Cindy won $50.00 for getting first place in the Science Fair. She spent her winnings on robotics kits and chemistry sets. Each robotics kit (y) costs $10.00, while each chemistry set costs $8.00 (x). Which of the following is a viable solution to the number of robotics kits and chemistry sets Cindy can purchase, assuming she spends her entire winnings?(1 point)

1 answer

To determine a viable solution for how many robotics kits (y) and chemistry sets (x) Cindy can purchase with her $50 winnings, we can create an equation based on the total cost:

\[ 10y + 8x = 50 \]

In this equation:

  • \(y\) is the number of robotics kits,
  • \(x\) is the number of chemistry sets,
  • $10 is the cost of each robotics kit,
  • $8 is the cost of each chemistry set.

We are looking for non-negative integer solutions (0 or greater) for \(x\) and \(y\).

To find solutions, we can express \(y\) in terms of \(x\):

\[ 10y = 50 - 8x \] \[ y = \frac{50 - 8x}{10} \]

For \(y\) to be a non-negative integer, the right side of the equation must also be a non-negative integer. This means that \(50 - 8x\) must be non-negative and divisible by 10.

Let's explore some integer values for \(x\):

  1. If \(x = 0\): \[ y = \frac{50 - 8(0)}{10} = \frac{50}{10} = 5 \] Solution: (5 robotics kits, 0 chemistry sets)

  2. If \(x = 1\): \[ y = \frac{50 - 8(1)}{10} = \frac{42}{10} = 4.2 \quad \text{(not an integer)} \]

  3. If \(x = 2\): \[ y = \frac{50 - 8(2)}{10} = \frac{34}{10} = 3.4 \quad \text{(not an integer)} \]

  4. If \(x = 3\): \[ y = \frac{50 - 8(3)}{10} = \frac{26}{10} = 2.6 \quad \text{(not an integer)} \]

  5. If \(x = 4\): \[ y = \frac{50 - 8(4)}{10} = \frac{18}{10} = 1.8 \quad \text{(not an integer)} \]

  6. If \(x = 5\): \[ y = \frac{50 - 8(5)}{10} = \frac{10}{10} = 1 \] Solution: (1 robotics kit, 5 chemistry sets)

  7. If \(x = 6\): \[ y = \frac{50 - 8(6)}{10} = \frac{2}{10} = 0.2 \quad \text{(not an integer)} \]

From these possibilities, we find two viable integer solutions:

  1. 5 robotics kits and 0 chemistry sets: \((y, x) = (5, 0)\)
  2. 1 robotics kit and 5 chemistry sets: \((y, x) = (1, 5)\)

These combinations are valid as they satisfy the original monetary constraint.

Thus, two possible viable combinations are:

  • 5 robotics kits and 0 chemistry sets.
  • 1 robotics kit and 5 chemistry sets.

You can choose either of these as a viable solution.