CAT 2024 Slot 2 · DILR · Set 3: Daily Product Ratings of an E-commerce Firm
The full set and all 4 questions (Q10–Q13) from CAT 2024 Slot 2, with answers and solutions.
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.
Distribution of Ratings on Day 2
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 10 · Reasoning Based DI
How many buyers gave ratings on Day 1?
Answer & solution
Answer: 150
Solving the set
Step 1 – Day 2 data from the chart.
|
Rating |
Buyers on Day 2 |
Rating × Buyers |
|
1 |
5 |
5 |
|
2 |
10 |
20 |
|
3 |
5 |
15 |
|
4 |
20 |
80 |
|
5 |
10 |
50 |
|
Total |
50 |
170 |
Step 2 – Day 1. Let n buyers rate on Day 1. Their average is 3, so their total is 3n. The cumulative average after Day 2 is 3.1: (3n + 170)/(n + 50) = 3.1 ⇒ 3n + 170 = 3.1n + 155 ⇒ 0.1n = 15 ⇒ n = 150. Day 1 total = 450.
Step 3 – Day 3. Let ratings 1 and 2 each have x buyers; rating 3 then has 2x (fact 4). Ratings 4 and 5 are both modes, so they have the same number y, and y must be more than 2x. Total: x + x + 2x + y + y = 100 ⇒ 4x + 2y = 100. All counts are non-zero multiples of 10.
|
Buyers giving 1 (= giving 2) |
Buyers giving 3 |
Buyers giving 4 (= giving 5) |
Are 4 and 5 the modes? |
|
10 |
20 |
(100 − 40)/2 = 30 |
Yes (30 > 20 > 10) |
|
20 |
40 |
(100 − 80)/2 = 10 |
No (3 has 40 buyers) |
Only x = 10, y = 30 works.
|
Rating |
Buyers on Day 3 |
Rating × Buyers |
Cumulative buyers |
|
1 |
10 |
10 |
1–10 |
|
2 |
10 |
20 |
11–20 |
|
3 |
20 |
60 |
21–40 |
|
4 |
30 |
120 |
41–70 |
|
5 |
30 |
150 |
71–100 |
|
Total |
100 |
360 |
Step 4 – Cumulative averages.
|
Day |
Buyers |
Sum of ratings |
Cumulative buyers |
Cumulative sum |
Cumulative average |
|
1 |
150 |
450 |
150 |
450 |
3.00 |
|
2 |
50 |
170 |
200 |
620 |
3.10 |
|
3 |
100 |
360 |
300 |
980 |
3.27 |
This question
From Step 2, n = 150. Check: (450 + 170)/(150 + 50) = 620/200 = 3.1 ✓.
Hence, the answer is 150.
Question 11 · Reasoning Based DI
What is the daily average rating of Day 3?
- A
3.2
- B
3.6
- C
3.0
- D
3.5
Answer & solution
Answer: Option B
Solving the set
Step 1 – Day 2 data from the chart.
|
Rating |
Buyers on Day 2 |
Rating × Buyers |
|
1 |
5 |
5 |
|
2 |
10 |
20 |
|
3 |
5 |
15 |
|
4 |
20 |
80 |
|
5 |
10 |
50 |
|
Total |
50 |
170 |
Step 2 – Day 1. Let n buyers rate on Day 1. Their average is 3, so their total is 3n. The cumulative average after Day 2 is 3.1: (3n + 170)/(n + 50) = 3.1 ⇒ 3n + 170 = 3.1n + 155 ⇒ 0.1n = 15 ⇒ n = 150. Day 1 total = 450.
Step 3 – Day 3. Let ratings 1 and 2 each have x buyers; rating 3 then has 2x (fact 4). Ratings 4 and 5 are both modes, so they have the same number y, and y must be more than 2x. Total: x + x + 2x + y + y = 100 ⇒ 4x + 2y = 100. All counts are non-zero multiples of 10.
|
Buyers giving 1 (= giving 2) |
Buyers giving 3 |
Buyers giving 4 (= giving 5) |
Are 4 and 5 the modes? |
|
10 |
20 |
(100 − 40)/2 = 30 |
Yes (30 > 20 > 10) |
|
20 |
40 |
(100 − 80)/2 = 10 |
No (3 has 40 buyers) |
Only x = 10, y = 30 works.
|
Rating |
Buyers on Day 3 |
Rating × Buyers |
Cumulative buyers |
|
1 |
10 |
10 |
1–10 |
|
2 |
10 |
20 |
11–20 |
|
3 |
20 |
60 |
21–40 |
|
4 |
30 |
120 |
41–70 |
|
5 |
30 |
150 |
71–100 |
|
Total |
100 |
360 |
Step 4 – Cumulative averages.
|
Day |
Buyers |
Sum of ratings |
Cumulative buyers |
Cumulative sum |
Cumulative average |
|
1 |
150 |
450 |
150 |
450 |
3.00 |
|
2 |
50 |
170 |
200 |
620 |
3.10 |
|
3 |
100 |
360 |
300 |
980 |
3.27 |
This question
From Step 3, the Day 3 total is 360 from 100 buyers, so the daily average = 360/100 = 3.6.
Hence, the answer is option 2.
Question 12 · Reasoning Based DI
What is the median of all ratings given on Day 3?
Answer & solution
Answer: 4
Solving the set
Step 1 – Day 2 data from the chart.
|
Rating |
Buyers on Day 2 |
Rating × Buyers |
|
1 |
5 |
5 |
|
2 |
10 |
20 |
|
3 |
5 |
15 |
|
4 |
20 |
80 |
|
5 |
10 |
50 |
|
Total |
50 |
170 |
Step 2 – Day 1. Let n buyers rate on Day 1. Their average is 3, so their total is 3n. The cumulative average after Day 2 is 3.1: (3n + 170)/(n + 50) = 3.1 ⇒ 3n + 170 = 3.1n + 155 ⇒ 0.1n = 15 ⇒ n = 150. Day 1 total = 450.
Step 3 – Day 3. Let ratings 1 and 2 each have x buyers; rating 3 then has 2x (fact 4). Ratings 4 and 5 are both modes, so they have the same number y, and y must be more than 2x. Total: x + x + 2x + y + y = 100 ⇒ 4x + 2y = 100. All counts are non-zero multiples of 10.
|
Buyers giving 1 (= giving 2) |
Buyers giving 3 |
Buyers giving 4 (= giving 5) |
Are 4 and 5 the modes? |
|
10 |
20 |
(100 − 40)/2 = 30 |
Yes (30 > 20 > 10) |
|
20 |
40 |
(100 − 80)/2 = 10 |
No (3 has 40 buyers) |
Only x = 10, y = 30 works.
|
Rating |
Buyers on Day 3 |
Rating × Buyers |
Cumulative buyers |
|
1 |
10 |
10 |
1–10 |
|
2 |
10 |
20 |
11–20 |
|
3 |
20 |
60 |
21–40 |
|
4 |
30 |
120 |
41–70 |
|
5 |
30 |
150 |
71–100 |
|
Total |
100 |
360 |
Step 4 – Cumulative averages.
|
Day |
Buyers |
Sum of ratings |
Cumulative buyers |
Cumulative sum |
Cumulative average |
|
1 |
150 |
450 |
150 |
450 |
3.00 |
|
2 |
50 |
170 |
200 |
620 |
3.10 |
|
3 |
100 |
360 |
300 |
980 |
3.27 |
This question
With 100 ratings in order, the median is the average of the 50th and 51st ratings. From the cumulative column of the Day 3 table, ratings 41 to 70 are all 4.
Median = 4.
Hence, the answer is 4.
Question 13 · Reasoning Based DI
Which of the following is true about the cumulative average ratings of Day 2 and Day 3?
- A
The cumulative average of Day 3 increased by more than 8% from Day 2.
- B
The cumulative average of Day 3 increased by a percentage between 5% and 8% from Day 2.
- C
The cumulative average of Day 3 increased by less than 5% from Day 2.
- D
The cumulative average of Day 3 decreased from Day 2.
Answer & solution
Answer: Option B
Solving the set
Step 1 – Day 2 data from the chart.
|
Rating |
Buyers on Day 2 |
Rating × Buyers |
|
1 |
5 |
5 |
|
2 |
10 |
20 |
|
3 |
5 |
15 |
|
4 |
20 |
80 |
|
5 |
10 |
50 |
|
Total |
50 |
170 |
Step 2 – Day 1. Let n buyers rate on Day 1. Their average is 3, so their total is 3n. The cumulative average after Day 2 is 3.1: (3n + 170)/(n + 50) = 3.1 ⇒ 3n + 170 = 3.1n + 155 ⇒ 0.1n = 15 ⇒ n = 150. Day 1 total = 450.
Step 3 – Day 3. Let ratings 1 and 2 each have x buyers; rating 3 then has 2x (fact 4). Ratings 4 and 5 are both modes, so they have the same number y, and y must be more than 2x. Total: x + x + 2x + y + y = 100 ⇒ 4x + 2y = 100. All counts are non-zero multiples of 10.
|
Buyers giving 1 (= giving 2) |
Buyers giving 3 |
Buyers giving 4 (= giving 5) |
Are 4 and 5 the modes? |
|
10 |
20 |
(100 − 40)/2 = 30 |
Yes (30 > 20 > 10) |
|
20 |
40 |
(100 − 80)/2 = 10 |
No (3 has 40 buyers) |
Only x = 10, y = 30 works.
|
Rating |
Buyers on Day 3 |
Rating × Buyers |
Cumulative buyers |
|
1 |
10 |
10 |
1–10 |
|
2 |
10 |
20 |
11–20 |
|
3 |
20 |
60 |
21–40 |
|
4 |
30 |
120 |
41–70 |
|
5 |
30 |
150 |
71–100 |
|
Total |
100 |
360 |
Step 4 – Cumulative averages.
|
Day |
Buyers |
Sum of ratings |
Cumulative buyers |
Cumulative sum |
Cumulative average |
|
1 |
150 |
450 |
150 |
450 |
3.00 |
|
2 |
50 |
170 |
200 |
620 |
3.10 |
|
3 |
100 |
360 |
300 |
980 |
3.27 |
This question
From Step 4, the cumulative average went from 3.10 (Day 2) to 980/300 ≈ 3.267 (Day 3).
Percentage increase = (3.267 − 3.1)/3.1 × 100 ≈ 5.4%, which is between 5% and 8%.
Shortcut: 5% of 3.1 is 0.155 and 8% is 0.248; the increase of 0.167 lies between them.
Hence, the answer is option 2.
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