In a race of 1000 m, A beats B by 150 m, while in another race of 3000 m, C beats D by 400 m. Speed of B is equal to that of D. (Assume that A, B, C and D run with uniform speed in all the events). If A and C participate in a race of 6000 m, then which one of the following is correct?

Question

Prepp Team · Accepted Answer

Understanding the Race Scenario and Relative Speeds

The question describes a series of races involving four runners: A, B, C, and D, who run at uniform speeds. We are given the results of two races and a condition about the speeds of B and D. We need to determine the outcome of a race between A and C over a specific distance.

Let's denote the speeds of A, B, C, and D as \(S_A\), \(S_B\), \(S_C\), and \(S_D\) respectively.

Analyzing the Given Race Information

We have information from two different races:

Race 1: 1000 m race where A beats B by 150 m.
Race 2: 3000 m race where C beats D by 400 m.

We are also given that the speed of B is equal to the speed of D, i.e., \(S_B = S_D\).

Calculating Speed Ratios from Race Results

In a race where one person beats another by a certain distance, when the winner finishes the full distance, the loser has covered the full distance minus the margin. Since both run for the same amount of time until the winner finishes, the ratio of distances covered is equal to the ratio of their speeds.

From Race 1 (1000 m):

When A covers 1000 m, B covers \(1000 \, \text{m} - 150 \, \text{m} = 850 \, \text{m}\).

The ratio of A's speed to B's speed is:

\begin{equation*} \frac{S_A}{S_B} = \frac{\text{Distance covered by A}}{\text{Distance covered by B}} = \frac{1000}{850} = \frac{100}{85} = \frac{20}{17} \end{equation*}

So, \(S_A = \left(\frac{20}{17}\right) S_B\).

From Race 2 (3000 m):

When C covers 3000 m, D covers \(3000 \, \text{m} - 400 \, \text{m} = 2600 \, \text{m}\).

The ratio of C's speed to D's speed is:

\begin{equation*} \frac{S_C}{S_D} = \frac{\text{Distance covered by C}}{\text{Distance covered by D}} = \frac{3000}{2600} = \frac{30}{26} = \frac{15}{13} \end{equation*}

So, \(S_C = \left(\frac{15}{13}\right) S_D\).

Using the Condition \(S_B = S_D\) to Compare A and C

We are given \(S_B = S_D\). Let's use this to compare \(S_A\) and \(S_C\). If we assume \(S_B = S_D = S\) (some arbitrary speed unit), then:

\(S_A = \left(\frac{20}{17}\right) S\)
\(S_C = \left(\frac{15}{13}\right) S\)

To see who is faster, we compare the fractions \(\frac{20}{17}\) and \(\frac{15}{13}\).

\begin{align*} \frac{20}{17} &\quad? \quad \frac{15}{13} \\ 20 \times 13 &\quad? \quad 15 \times 17 \\ 260 &\quad? \quad 255 \end{align*}

Since \(260 > 255\), we have \(\frac{20}{17} > \frac{15}{13}\).

This means \(S_A > S_C\). A is faster than C.

Predicting the Outcome of A vs C in a 6000 m Race

Since A is faster than C, A will win the 6000 m race against C. We need to find by what distance A beats C.

When A finishes the 6000 m race, the time taken by A is:

\begin{equation*} T_A = \frac{\text{Distance}}{\text{Speed}} = \frac{6000}{S_A} \end{equation*}

In this same amount of time \(T_A\), C will cover a distance \(D_C\).

\begin{equation*} D_C = S_C \times T_A = S_C \times \frac{6000}{S_A} \end{equation*}

We know \(S_A = \left(\frac{20}{17}\right) S_B\) and \(S_C = \left(\frac{15}{13}\right) S_D\). Since \(S_B = S_D\), let's use \(S_B\) for both.

\begin{equation*} D_C = \left(\frac{15}{13} S_B\right) \times \frac{6000}{\left(\frac{20}{17} S_B\right)} \end{equation*}

We can cancel \(S_B\) from the equation:

\begin{equation*} D_C = \frac{15}{13} \times \frac{6000}{\frac{20}{17}} = \frac{15}{13} \times \frac{6000 \times 17}{20} \end{equation*}

Simplify the calculation:

\begin{equation*} D_C = \frac{15}{13} \times \frac{300 \times 17}{1} = \frac{15 \times 5100}{13} = \frac{76500}{13} \end{equation*}

Now, calculate the value of \(D_C\):

\begin{equation*} D_C \approx 5884.61538 \, \text{m} \end{equation*}

When A finishes the 6000 m race, C has covered approximately 5884.61538 m. The distance by which A beats C is the difference between the race distance and the distance covered by C.

Distance A beats C by = \(6000 \, \text{m} - D_C\)

\begin{equation*} \text{Margin} = 6000 - \frac{76500}{13} = \frac{6000 \times 13 - 76500}{13} = \frac{78000 - 76500}{13} = \frac{1500}{13} \end{equation*}

Calculate the value:

\begin{equation*} \frac{1500}{13} \approx 115.3846 \, \text{m} \end{equation*}

So, A beats C by approximately 115.38 m.

Conclusion

Based on the calculations:

A is faster than C.
In a 6000 m race, A beats C by approximately 115.38 m.

Let's summarize the speeds relative to \(S_B = S_D = S\):

Runner	Speed (relative to S)	Approximate Value
A	\(\frac{20}{17} S\)	\(1.176 S\)
B	\(S\)	\(1.000 S\)
C	\(\frac{15}{13} S\)	\(1.154 S\)
D	\(S\)	\(1.000 S\)

Comparing \(S_A\) and \(S_C\): \(\frac{20}{17} > \frac{15}{13}\), confirms A is faster.

The distance difference calculated is approximately 115.38 m, with A being the winner.

Revision Table: Key Concepts in Race Problems

Concept	Explanation	Formula/Relation
Uniform Speed	Speed remains constant throughout the race.	Speed = Distance / Time
Relative Speed in Races	When comparing two runners over the same time, the ratio of distances covered equals the ratio of their speeds.	\(D_1 / D_2 = S_1 / S_2\) (when time is constant)
Beating by Distance	If runner X beats runner Y by 'm' meters in a 'D' meter race, when X covers D meters, Y covers (D - m) meters.	\(S_X / S_Y = D / (D-m)\)

Additional Information on Race Calculations

Race problems often involve understanding the relationship between speed, distance, and time. When comparing two runners in a race, there are two main scenarios:

Same Time: If runners start together and the comparison is made when the first runner finishes or at any point when both have been running for the same duration, the ratio of distances covered is directly proportional to the ratio of their speeds. This is what we used in this problem to find speed ratios from the initial races.
Same Distance: If runners cover the same distance (e.g., finishing the same race), the ratio of their times is inversely proportional to the ratio of their speeds (\(T_1/T_2 = S_2/S_1\)).

In our problem, to find the margin by which A beats C in the 6000 m race, we calculated how much distance C covers in the time A takes to complete 6000 m. The difference between 6000 m and the distance covered by C is the winning margin.

It's crucial to correctly set up the speed ratios from the initial conditions. The condition \(S_B = S_D\) serves as a link between the information from the two separate races (A vs B and C vs D), allowing us to compare A's speed directly to C's speed.

The calculations involve fractions and can result in decimals. It's important to maintain precision or use fractional calculations until the final step to get an accurate answer matching the options.

Prepp Team · Answer

A beats C by 250 m

Prepp Team · Answer

C beats A by 250 m

Prepp Team · Answer

C beats A by 115.38 m

Understanding the Race Scenario and Relative Speeds

Analyzing the Given Race Information

Calculating Speed Ratios from Race Results

From Race 1 (1000 m):

From Race 2 (3000 m):

Using the Condition \(S_B = S_D\) to Compare A and C

Predicting the Outcome of A vs C in a 6000 m Race

Conclusion

Revision Table: Key Concepts in Race Problems

Additional Information on Race Calculations

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Important Questions from Relative Speed

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