-1)=1/3
which is fraction so
1/3=33.3%

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