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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 for CAT 2024 DILR question 10 (Data Interpretation)

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.

How many buyers gave ratings on Day 1?

Entered answer:

Solution

✅ Correct Answer: 150
Slide 1/4

According to the chart, we can make the following table to show the distribution of ratings on day 2.

RatingsNumber of Buyers
15
210
35
420
510

From here we get the total number of buyers on day 2 was 50.

Average on day 2 =

1×5+2×10+3×5+4×20+5×1050=17050=3.4\small \dfrac{1 \times 5 + 2 \times 10 + 3 \times 5 + 4 \times 20 + 5 \times 10}{50} = \frac{170}{50} = 3.4

Therefore, we can make the following table:

DayNumber of BuyersDaily AverageCumulative Average
133
2503.43.1
3
DayNumber of BuyersDaily AverageCumulative Average
115033
2503.43.1
3

Day - 1

From the information given in the question, we know that cumulative averages on day 1 and day 2 were 3 and 3.1 respectively.

Let the number of buyers in day 1 = xx.

3x+170x+50=3.1\small \dfrac{3x + 170}{x + 50} = 3.1

Solving: 3x+170=3.1(x+50)3x + 170 = 3.1(x + 50);

3x+170=3.1x+1553x + 170 = 3.1x + 155;

0.1x=150.1x = 15;

x=150x = 150.

Therefore, the number of buyers in day 1 = 150.

Day - 3

RatingsNumber of Buyers
1A
2A
32A
4B (Mode)
5B (Mode)

As per statement 1: Total number of buyers in day 3 = 100.

Also using statements 2, 3, and 4:

The numbers of buyers giving each product rating are non-zero multiples of 10.

Number of buyers giving rating 1 = Number of buyers giving rating 2 = Half of number of buyers giving rating 3.

The modes of the product ratings were 4 and 5.

Adding all of them we get:

Sum of the number of buyers = 4A+2B=1004A + 2B = 100.

The only possible solution for the equation with B value being the mode is A=10A = 10 and B=30B = 30.


Average on day 3:

1×10+2×10+3×20+4×30+5×30100\small \dfrac{1 \times 10 + 2 \times 10 + 3 \times 20 + 4 \times 30 + 5 \times 30}{100}

=3.6 =3.6


Cumulative average:

=(150×3)+(50×3.4)+(100×3.6)150+50+100= \small \dfrac{(150 \times 3) + (50 \times 3.4) + (100 \times 3.6)}{150 + 50 + 100}

=3.266= 3.266

DayNumber of BuyersDaily AverageCumulative Average
115033
2503.43.1
31003.63.266

We get the above final table.

From the table we can see that 150 buyers gave rating on day 1.

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