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) cost $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?

Question

(0,5)
(3, 2.6)
(-5, 9)
(5,1)

Answer 1

To determine which pairs of values (x, y) represent a viable solution to Cindy's spending on robotics kits and chemistry sets, we can set up the equation based on her total winnings of $50.00:

Let:

• \( x \) = number of chemistry sets
• \( y \) = number of robotics kits

The cost for chemistry sets is \( 8x \) and for robotics kits is \( 10y \). Therefore, the equation based on her total amount spent will be:

\[ 8x + 10y = 50 \]

Now we can evaluate the given options:

1. (0, 5):

   • \( x = 0 \), \( y = 5 \)
   • Substituting into the equation: \[ 8(0) + 10(5) = 0 + 50 = 50 \quad \text{(Valid solution)} \]

2. (3, 2.6):

   • \( x = 3 \), \( y = 2.6 \)
   • Substituting into the equation: \[ 8(3) + 10(2.6) = 24 + 26 = 50 \quad \text{(Valid solution)} \]

3. (-5, 9):

   • \( x = -5 \), \( y = 9 \)
   • Substituting into the equation: \[ 8(-5) + 10(9) = -40 + 90 = 50 \quad \text{(The negative value for x is not feasible)} \]

4. (5, 1):

   • \( x = 5 \), \( y = 1 \)
   • Substituting into the equation: \[ 8(5) + 10(1) = 40 + 10 = 50 \quad \text{(Valid solution)} \]

Valid solutions are:

• (0, 5)
• (3, 2.6)
• (5, 1)

The only invalid option is (-5, 9), as having a negative quantity of chemistry sets is not feasible.

So, the viable solutions Cindy can use are:

• (0, 5)
• (3, 2.6)
• (5, 1)