punwish wrote:

This is the longer version of solution , there is a quick way to do it also :

Lets say As is the speed of A , Bs is the speed of B and similarly Cs is for C & Ds is for D.

if A took T1 time to finish the 400 yards race , so in the same time B will cover 340 yards

Therefore, we can say that T1 = 400/As and T1 = 340/Bs (Speed = Distance/Time or Time = Distance/speed ).

T1 = 400/As = 340/Bs => As/Bs = 400/340 or As/Bs = 20/17

Similarly , we can get

Bs/Cs = 17/15 & Cs/Ds= 15/14

therefore we can say , As/Cs = 4/3 ( As = 20/17 * Bs = 20/17 * 17/15 * Cs = 20/15 Cs

Lets assume to complete a 200 yards race A took time T , then

As=200/T or T = 200/As , in the same time C will cover X distance , so X = Cs * T (Distance = Speed * time)

Therefore X = Cs * 200/As ( from above we knw that As/Cs = 4/3 )

X = 200 / (As/Cs) = 200 / (4/3) = 150.

Thereby we can say that A will beat C in a 200 yards race by 50 yards.

Similarly we can get that B will beat D in a 200 yards race by 35 yards.

Hi punwish,

That's the correct answer. Yes, you are right this is a longer method to approach the question. I think the approach below is smarter one.

We can find out the continued ratios of all 4 players.

--A--:---B--:---C--:---D----

400 : 340

------:850 : 750

------:-----: 600 : 560

------------------------------------

400 : 340 : 750/850*340 : --

400 : 340 : 300 : 560/600*300

400 : 340 : 300 : 280So. we can infer that A beats C by 100 yards in a 400 yard race, so it will beat C by 50 yards in a 200 yard race.

Similarly, we can infer that B beats D by 60 yards in 340 yard race, so it will beat D by (340-280)/340*200 = ~35 yards in a 200 yard race.

Meanwhile, don't forget to attend the SC session on Saturday. Please find the link below:

-Shalabh

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