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Expert replies

by carlex_mx » Tue Feb 21, 2012 1:42 pm
Stuart Kovinsky wrote:
hman768 wrote:Wow. I was having the same problem at first and I kept coming up with E. The question is poorly worded. I originally read "z is closer to 10 than x" to mean that if X is 1 then Z only need be 2 or above because on the number line 2 is closer to 10 than 1 is to 10. But after reading the posts I see that this question means that Z must be closer to 10 than Z is to X. It's just a bad question. I did the official guide and I don't remember seeing it. I wonder where the question came from.
Hi,

if the question read as you reposted it, "z is closer to 10 than x", your critique would be valid.

However, if you look at the OP the question reads "z is closer to 10 than TO x", which only has one interpretation - the one that makes (1) sufficient all by itself.

As an aside, nowhere does it say that x and z are integers. So we could also satisfy (2) with x=.1 and z=.5 (not that it matters in this case, since x=1 and z=5 also give us the "no" answer that renders (2) insufficient alone).

As the first responder posted, (1) is clearly sufficient alone. Just picture a number line:

x ------- average of x and 10 -------- 10

The average of two terms is always dead centre of the two terms. In order for z to be closer to 10 than z is to x, z must be to the right of the mid point (if z were to the left, then it would be closer to x; if z were right in the middle, then it would be equidistant from x and 10). Since z must be to the right of the midpoint, which is the average of x and 10, z must be greater than the average.
Thanks Stuart!

I skipped the ""z is closer to 10 than TO x". Really important to read carefully :)
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by dario.brignone » Tue Mar 27, 2012 8:45 am
From the statement:
Z > ((10+x)/2)

From 2)
5x > ((10+x)/2) -> 10x > 10+x -> x > 10/9
That is not true as we only know that x > 0

Does the above make sense? Can we eliminate 2) base on that?

Thanks.
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