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Explanation:

This questions was discarded by IIM Bangalore.

A cannot be the last city to be visited while satisfying all the conditions given in the caselet.

Explanation:

Demand
A – 50 (40%); 70 (60%)
B – 40 (30%); 60 (70%)
C – 70 (30%); 100 (70%)
D – 30 (40%); 50 (60%)

For Ahmednagar to be last, it should have the least demand of the 4 cities.
⇒ The only way Ahmednagar’s demand can be the least of the 4 cities is when its demand is 50.

Now, demand of all other cities should be greater than or equal to 50.

⇒ Demand at
B = 60
C = 70 or 100
D = 50

∴ Sequence of cities according to demand will be C → B → D → A

Distance travelled from
Warehouse → C = 12
C → B = 4
B → W → D = 12
D → W → A = 7 [shortest route from D to A is through Warehouse and not the direct route]

∴ Total distance travelled = 12 + 4 + 12 + 7 = 35.

Ambiguity: There is some ambiguity in this question. Once you reach B, demand at both A and D is same (i.e., 50). You would go the nearest of A and D which is A and hence A cannot be the last city to be visited then.

Hence, this question was discarded.

Note: The answer given by IIM-B in the cadidate response sheet was 35.

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