**Concept:The question tests on the application of Rates(Time-Speed-Distance) in a Race question**

__1__^{st}RaceLet Jerry take x sec

=>Jim would take (x+30) sec to complete the same race

=>Jerry’s speed = Distance of the race/His speed

=>Jerry’s speed = Distance of the race/His speed

= 2000 / x m/s

Jim’s speed= Distance of the race/His speed

Jim’s speed= Distance of the race/His speed

=1800 / (x+30) m/s

__2__^{nd}RaceLet Jerry take y sec to cover the race.

Jerry has not covered the 2000m; instead he is beaten by 1000m

=>He has covered 2000-1000 = 1000m in y sec

=> Jerry’s speed here = 1000/y m/s

Jim gets a start of 3min or 180sec

Jim gets a start of 3min or 180sec

=> He gets 180 sec additionally for the race and he also completes the 2000 m race

- Speed of Jim = 2000 / (180+y)m/s ----------(1)

- That means, 2000/x = 1000/y (Equating Jerry’s speed from both races)

=> y=x/2

=> 1800 / (x+30) = 4000 / (360 + y ) (Substitute y=x/2 in eq(1))

=> 1800 / (x+30) = 4000 / (360 + y ) (Substitute y=x/2 in eq(1))

=>2200x = 526000

=> x = 239 seconds.

Hence, Jerry will complete 2000m race in 2000/239

Hence, Jerry will complete 2000m race in 2000/239

= 3.98 ~= 4 minutes.

__(You may not need to solve ahead as the option b starts for Jerry with 4)__ Speed of Jim is 1800 / (239+30)

= 6.69 m/s

=> The time required for Jim to finish 2000m = 2000/6.69 = 298.95 s

Therefore, the time required in minutes = 298.95/60 = 4.98 ~= 5 minutes.

Therefore, the time required in minutes = 298.95/60 = 4.98 ~= 5 minutes.

**(B)**