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.
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-Shalabh
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