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1. The average of x and 10 will be at the same distance from x and 10. Since is closer to 10, and x < 10, z has to be greater than average of x and 10. SUFF.

2. If x = 1 => z = 5; Average of x and 10 = 11/2 = 5.5 > z
If x = 2 => z = 10; Average of x and 10 = 12/2 = 6 < z.
INSUFF. _________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

1. The average of x and 10 will be at the same distance from x and 10. Since is closer to 10, and x < 10, z has to be greater than average of x and 10. SUFF.

2. If x = 1 => z = 5; Average of x and 10 = 11/2 = 5.5 > z If x = 2 => z = 10; Average of x and 10 = 12/2 = 6 < z. INSUFF.

agree with giddi77, however z cannot be 10.

1. if st 1 is true, z is always greater than the average of x and 10.
2. if x = 0.2, z = 1 where as the avg of 10 and x is > than z. if x = 1.9, z = 9.55, which is > than the avg of x and 10.

1) z is closer to 10 on the number line than to x and x is less than 10
In all cases of x from 0-9 if z is closer to 10 , it is greater than the average of 10 and x

2) z=5x

When x = 8 , AV = 9 and z = 40 hence z> AV
When x = 1 , AV = 5.5 and z =5 hence z< AV
therefor not suff

Please help me. I do not understand how the answer is A

1) Consider the example of X =9 and Z = 9.1

Z is closer to 10 compared to X

Avg of X and 10 = 19/2 = 9.5 > Z

2) Consider the example of X = 9 and Z = 9.6

Z is closer to 10 compared to X

Avg of X and 10 = 19/2 = 9.5 < Z

Contradictory results. How can the answer be A?

The first examplae has a wrong value for z the statement z is closer to as compared to x means that 10 - z<z-x
Therefor on a number line z will always be on right hand side of the mid pt of segment x-10 therefore z will be greater than the average.

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