Question

Karan runs 4/3 times as fast as Sanjay. If Karan gives Sanjay a head start of 20 meters, how far must the winning post be so that both of them reach at the same time?

• A. 80 m
• B. 60 m
• C. 140 m
• D. 100 m

Answer 1

The correct option is A

80 m

Here, we could say that, if the speed of Sanjay is 'S', then speed of Karan is '$\frac{4}{3}\times S$

Or, by using the ratio concept, their speed ratio = $\frac{SpeedofKaran}{SpeedofSanjay}=\frac{4}{3}$

ie, if Karan runs as 4 m/s, then Sanjay will be 3 m/s.

And its said that they reach at the same time means the time taken is constant, so, the ratio of speed = ratio of the distance covered!

Now, with the concepts of Relative speed, as they are running in the same direction, their relative speed = 4 -3 =1 m/s.

Also, its given that Karan gives a head start of 20 meters to Sanjay implies that Sanjay is ahead of Karan by 20 m.

So, we can say the time taken for them to meet = $\frac{20m}{1m/s}=20seconds$ ------(Time = Distance/speed)

So, in 20 seconds, Karan would have covered $20seconds\times 4m/s=80m$

And in the same 20 seconds, Sanjay would have covered $20seconds\times 3m/s=60m$

So, we can say that the race is for a total of 80 meters. As Karan has given 20 m head start to Sanjay, he will be covering only 60 m.