IMO E


Option a :
1. lets say x = 4, z=5 ( closer to 10)
avg : 10+4 =14/2 =7 hence, z is less than avg
2. lets say x=4 z= 9 (closer to 10). in this case, z is higher than avg.

INSUFFICIENT

Option b:
1. lets say x =1. Hence, z = 5(1) = 5
avg : 10+1 = 11/2 =5.5..hence, z is less than avg
2. lets say x=2. hence, z = 5(2) =10
avg = 10+2 =12/2 =6. hence, z is more.

INSUFFICIENT

combining both,
again, x = 1, z = 5 (closer to 10). z is less than avg
also, x=2, z=10 (closer to 10). z is more than avg.

INSUFFICIENT

1. lets say x = 4, z=5 ( closer to 10)
avg : 10+4 =14/2 =7 hence, z is less than avg

In this case, z is not closer to 10 than it is to x.
In fact, by definition, if x<10 and z is closer to 10 than it is to x, it must be above the average of 10 and x, since the average of 10 and x is exactly halfway between 10 and x.
So Statement 1 is sufficient.

Statement 2 can also be approached algebraically:
5x > (x+10)/2 ?
10x > (x+10) ?
9x > 10 ?
x > 10/9 ?
[spoiler]Since we don't know whether this is true or not, inssufficient[/spoiler]

Given x<10, question is (x+10)/2 < Z


1> z is closer to 10 than it is to x can be written in an equation as -> z-x > 10-z

10-z < z-x
x+10 < 2z
(x+10)/2 < Z sufficient to answer a YES.

also if u visualize on a number line .

0-------x------------------------------------(x+10)/2---------z----------10

Z is more towards 10, therefore the weighted average of 10 and x will always be less than z as shown above.


2>(x+10)/2< Z (from question)
put z = 5x
(x+10)/2 < 5x
x+10 <10x
9x > 10 . multiple value of x will satisfy of dis-satisfy this with x<10 -> insufficeint.


pick A

ooopss...you are absolutely right...i missed the 'than x' part in the line one...
as stated above, statement 1 is sufficient..thnx for pointing out guys..