CAT 2023 Slot 1 Solved Paper

Given that 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 ratings given by R1,R2,R3R4,R5 are 5⋆ means =17,11,19,14, and 17 respectively.
Similarly the sum of ratings received by U,V,W,X and Y are 5⋆ means =11,19,17,18, and 13 respectively.
Also capturing the absolute data given in the partial information (a) and (b) and representing as a table, we get:Now,
Consider U
Given median =2, mode =2 and range =3
⇒ His ratings should be of the form 1,a,2,b,4⇒1+2+4+a+b=11⇒a+b=4. For mode =2⇒a=b=2
⇒ U's ratings are 1,2,2,2,4.
Consider V
Given median =4, mode =4 and range =3
⇒ His ratings should be of the form 2,a,4,b,5⇒2+4+5+a+b=19⇒a+b=8⇒ For mode =4⇒a=b= 4
⇒ V's ratings are 2, 4, 4, 4, 5 .
Consider W
Given median =4, mode =5 and range =4
⇒ His ratings should be of the form 1,a,4,5,5⇒1+a+4+5+5=17⇒a=2
⇒ W's ratings are 1, 2, 4, 5 , 5 .
Consider X
Given median =4, mode =5 and range =4
⇒ His ratings should be of the form 1,a,4,5,5⇒a+1+4+5+5=18⇒a=3
⇒ X's ratings are 1, 3, 4, 5, 5
Consider Y
Given median =3, mode =1&4, Range =3
⇒ His ratings are 1,1,3,4,4.
Capturing this data in the table, we get:Given median =4, mode =5 and range =4
⇒ His ratings should be of the form 1,a,4,5,5⇒a+1+4+5+5=18⇒a=3
⇒ X's ratings are 1, 3, 4, 5, 5
Consider Y
Given median =3, mode =1&4, Range =3
⇒ His ratings are 1,1,3,4,4.
Capturing this data in the table, we get:
Now, consider column R3 ⇒ The two missing entries should add up to 19−1−5−5=8, (only possibility is 4+4 )⇒ We can fill the row "U" and 4 in the row "V"Now, consider column R2 ⇒ Missing entry should be11−2−1−5−1=2
Consider column R1, the missing elements should add up to 17−5−4−1=7(3+4or4+3) ----(1)Consider R5, the missing elements should add up to 10⇒2+4+4or4+3+3 (not possible) as (1) requires a 3.
Now, we can fill column R1 as 3 + 4 and the remaining in column R4 and we can get the complete table
⇒ Ratings give by R3 are 1,4,4,5,5⇒ Median =4.