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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
In every month, both online and offline registration numbers were multiples of 10.
From (2), in Jan, the number of offline registrations was double that of online registrations.
Let x be the number of online registrations
2x be the number of offline registrations
Total number of registrations be x + 2x = 3x
According to the table data:
3x should lie between the minimum and maximum total number of registrations.
x = 40 (as x should also be a multiple of 10)
In January, 40 and 80 are the online and offline registrations respectively.
From 5, we can say that the number of online registrations is highest in may.
In may, there are total 100 online registrations.
Maximum possible total registrations is 130
Lowest possible number of offline registrations is 30
Let x be the number of offline registrations in May which is equal to number of online registrations in March.
Let's arrange the data in table:
From the table mentioned in question, 50 is the median for Offline data
Now x should lie between 50 and 80
For 80 to be the median for the online data
y should lie between 80 and 100
Now, let's consider the following:
Feb ⇒ Minimum value of y + x
= 80 + 50 = 130
Therefore, x = 50 and y = 80
Since, 110 is the minimum number of total registrations, the only possibility is in March:
50 + z = 110
z = 60
Let's complete the table:
The correct answer is Only I.