Discussion

Explanation:

Let us tabulate the information given in the bar chart. As per condition (i) we get the following information

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From this we can conclude that Dorm 6 would require light repairs and that Dorm 3 and 9 would require moderate repairs. Now following condition (2), we get to know that Dorm 7 would require 6 cr for repair. and Dorm 8 would required 1 cr. This means that in light repairs, Dorm 8 requires 1 crore for repair and Dorm 2 requires 6 crores for repair. Now as each of Dorms to 9 requires a different amount for repairs, and no even numbered dorm requires ,moderate repair, it would imply that Dorm no 4 requires heavy repairs. Already Dorm 7 needs the maximum amount for repairs, (i.e, 6 crores) which means that Dorm4 would require 5 crores for repair. Now Dorm 9 would require 3 or 4 cr for repair. This leads us to 2 possible cases.

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In case 1, as 4 cr is required for Dorm no 9, 3 cr will be required for Dorm no. 5. Also, as no even numbered dorm can have a moderate expenditure, out of the remaining 4 dorms, 1 and 3 (which are also the only odd numbered dorms out of the remaining 4 dorms) will have an expenditure of 3 crores (also since there is only 1 dorm which has a expense of 4 cr which in this case is the expense for Dorm 9). This leaves out only Dorm 2 and Dorm 10, one of these 2 dorms will have an expense of 1 cr and the other an expense of 6 crores.
In case 2, 3 cr is the expense for Dorm No. 9 and 4 cr will be the expense for Dorm no 5. Like in case 1, Dorm no 1 and 3 will have an expense of 1 crore and the other an expense of 6 crores. So our final table for both cases will be as below

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Dorms 1 and 3 cost 3 crores each to repair.

One of Dorm 5 or Dorm 9 costs 3 crores to repair and the other costs 4 crores repair.

Dorm 7 costs 6 crores to repair.

Total cost of repairing odd numbered dorms = 3 + 3 + 3 + 4 + 6 = 19 crores

Answer: 19

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