How to solve XAT 2018 Question 40?

This question can be solved by applying the core concept explained in the solution above, followed by step-by-step calculation and elimination of options.

Which concept is tested in this question?

This question tests fundamental concepts commonly asked in XAT exams.

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XAT 2018 Question 40

This is a detailed step-by-step solution for XAT 2018 Question 40. Understand the concept, common mistakes, and expert tips to solve similar questions in XAT exam.

XAT

T201: Howrah station to Kharagpur station (6:30 - 8:15) --> Delivery (8:16 - 9:45) --> Kharagpur to Jamshedpur via T202 (10:40 - 12:35) -- > Delivery (12:35 - 13:25). Thus, a total of 6hr 55mins.
T202: Howrah station to Kharagpur station (8:30 - 10:45) --> Delivery (10:46 - 12:10) --> Kharagpur to Jamshedpur via T203 (12:15 - 1400) --> Delivery (14:01 - 14:50). Thus, a total of 6hr 20mins.
T204: Howrah station to Kharagpur station (14:30 - 16:15) --> Delivery (16:16 - 17:45) --> Kharagpur to Jamshedpur via T205 (17:55 - 20:00) --> Delivery (20:01 - 20:50). Thus, a total of 6hr 20mins.
Thus, T202 and T204 require the least amount of time.
Hence, option D is the answer.

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