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.
- In every month, both online and offline registration numbers were multiples of 10.
- In January, the number of offline registrations was twice that of online registrations.
- In April, the number of online registrations was twice that of offline registrations.
- The number of online registrations in March was the same as the number of offline registrations in February.
- The number of online registrations was the largest in May.
Solution
understanding the set :-
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There is some information regarding a coaching class, where some students register online, and some others register offline.
-
the total registration number is the sum of online and offline registrations
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There’s a table which shows the minimum, maximum, median registration numbers from January to May of 2023
In this question we will be finding number of students online and offline in the coaching class in all the 5 months, for which we can make a normal simple table.
| 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 |
From the table we can see that both jan and march had equal number of registrations => 110
and both feb and may had equal number of registrations => 130