CAT 2004 | Question: 41

Question

Karan and Arjun run a $100$-metre race, where Karan beats Arjun by $10$ metres. To do a favour to Arjun, Karan starts $10$ meters behind the standing 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 and Arjun reach the finishing line simultaneously
• B. Arjun beats Karan by $1$ metre
• C. Arjun beats Karan by $11$ metre
• D. Karan beats Arjun by $1$ metre

Answer 1

D
$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$
So, $\frac{d}{v_K}= \frac{d-10}{v_A}$

$\implies v_K=\frac{10v_A}{9}$

in 2nd race

$\frac{d+10}v_K=\frac{d}{v_A}$

$\frac{d+10) 9}{10v_A}=\frac{d}{v_A} \\ \implies 9d + 90 = 10d \\\implies d = 90$

So, they meet at a distance of 90m and since Karan runs faster, he reaches the destination first. So naswer must be D. Still to confirm we can do

$\frac{110}{v_K} = \frac{d}{v_A} \\ \implies \frac{110 \times 9}{10v_A} = \frac{d}{v_A} \\\implies d = 99$.

So, Karan wins Arjun by $100 - 99 = 1m.$