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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
We can represent the money raised each year for each company as follows.
For each or these companies we will try to make cases satisfying the given conditions.
We will write the amount of money raised by them in consecutive years from starting from their 1st year till their last year.
For A, the sum is 21 where first and last values must be 1.
This is possible for the following order of numbers.
Case 1: 1 → 2 → 3 → 4 → 5 → 3 → 2 → 1
Case 2: 1 → 2 → 3 → 5 → 4 → 3 → 2 → 1
For B, the only two possible order of numbers is
Case 1: 1 → 2 → 3 → 1
Case 2: 1 → 3 → 2 → 1
For C, the sum is 9 where first and last values must be 1. This is only possible when the order of numbers is
1 → 2 → 3 → 2 → 1
For D, the sum is 10 where first and last values must be 1. This is only possible when the order of numbers is
1 → 2 → 4 → 2 → 1
For E, the sum is 13 where first and last values must be 1. This is possible for the following order of numbers.
Case 1: 1 → 2 → 4 → 3 → 2 → 1
Case 2:1 → 2 → 3 → 4 → 2 → 1
Case 3: 1 → 3 → 5 → 3 → 1
Hence, we get the following table with all possible cases.
Highest amoun of money that can be raised in 2013 by:
A = 5, B = 3, C = 1, D = 4, E = 4
∴ Highest total amount of money that can be raised by them in 2013 = 5 + 3 + 1 + 4 + 4 = 17 crores