CAT 2023 Slot 3 Solved Paper

Given that in every month, both online and offline registration numbers were multiples of 10 .
From (2), in Jan, the number of offline registrations was twice that of online registrations.
⇒ If $\mathit{x}$ is number of online registrations $\Rightarrow 2\mathit{x}$ is the number of offline registrations $\Rightarrow 3\mathit{x}$ is the total number of registrations.
According to the data given in the table $\Rightarrow 3\mathit{x}$ should lie between the minimum and maximum total number of registrations. $\Rightarrow \mathit{x}=40$ (as $\mathit{x}$ should also be a multiple of 10 )
⇒ ln Jan $\Rightarrow \left(40,80\right)$ are the online and offline registrations respectively.
Similarly from (3) ⇒ ln Apr $\left(80,40\right)$ are the online and offline registrations respectively.
From-5, the number of online registrations is highest in may ⇒ In may there are 100 online registrations. The lowest possible number of offline registrations is 30 and maximum possible total registrations is $130$ ⇒ ln May $\left(100,30\right)$ are the online and offline registrations respectively.
Let us assume, ' $\mathit{x}$ ' to be the number of offline registrations in May = number of online registrations in March.
Let us capture all this data in a table:From the table given in the question, 50 is the median for Offline data
$\Rightarrow \mathit{x}$ should lie between 50 and 80 (included)
For 80 to be the median for the online data $\Rightarrow \mathit{y}$ lie between 80 and 100 (included).
Now, consider Feb ⇒ Minimum value of $\mathit{y}+\mathit{x}=80+50=130$ (which is the maximum value possible of the total possible registrations)
$\Rightarrow \mathit{x}=50$ and $\mathit{y}=80$
Since, 110 is the minimum number of total registrations, the only possibility is in March $\Rightarrow 50+\mathit{z}=110\Rightarrow \mathit{z}=$ 60.
Now, filling the complete table we get,The number of offline registrations in Feb is 50.