Points $A, P, Q$ and $B$ lie on the same line such that $P, Q$ and $B$ are, respectively, $100$ km, $200$ km and $300$ km away from $A$. Cars $1$ and $2$ leave $A$ at the same time and move towards $B$. Simultaneously, car $3$ leaves $B$ and moves towards $A$. Car $3$ meets car $1$ at $Q$, and car $2$ at $P$. 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 Answer

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 $\text{car1, car2}$ and $\text{car3}$ be $x,y$ and $z \; \text{km/hr}.$

After some time, $\text{car3}$ meets $\text{car1}$ 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{car1} = 200 \; \text{km} $
- Distance traveled by $ \text{car3} = 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, $\text{car3}$ meets $\text{car2}$ at $P.$

- Distance traveled by $\text{car2} = 100 \; \text{km} $
- Distance traveled by $ \text{car3} = 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 $: \text{C}$