x articles are bought at Rs `(x-8)` each and `(x-2)` some other articles are bought at Rs `(x-3)` each. If the total cost of all these articles is Rs 76, how many articles of first kind were bought?

Answer

To solve the problem step by step, we will follow the given information and set up equations based on the costs of the articles.

### Step 1: Define the variables
Let:
- \( x \) = number of articles of the first kind.
- \( x - 2 \) = number of articles of the second kind.

### Step 2: Calculate the cost of the first kind of articles
The cost of each article of the first kind is \( x - 8 \). Therefore, the total cost for \( x \) articles is:
\[ \text{Cost of first kind} = x \times (x - 8) = x^2 - 8x \]

### Step 3: Calculate the cost of the second kind of articles
The cost of each article of the second kind is \( x - 3 \). Therefore, the total cost for \( x - 2 \) articles is:
\[ \text{Cost of second kind} = (x - 2) \times (x - 3) = (x - 2)(x - 3) = x^2 - 3x - 2x + 6 = x^2 - 5x + 6 \]

### Step 4: Set up the equation for total cost
According to the problem, the total cost of all articles is Rs 76. Therefore, we can write the equation:
\[ \text{Total Cost} = \text{Cost of first kind} + \text{Cost of second kind} \]
\[ x^2 - 8x + x^2 - 5x + 6 = 76 \]

### Step 5: Simplify the equation
Combine like terms:
\[ 2x^2 - 13x + 6 = 76 \]
Subtract 76 from both sides to set the equation to zero:
\[ 2x^2 - 13x + 6 - 76 = 0 \]
\[ 2x^2 - 13x - 70 = 0 \]

### Step 6: Factor the quadratic equation
Now we will factor the quadratic equation \( 2x^2 - 13x - 70 = 0 \). We look for two numbers that multiply to \( 2 \times -70 = -140 \) and add to \( -13 \). The numbers are \( -20 \) and \( 7 \).
\[ 2x^2 - 20x + 7x - 70 = 0 \]
Group the terms:
\[ (2x^2 - 20x) + (7x - 70) = 0 \]
Factor by grouping:
\[ 2x(x - 10) + 7(x - 10) = 0 \]
Factor out \( (x - 10) \):
\[ (x - 10)(2x + 7) = 0 \]

### Step 7: Solve for \( x \)
Set each factor to zero:
1. \( x - 10 = 0 \) → \( x = 10 \)
2. \( 2x + 7 = 0 \) → \( x = -\frac{7}{2} \) (not valid since \( x \) must be positive)

Thus, the only valid solution is:
\[ x = 10 \]

### Conclusion
The number of articles of the first kind bought is \( \boxed{10} \).