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IBPS PO Mains 5 Nov 2023 Paper

Question Numbers: 137-140

Directions: The below table shows the four points based on starting/ Ending and distance in km between three points.

Question : 140 of 155

Marks: +1, -0

Solution:

Concept:

Minimum time is obtained by taking the fastest direct path, while maximum time is obtained by taking the slowest possible route.

Explanation:

Minimum time from A to D is on the direct distance of

260

km at

40

km/h:

\text{Minimum time} = \dfrac{260}{40} = \dfrac{13}{2}

hours.

Maximum time is via A to B and then B to D:

\text{Maximum time} = \dfrac{120}{40} + \dfrac{200}{25} = 3 + 8 = 11

hours.

Required sum =

\dfrac{13}{2} + 11 = \dfrac{13 + 22}{2} = \dfrac{35}{2}

hours.

Answer:

\dfrac{35}{2}

hours, i.e., Option B.

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