A plane left half an hour later than the scheduled time and in order to reach its destination 1500 kilometers away in time, it had to increase its speed by 33.33 percent over its usual speed. Find its increased speed.
A. 250 kmph
B. 500 kmph
C. 750 kmph
D. 1000 kmph
E. 1200 kmph
Answer: Option D
Solution (By Examveda Team)
By increasing the speed by 33.33%, it would be able to reduce the time taken for traveling by 25%. But since this is able to overcome the time delay of 30 minutes, 30 minutes must be equivalent to 25% of the time originally taken. Hence, the original time must have been 2 hours and the original speed would be 750 kmph. Hence, the new speed would be 1000 kmph.This Question Belongs to Arithmetic Ability >> Speed Time And Distance
In MD Rakib's Comment ,
in 2nd approach: 1500/x - 1500/1.3333x=1/2 (1.33 is not correct)
let original speed x km
1500/x - 1500/4x/3 =1/2
x=750
increased speed 4x/3= 1000
1st approach :
speed=100:133.33 =3:4 2nd approach: 1500/x+1500/1.33x=1/2
time=4:3[gap 1] x=750*1.33=1000kmph
but we need to gap 30
so time=120:90
so increased speed=1500/30/60=1000kmph...(ans)
how did we get 2 hrs???
Good solve
This is a fact of the percentage that if speed increases 33.33% then time must be reduce by 25%. Let is understand it with an example.
Say, Let speed of a train is 100 kmph. Now,
Speed of the Train increases by 33.33%. Then how much time train will take to cover 100 km?
New Speed = 100 + 33.33% of 100 = 133.33 kmph.
Now,
Train travels 133.33 km in 60 minutes.
Then, to travel 1 km it will take = 60/133.33 minutes
And, to travel 100 km it will take = (60 * 100)/133.33 = 45 minutes.
Decrease in Time = 15 minutes.
% Decrease in Time = (15 * 100)/60 = 25%.
Also,
** If time increases by 25% then speed will decrease by 33.33%. This is Because,
ST = D
When D = constant,
S is inversely proportional to Time.
Hope, this is clear to you both @Sachin and @Divyani.
how u get 25% to reduce..please define..
Time 25% kese kam ho gya...????