Another Triangle question - Kaplan - 700+

[imhimanshu]

IF the sides of the triangle have lengths x,y and z,x+y =30 and y+z = 20, then which of the following could be the perimeter of the triangle.
1- 28
2- 36
3- 42
options are -
a)1 only
b)2 only
c)1 and 2 only
d)1 and 3 only
e)1,2 and 3 only
OA to follow
Thanks

[GMATGuruNY]

IF the sides of the triangle have lengths x,y and z,x+y =30 and y+z = 20, then which of the following could be the perimeter of the triangle.
1- 28
2- 36
3- 42
options are -
a)1 only
b)2 only
c)1 and 2 only
d)1 and 3 only
e)1,2 and 3 only
OA to follow
Thanks

I: x+y+z = 28.
Since x+y=30, it is not possible that x+y+z = 28.
Eliminate any answer choice that includes I.
Eliminate A,C,D and E.

The correct answer is B.

[sl750]

Both options 2 and 3 are possible. Are those answer choices correct?

[GMATGuruNY]

Both options 2 and 3 are possible. Are those answer choices correct?

Option III is not possible.

IF the sides of the triangle have lengths x,y and z,x+y =30 and y+z = 20, then which of the following could be the perimeter of the triangle.
1- 28
2- 36
3- 42

The third side of a triangle must be less than the sum of the other 2 sides.

III: x+y+z = 42
Since x+y=30, z=12.
Since z=12 and y+z=20, y=8.
Since y=8 and x+y=30, x=22.
Thus, the 3 sides are x=22, y=8, and z=12.
Such a triangle is not possible.
Given that y+z=20, the third side (x) must be less than 20.
Thus, it is not possible that x=22.