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Question

don't agree with OA

tkirk32 · Answer

Should Be E.

.xyz

x=8 and y =5 or y =5 and x = 8

.589 or .856

Dan · Answer

I agree on C.

1. .xy could be: .96, .87, .78, .69, .68, .59,.95 not suff (.59 less, others > 2/3)
2. .xyz could be: .7y8, .8y7, .9y6, .6y9 not suff (.6y9 could be less or more than 2/3)

both, .59 is out since it does not meet both conditions. All others are ok/

cloudz9 · Answer

for statement I to be true where x+y>13, there are several possibilities...however since we only need to worry about the lower ones, they are
0.59z, 0.68z ... (0.68z is greater than 2/3 so the only one to worry about is 0.59z
however this in itself doesn't provide enough information

for statement II to be true where x+z>14 again we only need to find the lowest possible numbers...
which are 0.6y9, 0.7y8 (0.7z8 is definitely greater than 2/3 so the only one to worry about is 0.6y9). 
however this alone doesn't provide enough information

But if both statements are taken together...
from the first statement, the one number that was lower than 2/3 was 0.59z...however using the second statement this can be eliminated.
from the second statement the number that could be lower than 2/3 is 0.6y9....however from the first statement, the lowest value of y if x is 6 is 8...which makes the lowest possible value of 0.xyz to be 0.689 which is greater than 2/3
this gives us enough info to determine that given the two statements, 0.xyz will always be greater than 2/3

Thus, answer is C

tkirk32 · Answer

haha. For some reason I totaly ignored the ">" sign.

HongHu · Answer

1) x+y>13
.95>2/3
.5914 
zx>=6
Not knowing y
.612/3
Insufficient

Combined
x>=6
When x>6 .xyz>2/3
When x=6, y>7, .xyz>2/3
Sufficient

(C)

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