XAT 2015 — QA & DI Question 15
Devanand’s house is 50 km West of Pradeep’s house. On Sunday morning, at 10 a.m., they leave their respective houses.
Under which of the following scenarios, the minimum distance between the two would be 40 km?
Scenario I: Devanand walks East at a constant speed of 3 km per hour and Pradeep walks South at a constant speed of 4 km per hour.
Scenario II: Devanand walks South at a constant speed of 3 km per hour and Pradeep walks East at a constant speed of 4 km per hour.
Scenario III: Devanand walks West at a constant speed of 4 km per hour and Pradeep walks East at a constant speed of 3 km per hour.
Answer & solution
Scenario I only
- B
Scenario II only
- C
Scenario III only
- D
Scenario I and II
- E
None of the above
Scenario 1: Devanand walks East at a constant speed of 3 km per hour and Pradeep towards South at a constant speed of 4 km per hour,
Let the two walk for x hours.
Distance travelled by Devanand and Pradeep is 3x km and 4x km respectively.
Thus, we have
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∴ (AP)2 + (CP)2 = (50 – 3x)2 + (4x)2 = 402
Solving this, we get x = 6
Thus, after 6 hours, the minimum distance between the two would be 40 km.
The distance between the two will increase in other two scenarios.
Hence, option (a).