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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

1. $500$
2. $475$
3. $550$
4. $450$

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

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

$\textbf{Case 1:}$

$\textbf{Case 2:}$

$\textbf{Case 3:}$

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 ratios of speed:

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

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

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

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

$\Rightarrow \frac{x}{x-90} = \frac{x^{2}}{(x-45)(x-50)} \quad [\because \text{From equation (3)}]$

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

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

$\Rightarrow – 95x + 90x = – 2250$

$\Rightarrow -5x = -2250$

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

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

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

Correct Answer$: \text{D}$

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