A motorbike leaves a point X at 1 P.M. and moves towards the point Y at a uniform speed. A car leaves the point Y at 2 P.M. and moves towards the point X at a uniform speed which is double that of the motorbike. They meet at 3.40 P.M. at a point which is 168 km away from X. What is the distance between the points X and Y ?

Given:

The motorbike leaves point X at 1 P.M. and moves towards point Y at a uniform speed.

The car leaves point Y at 2 P.M. and moves towards point X at a speed double that of the motorbike.

They meet at 3:40 P.M. at a point 168 km away from X.

Concept Used:

Let the speed of the motorbike be S km/hr.

The speed of the car is 2S km/hr.

We need to calculate the total distance between points X and Y.

Calculation:

Time taken by the motorbike from 1 P.M. to 3:40 P.M. = 2 hours 40 minutes = 2 + 40/60 = 2²/₃ hours.

Distance covered by the motorbike till they meet = Speed × Time

⇒ Distance covered by the motorbike = S × 2²/₃ = (8S/3) km

It is given that this distance is 168 km.

⇒ 8S/3 = 168

⇒ S = (168 × 3)/8

⇒ S = 63 km/hr

Speed of the car = 2S = 2 × 63 = 126 km/hr

Time taken by the car from 2 P.M. to 3:40 P.M. = 1 hour 40 minutes = 1²/₃ hours.

Distance covered by the car = 126 × 1²/₃ = 210 km

Total distance between X and Y = Distance covered by the motorbike + Distance covered by the car

⇒ Total distance = 168 + 210 = 378 km

∴ The distance between points X and Y is 378 km.