Solution for free throws, Problem for probability

Thanks everyone for trying out the problem, and congrats to those who got the problem right!

Problem:

The players on a basketball team made some three-point shots, some two-points shots, and some one-point free throws. They scored as may points with two-points shots as with three-points shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was 61 points. How many free throws did they make?

Solution:

First of all, we will need to set up a system of equations to solve this problem. Let's make x= # of free throws, y= # of two-point shots, and z= # of three-point shots. Now, we have x + 2y + 3z = 61. If they scored as many points with two-point shots as three-point shots, then 2y=3z. Their number of successful three throws was one more than their number of successful two-point shots, which means x = y + 1. Now we can substitute these into the original equation.

x + 2y + 3z = 61

(y + 1) + 2y + (2y) = 61

y + 1 + 2y + 2y = 61

5y + 1 = 61

5y = 60

y = 12

We can now plug in 12 in for y into x = y + 1 to solve for the number of free throws.

x = y + 1

x = 12 + 1

x = 13

The answer is 13 free throws.

Here's the next problem

Problem:

A pair of standard 6-sided dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference.

Good luck! and remember, calculators are not allowed. :)