Latest Tech Mahindra Aptitude Question SOLUTION: A runs 5/2 time as fast as B, and gives B and a start of 40m. How long should the racecourse be so that A and B might reach in the same time? a. 150 m b. 100 m c. 75 m d. 80 m

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A runs 5/2 time as fast as B, and gives B and a start of 40m. How long should the racecourse be so that A and B might reach in the same time?

a. 150 m
b. 100 m
c. 75 m
d. 80 m

Asked In Tech Mahindra Aashu (8 years ago)

Unsolved Read Solution (3)

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Read Solution (3) :

kishore  (8 years ago)

verification of options..... (2/5)*100 =40
So answer is B).100m

Is this solution Helpfull? Yes (3) | No   

G Vijay Bhaskar  (7 years ago)

let us consider speed of B is 10 m/s ,from given data A's speed is 5/2 of B's speed, so 5/2*10=20 m/s, as time is constant let us calculate the distance travelled by B in 3 seconds,10*3=30m,as he has a start of 40m ,distance travelled by B is 40+30 =70m,mean while distance travelled by A is 20*3=60m,still A lags behind B, now let us consider 4 seconds ,distance travelled by B is 10*40=40m,additional i.e. 40+40=80m,distance travelled by A is 20*4=80m,so at 80m both A and B meet each other

Is this solution Helpfull? Yes (1) | No (5)  

G Vijay Bhaskar  (7 years ago)

let us consider speed of B is 10 m/s ,from given data A's speed is 5/2 of B's speed, so 5/2*10=20 m/s, as time is constant let us calculate the distance travelled by B in 3 seconds,10*3=30m,as he has a start of 40m ,distance travelled by B is 40+30 =70m,mean while distance travelled by A is 20*3=60m,still A lags behind B, now let us consider 4 seconds ,distance travelled by B is 10*40=40m,additional i.e. 40+40=80m,distance travelled by A is 20*4=80m,so at 80m both A and B meet each other

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