A shopkeeper sells article A at 8% profit and article B at 10% loss, thereby getting a sum of 1,008. If he sells the article A at 10% profit and article B at 8% loss, he would have 1,028. Find the cost price of article A and B to the shopkeeper.

Answer

To find the cost price of articles A and B, we can set up the problem as follows:

Let:
- \( X \) = Cost price of article A
- \( Y \) = Cost price of article B

### Step 1: Set up the equations based on the selling prices.

1. **First Scenario:**
- Article A is sold at 8% profit.
- Article B is sold at 10% loss.
- The total selling price is 1008.

The selling price of article A:
\[ SP_A = X + \frac{8}{100}X = \frac{108}{100}X = \frac{108X}{100} \]

The selling price of article B:
\[ SP_B = Y - \frac{10}{100}Y = \frac{90}{100}Y = \frac{90Y}{100} \]

Therefore, we can write the first equation:
\[ \frac{108X}{100} + \frac{90Y}{100} = 1008 \]
Multiplying through by 100 to eliminate the fraction:
\[ 108X + 90Y = 100800 \quad \text{(Equation 1)} \]

2. **Second Scenario:**
- Article A is sold at 10% profit.
- Article B is sold at 8% loss.
- The total selling price is 1028.

The selling price of article A:
\[ SP_A = X + \frac{10}{100}X = \frac{110}{100}X = \frac{110X}{100} \]

The selling price of article B:
\[ SP_B = Y - \frac{8}{100}Y = \frac{92}{100}Y = \frac{92Y}{100} \]

Therefore, we can write the second equation:
\[ \frac{110X}{100} + \frac{92Y}{100} = 1028 \]
Multiplying through by 100:
\[ 110X + 92Y = 102800 \quad \text{(Equation 2)} \]

### Step 2: Solve the system of equations.

Now we have a system of linear equations:
1. \( 108X + 90Y = 100800 \) (Equation 1)
2. \( 110X + 92Y = 102800 \) (Equation 2)

We can solve these equations using the elimination method. First, we can multiply Equation 1 by 110 and Equation 2 by 108 to align the coefficients of \( X \):

\[ 110(108X + 90Y) = 110(100800) \implies 11880X + 9900Y = 11088000 \quad \text{(Equation 3)} \]
\[ 108(110X + 92Y) = 108(102800) \implies 11880X + 9936Y = 11078400 \quad \text{(Equation 4)} \]

### Step 3: Subtract Equation 4 from Equation 3.

\[ (11880X + 9900Y) - (11880X + 9936Y) = 11088000 - 11078400 \]
\[ 9900Y - 9936Y = 9600 \]
\[ -36Y = 9600 \]
\[ Y = -\frac{9600}{-36} = 266.67 \quad \text{(Cost price of article B)} \]

### Step 4: Substitute \( Y \) back to find \( X \).

Using \( Y = 266.67 \) in Equation 1:
\[ 108X + 90(266.67) = 100800 \]
\[ 108X + 24000 = 100800 \]
\[ 108X = 100800 - 24000 \]
\[ 108X = 76800 \]
\[ X = \frac{76800}{108} = 711.11 \quad \text{(Cost price of article A)} \]

### Final Answer:
- Cost price of article A \( X \) = 711.11
- Cost price of article B \( Y \) = 266.67