x ≥y/3 and y + z ≥ 65

So, since X is an integer, Y will be multiple of 3. Taking Y=3, x could be anything >= 1.

Y=3 also satisfies the second equation if Z is anything >= 62

So from the answer choices, it looks like it will be c

Somehow, I don't like my solution. Am I missing something here?

If x,y,z are positive integers, x ≥y/3 , y ≥ z/5, and y+z ≥ 65 then the smallest possible value of x is
(a) 6
(b) 5
(c) 2
(d) 4
(e) 3

Sorry Guys. I left out y ≥ z/5 and just realized that.

y>z/5 --> 5y - z>0
Combinining with y + z > 65:
6y > 65 --> y > 11
Since x>y/3, we have x> 4

Answer is D

agoyal2 wrote:x>y/3 and y + z > 65

So, since X is an integer, Y will be multiple of 3. Taking Y=3, x could be anything >= 1.

watch out. y does not need to be a multiple of 3. It would if the first inequality was an equality (x = y/3) but not in this case

Thanks avenus, I see your point...