A and B take part in 100 meters race, where A beats B by 10 meters. To favour B, A starts 10 meters behind the starting line in a second 100 meters race, running at their earlier speeds. Which one among the following is true in view of the second race?
In the first race, A runs 100m and B runs 90m in the same time, say T.
Speed of A, $V_A = 100 / T$.
Speed of B, $V_B = 90 / T$.
Ratio of speeds: $V_A / V_B = (100/T) / (90/T) = 100/90 = 10/9$. So, $V_A = (10/9) V_B$.
In the second race, the finish line is 100 meters from the standard starting line.
A starts 10 meters behind the starting line, so A has to run $100 + 10 = 110$ meters to reach the finish line.
B starts at the regular starting line, so B has to run 100 meters to reach the finish line.
Time taken by A to cover 110 meters: $T_A = \text{Distance} / \text{Speed} = 110 / V_A$.
Time taken by B to cover 100 meters: $T_B = \text{Distance} / \text{Speed} = 100 / V_B$.
Substitute $V_A = (10/9) V_B$ into the expression for $T_A$:
$T_A = 110 / ((10/9) V_B) = (110 \times 9) / (10 \times V_B) = 990 / (10 V_B) = 99 / V_B$.
Now compare $T_A$ and $T_B$:
$T_A = 99 / V_B$ and $T_B = 100 / V_B$.
Since $99 < 100$ and $V_B$ is positive, $T_A < T_B$. A reaches the finish line faster than B. A beats B. To find by how much A beats B, calculate the distance covered by B when A finishes the race (at time $T_A$). Distance covered by B at time $T_A$ is $D_B = V_B \times T_A = V_B \times (99 / V_B) = 99$ meters. When A crosses the finish line, B has covered 99 meters from their starting line. The finish line is at 100 meters for B. So, B is $100 - 99 = 1$ meter behind the finish line when A finishes. A beats B by 1 meter.