Karan and Arjun run a 100-metre race, where Karan beats Arjun by 10 me

Question

Karan and Arjun run a 100-metre race, where Karan beats Arjun by 10 metres. To do a favour to Arjun, Karan starts 10 metres behind the starting line in a second 100-metre race. They both run at their earlier speeds. Which of the following is true in connection with the second race?

(A) Karan & Arjun reach finishing line together. 
(B) Arjun beats Karan by 1 metre.
(C) Arjun beats Karan by 11 metres. 
(D) Karan beats Arjun by 1 metre.
(E) Karan beats Arjun by 2 metre.

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nick1816 · Answer

\frac{V_K}{V_A}= \frac{100}{(100-10)}= \frac{10}{9}

Time taken to cover 110m by Karan= 110/10x= 11/x

Distance covered by Arjun in 11/x time= 9x*(11/x)= 99m

Karan beats Arjun by (100-99)= 1m

gracie · Answer

k/a rates=10/9
10/9=110/a→
a=99 meters
D

sanjitscorps18 · Answer

Let Karan's speed be K
Let Arjun's speed be A

Since time taken to finish the race as given in the situation is the same, we can infer that 
Time for 100 meters(K) = Time for 90 meters(A)
100/K = 90/A
=> A/K = 9/10

So, A could be 9x and K could be 10x.

Now in the second race Karan is running 110 m hence time take by Karan is 110/10x = 11/x
In the same time Arjun would run 9x * 11/x = 99 m

Since the race actually is for 100 m, we can say that Arjun covered 99 meters while Karan covered 100 meters, hence Karan beat Arjun by 1 meter.

Option D

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Karan and Arjun run a 100-metre race, where Karan beats Arjun by 10 me

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Karan and Arjun run a 100-metre race, where Karan beats Arjun by 10 me

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