Cost of 8pencils, 5pens and 3erasers is rs.111. Cost of 9pencil, 6pens and 5eraser is rs. 130. Cost of 16pencils, 11pens and 3erasers is rs.221. what is the cost of 39pencil, 26pen and 13 erasers.

Options

A. a. 316
B. b. 546
C. c. 624
D. d.can't be determined

Answer

To solve the problem, we need to set up a system of equations based on the information provided about the costs of pencils, pens, and erasers.

Let:
- Cost of 1 pencil = y
- Cost of 1 pen = x
- Cost of 1 eraser = z

From the problem, we have the following equations based on the costs:

1. From the first condition:
\[ 8y + 5x + 3z = 111 \quad \text{(Equation 1)} \]

2. From the second condition:
\[ 9y + 6x + 5z = 130 \quad \text{(Equation 2)} \]

3. From the third condition:
\[ 16y + 11x + 3z = 221 \quad \text{(Equation 3)} \]

Next, we will solve these equations step by step.

### Step 1: Add the equations
We will add all three equations together to simplify our calculations:
\[ (8y + 5x + 3z) + (9y + 6x + 5z) + (16y + 11x + 3z) = 111 + 130 + 221 \]
This simplifies to:
\[ 33y + 22x + 11z = 462 \quad \text{(Equation 4)} \]

### Step 2: Simplify Equation 4
Now, we can divide the entire equation by 11 to make it simpler:
\[ 3y + 2x + z = 42 \quad \text{(Equation 5)} \]

### Step 3: Find the cost of 39 pencils, 26 pens, and 13 erasers
We need to find the cost of:
\[ 39y + 26x + 13z \]
Notice that \(39y + 26x + 13z\) can be factored out from Equation 5:
\[ 39y + 26x + 13z = 13(3y + 2x + z) \]
Substituting Equation 5 into this gives:
\[ 39y + 26x + 13z = 13 \times 42 = 546 \]

### Conclusion
The cost of 39 pencils, 26 pens, and 13 erasers is Rs. 546.

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