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rohit929
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Ive got a question: for stat 2, can you not start off by plugging in z=5x to:

(x+10)/2 > z ? --> x+10>2x --> x+10>10x --> 10/9 > x ...
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rohit929
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greenoak
It's A. Here is the explanation:

Quote:
If x>0 and x<10, is z greater than the average of x and 10?

1) z is closer to 10 than it is to x
It means two possibilities for z:
i. z is between x and 10, but closer to 10 => (x+10)/2 < z
ii. z is greater than 10 => (x+10)/2 <z
So, using this statement, we can answer the question => 1) is sufficient

2) z=5x
Not sufficient. Let’s consider two examples:
Let x=2 => z=10. Average of 2 and 10 is 6, so z is greater than the average.
Let x=0.001 (smth. really small) => z=0.005. Average of 0.001 and 10 is slightly greater than 5, so clearly it is greater than z.
hi greenoak,
for statement 1:

let x=9.8 since x<10 and positive number it may or may not be an integer
and z=9.9 as z is closer to 10
then avg of x and 10 =9.9
which is equal to z.

please correct me where i am going wrong
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durgesh79
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rohit929


let x=9.8 since x<10 and positive number it may or may not be an integer
and z=9.9 as z is closer to 10
then avg of x and 10 =9.9
which is equal to z.

please correct me where i am going wrong

when x = 9.8, z = 9.9
z is actally in middle of x and 10. As per question stem z should be more close to 10 than to x.



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