Case Study-2: A company manufactores two types of sanitizers Alpha and Beta. The cost of the small bottle of Alpha sanitizer is Rs. 10 and for beta sanitizer is Rs. 12. In the month of June, the company sold total 1000 bottles and makes a total sale of Rs. 10,820. Seeing the great demand and short of supply, company decided to increase the price of both the sanitizer by Rs. 1. In the next month i.e. July, the company sold 2,500 bottles and total sales of Rs. 29,200.
Answer the following questions:
How many sanitizers of each type was sold in June?

Options

A. 460, 510
B. 540, 460
C. 410, 590
D. 590, 410

Answer

To solve the problem, we need to find out how many bottles of Alpha and Beta sanitizers were sold in June. We can set up a system of equations based on the information given.

### Step-by-Step Solution:

1. **Define Variables:**
Let:
- \( x \) = number of Alpha sanitizers sold
- \( y \) = number of Beta sanitizers sold

2. **Set Up Equations:**
From the problem, we know:
- The total number of bottles sold in June is 1000:
\[ x + y = 1000 \quad \text{(Equation 1)} \]
- The total sales amount in June is Rs. 10,820. The price of Alpha is Rs. 10 and the price of Beta is Rs. 12:
\[ 10x + 12y = 10820 \quad \text{(Equation 2)} \]

3. **Solve the Equations:**
We can solve these equations simultaneously. Start with Equation 1:
\[ y = 1000 - x \quad \text{(Substituting for } y \text{)} \]

Substitute \( y \) in Equation 2:
\[ 10x + 12(1000 - x) = 10820 \]
Simplifying this:
\[ 10x + 12000 - 12x = 10820 \]
\[ -2x + 12000 = 10820 \]
\[ -2x = 10820 - 12000 \]
\[ -2x = -1180 \]
\[ x = \frac{1180}{2} = 590 \]

4. **Find the Value of \( y \):**
Now substitute \( x \) back into Equation 1 to find \( y \):
\[ y = 1000 - 590 = 410 \]

5. **Conclusion:**
Therefore, the number of sanitizers sold in June is:
- Alpha: 590
- Beta: 410

### Final Answer:
- Alpha sanitizers sold: **590**
- Beta sanitizers sold: **410**