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 (denoted as y) costs \$10.00, while each chemistry set (denoted as x) costs \$8.00. 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?

A. (3, 2.6)

B. (0, 5)

C. (5, 1)

D. (-5, 9)

1 answer

To determine which of the provided options represents a viable solution for the number of robotics kits (y) and chemistry sets (x) that Cindy can purchase with her $50.00 winnings, we can set up the equation based on the costs of the items:

The total cost equation is: \[ 10y + 8x = 50 \]

Now we will evaluate each option:

A. \((3, 2.6)\)

  • Plugging in \(y = 3\) and \(x = 2.6\): \[ 10(3) + 8(2.6) = 30 + 20.8 = 50.8 \] This exceeds $50. Not viable.

B. \((0, 5)\)

  • Plugging in \(y = 0\) and \(x = 5\): \[ 10(0) + 8(5) = 0 + 40 = 40 \] This is under $50. Not fully utilizing the money, but it's a viable option for spending only part of the winnings.

C. \((5, 1)\)

  • Plugging in \(y = 5\) and \(x = 1\): \[ 10(5) + 8(1) = 50 + 8 = 58 \] This exceeds $50. Not viable.

D. \((-5, 9)\)

  • Plugging in \(y = -5\) and \(x = 9\): \[ 10(-5) + 8(9) = -50 + 72 = 22 \] Negative number of robotics kits is not possible. Not viable.

Now, checking the options again, we recognize:

  • Only option B results in a valid outcome, where Cindy purchases 0 robotics kits and 5 chemistry sets, although it leaves her with some remainder of her winnings.

The only viable option that Cindy can fully or partially spend in whole numbers of items is: B. (0, 5)