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by lilu » Mon Apr 27, 2009 8:14 pm
If x is a positive number less than 10, is z greater than the average (arithmetic mean) of x and 10?

1. On the number line z is closer to 10 than to x
2. z=5x
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by Feep » Mon Apr 27, 2009 8:32 pm
Looks like A to me.

(1) tells us that z is closer to 10 than to x. Since the average of x and 10 will be exactly at the midpoint of those two numbers, z will be to the right of the midpoint on the number line, and is therefore greater. Sufficient.

(2) gives us z = 5x. If x = 1, z = 5, and is NOT greater than the average of x and 10 (5.5). If x = 10, z = 50, and is obviously greater than the average of x and 10 (10). Since we can see either case, this is insufficient.

A.
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by bulla4life » Thu Apr 30, 2009 5:57 pm
Hey Feep,

Option 2 is insufficient too

a.
What if x is 6 and z is 7.

Z is closer to 10.

the average of 6 and 10 is 8.

it means z is less then the average

b.
what if x is 7 and z is 9
average of 7 and 10 is 8.5
it means z is greater

Insufficient. it should be E
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by Vemuri » Thu Apr 30, 2009 6:30 pm
IMO E is the answer. What is the OA?
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by Feep » Thu Apr 30, 2009 11:41 pm
bulla4life wrote:Hey Feep,

Option 2 is insufficient too

a.
What if x is 6 and z is 7.

Z is closer to 10.

the average of 6 and 10 is 8.

it means z is less then the average

b.
what if x is 7 and z is 9
average of 7 and 10 is 8.5
it means z is greater

Insufficient. it should be E
First of all, I assume you were talking about statement (1), since I stated in my original post that (2) was indeed insufficient.

Your test case for statement (1) is invalid: if x is 6, z cannot be 7. 7 is closer to 6 than 10, and so z would be closer to x than to 10.
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by mike22629 » Fri May 01, 2009 8:25 am
IMO A.
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by aj5105 » Sat May 02, 2009 7:06 pm
(A)
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by awesomeusername » Sat May 02, 2009 10:56 pm
bulla4life wrote:Hey Feep,

Option 2 is insufficient too

a.
What if x is 6 and z is 7.

Z is closer to 10.

the average of 6 and 10 is 8.

it means z is less then the average
If z is closer to 10 than 6, then z is greater than 8.
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by Objectivo » Sun May 03, 2009 5:33 am
To me, it's C

Since x is a positive integer, it is either 1,2,3,4,5,6,7,8 or 9, the average between x and 10 is either 5,5; 6; 6,5; 7; 7,5; 8; 8,5; 9 or 9,5

(1) tells us that z is closer to 10 than to x.
(2) tells us that z=5x

Let's start with (2): this tells us that, since x is either one of the number above, z must be 5,10,15,20,25,30, 35,40 or 45.

Since (1) tells us that z is closer to 10 than to x, this excludes x being 1 and z being 5 (with 5 being closer to one than to 10).

In all other cases, z is greater than the average between x and 10.

Hence, (1) and (2) together are sufficient.
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by Feep » Sun May 03, 2009 7:24 pm
Objectivo wrote:To me, it's C

Since x is a positive integer, it is either 1,2,3,4,5,6,7,8 or 9, the average between x and 10 is either 5,5; 6; 6,5; 7; 7,5; 8; 8,5; 9 or 9,5

(1) tells us that z is closer to 10 than to x.
(2) tells us that z=5x

Let's start with (2): this tells us that, since x is either one of the number above, z must be 5,10,15,20,25,30, 35,40 or 45.

Since (1) tells us that z is closer to 10 than to x, this excludes x being 1 and z being 5 (with 5 being closer to one than to 10).

In all other cases, z is greater than the average between x and 10.

Hence, (1) and (2) together are sufficient.
Sure they are, but you need to look at them individually first. (1) is sufficient by itself, and so the answer is A.

OP, can you post the OA so we can put this to rest?
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by lilu » Sun May 03, 2009 8:01 pm
Sorry I haven't replied right away.
I just double-checked the exam and the OA is A
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by hman768 » Mon May 31, 2010 5:48 pm
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.
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by akhpad » Thu Jul 15, 2010 8:34 am
I re-checked A is correct.

Thing is that x is less than 10
Last edited by akhpad on Thu Jul 15, 2010 6:41 pm, edited 1 time in total.
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by outreach » Thu Jul 15, 2010 10:12 am
but VISUAL approach gives the correct solution
akhp77 wrote:I can also say that A is not sufficient.

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by Stuart@KaplanGMAT » Thu Jul 15, 2010 1:06 pm
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.
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