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The distance from $\text{A}$ to $\text{B}$ is $60$ km. Partha and Narayan start from $\text{A}$ at the same time and move towards $\text{B}$. Partha takes four hours more than Narayan to reach $\text{B}$. Moreover, Partha reaches the mid-point of $\text{A}$ and $\text{B}$ two hours before Narayan reaches $\text{B}$. The speed of Partha, in km per hour, is

1. $4$
2. $6$
3. $5$
4. $3$

Given that, the distance from $\text{A}$ to $\text {B}=60\; \text{km}$

Partha and Narayan start from $\text {A}$ at the same time and move towards $\text {B}$.

Let the time taken by Narayan to reach from $\text {A}$ to $\text {B}$ be $x$ hours.

Partha takes four hours more than Narayan to reach $\text {B}$.

• The time taken by partha to reach from $\text {A}$ to $\text{B} = (x+4)$ hours.
• The time taken by Partha to complete half of the distance $= \left ( \frac {x+4}{2} \right)$ hours.

Partha reaches the mid point of $\text {A}$ and $\text {B}$ two hours before Narayan reaches $\text {B}:$

$x – \left(\frac {x+4}{2}\right) = 2$

$\Rightarrow \frac {2x-(x+4)}{2} = 2$

$\Rightarrow 2x-x-4 = 4$

$\Rightarrow x=8$ hours.

The time taken by Partha and Narayan to complete cover the distance from $\text {A}$ to $\text {B}:$

• Narayan time $= x = 8\; \text{hours}$
• Partha time $= x+4 = 8+4 = 12\; \text{hours}$

Let $S_{p}$ be the speed of partha.

Apply the formula for the speed $: \boxed {\text {Speed} =\frac{\text{Distance}} { \text{Time}}}$

$\therefore$ The speed of partha $S_{p} = \frac {60}{12} = 5 \frac { \text {km}}{ \text {h}}$

Correct Answer $: \text C$

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