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 bottles of each type were sold in the next month when rate was increased?

Options

A. 1200, 1300
B. 1300, 1200
C. 1550, 950
D. 1650, 850

Answer

To solve the problem, we need to set up a system of equations based on the information provided.

### Step 1: Define Variables
Let:
- \( x \) = number of Alpha sanitizer bottles sold in July
- \( y \) = number of Beta sanitizer bottles sold in July

### Step 2: Set Up Equations
From the problem, we know:
1. The total number of bottles sold in July is 2500:
\[ x + y = 2500 \quad \text{(Equation 1)} \]

2. The selling price of Alpha sanitizer in July is Rs. 11 and for Beta sanitizer is Rs. 13. The total sales amount to Rs. 29,200:
\[ 11x + 13y = 29200 \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations
Now, we will solve these two equations simultaneously.

**From Equation 1:**
\[ y = 2500 - x \]

**Substituting \( y \) in Equation 2:**
\[ 11x + 13(2500 - x) = 29200 \]

### Step 4: Simplify the Equation
Distributing the 13:
\[ 11x + 32500 - 13x = 29200 \]

Combining like terms:
\[ -2x + 32500 = 29200 \]

### Step 5: Isolate \( x \)
Subtract 32500 from both sides:
\[ -2x = 29200 - 32500 \]
\[ -2x = -3300 \]

Dividing by -2:
\[ x = 1650 \]

### Step 6: Find \( y \)
Now substitute \( x \) back into Equation 1 to find \( y \):
\[ y = 2500 - 1650 \]
\[ y = 850 \]

### Conclusion
The number of bottles sold in July is:
- Alpha sanitizer: **1650 bottles**
- Beta sanitizer: **850 bottles**