Question

Eve is organizing a race to raise money for a local children's hospital. She makes $20 for each 5K participant and $45 for each 10K participant. Her goal is to raise $9,500. If x represents the number of participants in the 5K race and y represents the number of participants in the 10K race, the linear equation for the given scenario is ______________. If Eve raises exactly $9,500, is it possible for there to be 220 participants in the 5K race? (yes or no)

Answer 1

Answer: $20x +45y=9500$ is the linear equation.

yes, it is possible.

Step-by-step explanation:

given data:

we were told that she would make $20 for each 5K participant and $45 for each 10K participant.

cost of each 5k participant = $20

cost of each 10k participant = $45.

her target = $9500

solution:

let y represent the 10k

let x represent the 5k participant

the linear equation for this scenarios is

$20x +45y=9500$

(b.) possibility of having 220 participant in the 5k race.

$\\20(220) +45y=9500\\4400 + 45y = 9500\\45y = 9500- 4400\\45y = 5100\\y = \frac{5100}{45} \\y = 113$

Answer 2

Answer:

Step-by-step explanation:

It is given that x is the number of participants in the 5K race and y is the number of participants in the 10K race.

Eve makes $20 for each 5K participant and $45 for each 10K participant. She wants to raise $9,500. The linear equation below represents the given scenario.

If Eve raises exactly $9,500 and there are 220 5K participants, the number of 10K participants is found as shown below.

It is not possible to have 113.3 participants in the 10K race. So, no, it is not possible for Eve to raise exactly $9,500 if there are 220 participants in the 5K race.