CAT 2024 Slot 2 — DILR Question 11

Mixed PracticeEasy

Passage / Data

Answer the following questions based on the information given 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.

Distribution of ratings on Day 2

Rating (Day 2)	1	2	3	4	5
Number of buyers	5	10	5	20	10

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.

What is the daily average rating of Day 3?

Answer & solution

A
3.0
B
3.5
C
3.2
3.6

Solution

Medium

Pin the Day-3 counts. With 100 buyers, all counts non-zero multiples of 10, $n_1=n_2$ and $n_3=2n_1$ , and modes 4 and 5 (so $n_4=n_5$ are the joint maximum), only one distribution fits. Then take the mean.

Day 3: 100 buyers; counts are non-zero multiples of 10; $n_1=n_2$ , $n_3=2n_1$ ; modes 4 and 5 (so $n_4=n_5$ and these are the largest).

Set up the counts. Let $n_1=n_2=x$ and $n_3=2x$ , with $n_4=n_5=m$ .

\begin{aligned} &x+x+2x+m+m=100\\ &\Rightarrow\ 2x+m=50 \quad\text{(divide by 2)} \end{aligned}

Fit the multiples of 10 with $m$ the unique max. $x=10$ gives $m=30>2x$ ; $x=20$ gives $m=10$ , not a maximum.

\begin{aligned} &x=10\Rightarrow m=30 \quad\text{(}30>20=n_3\text{)}\\ &\Rightarrow\ (n_1,n_2,n_3,n_4,n_5)=(10,10,20,30,30) \end{aligned}

Daily average. Weighted mean of the Day-3 ratings.

\begin{aligned} &\frac{1(10)+2(10)+3(20)+4(30)+5(30)}{100}\\ &\Rightarrow\ \frac{10+20+60+120+150}{100}=\frac{360}{100}\\ &\Rightarrow\ 3.6 \end{aligned}

3.6