Question

In a 1000 metre race, Rahul reaches the finishing line 5 seconds before than Raj and beats Raj by 50 metre. What is Rahul's speed (in m/s)?

• A. $ 11 \frac9{19}$
• B. $ 9 \frac9{19}$
• C. $ 10 \frac{10}{19}$
• D. $ 9 \frac{10}{19}$

Answer 1

The Correct Option is C

To solve this problem, we need to find Rahul's speed in a 1000-meter race. We know from the problem that Rahul reaches the finish line 5 seconds before Raj and he beats Raj by a distance of 50 meters.

The given conditions can be broken down as follows:

1. Rahul finishes the race while covering 1000 meters. 
2. Raj finishes the race while still needing to cover an additional 50 meters when Rahul has finished.
3. The time difference between Rahul finishing and Raj finishing the race is 5 seconds.

Let's denote:

• \(T_r\) as the time taken by Rahul to finish the race (in seconds).
• \(T_j = T_r + 5\) as the time taken by Raj to finish the 1000 meter race.
• \(S_r\) as Rahul's speed in m/s.
• \(S_j\) as Raj's speed in m/s.

From the data given:

1. Rahul's speed: \(S_r = \frac{1000}{T_r}\).
2. In \(T_r\) seconds, Raj runs 950 meters (\(1000 - 50\)). So, Raj's speed: \(S_j = \frac{950}{T_r}\).
3. Since Raj finishes the race 5 seconds after Rahul, we have: \(T_j = T_r + 5\).

Considering Raj covers 1000 meters in \((T_r + 5)\) seconds, we write:

\(S_j = \frac{1000}{T_r + 5}\).

By equating the two expressions for Raj's speed, we have:

\(\frac{950}{T_r} = \frac{1000}{T_r + 5}\).

Cross-multiplying gives:

\(950(T_r + 5) = 1000T_r\).

Expanding and simplifying:

\(950T_r + 4750 = 1000T_r\).

Subtracting \(950T_r\) from both sides:

\(4750 = 50T_r\).

Thus, we find \(T_r = 95\) seconds.

Rahul's speed \(S_r\) is then:

\(S_r = \frac{1000}{95} \approx 10.5263\) m/s.

Converting this to mixed fraction:

\(S_r = 10 \frac{10}{19}\) m/s.

Therefore, the correct answer is \(10 \frac{10}{19}\) m/s.