This problem involves calculating relative distances covered by runners in a race based on given margins.
When runner A completes the 100 m race, runner B is 10 m behind.
This implies the ratio of distances covered by A and B in the same time is $\frac{\text{Distance}_A}{\text{Distance}_B} = \frac{100}{90}$.
When runner B completes a race (equivalent to 100 m for comparison), runner C is 5 m behind.
This implies the ratio of distances covered by B and C in the same time is $\frac{\text{Distance}_B}{\text{Distance}_C} = \frac{100}{95}$.
We need to determine how far C runs when A finishes the 100 m race.
From the A vs B analysis, when A runs 100 m, B runs 90 m.
Now, we use the B vs C ratio to find C's distance when B runs 90 m:
The ratio $\frac{\text{Distance}_C}{\text{Distance}_B} = \frac{95}{100}$.
Therefore, when $\text{Distance}_B = 90 \text{ m}$, the distance C covers is:
$\text{Distance}_C = 90 \text{ m} \times \frac{95}{100}$
$\text{Distance}_C = 90 \times 0.95 = 85.5 \text{ m}$
When A completes the 100 m race, C has run 85.5 m.
The distance A beats C by is the difference between the race distance and C's covered distance:
$100 \text{ m} - 85.5 \text{ m} = 14.5 \text{ m}$
A and B have to travel from place P to place Q following the same route in their respective cars. A drives at $60$ kmph while B drives at $80$ kmph. Find the time taken by B to reach place Q if A takes $12$ hrs.
On a straight road, a bus is $60$ km ahead of a car running in the same direction. After $3$ hours, the car is $90$ km ahead of the bus. If the speed of the bus is $45$ km/h, then what is the speed of the car (in km/h)?
A train running at the speed of $90$ kmph crosses a $250$ m long platform in $26$ seconds. What is the length of the train (in m)?
A car covers 4 successive stretches of 3 km each at speed of 10 kmph, 20 kmph, 30 kmph and 60 kmph respectively. The average speed of the car for the entire journey is: