QJEEM21Ma3002: JEE Main 2021 Mathematics - Class 11 JEEM21Ma3002 Solution: Let  A = {n ∈ N: n is a 3-digit number}…

In this problem, we're dealing with 3-digit numbers in set $A$ , and subsets $B$ and $C$ which represent numbers of specific forms.

1. First, we consider the numbers of the form $9k + 2$ (Set $B$ ) within the 3-digit range, which starts at 101 and ends at 992.

2. We calculate the sum of these numbers, denoted as $s_1$ . To calculate $s_1$ , you use the formula for the sum of an arithmetic series :

$$(n/2)\times (\text{{first term}}+\text{{last term}})$$

Here, $n$ is the total count of such numbers. These are 3-digit numbers of the form $9k + 2$ , and we can find the total count by subtracting the smallest such number (101) from the largest (992), dividing the result by 9 (because we're considering numbers with a difference of 9), and then adding 1.

The sum $s_1$ is calculated as follows :

$$(100/2)\times (101+992)=54650$$

3. According to the problem, the sum of all elements of the set $A \cap (B \cup C)$ is $274 \times 400 = 109600$ .

Since the set $A \cap (B \cup C)$ is the union of two disjoint sets (the set of all three-digit numbers of form $9k + 2$ and the set of all three-digit numbers of form $9k + l$ ), we can write this sum as :

$$s_1$$(for numbers of the form 9k + 2) + $$s_2$$ (for numbers of the form 9k + l) = 109600

4. Solving this equation for $s_2$ (the sum of numbers of the form $9k + l$ ), we get :

$$s_2=109600-s_1=109600-54650=54950$$

5. The sum $s_2$ can be expressed as $(n/2) \times (\text{{first term}} + \text{{last term}})$ , where $n$ is the count of numbers of the form $9k + l$ . The first term here is the smallest 3-digit number of this form, which is $99 + l$ , and the last term is the largest such number, which is $990 + l$ .

We equate this to $s_2$ to solve for $l$ :

$$54950=(100/2)[(99+l)+(990+l)]$$

6. Simplifying this equation, we get :

$$2l+1089=1099$$

Solving for $l$ , we find :

$$l=5$$

So, the correct answer is 5.