CAT 2023 Slot 1 · DILR · Question 12
CAT 2023 Slot 1 previous year question on Caselet DI, with the answer and a step-by-step solution.
Set 3: Restaurant Ratings of Gig Workers — this question belongs to a set. Read the full set with all 5 questions →
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 | |
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.
Answer & solution
Answer: 0
Solving the set
Step 1 – Each worker's five ratings. Sum = 5 × mean. Use the median, mode, range and the known 1s and 5s.
• Ullas: sum 11, has a 1 (R1), range 3 ⇒ maximum 4. Median 2 and mode 2 ⇒ 1, 2, 2, x, 4 with 1 + 2 + 2 + x + 4 = 11 ⇒ x = 2.
• Vasu: sum 19, has a 5 (R5), range 3 ⇒ minimum 2. Median 4 ⇒ 2, a, 4, b, 5 with a + b = 8 ⇒ a = b = 4.
• Waman: sum 17, two 5s (R1, R3) and a 1 (R2), median 4 ⇒ 1, x, 4, 5, 5 with x = 2.
• Xavier: sum 18, two 5s (R2, R3), range 4 ⇒ minimum 1. Median 4 ⇒ 1, x, 4, 5, 5 with x = 3.
• Yusuf: sum 13, two 1s (R2, R3), modes 1 and 4, median 3 ⇒ 1, 1, 3, 4, 4.
Worker | Sum (5 × mean) | Known 1s/5s | Ratings (from median, mode, range) |
Ullas | 11 | 1 (R1) | 1, 2, 2, 2, 4 |
Vasu | 19 | 5 (R5) | 2, 4, 4, 4, 5 |
Waman | 17 | 5 (R1, R3), 1 (R2) | 1, 2, 4, 5, 5 |
Xavier | 18 | 5 (R2, R3) | 1, 3, 4, 5, 5 |
Yusuf | 13 | 1 (R2, R3) | 1, 1, 3, 4, 4 |
Step 2 – Restaurant totals (5 × mean): R1 = 17, R2 = 11, R3 = 19, R4 = 14, R5 = 17.
• R2: Waman 1 + Xavier 5 + Yusuf 1 = 7, so Ullas + Vasu = 4. Both are at least 2, so both are 2. Vasu's single 2 is at R2, so Vasu gets 4 from R1, R3 and R4.
• R3: Vasu 4 + Waman 5 + Xavier 5 + Yusuf 1 = 15, so Ullas = 4. Ullas then gets 2 from R4 and R5.
• R1: Ullas 1 + Vasu 4 + Waman 5 = 10, so Xavier + Yusuf = 7: (3, 4) or (4, 3).
• If Xavier = 4 and Yusuf = 3 at R1, Yusuf has 4 at both R4 and R5. R4 needs Waman + Xavier = 14 − 2 − 4 − 4 = 4, but Waman's options (2 or 4) and Xavier's (1 or 3) never add to 4. Not possible.
• So Xavier = 3 and Yusuf = 4 at R1. R4 needs Waman + Xavier + Yusuf = 8: only 4 + 1 + 3 works. R5 then gets Waman 2, Xavier 4, Yusuf 4 (total 17 ✓).
Restaurant | Ullas | Vasu | Waman | Xavier | Yusuf | Sum | Mean |
R1 | 1 | 4 | 5 | 3 | 4 | 17 | 3.4 |
R2 | 2 | 2 | 1 | 5 | 1 | 11 | 2.2 |
R3 | 4 | 4 | 5 | 5 | 1 | 19 | 3.8 |
R4 | 2 | 4 | 4 | 1 | 3 | 14 | 2.8 |
R5 | 2 | 5 | 2 | 4 | 4 | 17 | 3.4 |
This question
From the grid, R2 gave 2 (Ullas), 2 (Vasu), 1 (Waman), 5 (Xavier) and 1 (Yusuf). No worker got a 4 from R2.
Hence, the answer is 0.
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