If x,y,z are positive integers, x ≥y/3 and y + z ≥ 65 then the smallest possible value of x is
(a) 6
(b) 5
(c) 2
(d) 4
(e) 3[/list][/u]
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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?
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.
(a) 6
(b) 5
(c) 2
(d) 4
(e) 3
Sorry Guys. I left out y ≥ z/5 and just realized that.
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 caseagoyal2 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.