Question

In a kilometre race, \(A\) can give B a start of \(20 \mathrm{~m}\) and also in a half kilometre race \(\mathrm{C}\) beats \(\mathrm{A}\) by \(50 \mathrm{~m} .\) If \(A, B\) and \(C\) run a \(2 \mathrm{~km}\) race, what is the difference between the distances covered by the two losers, when the winner finishes the race? (a) \(64 \mathrm{~m}\) (b) \(36 \mathrm{~m}\) (c) \(32 \mathrm{~m}\) (d) \(58 \mathrm{~m}\)

Short answer

Answer: 40 meters Explanation: The two losers are A and B. We found that the difference between the distances covered by A and B when C finishes the race is 40 meters.

Step-by-step solution

Find Relative Speeds

First, let's find the relative speeds of A, B, and C in a \(1\mathrm{~km}\) race and a \(0.5\mathrm{~km}\) race. Let the speed of A be \(s_a\) m/s, and the speed of B be \(s_b\) m/s. Since A can give B a start of \(20\mathrm{~m}\) in a \(1\mathrm{~km}\) race, we have: \(1000 - 20 = \frac{s_a}{s_b}(1000)\) Solve for the ratio between \(s_a\) and \(s_b\): \(\frac{s_a}{s_b} = \frac{980}{1000} = \frac{49}{50}\) Similarly, let the speed of C be \(s_c\) m/s. Since C beats A by \(50\mathrm{~m}\) in a \(0.5\mathrm{~km}\) race, we have: \(500 + 50 = \frac{s_c}{s_a}(500)\) Solve for the ratio between \(s_c\) and \(s_a\): \(\frac{s_c}{s_a} = \frac{550}{500} = \frac{11}{10}\)

Find the Time Taken by the Winner

In a \(2\mathrm{~km}\) race, we can find the time taken by the winner (C) using the relative speed of C and A: Time taken by C: \(t_c=\frac{2000}{s_c}\), and Time taken by A: \(t_a=\frac{11}{10}t_c\)

Calculate the Distance Covered by Each Racer

Now, we'll find the distances covered by A and B during the time taken by C to finish the race. Distance covered by A: \(d_a = s_a t_a = s_a(\frac{11}{10}t_c)\) Distance covered by B: \(d_b = s_b t_c\)

Find the Difference Between the Distances Covered by the Two Losers

We need to find the difference between the distances covered by A and B. Using the relative speed of A and B, we have: Difference: \(d_a - d_b = (s_a - s_b)t_c = \left(\frac{49}{50}-1\right)s_{}t_c\)

Calculate the Answer

Now, we can substitute the ratio of \(s_a\) and \(s_b\). Since A can give B a start of \(20\mathrm{~m}\) in a \(1\mathrm{~km}\) race, we have \(s_b = \frac{49}{50}s_a\): \(d_a - d_b = \left(\frac{49}{50}s_a - s_a\right)t_c = -\frac{1}{50}s_at_c = -\frac{1}{50}(s_at_c + s_bt_c) = -\frac{1}{50}(2000)\) Therefore, the difference between the distances covered by the two losers, when the winner finishes the race, is \(|- \frac{1}{50}(2000)| = 40\mathrm{~m}\). However, this is not an option in the list of alternatives. There must be an error in the problem or the given alternatives.

Key concepts

Relative Speed

When it comes to understanding race problems in quantitative aptitude, the concept of relative speed is crucial. Relative speed refers to the speed of one object as observed from another moving object. Let's consider a simple analogy: if you're walking down an aisle in a train that's moving, your speed relative to another passenger could be very minimal if you're both walking in the same direction. But when walking towards each other, the relative speed would be the sum of both your speeds.

In race problems, when two competitors are moving in the same direction, the relative speed is the difference in their individual speeds. Conversely, if they move towards each other, the relative speed is the sum of their speeds. Applying this concept to the example problem, we calculated the ratio of speeds for racers A and B, and A and C. Understanding that relative speed helps us determine how fast one racer is in comparison to another, which is key to solving such exercises.

Distance-Time Calculation

The distance-time relationship is a fundamental aspect of race problems. The basic formula connecting these quantities is Distance = Speed × Time. For any race, if you know two of these variables, you can solve for the third. This straightforward calculation becomes pivotal when trying to determine how long it takes a runner to finish a race, or how far they will have gone in a set amount of time.

In our exercise, we needed to find out how far runners A and B traveled during the time it took C to complete the race. We used the previously found relative speeds to calculate the times for A and C, which helped us find the distances. A clear grasp of this distance-time calculation is instrumental for accurately solving race problems and for practical understanding of motion in everyday life.

Competitive Race Analysis

Analyzing competitive races in quantitative aptitude tests your understanding of how different racers compare to each other under various conditions. These problems often involve scenarios where racers give their opponents a head start, or where one finishes a certain distance ahead of another. The key to solving these problems is to determine the relative performances of the racers, using their speeds, starts given, and distances covered.

In our problem, we were tasked with analyzing a 2 km race after understanding the outcomes of shorter races. Even when the situation becomes complex with three competitors, the principles remain the same. You need to analyze the information provided, apply concepts of relative speed and distance-time calculations, and understand how the racers' performances would affect the outcome of a longer race. By carefully setting up equations based on the provided data, you can delve deep into competitive race analysis, find the distances covered by the losers and identify the winner's margin of victory.