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

$364$

$378$

$380$

$388$

✅ Correct Option: 2

We first map out what happens when:

1:00 PM: Motorbike leaves point A

2:00 PM: Car leaves point B

3:40 PM: They meet at a point 168 km from A

Motorbike's travel time: From 1:00 PM to 3:40 PM

3:40 PM - 1:00 PM = 2 hours 40 minutes = 2 $\tfrac{2}{3}$ hours = $\tfrac{8}{3}$ hours

Car's travel time: From 2:00 PM to 3:40 PM

3:40 PM - 2:00 PM = 1 hour 40 minutes = 1 $\tfrac{2}{3}$ hours = $\tfrac{5}{3}$ hours

We know the motorbike traveled 168 km in $\tfrac{8}{3}$ hours.

Using Speed = Distance ÷ Time:

Motorbike's speed = $168 \div \tfrac{8}{3}$

$= 168 \times \tfrac{3}{8}$

$= 63$ km/h

The problem states the car's speed is double that of the motorbike.

Car's speed = $2 \times 63 = 126$ km/h

The car traveled for $\tfrac{5}{3}$ hours at 126 km/h.

Distance traveled by car = $126 \times \tfrac{5}{3} = 210$ km

Since they meet at a point between A and B:

Distance from A to meeting point = 168 km

Distance from B to meeting point = 210 km

Total distance between A and B = $168 + 210 = 378$ km

When two objects move towards each other and meet, the total distance they cover together equals the distance between their starting points. This is a fundamental concept in relative motion problems that will help you solve similar questions quickly!

Answer: 378 km

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