Question:

In a locality, there are five small cities: A, B, C, D, and E. The distances of these cities from each other are as follows. AB = 2 km; AC = 2 km; AD>2 km; AE>3 km; BC = 2 km; BD = 4 km; BE = 3 km; CD = 2 km; CE = 3 km; DE>3 km. If a ration shop is to be set up within 3 km of each city, how many ration shops will be required?

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When solving coverage problems, identify overlapping regions where resources can be concentrated to minimize the number of required resources.
Updated On: Jul 24, 2025
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The Correct Option is C

Solution and Explanation

We are given the distances between various cities. The objective is to set up ration shops such that each city is within 3 km of at least one ration shop. Here are the distances provided: \[ AB = 2 \, \text{km}, \quad AC = 2 \, \text{km}, \quad AE>3 \, \text{km}, \quad BC = 2 \, \text{km}, \quad BD = 4 \, \text{km}, \quad BE = 3 \, \text{km} \] \[ CD = 2 \, \text{km}, \quad CE = 3 \, \text{km}, \quad DE>3 \, \text{km} \] We need to set up ration shops such that they cover cities that are within 3 km of each other. We observe that:
- Cities \(A, B, C, D\) are within 3 km of each other based on the provided distances.
- City \(E\) is only within 3 km of \(B\) and \(C\), so a ration shop can be placed at one of these cities.
Thus, we need to place ration shops at \(B\), \(C\), and \(A\) to cover all cities, as \(B\) and \(C\) can cover \(E\), and \(A\) can cover the remaining cities within the required distance.
Therefore, the answer is c. 3.
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