CAT 2023 Slot 1 · DILR · Set 3: Restaurant Ratings of Gig Workers
The full set and all 5 questions (Q11–Q15) from CAT 2023 Slot 1, with answers and solutions.
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.
Question 11 · Caselet DI
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
The grid in Step 2 is completely determined. So no rating is undetermined.
Hence, the answer is 0.
Question 12 · Caselet DI
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.
Question 13 · Caselet DI
Answer & solution
Answer: 3
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 Step 2, R1 gave Xavier a rating of 3.
Hence, the answer is 3.
Question 14 · Caselet DI
Answer & solution
Answer: 4
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
R3 gave 4, 4, 5, 5, 1. Arranged in order: 1, 4, 4, 5, 5. The median is 4.
Hence, the answer is 4.
Question 15 · Caselet DI
- AR2
- BR5
- CR4
- DR3
Answer & solution
Answer: Option C
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
Restaurant | Ratings in order | Median | Workers given the median |
R2 | 1, 1, 2, 2, 5 | 2 | 2 (Ullas, Vasu) |
R3 | 1, 4, 4, 5, 5 | 4 | 2 (Ullas, Vasu) |
R4 | 1, 2, 3, 4, 4 | 3 | 1 (Yusuf) |
R5 | 2, 2, 4, 4, 5 | 4 | 2 (Xavier, Yusuf) |
R4's median is 3 and only Yusuf got a 3 from R4.
Hence, the answer is option 3.
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