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Point $P$ lies between points $A$ and $B$ such that the length of $BP$ is thrice that of $AP$. Car $1$ starts from $A$ and moves towards $B$. Simultaneously, car $2$ starts from $B$ and moves towards $A$. Car $2$ reaches $P$ one hour after car $1$ reaches $P$. If the speed of car $2$ is half that of car $1$, then the time, in minutes, taken by car $1$ in reaching $P$ from $A$ is _________.
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Given that, point $P$ lies between $A$ and $B$ such that the length of $BP$ is thrice that of $AP$. It means $\boxed{BP = 3 \; AP}$

Let the total distance be $`4d\text{’} \; \text{km}.$ 



Let $\text{car 1}$ takes $`t\text{’} \; \text{hours}$ to reach point $P.$

So, $ \text{car 2}$ takes $`(t+1)\text{’} \; \text{hours}$ to reach point $P.$

The speed of $\text{car 2}$ is half that of $ \text{car 1.}$

Thus$, \text{S}_{\text{car 2}} = \frac{1}{2} \text{S}_{\text{car 1}}$

$ \Rightarrow \boxed {\text{S}_{\text{car 1}} = 2 \text{S}_{\text{car 2}}} \quad \longrightarrow (1)$

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

Now, the speed of $\text{car 1,}$ and $\text{car 2.}$

  • $\text{S}_{\text{car 1}} = \frac{d}{t}$
  • $\text{S}_{\text{car 2}} = \frac{3d}{t+1}$

Put the values in equation ${(1),}$ we get

$ \frac{d}{t} = 2 \left( \frac{3d}{t+1} \right)$

$ \Rightarrow \frac{1}{t} = \frac{6}{t+1}$

$ \Rightarrow t+1 = 6t $

$ \Rightarrow 5t = 1 $

$ \Rightarrow t = \frac {1}{5} \; \text{hours}$

$ \Rightarrow t = \frac{1}{5} \times 60 $

$ \Rightarrow t = 12 \; \text{minutes}$

$ \therefore$ The time taken by $ \text{car 1}$ to reach $P$ from $A$ is $12 \; \text{minutes}.$

Correct Answer $: 12$

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