CAT 2023 Slot 1 — DILR Question 16

Mixed PracticeEasy

Passage / Data

Answer the following questions based on the information given below:

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.

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* 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.

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

Answer & solution

Answer: 0

Solution

Easy

This one set powers all five questions. The trick is to first reconstruct each worker's five ratings from the summary statistics, then use the restaurant row-sums (mean $\times 5$ ) plus the given 1's and 5's to lock every cell of the $5\times5$ grid. Once the grid is unique, every question is just a lookup.

Summary statistics (transcribed from the table). The five restaurants R1–R5 rate five workers on $1$ – $5$ . Restaurant means $3.4,\,2.2,\,3.8,\,2.8,\,3.4 \Rightarrow$ row-sums $17,\,11,\,19,\,14,\,17$ .

	Ullas	Vasu	Waman	Xavier	Yusuf
Mean	2.2	3.8	3.4	3.6	2.6
Median	2	4	4	4	3
Mode	2	4	5	5	1 & 4
Range	3	3	4	4	3
Sum	11	19	17	18	13

Given 1's & 5's: 5's — R1&R3→Waman, R2&R3→Xavier, R5→Vasu. 1's — R1→Ullas, R2→Waman, R2&R3→Yusuf.

Reconstruct each worker's five ratings. Use range (with a known max or min), then force median, mode and sum to match.

\begin{aligned} &\textbf{Ullas: } \min=1,\ \text{range }3\Rightarrow\max=4;\ \text{med }2,\ \text{mode }2,\ \text{sum }11 \Rightarrow \{1,2,2,2,4\}\\ &\textbf{Vasu: } \max=5,\ \text{range }3\Rightarrow\min=2;\ \text{med }4,\ \text{mode }4,\ \text{sum }19 \Rightarrow \{2,4,4,4,5\}\\ &\textbf{Waman: } \max=5,\ \min=1;\ \text{med }4,\ \text{mode }5,\ \text{sum }17 \Rightarrow \{1,2,4,5,5\}\\ &\textbf{Xavier: } \max=5,\ \text{range }4\Rightarrow\min=1;\ \text{med }4,\ \text{mode }5,\ \text{sum }18 \Rightarrow \{1,3,4,5,5\}\\ &\textbf{Yusuf: } \min=1,\ \text{range }3\Rightarrow\max=4;\ \text{med }3,\ \text{modes }1\&4,\ \text{sum }13 \Rightarrow \{1,1,3,4,4\} \end{aligned}

Place the known 5's and 1's, then use row-sums. Waman's column already has $5,5$ from R1,R3 and a $1$ from R2; the remaining $\{2,4\}$ go to R4,R5. Xavier has $5,5$ from R2,R3. For R3 (sum $19$ ): R3 gives $5$ to Waman and Xavier and $1$ to Yusuf, leaving $19-5-5-1=8$ for Ullas+Vasu $\Rightarrow$ both get $4$ .

Pin down R2 (sum 11). R2 gives $1$ to Waman and Yusuf and $5$ to Xavier; Ullas' three remaining $2$ 's must come from R2,R4,R5 so R2→Ullas $=2$ . Then Vasu's R2 cell $=11-2-1-5-1=2$ . So Vasu's two remaining $4$ 's go to R1 and R4.

Xavier's lone 1 must be R4. If R1 or R5 (each row-sum $17$ ) gave Xavier a $1$ , that row could not reach $17$ . So R4→Xavier $=1$ . Now R4 (sum $14$ ): already $2$ (Ullas) $+4$ (Vasu) $+1$ (Xavier) $=7$ , leaving $7$ for Waman+Yusuf $\Rightarrow$ Waman $=4$ , Yusuf $=3$ . Hence Waman's last cell (R5) $=2$ , and Yusuf's two remaining $4$ 's go to R1 and R5.

Finish with R1 (sum 17). R1 has $1$ (Ullas) $+4$ (Vasu) $+5$ (Waman) $+4$ (Yusuf) $=14$ , so R1→Xavier $=17-14=3$ . The last empty cell — Xavier from R5 — is then $4$ . The grid is now completely and uniquely determined:

	Ullas	Vasu	Waman	Xavier	Yusuf	Sum
R1	1	4	5	3	4	17
R2	2	2	1	5	1	11
R3	4	4	5	5	1	19
R4	2	4	4	1	3	14
R5	2	5	2	4	4	17
Sum	11	19	17	18	13	78

Since every one of the $25$ ratings is forced, no individual rating is left undetermined.

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