There are 20 balls in the virtual box. 8 of the balls are colored blue. Math Question, Prealgebra? Multiple Choice, how to solve?

At random some of the remaining 12 balls are green and some are red and some will be yellow. There will always be 3 more green balls than red regardless of the scenario. Which is feasible?

a. 4 red, 1 green, 7 yellow

b. 2 red, 5 green, 5 yellow

c. 3 red, 0 green, 9 yellow

d. 2 red, 8 green, 2 yellow

e. 6 red, 6 green, 0 yellow

Relevance
• 1 month ago

a. 4 red, 1 green, 7 yellow

Not Feasible - must have 3 more green balls than red

b. 2 red, 5 green, 5 yellow  <–––––––

Feasible 3 more green balls than red

c. 3 red, 0 green, 9 yellow

Not Feasible - must have 3 more green balls than red

d. 2 red, 8 green, 2 yellow

Not Feasible - must have 3 more green balls than red

e. 6 red, 6 green, 0 yellow

Not Feasible - must have 3 more green balls than red

• JJ
Lv 7
1 month ago

It's B because that's the only choice where green = 3+red

• 1 month ago

"There will always be 3 more green balls than red".  The only option where there are three more green than red is B.  It's not math, it's being able to read.