if x = 1 then avg of 1 & 10 = 5.5......I
if x = 3 then avg of 3 & 10 = 6.5.......II

a)
x = 1 & let z = 9 greater than avg ...... from I
x = 1 & let z = 2 smaller than avg ....... from I
Insuff

b)
if x = 1 then z = 5 smaller than avg ....... from I
if x = 3 then z = 15 greater than avg ........ from II
Insuff

Both together Insuff...

Ans E.

The correct answer is A

Suyog wrote: x = 1 & let z = 9 greater than avg ...... from I
x = 1 & let z = 2 smaller than avg ....... from I

The second option is not possible as Z will be closer to X rather than 10.

Now look at it this way, if Z is equidistant from X and 10, then Z is equal to the average of 10 & X. You need to visualize this....
Now if Z is closer to X, Z is less than average of X & 10, and if Z is closer to 10, Z is greater than the average of X & 10

simplyjat that was indeed a great great explanation and approach!!

...I am not a fan of picking numbers (even though some times we must)because its both time consuming and you never know if it will work for other numbers or if there ia an exception..so this explanation of why A is SUFF is really the one!.

:D

Was just looking at this one, I think it should be solved like this

1) 10-z < z-x which resolves to 5+x/2 <z> average

SUFF

2) You'd have to sub # in there, cuz you can't use the condition given in 1) for 2)
x =1 , z = 5, A = 5.5, then z <A> A
INSUF

so answer is A

helen123 how did you get 10-z < z-x?

did you mean 10-z<10-x?

Z is closer to 10 than it is to x ..so the difference between 10 and z is less than the difference between z and x ..hence
10-z < z-x?