please explain gmat prep question

ORIGINAL QUESTION WITH ANSWER CHOICES

From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the least number of yellow balls that must be in the bag?
A. 17
B. 18
C. 19
D. 20
E. 21

If you're not sure where to begin here, TESTING the answer choices will reveal the correct answer in no time.

A) 17
If there are 17 yellow balls, then the TOTAL number of balls = 12 + 17 = 29
So, P(selected ball is blue) = 12/29
12/29 is GREATER THAN 2/5, so we can eliminate A

B) 18
If there are 18 yellow balls, then the TOTAL number of balls = 12 + 18 = 30
So, P(selected ball is blue) = 12/30
12/30 is EQUAL TO 2/5, so we can eliminate B

At this point, we can see that adding 1 more yellow ball (i.e., 19 yellow balls) will make P(selected ball is blue) LESS THAN 2/5
So, the correct answer is C

Cheers,
Brent

yvonne12 wrote:from a bag containing 12 identical balls, y identical yellow balls and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the best number of yellow balls that must be in the bag?

please explain how I can work out this problem

We can create the following inequality:

12/(y+12) < 2/5

60 < 2(y + 12)

30 < y + 12

18 < y

Since y is greater than 18, the least number of yellow balls is 19.