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1 vote

A motorbike leaves point $\text{A}$ at $1$ pm and moves towards point $\text{B}$ at a uniform speed. A car leaves point $\text{B}$ at $2$ pm and moves towards point $\text{A}$ at a uniform speed which is double that of the motorbike. They meet at $3:40$ pm at a point which is $168$ km away from $\text{A}$. What is the distance, in km, between $\text{A}$ and $\text{B}?$

- $364$
- $378$
- $380$
- $388$

1 vote

Let the distance between $A$ and $B$ be $D$ km, and the speed of the motorbike be $x$ km /hour. Then the speed of the car is $2x$ km/hour.

Let's say they met at point $C$.

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

__For motorbike:__

$x = \dfrac{168}{\frac{8}{3}} \quad \left[\because 2 \;\text{hours}\; 40 \;\text{minutes} = 2 + \frac{40}{60} = 2 + \frac{2}{3} = \frac{8}{3} \;\text{hours}\right]$

$\Rightarrow \boxed{x = 63}$ km/hours

__For car:__

$2x = \dfrac{D-168}{\frac{5}{3}}$

$\Rightarrow2(63) = \dfrac{3(D-168)}{5} \quad \left[\because 1\;\text{hours}\; 40 \;\text{minutes} = 1 + \frac{40}{60} = 1 + \frac{2}{3} = \frac{5}{3}\;\text{hours} \right]$

$\Rightarrow 42\times5 = D-168$

$\Rightarrow 168 + 210 = D$

$\Rightarrow \boxed{D = 378\;\text{km}}$

$\therefore$ The distance between $A$ and $B$ is $378\;\text{km}.$

Correct Answer $:\text{B}$