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1) As z is closer to 10 than it is to x, we can have z > 10

or z < 10 on the condition that it stays closer to 10 than to x.

Case

1.

If z > 10, then the value of z will always be greater than average of x & 10.

For ex. z = 12 & x = 7 ---> Will satisfy option 1.

Now avg.(x+10) = 17/2 = 8.5, which is less than 12

Case

2.

Now let us see if z < 10.

If we consider z=9 & x=7, then avg.(x+10) = 17/2 = 8.5, which is less than 9. You can check by taking any legal values.

Another way of solving option 1:

We know that z > 10 will always satisfy, so we will consider only for z < 10.

As 10-z < z-x,

Therefore, 2z -x > 10

or, z > (x+10)/2

So, option

1 is

sufficient.

-------------------------------------

2) z = 5x

So possible values for x = 1,2,...9

For x=1, z=5 So, avg(x+10) = 11/2 = 5.5, which is greater than z

For x=2, z=10 So, avg(x+10) = 12/2 = 6, which is less than z

So, option

2 is

not sufficient.

-------------------------------------

So, only option

1 is sufficient to answer the question.

HTH

_________________

+++ Believe me, it doesn't take much of an effort to underline SC questions. Just try it out. +++

+++ Please tell me why other options are wrong. +++

~~~ The only way to get smarter is to play a smarter opponent. ~~~