From a jar containing 50 pieces of candy, of which 25 are red and 25 are green, Ari has taken 3 red and 4 green pieces. He takes an additional 13 pieces from the jar. What is the least number of these additional pieces that must be red in order for Ari to have more red candies than green candies among all the pieces he has taken?

Algebraically, here’s how you solve:

3 + *r* > 4 + *g*

*r* + *g* = 13

*g* = 13 – *r*

3 + *r* > 4 + (13 – *r*)

3 + *r* > 14 – *r*

2*r* > 14

*r* > 7

Therefore, at least 8 of the candies must be red.

You also, maybe, could calculate this in your head without the algebra. Ari will have 20 candies once he adds 13 more to the 7 he took. To have more red than green, he’ll need 11 red. He already has 3, so he needs 8 more.

## Comments (1)

Where did you get “14” in 3 + r > 14 – r ?