Question

Keenan and Dalton decided to shoot arrows at a simple target with a large outer ring and a smaller bull's-eye. Keenan went first and landed 3 arrows in the outer ring and 1 arrow in the bull's-eye, for a total of 101 points. Dalton went second and got 3 arrows in the outer ring and 4 arrows in the bull's-eye, earning a total of 251 points. How many points is each region of the target worth? The outer ring is worth points, and the bull's-eye is worth points.

Answer 1

To find the point values for each region of the target, we set up a system of equations. Subtracting one equation from the other reveals that the bull's-eye is worth 50 points and the outer ring is worth 17 points.

Step-by-step explanation:

We can solve the problem by setting up a system of equations based on the information given. Let's denote the points for the outer ring as x and the points for the bull's-eye as y.

For Keenan, the equation based on his score is:
3x + 1y = 101

For Dalton, the equation based on his score is:
3x + 4y = 251

To solve the system of equations, we subtract the first equation from the second to eliminate x and find the value of y.
So, (3x + 4y) - (3x + 1y) = 251 - 101,
which simplifies to 3y = 150, meaning y = 50.

Now that we know the value of y, we can substitute it back into the first equation:
3x + 1(50) = 101,
which simplifies to 3x + 50 = 101 and then to 3x = 51. Therefore, x = 17.

The outer ring is worth 17 points, and the bull's-eye is worth 50 points.