Perimeter of a triangle with unknown side

GRE Quant Problem Solving
1

from manhattan 5lb… chp 27, problem 18

if p is the peremiter of a triangle with one side of 7 and another side of 9, what is the range of possible values for p? the answer is 18 < p < 32

my understanding was to obtain the range for the other side of the triangle… to get the minimum 9 - 7 = 2 // to get the maximum 9 + 7 = 16. I thought this meant that 2 < x < 16. meaning that x must be greater than 2 and less than 16. not greater than or equal too, less than or equal to.
so the range i obtained was 9 +7 + 3 = 19 and 7 + 9 + 15 = 31 meaning 19 < p < 31.

In Greg’s triangle 2 video under week 3 day 2 he states x must be greater or smaller, not equal to, the ranges calculated. am i missing something? or can x equal 2 & 16?

2

18 < p < 32
Implies that p>18, but p can’t be equal to 18
By, 9-7, you got x > 2
So, p > 18
If you add 1 to it, you are excluding x = 2.1,2.01,2.000001, 2.5, etc

3

ahhhh i see now. i was tunnel vision, the range is 19 to 31 meaning 18 < p < 32. thank you for clearing that up for me.

4

No, p can also be 18.2 or 31.75
19 to 31 excludes few values

5

i see. i was refering to integers but the question does not state that. so, you’re correct. thank you