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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
Read Solution (Total 3)
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- verification of options..... (2/5)*100 =40
So answer is B).100m - 8 years agoHelpfull: Yes(3) No(0)
- 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
- 8 years agoHelpfull: Yes(1) No(5)
- 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
- 8 years agoHelpfull: Yes(0) No(4)
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