Question

A fruit company delivers its fruit in two types of boxes: large and small. A delivery of 3 large boxes and 5 small boxes has a total weight of 116 kilograms. A dellvery of 9 large boxes and 7 small boxes has a total weight of 238 kilograms. How much does each type of box weigh?

Answer 1

Let \( x \) be the weight of a large box and \( y \) be the weight of a small box. We can set up the following system of equations based on the information provided:

1. For the delivery of 3 large boxes and 5 small boxes:
\[
3x + 5y = 116 \quad \text{(Equation 1)}
\]

2. For the delivery of 9 large boxes and 7 small boxes:
\[
9x + 7y = 238 \quad \text{(Equation 2)}
\]

We can solve this system of equations using the method of substitution or elimination. Here, we will use the elimination method.

First, we will multiply Equation 1 by 3 to match the coefficient of \( x \) in Equation 2:

\[
3(3x + 5y) = 3(116)
\]
This simplifies to:
\[
9x + 15y = 348 \quad \text{(Equation 3)}
\]

Now we have:

- Equation 2: \( 9x + 7y = 238 \)
- Equation 3: \( 9x + 15y = 348 \)

Next, we subtract Equation 2 from Equation 3 to eliminate \( x \):

\[
(9x + 15y) - (9x + 7y) = 348 - 238
\]
This simplifies to:
\[
15y - 7y = 110
\]
\[
8y = 110
\]
\[
y = \frac{110}{8} = 13.75
\]

Now that we have \( y \), we can substitute back into Equation 1 to find \( x \):

\[
3x + 5(13.75) = 116
\]
\[
3x + 68.75 = 116
\]
\[
3x = 116 - 68.75
\]
\[
3x = 47.25
\]
\[
x = \frac{47.25}{3} = 15.75
\]

Now we have the weights of the boxes:
- The weight of a large box \( x \) is 15.75 kilograms.
- The weight of a small box \( y \) is 13.75 kilograms.

Thus, the final answer is:
- **Large box weight:** 15.75 kg
- **Small box weight:** 13.75 kg