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Numerical Ability
Time Distance and Speed
A person run from A to B.He took ΒΌ 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
Read Solution (Total 9)
-
- 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% - 9 years agoHelpfull: Yes(7) No(0)
- 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. - 9 years agoHelpfull: Yes(4) No(0)
- 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....... - 9 years agoHelpfull: Yes(1) No(0)
- 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. - 9 years agoHelpfull: Yes(1) No(1)
- ratio b/w time is: 1: 3/4
speed: 3/4: 1
therefore, ((3/4-1)/ 3/4) *100 =33.3 % - 9 years agoHelpfull: Yes(1) No(0)
- 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% - 9 years agoHelpfull: Yes(0) No(0)
- (v+(v*x/100))/v=100/75 => x=33.33%
- 9 years agoHelpfull: Yes(0) No(0)
- 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 - 9 years agoHelpfull: Yes(0) No(0)
- speed of a is 1/4 less than b
than 1/(4-1)=1/3
which is fraction so
1/3=33.3% - 8 years agoHelpfull: Yes(0) No(0)
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