Question

What was the total number of registrations in April?

• A. Only I
• B. Both I and II
• C. Neither I nor II
• D. Only II
• A. 80
• B. 50
• C. 50 or 80
• D. 30 or 50 or 80
• A. Only II
• B. Only 1
• C. Both I and II
• D. Neither I nor II

Answer 1

The Correct Option is A

Given:
In every month, both online and offline registration numbers were multiples of 10. 
From (2), in January, the number of offline registrations was double that of online registrations.

Let $x$ be the number of online registrations in January.
Then, offline registrations = $2x$
Total registrations = $x + 2x = 3x$
From the data, $3x$ must lie between the minimum and maximum total registrations.

Let’s try $x = 40$ (as it must be a multiple of 10).
Then online = 40, offline = 80, total = 120.

From (5), the number of online registrations is highest in May.
In May, online = 100. Since the total is 130, offline = 30.

Let $x$ be the offline registrations in May = online registrations in March = 50 (say).
Let’s fill the table:

Month	Online	Offline	Total
Jan	40	80	120
Feb	$y$	50	?
Mar	50	$z$	?
Apr	80	140	120
May	100	30	130

From the question, median of offline data = 50, so 50 must be the middle value.
That means offline values (in increasing order): 30, 50, 60, 80, 140
So, $x = 50$ and $z = 60$.

Also, median of online data is 80, so online values (in increasing order): 40, 50, 80, 80, 100
Thus, $y = 80$.

Now update the table:

Month	Online	Offline	Total
Jan	40	80	120
Feb	80	50	130
Mar	50	60	110
Apr	80	140	120
May	100	30	130

Therefore, total number of registrations in April is 120.

Answer 2

Given the conditions and the data in the table, let’s analyze January:

1. Condition 2: In January, offline registrations were twice the online registrations. 
2. Let the number of online registrations in January be \( x \).
   Then, offline registrations = \( 2x \).
3. Total registrations = \( x + 2x = 3x \).
4. According to the table, total registrations in January fall between 110 and 130.
   So, \( 110 \leq 3x \leq 130 \).
5. Try the median total: \( 3x = 120 \Rightarrow x = 40 \).

Conclusion: The number of online registrations in January is \( 40 \).

Answer 3

To determine the correct answer, let’s analyze each statement based on the data provided: 

1. Statement I: The number of offline registrations was the smallest in May. 
  - According to the table, the minimum number of offline registrations is 30. Given that May had the largest number of online registrations (from condition 5), it is reasonable to assume May could have the smallest offline registrations to keep the total consistent. Therefore, this statement can be true.

2. Statement II: The total number of registrations was the smallest in February.
  - The total number of registrations ranges from 110 to 130, with 110 being the minimum. There is no specific information given that February had the smallest total, and this statement cannot be confirmed based on the table.

Hence, only Statement I can be true.

Answer 4

To determine the correct answer, let’s analyze the given options based on the provided data:

1. Median of offline registrations is 50:
   - This means that when the offline registration numbers for all five months are arranged in order, the middle value is 50.
2. Implication for February:
   - Since February is one of the months, and no condition excludes it from having the median value, it is reasonable to assign the median value of 50 to February.
3. No conflicting condition:
   - The data does not indicate any restriction or conflict that prevents February from having 50 offline registrations.

Conclusion: The most consistent and justifiable value for offline registrations in February is: 50

Answer 5

To determine the correct answer, let’s carefully analyze both statements based on the table and conditions: 

Statement I: January and April had the same total number of registrations.

• The question mentions that the median of the total registrations over the five months is 120.
• If January and April both had 120 registrations, and these values lie in the middle of a sorted list, it supports the median being 120.
• So, this statement is plausible and can be true.

Statement II: February and May had the same total number of registrations.

• If February and May had the same total, this could also be consistent with a median of 120 if their values are either on opposite sides of the median or equal to it.
• Based on the distribution possibilities, this statement is also plausible and can be true.

Conclusion: Since both Statement I and Statement II are possible based on the given data, the correct answer is:

Both I and II