Multiple choice

Directions : Answer the questions independently of each other. Well trodden paths connecting points A, B and C form a right-angled isosceles triangle with A and C being equidistant from B. Ravi decides to travel from point A to point C using the shortest path. Exactly halfway through the journeyhe meets with an accident and has to be rushed to the hospital located at point B. On path AB, the maximum speed the vehicle can travel at is 40 km/hr, on path BC it is 60 km/hr and on path CA it is 60√2 km/hr. There is also an untrodden path from the accident spot to the hospital at point B (which is also the shortest possible route between the two points). The maximum speed with which a vehicle can travel on any untrodden path is 30√2 km/hr. The distance between A and B is 2 km. What is the difference in time taken between the fastest two routes to reach the hospital? (Assume vehicles can travel only on the routes mentioned above).

  1. 1 min

  2. 20 min

  3. 1 hour

  4. 2 hours

  5. 5

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The triangle is right-angled isosceles with AB = BC = 2 km and AC = 2*sqrt(2) km. The shortest path from A to C is the hypotenuse (2*sqrt(2) km). Halfway is sqrt(2) km. The accident spot is on the hypotenuse. The distance from the accident spot to B is 1 km. Calculating the time for the two fastest routes yields a difference of 1 minute.