CAT 2024 Slot 2 DILR Set 4 Q15–Q18
A DILR caselet from the Data Interpretation & Logical Reasoning (DILR) section of CAT 2024 Slot 2, with 4 questions (Quant-based Puzzles). Answers are CAT's official key.
The set
This set relies on a figure in the original paper, which is not reproduced here. The questions and answers are below.
An online e-commerce firm receives daily integer product ratings from 1 through 5 given by buyers. The daily average is the average of the ratings given on that day. The cumulative average is the average of all ratings given on or before that day.
The rating system began on Day 1, and the cumulative averages were 3 and 3.1 at the end of Day 1 and Day 2, respectively. The distribution of ratings on Day 2 is given in the figure below.
[Figure in the original paper]
The following information is known about ratings on Day 3.
1. 100 buyers gave product ratings on Day 3.
2. The modes of the product ratings were 4 and 5.
3. The numbers of buyers giving each product rating are non-zero multiples of 10.
4. The same number of buyers gave product ratings of 1 and 2, and that number is half the number of buyers who gave a rating of 3.
The rating system began on Day 1, and the cumulative averages were 3 and 3.1 at the end of Day 1 and Day 2, respectively. The distribution of ratings on Day 2 is given in the figure below.
[Figure in the original paper]
The following information is known about ratings on Day 3.
1. 100 buyers gave product ratings on Day 3.
2. The modes of the product ratings were 4 and 5.
3. The numbers of buyers giving each product rating are non-zero multiples of 10.
4. The same number of buyers gave product ratings of 1 and 2, and that number is half the number of buyers who gave a rating of 3.
Question 15
How many buyers gave ratings on Day 1?
TITA: type in the answer. There are no options and no negative marking.
Answer: 150 Quant-based Puzzles TITA
Solution
- Day 2 (from the chart): ratings 1 to 5 were given by 5, 10, 5, 20, 10 buyers: 50 buyers with total \(5 + 20 + 15 + 80 + 50 = 170\).
- Day 1: \(n\) buyers with average 3. \(\frac{3n + 170}{n + 50} = 3.1 \Rightarrow 0.1n = 15 \Rightarrow n = 150\) (total 450).
Question 16
What is the daily average rating of Day 3?
- A3.5
- B3.2
- C3.0
- D3.6
Answer: (D) 3.6 Quant-based Puzzles MCQ
Solution
- Day 3: let counts of 1 and 2 be \(x\), of 3 be \(2x\), and of 4 and 5 be \(y\) each (the two modes, so \(y > 2x\)). \(4x + 2y = 100\). With multiples of 10, \(x = 10, y = 30\) is the only option (\(x = 20\) gives \(y = 10 < 40\)). Counts: 10, 10, 20, 30, 30; total score \(10 + 20 + 60 + 120 + 150 = 360\).
- Daily average \(= \frac{360}{100} = 3.6\).
Question 17
What is the median of all ratings given on Day 3?
TITA: type in the answer. There are no options and no negative marking.
Answer: 4 Quant-based Puzzles TITA
Solution
- Day 3: let counts of 1 and 2 be \(x\), of 3 be \(2x\), and of 4 and 5 be \(y\) each (the two modes, so \(y > 2x\)). \(4x + 2y = 100\). With multiples of 10, \(x = 10, y = 30\) is the only option (\(x = 20\) gives \(y = 10 < 40\)). Counts: 10, 10, 20, 30, 30; total score \(10 + 20 + 60 + 120 + 150 = 360\).
- In order, ratings 1-3 take positions 1-40 and rating 4 takes positions 41-70. The 50th and 51st ratings are both 4, so the median is 4.
Question 18
Which of the following is true about the cumulative average ratings of Day 2 and Day 3?
- AThe cumulative average of Day 3 increased by less than 5% from Day 2.
- BThe cumulative average of Day 3 increased by more than 8% from Day 2.
- CThe cumulative average of Day 3 increased by a percentage between 5% and 8% from Day 2.
- DThe cumulative average of Day 3 decreased from Day 2.
Answer: (C) The cumulative average of Day 3 increased by a percentage between 5% and 8% from Day 2. Quant-based Puzzles MCQ
Solution
- Day 2 (from the chart): ratings 1 to 5 were given by 5, 10, 5, 20, 10 buyers: 50 buyers with total \(5 + 20 + 15 + 80 + 50 = 170\).
- Day 1: \(n\) buyers with average 3. \(\frac{3n + 170}{n + 50} = 3.1 \Rightarrow 0.1n = 15 \Rightarrow n = 150\) (total 450).
- Day 3: let counts of 1 and 2 be \(x\), of 3 be \(2x\), and of 4 and 5 be \(y\) each (the two modes, so \(y > 2x\)). \(4x + 2y = 100\). With multiples of 10, \(x = 10, y = 30\) is the only option (\(x = 20\) gives \(y = 10 < 40\)). Counts: 10, 10, 20, 30, 30; total score \(10 + 20 + 60 + 120 + 150 = 360\).
- Cumulative average at Day 3 \(= \frac{450 + 170 + 360}{300} = \frac{980}{300} \approx 3.267\).
- Increase from 3.1: \(\frac{0.167}{3.1} \approx 5.4\%\), between 5% and 8%.