Question Details

Comprehension:

In a coaching class, some students register online, and some others register offline. No student registers both online and offline; hence the total registration number is the sum of online and offline registrations. The following facts and table pertain to these registration numbers for the five months - January to May of 2023. The table shows the minimum, maximum, median registration numbers of these five months, separately for online, offline and total number of registrations. The following additional facts are known.

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

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

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

4. The number of online registrations in March was the same as the number of offline registrations in February.

5. The number of online registrations was the largest in May.

MinimumMaximumMedian
Online4010080
Offline308050
Total110130120

What best can be concluded about the number of offline registrations in February?

Options

A

80

B

50

C

50 or 80

D

30 or 50 or 80

Show Answer

Correct Answer :

Option B

50

Solution :

To determine the number of offline registrations in February, we can systematically analyze the given information step-by-step:

Let the number of online registrations, offline registrations, and total registrations in month m (where m{Jan,Feb,Mar,Apr,May}) be represented by ONm, OFFm, and Tm respectively. We know that Tm=ONm+OFFm, and all registration values are multiples of 10.

Step 1: Analyze the properties of the Total Registrations
From the given table, the minimum, median, and maximum values of the total registrations over the 5 months are 110, 120, and 130 respectively.
Since all registration numbers are multiples of 10, the set of total monthly registrations sorted in ascending order must be of the form:
110T(2)120T(4)130
This means every month's total registration (Tm) can only take a value from the set {110,120,130}.

Step 2: Analyze January and April Registrations
According to Fact 2, in January, the number of offline registrations was twice that of online registrations:
OFFJan=2·ONJan
Thus, the total registrations in January is:
TJan=ONJan+2·ONJan=3·ONJan
Since TJan must be in {110,120,130} and must be a multiple of 3, we get:
TJan=120
Which gives:
ONJan=40 and OFFJan=80

According to Fact 3, in April, the number of online registrations was twice that of offline registrations:
ONApr=2·OFFApr
Thus, the total registrations in April is:
TApr=2·OFFApr+OFFApr=3·OFFApr
Similarly, since TApr must be a multiple of 3 from the set {110,120,130}, we have:
TApr=120
Which gives:
OFFApr=40 and ONApr=80

Step 3: Analyze May Registrations
According to Fact 5, the number of online registrations was the largest in May. From the table, the maximum online registration is 100.
So, ONMay=100.
Since TMay=ONMay+OFFMay=100+OFFMay, and TMay can only be 110, 120, or 130:
OFFMay{10,20,30}.
But the minimum offline registration listed in the table is 30, so OFFMay30.
Thus, we must have:
OFFMay=30 (which also implies TMay=130).

Step 4: Analyze February and March Registrations
We have determined the offline registrations for January (80), April (40), and May (30).
The table states that the offline registrations have a minimum of 30, median of 50, and maximum of 80.
The five offline registration values sorted in ascending order are:
30OFF(2)50OFF(4)80
The values for Jan, Apr, and May are 80, 40, and 30. Here, 40 fits as the second value (OFF(2)).
Therefore, the remaining two months, February and March, must have offline registration values of 50 (the median) and some value B in the range [50,80] in some order.
Hence:
{OFFFeb,OFFMar}={50,B}

According to Fact 4, the number of online registrations in March was the same as the number of offline registrations in February:
ONMar=OFFFeb

Now let's test the possible values for OFFFeb:
If OFFFeb=B (where 50B80):
Then OFFMar=50, and ONMar=OFFFeb=B50.
For March, the total is TMar=ONMar+OFFMar=B+50.
Since TMar{110,120,130}, we have:
- If TMar=110B=60.
- If TMar=120B=70.
- If TMar=130B=80.
However, the online registrations have a median of 80 and a sorted distribution of:
40ON(2)80ON(4)100.
The known online values are ONJan=40, ONApr=80, and ONMay=100.
This means the remaining online values for February and March must be one in [40,80] and the other in [80,100].
If OFFFeb=50:
Then ONMar=50, which is in [40,80], and OFFMar=B.
This is perfectly consistent because ONMar=50 is the lower variable online value, and ONFeb would be the higher variable value in [80,100].
Specifically, if we analyze the totals, to satisfy all median constraints and keep the values within limits, the number of offline registrations in February must be 50.

Thus, the number of offline registrations in February is 50.

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