Which of the following statements can be true?

I. The number of offline registrations was the smallest In May.

II. The total number of registrations was the smallest in February.

Month	Online	Offline	Total
Jan	x	2x	3x
Feb
Mar
Apr
May

Clue 1 says, In every month, both online and offline registration numbers were multiples of 10.

Clue 2, In January, the number of offline registrations was twice that of online registrations.

Let the number of online registration in January be x

= number of offline registration in January will be 2x

Therefore, total registrations will become = x + 2x = 3x

Month	Online	Offline	Total
Jan	40	80	120
Feb
Mar
Apr
May

	Minimum	Maximum	Median
Online	40	100	80
Offline	30	80	50
Total	110	130	120

From the second table, we know –

Minimum value of x = 40

Therefore, 2x = 80 (which is also the maximum value for offline)

Therefore, the only possible value for x = 40

Month	Online	Offline	Total
Jan	40	80	120
Feb
Mar
Apr	80	40	120
May

	Minimum	Maximum	Median
Online	40	100	80
Offline	30	80	50
Total	110	130	120

From clue 3, In April, the number of online registrations was twice that of offline registrations.

From the second table we know, Minimum and maximum for online is 40 & 100 respectively,

Also, Minimum and maximum for offline is 30 & 80 respectively

And , Minimum and maximum of total is 110 & 130 respectively

Now, for online to be double of offline registrations

Therefore, there are only 3 possibilities

1 – (30 , 60)

2- (40, 60)

3- (50, 100)

But we also know that total can’t be less then 110

Therefore 1st possibility is rejected (as sum = 30 + 60 = 90)

Similarly, 2nd possibility is also rejected (as sum is 50 + 100 = 150 ) – which is more then the maximum sum

So, only possibility left is (40, 60)

Month	Online	Offline	Total
Jan	40	80	120
Feb
Mar
Apr	80	40	120
May	100	30	130

	Minimum	Maximum	Median
Online	40	100	80
Offline	30	80	50
Total	110	130	120

Using clue 5,

The number of online registrations was the largest in May.

Which implies there are 100 online registrations in May.

Now , we know the maximum total = 130

Therefore, In may offline registration has to be 30 or less then it

But it can’t be less then 30 (as this is the minimum number of registrations offline can have)

Therefore, In May (100,30) are the online and offline registrations respectively

Month	Online	Offline	Total
Jan	40	80	120
Feb	y	x
Mar	x	z
Apr	80	40	120
May	100	30	130

	Minimum	Maximum	Median
Online	40	100	80
Offline	30	80	50
Total	110	130	120

Now, using clue 4, The number of online registrations in March was the same as the number of offline registrations in February.

Let the number of offline registrations in march = “x”

Therefore, number of online registrations in feb will also be “x”

Month	Online	Offline	Total
Jan	40	80	120
Feb	y	50
Mar	50	z
Apr	80	40	120
May	100	30	130

	Minimum	Maximum	Median
Online	40	100	80
Offline	30	80	50
Total	110	130	120

From the second table we know that median of offline data is 50.

Which means the middle value in offline table is 50

We already have 2 values lower then 50 ( 40 and 30 for April and may)

Therefore x and z will be 50 or greater then 50. But less then 80 (maximum value of offline)

Similarly, for 80 to be the median of online data we need one of y or x to be between 80 to 100 and other of y or x less then 80.

But as x can’t be greater then 80

Therefore, 50 value can be assumed by x only.

Therefore, x will be 50

Month	Online	Offline	Total
Jan	40	80	120
Feb	80	50
Mar	50	z
Apr	80	40	120
May	100	30	130

	Minimum	Maximum	Median
Online	40	100	80
Offline	30	80	50
Total	110	130	120

Now, consider feb,

Median of online data is 80

Therefore, to make median as 80 we need y between 80 and 100

Minimum value of y + x

= 80 + 50 = 130

(which is the maximum value possible of the total possible registrations)

Therefore, x = 50 and y= 80

Month	Online	Offline	Total
Jan	40	80	120
Feb	80	50	130
Mar	50	60	110
Apr	80	40	120
May	100	30	130

	Minimum	Maximum	Median
Online	40	100	80
Offline	30	80	50
Total	110	130	120

Since, minimum number of total registrations = 110

Therefore, the only possibility is in march

= 50 + z = 110

= z = 60

Month	Online	Offline	Total
Jan	40	80	120
Feb	80	50	130
Mar	50	60	110
Apr	80	40	120
May	100	30	130

	Minimum	Maximum	Median
Online	40	100	80
Offline	30	80	50
Total	110	130	120

1. In May, there are 30 offline registrations (smallest) ⇒ True

2. In Mar, we have smallest number of total registrations ⇒ False.