Lizzy: Carter has a jar with 20 green M&Ms and 20 red M&Ms. He eats 12 of the green M&Ms, then his sister comes and eats half the red M&Ms and adds 14 yellow M&Ms. If Carter picks an M&M at random now, what is the percentage chance he'll get a green M&M?.
Me: Hmmm, let me think. I think this is the detailed solution:
First find the final number of green M&Ms by subtracting 12 from the starting number: 20 M&Ms - 12 M&Ms = 8 green M&Ms. Then find the final number of red M&Ms by dividing the starting number by half: 20 M&Ms / 2 = 10 red M&Ms. Then find the total number of M&Ms at the end by adding the number of red, green and yellow M&Ms: 8 M&Ms + 10 M&Ms + 14 M&Ms = 32 M&Ms. Then divide the number of green M&Ms by the total number of M&Ms and multiply by 100 to find the percentage chance Carter gets a green M&M: 8 M&Ms / 32 M&Ms * 100 = 25%. Final answer: 25.