Question 28.
How many different valid combinations of the missing ratings are possible?
Question Explanation
The first thing we should do while solving this set is we should replace the average with the sum of the ratings as we have done in the following table:
We know that the ratings can be 1/2/3/4/5 only.
R1 has given 4 and 3 to B and D respectively and the sum of the ratings given by R1 in total is 16.
So, the sum of the ratings given by R1 to A and C is (16-4-3 = 9)
Now, this sum is only possible when he has given 4 to one of them and 5 to another.
Let's make both the cases:
Now, with this, we can derive the score given by R3 to C.
In case 1, it will be (16-4-4-5) = 3
In case 2, it will be (16-5-4-5) = 2
Now, R2 gave 3 and 4 score to A and C respectively.
Hence, he would have given a sum of (16 - 3 - 4 = 9) to B and D.
It could be possible when he gave 4 to one and 5 to the other one.
Hence, we can make the possible cases and sub cases:
With this only, we can fill the complete table.
Hence, we get:
However, Case - 2 is not possible as there no reviewer can give 6 points to any gadget.
Hence, there are only three possible combinations.
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