Question

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 km away from A. What is the distance, in km, between A and B?

• A. 364
• B. 378
• C. 380
• D. 388

Answer 1

The Correct Option is B

To solve this problem, we first need to determine the speeds of the motorbike and the car. Let's denote the speed of the motorbike as \( v \) km/h. The car's speed is twice that of the motorbike, so its speed will be \( 2v \) km/h.

Step 1: Calculate the total time traveled by each vehicle until they meet.

• The motorbike leaves A at 1 pm and meets the car at 3:40 pm. Thus, it travels for a total of 2 hours and 40 minutes, which is \( \frac{8}{3} \) hours.
• The car leaves B at 2 pm and meets the motorbike at 3:40 pm, traveling for 1 hour and 40 minutes, which is \( \frac{5}{3} \) hours.

Step 2: Establish the equations for the distance traveled by each vehicle.

• The motorbike travels a distance of \( v \times \frac{8}{3} \) km.
• The car travels a distance of \( 2v \times \frac{5}{3} \) km.

Since they meet at the point 168 km from A, and since the sum of the distances traveled by both vehicles equals the total distance between A and B, we have:

\[ v \times \frac{8}{3} + 2v \times \frac{5}{3} = \text{Total Distance between A and B} \]

The motorbike covers 168 km, thus:

\[ v \times \frac{8}{3} = 168 \]

Solving for \( v \):

\[ v = \frac{168 \times 3}{8} = 63 \text{ km/h} \]

Substitute \( v = 63 \) km/h into the equation for the car's distance:

\[ 63 \times \frac{8}{3} + 2 \times 63 \times \frac{5}{3} = D \]

Simplifying:

\[ 168 + 210 = D \]

\[ D = 378 \text{ km} \]

Therefore, the distance between A and B is 378 km.

Options	Values (km)
364
378	✓
380
388