In a car race, car $\text{A}$ beats car $\text{B}$ by $45 \; \text{km},$ car $\text{B}$ beats car $\text{C}$ by $50 \; \text{km},$ and car $\text{A}$ beats car $\text{C}$ by $90\;\text{km}.$ The distance $\text{(in km)}$ over which the race has been conducted is

- $500$
- $475$
- $550$
- $450$

## 1 Answer

Given that, in a car race, car$\text{A}$ beats car$\text{B}$ by $45 \; \text{km},$ and car$\text{B}$ beats car$\text{C}$ by $50 \; \text{km},$ and car$\text{A}$ beats car$\text{C}$ by $90 \; \text{km}.$

Let the distance over which the race has been conducted be $’x’ \; \text{km}.$

Case $1:$

DIAGRAM

Case $2:$

DIAGRAM

Case $3:$

DIAGRAM

We know that, $ \text{Speed} = \frac{\text{Distance}}{\text{Time}} $

$\Rightarrow \boxed {\text{Speed} \propto \text{Distance}} \; (\text{Time = constant})$

Now, we can calculate the ratio of speeds:

- In Case $1: \; \frac{S_{A}}{S_{B}} = \frac{x}{x – 45} \; \longrightarrow (1) $
- In Case $2: \; \frac{S_{B}}{S_{C}} = \frac{x}{x – 50} \; \longrightarrow (2) $
- In Case $3: \; \frac{S_{A}}{S_{C}} = \frac{x}{x – 90} \; \longrightarrow (3) $

Now, multiply equation $(1) \& (2),$ we get

$\left( \frac{S_{A}}{S_{B}} \right) \times \left( \frac{S_{B}}{S_{C}} \right) = \left( \frac{x}{x-45} \right) \left( \frac{x}{x-50} \right) $

$ \Rightarrow \frac{S_{A}}{S_{C}} = \frac{x^{2}}{(x-45)(x-50)} $

$ \Rightarrow \frac{x}{x-90} = \frac{x^{2}}{(x-45)(x-50)} $

$ \Rightarrow (x-45)(x-50) = x(x-90) $

$ \Rightarrow x^{2} – 50x – 45x + 2250 = x^{2} – 90x $

$ \Rightarrow – 95x + 90x = – 2250 $

$ \Rightarrow -5x = -2250 $

$ \Rightarrow x = \frac{2250}{5} $

$ \Rightarrow \boxed{x = 450 \; \text{km}}$

$\therefore$ The distance (in km) over which the race has been conducted is $450.$

Correct Answer$: \text{D}$