A bag contains balls that are red, green, or blue. One third of the balls are red, and 2/7 are blue. The number of green balls is eight fewer than the twice the number of blue balls. How many green balls are in the bag?

16 green balls

Explanation:

x = total balls in bag

Red = 1/3x

Blue = 2/7x

Green = -8+2(2/7x)

Since these are balls, the answers have to be whole numbers. The equations representing the different colors contain fractions and have denominators 3 and 7. So the total number of balls, represented by x, needs to be a factor of 3 and 7. So I did a trial and error thing here:

First possible common denominator = 21, which is 3x7

We test that there are 21 balls in the bag. Substituting 21 into the equations for red, blue and green we get:

red =  1/3(21) = 7

blue = 2/7(21) = 6

green = -8+2(2/7)(21) = 4

But if you add up the colors: 7+6+4 = 14 which is < 21

So we test another common denominator

Next possible common denominator = 42, which is 3x7x2

So we test that there are 42 balls in the bag. Substituting 42 into the equations for red, blue and green we get:

red =  1/3(42) = 14

blue = 2/7(42) = 12

green = -8+2(2/7)(42) = 16

If you add up the colors: 14+12+16 = 42

if im not wrong, i believe you just $7 on the first year, and then 10% of 77 as so on