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Question:

How many individual ratings cannot be determined from the above information?

Explanation:

The means of the ratings given by R1, R2, R3, R4 and R5 were 3.4, 2.2, 3.8, 2.8 and 3.4 respectively.
∴ The sum of rating given by R1, R2, R3, R4 and R5 were 17, 11, 19, 14 and 17 respectively. (Multiplied with 5)

Similarly, we can find the sum of rating of all the 5 people.

Filling the data given in point 1 and 2, we get the following table:

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Range of ratings:
Ullas: lowest is given as 1, hence his highest = 1 + 3 = 4.
Vasu: highest is given as 5, hence his lowest = 5 – 3 = 2.
Waman: highest is 5 and lowest is 1.
Xavier: highest is given as 5, hence his lowest = 5 – 4 = 1.
Yusuf: lowest is given as 1, hence his highest = 1 + 3 = 4.

Ullas: Highest = 4, Lowest = 1, Median = 2 and Mode = 2

∴ Since median is 2, his reading in ascending order can be

Sum of reading of Ullas = 11, and his mode is 2
This is possible only when his readings in ascending order must be 1, 2, 2, 2, 4.

Vasu: Highest = 5, Lowest = 2, Median = 4 and Mode = 4.

∴ Since median is 4, his reading in ascending order can be

Sum of reading of Vasu = 19, and his mode is 2 
This is possible when his readings in ascending order must be 2, 4, 4, 4, 5.

Waman: Highest = 5, Lowest = 1, Median = 4 and Mode = 5.

∴ Since median is 4, his reading in ascending order can be

Sum of reading of Waman = 17, and his mode is 5 
This is possible when his readings in ascending order must be 1, 2, 4, 5, 5.

Xavier: Highest = 5, Lowest = 1, Median = 4 and Mode = 5.

∴ Since median is 4, his reading in ascending order can be

Sum of reading of Xavier = 18, and his mode is 5 
This is possible when his readings in ascending order must be 1, 3, 4, 5, 5.

Yusuf: Highest = 4, Lowest = 1, Median = 3 and Mode = 1 and 4.

∴ Since median is 4, his reading in ascending order can be

Sum of reading of Yusuf = 17, and his mode is 1 and 4
This is possible when his readings in ascending order must be 1, 1, 3, 4, 4.

Now, we have the following table.

For R3, sum of ratings is 19, hence R3’s sum of rating for Ullas and Vasu = 19 – 5 – 5 – 1 = 8.
This is possible only when R3 gives a rating of 4 to both Ullas and Vasu.

For Ullas, the remaining 3 2’s would be given by R2, R4 and R5.

For R2, rating given to Vasu = 11 – 2 – 1 – 5 – 1 = 2.
∴ For Vasu the remaining 2 4s would be given by R1 and R4.

Xavier receives a rating of 1. This can only be given by R4. If R1 or R5 give a rating of 1 to Xavier, their sum of 17 each could not be achieved.

Now, the sum of ratings given by R4 to Waman and Yusuf = 14 – 2 – 4 – 1 = 7.
This is only possible when Waman gets a rating of 4 from R4 and Yusuf gets a rating of 3 from R4.

∴ The remaining rating of 2 for Waman must have come from R5.
∴ The remaining 2 4s for Yusuf must have come from R1 and R5.

For R1, rating given to Xavier = 17 – 1 – 4 – 5 – 4 = 3
⇒ The remaining rating of 4 for Xavier must have been given by R5.

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Ratings of all can be uniquely determined.

Hence, 0.