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CAT 2023 Slot 1 Solved Paper

Instructions [35 - 39 ]

Five restaurants, coded R1, R2, R3, R4 and R5 gave integer ratings to five gig workers -
Ullas, Vasu, Waman, Xavier and Yusuf, on a scale of 1 to 5.
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 summary statistics of these ratings for the five workers is given below.

	Ullas	Vasu	Waman	Xavier	Yusuf h>
Mean rating	2.2	3.8	3.4	3.6	2.6
Median rating	2	4	4	4	3
Modal rating	2	4	5	5	1 and 4
Range of rating*	3	3	4	4	3

* Range of ratings is defined as the difference between the maximum and minimum ratings awarded to a worker.
The following is partial information about ratings of 1 and 5 awarded by the restaurants to the workers.
(a) R1 awarded a rating of 5 to Waman, as did R2 to Xavier, R3 to Waman and Xavier, and R5 to Vasu.
(b) R1 awarded a rating of 1 to Ullas, as did R2 to Waman and Yusuf, and R3 to Yusuf.

Question : 39

Total: 66

Which among the following restaurants gave its median rating to exactly one of the workers?

Validate

Solution:

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:

	R1	R2	R3	R4	R5	Total
U	1					11
V					5	19
W	5	1	5			17
X		5	5			18
Y		1	1			13
Total	17	11	19	14	17

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:

	R1	R2	R3	R4	R5	Total	Entries
U	1					11	2,2,2,4
V					5	19	2,4,4,4
W	5	1	5			17	2,4
X		5	5			18	1,3,4
Y		1	1			13	3,4,4
Total	17	11	19	14	17

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"

	R1	R2	R3	R4	R5	Total	Missing Entries
U	1	2	4	2	2	11
V			4		5	19	2,4,4
W	5	1	5			17	2,4
X		5	5			18	1,3,4
Y		1	1			13	3,4,4
Total	17	11	19	14	17

Now, consider column R2 ⇒ Missing entry should be 11−2−1−5−1=2

	R1	R2	R3	R4	R5	Total	Missing Entries
U	1	2	4	2	2	11
V	4	2	4	4	5	19
w	5	1	5			17	2,4
x		5	5			18	1,3,4
Y		1	1			13	3,4,4
Total	17	11	19	14	17

Consider column R1, the missing elements should add up to 17−5−4−1=7(3+4 or 4+3)--(1)
Consider R5, the missing elements should add up to 10⇒2+4+4 or 4+3+3 (not possible) as (1) requires a 3.

	R1	R2	R3	R4	R5	Total	Missing Entries
U	1	2	4	2	2	11
V	4	2	4	4	5	19
W	5	1	5	4	2	17
X		5	5		4	18	1,3
Y		1	1		4	13	3,4
Total	17	11	19	14	17

Now, we can fill column R1 as 3 + 4 and the remaining in column R4 and we can get the complete table

	R1	R2	R3	R4	R5	Total
U	1	2	4	2	2	11
V	4	2	4	4	5	19
W	5	1	5	4	2	17
X	3	5	5	1	4	18
Y	4	1	1	3	4	13
Total	17	11	19	14	17

⇒R2 median rating is 2⇒ given to 2 workers
⇒ R5 median rating is 4⇒ given to 2 workers
⇒ R4 median rating is 3⇒ given to only 1 worker.
⇒R3 median rating is 4⇒ given to 2 workers.

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