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Points $\text{A, P, Q}$ and $\text{B}$ lie on the same line such that $\text{P, Q}$ and $\text{B}$ are, respectively, $100$ km, $200$ km and $300$ km away from $\text{A}$. Cars $1$ and $2$ leave $\text{A}$ at the same time and move towards $\text{B}$. Simultaneously, car $3$ leaves $\text{B}$ and moves towards $\text{A}$. Car $3$ meets car $1$ at $\text{Q}$, and car $2$ at $\text{P}$. If each car is moving in uniform speed then the ratio of the speed of car $2$ to that of car $1$ is

- $1:2$
- $2:9$
- $1:4$
- $2:7$

1 vote

Given that, $A,P,Q$ and $B$ lie on the same line such that $P,Q$ and $B$ are respectively, $100 \; \text{km}, 200 \; \text{km}$ and $300 \; \text{km}$ away from $A.$

Let the speed of $\text{Car 1, Car 2}$ and $\text{Car 3}$ be $x,y$ and $z \; \text{km/hr}.$

After some time, $\text{Car 3}$ meets $\text{Car 1}$ at $Q.$

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

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

$ \Rightarrow \boxed{\frac{S_{1}}{S_{2}} = \frac{D_{1}}{D_{2}}} $

We can say that, if time is constant, speed is directly proportional to distance.

- Distance traveled by $\text{Car 1} = 200 \; \text{km} $
- Distance traveled by $ \text{Car 3} = 100 \; \text{km} $

So, $\frac{x}{z} = \frac{200}{100} $

$ \Rightarrow \frac{x}{z} = \frac{2}{1} $

$ \Rightarrow \boxed{x : z = 2 : 1} \quad \longrightarrow (1) $

After some time, $\text{Car 3}$ meets $\text{Car 2}$ at $P.$

- Distance traveled by $\text{Car 2} = 100 \; \text{km} $
- Distance traveled by $ \text{Car 3} = 200 \; \text{km} $

So, $\frac{y}{z} = \frac{100}{200} $

$ \Rightarrow \frac{y}{z} = \frac{1}{2} $

$ \Rightarrow \boxed{y : z = 1 : 2} \quad \longrightarrow (2) $

Now, combine the ratios, from equation $(1),$ and $(2),$ we get.

- $ x:z = (2:1) \times 2 = 4:2 $
- $ y:z = (1:2) \times 1 = 1:2 $
- $x:y:z = 4:1:2 $

$\therefore$ The ratio of the speed of $\text{car2}$ to that of $\text{car1} = y:x = 1:4.$

Correct Answer $: \text{C}$