Ten chairs and six tables together cost X 6200, three chairs and two tables together cost ? 1900. The cost of 4 chairs and 5 tables is

Options

A. 3000
B. X 3300
C. X 3500
D. 3800

Answer

To solve the problem step by step, we will define the variables, set up the equations based on the information given, and then solve for the unknowns.

### Step 1: Define Variables
Let:
- \( x \) = cost of one chair
- \( y \) = cost of one table

### Step 2: Set Up the Equations
From the problem statement, we have two pieces of information:
1. Ten chairs and six tables together cost 6200.
\[ 10x + 6y = 6200 \quad \text{(Equation 1)} \]
2. Three chairs and two tables together cost 1900.
\[ 3x + 2y = 1900 \quad \text{(Equation 2)} \]

### Step 3: Multiply Equation 2
To eliminate one of the variables when we subtract the equations, we can multiply Equation 2 by 3:
\[ 3(3x + 2y) = 3(1900) \]
This gives us:
\[ 9x + 6y = 5700 \quad \text{(Equation 3)} \]

### Step 4: Subtract Equation 3 from Equation 1
Now we will subtract Equation 3 from Equation 1:
\[ (10x + 6y) - (9x + 6y) = 6200 - 5700 \]
This simplifies to:
\[ 10x - 9x + 6y - 6y = 6200 - 5700 \]
\[ x = 500 \]

### Step 5: Substitute \( x \) back into Equation 1
Now that we have the value of \( x \), we can substitute it back into Equation 1 to find \( y \):
\[ 10(500) + 6y = 6200 \]
\[ 5000 + 6y = 6200 \]
Subtract 5000 from both sides:
\[ 6y = 6200 - 5000 \]
\[ 6y = 1200 \]
Now divide by 6:
\[ y = \frac{1200}{6} = 200 \]

### Step 6: Calculate the Cost of 4 Chairs and 5 Tables
Now we can find the cost of 4 chairs and 5 tables:
\[ \text{Cost} = 4x + 5y \]
Substituting the values of \( x \) and \( y \):
\[ = 4(500) + 5(200) \]
\[ = 2000 + 1000 \]
\[ = 3000 \]

### Final Answer
The cost of 4 chairs and 5 tables is **3000**.

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