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
x is number of online registrations
⇒2x is the number of offline registrations
⇒3x is the total number of registrations.
According to the data given in the table
⇒3x should lie between the minimum and maximum total number of registrations.
⇒x=40 (as
x should also be a multiple of 10 )
⇒ ln Jan
⇒(40,80) are the online and offline registrations respectively.
Similarly from (3) ⇒ ln Apr
(80,40) 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
(100,30) are the online and offline registrations respectively.
Let us assume, '
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
⇒x should lie between 50 and 80 (included)
For 80 to be the median for the online data
⇒y lie between 80 and 100 (included).
Now, consider Feb ⇒ Minimum value of
y+x=80+50=130 (which is the maximum value possible of the total possible registrations)
⇒x=50 and
y=80Since, 110 is the minimum number of total registrations, the only possibility is in March
⇒50+z=110⇒z= 60.
Now, filling the complete table we get,
The number of offline registrations in Feb is 50.