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If x is a positive number less than 10, is z less than the

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If x is a positive number less than 10, is z less than the average (arithmetic mean) of x and 10 ?

(1) z = 6x
(2) On the number line, z is closer to 10 than it is to x.

Actually, it's a 700+ 8-)

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If x is a positive number less than 10, is z less than the  [#permalink]

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New post 25 Apr 2012, 11:24
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If x is a positive number less than 10, is z less than the average (arithmetic mean) of x and 10 ?

Given: \(0<x<10\). Question: is \(z<\frac{x+10}{2}\)? --> is \(2z<x+10\)?

(1) z = 6x --> the question becomes: is \(2*6x<x+10\)? --> is \(x<\frac{10}{11}\)? We don't know that. Not sufficient.

(2) On the number line, z is closer to 10 than it is to x.

-----x-----average-----10----- (average of \(x\) and 10 is halfway between x and 10). So the question asks whether \(z\) is in the GREEN area.

Since, \(z\) is closer to 10 than it (\(z\)) is to \(x\) then this statement directly tells us that \(z\) is in the RED area. So, the answer to the question is NO. Sufficient.

Answer: B.

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Re: If x is a positive number less than 10, is z less than the  [#permalink]

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New post 26 Apr 2012, 12:06
Thank you Bunuel!, I have found another way to solve the stament (2), please confirm whether I am Ok.

(2) On the number line, z is closer to 10 than it is to x.

\(|10 - z| < z - x\)

1) \(- 10 + z < z - x\)
\(x > 10\)
Not a valid answer.

2) \(10 - z < z - x\)
\(10 + x < 2z\)

\(\frac{(10 + x)}{2} < z\)
This answers our question.

SUFF. B
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Re: If x is a positive number less than 10, is z less than the  [#permalink]

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New post 26 Apr 2012, 12:32
metallicafan wrote:
Thank you Bunuel!, I have found another way to solve the stament (2), please confirm whether I am Ok.

(2) On the number line, z is closer to 10 than it is to x.

\(|10 - z| < z - x\)

1) \(- 10 + z < z - x\)
\(x > 10\)
Not a valid answer.

2) \(10 - z < z - x\)
\(10 + x < 2z\)

\(\frac{(10 + x)}{2} < z\)
This answers our question.

SUFF. B


(1) On the number line, z is closer to 10 than it is to x --> \(|10-z|<|z-x|\) --> as z is closer to 10 than it is to x, then z>x, so \(|z-x|=z-x\) --> two cases for 10-z:

A. \(z\leq{10}\) --> \(|10-z|=10-z\) --> \(|10-z|<|z-x|\) becomes: \(10-z<z-x\) --> \(2z>10+x\). Answer to the question YES.

B. \(z>{10}\) --> in this case \(2z>20\) and as \(x<10\), then \(x+10<20\), hence \(2z>10+x\). Answer to the question YES.
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Re: If x is a positive number less than 10, is z less than the  [#permalink]

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New post 29 Apr 2014, 02:48
Bunuel wrote:
If x is a positive number less than 10, is z less than the average (arithmetic mean) of x and 10 ?

Given: \(0<x<10\). Question: is \(z<\frac{x+10}{2}\)? --> is \(2z<x+10\)?

(1) z = 6x --> the question becomes: is \(2*6x<x+10\)? --> is \(x<\frac{10}{11}\)? We don't know that. Not sufficient.

(2) On the number line, z is closer to 10 than it is to x.

-----x-----average-----10----- (average of \(x\) and 10 is halfway between x and 10). So the question asks whether \(z\) is in the GREEN area.

Since, \(z\) is closer to 10 than it (\(z\)) is to \(x\) then this statement directly tells us that \(z\) is in the RED area. So, the answer to the question is NO. Sufficient.

Answer: B.


Hi
I did not understand in statement 1..why x<10/11 ..not sufficient???Can you pls explain ??
Thanks
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Re: If x is a positive number less than 10, is z less than the  [#permalink]

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New post 29 Apr 2014, 03:03
drkomal2000 wrote:
Bunuel wrote:
If x is a positive number less than 10, is z less than the average (arithmetic mean) of x and 10 ?

Given: \(0<x<10\). Question: is \(z<\frac{x+10}{2}\)? --> is \(2z<x+10\)?

(1) z = 6x --> the question becomes: is \(2*6x<x+10\)? --> is \(x<\frac{10}{11}\)? We don't know that. Not sufficient.

(2) On the number line, z is closer to 10 than it is to x.

-----x-----average-----10----- (average of \(x\) and 10 is halfway between x and 10). So the question asks whether \(z\) is in the GREEN area.

Since, \(z\) is closer to 10 than it (\(z\)) is to \(x\) then this statement directly tells us that \(z\) is in the RED area. So, the answer to the question is NO. Sufficient.

Answer: B.


Hi
I did not understand in statement 1..why x<10/11 ..not sufficient???Can you pls explain ??
Thanks


(1) z = 6x --> the question becomes: is \(2*6x<x+10\)? --> is \(x<\frac{10}{11}\)? Can we answer this question? Do we know whether \(x<\frac{10}{11}\)? NO. Hence the statement is not sufficient.
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Re: If x is a positive number less than 10, is z less than the  [#permalink]

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New post 29 Apr 2014, 03:16
Sorry for this dumb question.

Can fraction be a number? Is it only whole numbers that are positive integers?

In this case, if X is whole number & x < 10, then A will be sufficient. If X is number, then Option A is insufficient. Please clarify.
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Re: If x is a positive number less than 10, is z less than the  [#permalink]

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New post 29 Apr 2014, 06:18
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maaadhu wrote:
Sorry for this dumb question.

Can fraction be a number? Is it only whole numbers that are positive integers?

In this case, if X is whole number & x < 10, then A will be sufficient. If X is number, then Option A is insufficient. Please clarify.


Fraction, integers, irrational numbers all are numbers.

You need to brush up fundamentals.

Number Theory chapter of Math Book: math-number-theory-88376.html
Best GMAT Quantitative Books: best-gmat-math-prep-books-reviews-recommendations-77291.html

Hope this helps.
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Re: If x is a positive number less than 10, is z less than the  [#permalink]

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New post 29 Apr 2014, 06:27
Bunuel wrote:
If x is a positive number less than 10, is z less than the average (arithmetic mean) of x and 10 ?

Given: \(0<x<10\). Question: is \(z<\frac{x+10}{2}\)? --> is \(2z<x+10\)?

(1) z = 6x --> the question becomes: is \(2*6x<x+10\)? --> is \(x<\frac{10}{11}\)? We don't know that. Not sufficient.

(2) On the number line, z is closer to 10 than it is to x.

-----x-----average-----10----- (average of \(x\) and 10 is halfway between x and 10). So the question asks whether \(z\) is in the GREEN area.

Since, \(z\) is closer to 10 than it (\(z\)) is to \(x\) then this statement directly tells us that \(z\) is in the RED area. So, the answer to the question is NO. Sufficient.

Answer: B.

First case is not sufficent no problem here.
But for the second case Lets suppose x=8 then average of 8 and 10 is 9. z can be 8.1 which is closer to 10 but less than 9 and also z can be 9.1 whihch is closer to 10 and greater than 9 so not sufficent is not it?
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Re: If x is a positive number less than 10, is z less than the  [#permalink]

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New post 29 Apr 2014, 06:43
agavaqif wrote:
Bunuel wrote:
If x is a positive number less than 10, is z less than the average (arithmetic mean) of x and 10 ?

Given: \(0<x<10\). Question: is \(z<\frac{x+10}{2}\)? --> is \(2z<x+10\)?

(1) z = 6x --> the question becomes: is \(2*6x<x+10\)? --> is \(x<\frac{10}{11}\)? We don't know that. Not sufficient.

(2) On the number line, z is closer to 10 than it is to x.

-----x-----average-----10----- (average of \(x\) and 10 is halfway between x and 10). So the question asks whether \(z\) is in the GREEN area.

Since, \(z\) is closer to 10 than it (\(z\)) is to \(x\) then this statement directly tells us that \(z\) is in the RED area. So, the answer to the question is NO. Sufficient.

Answer: B.

First case is not sufficent no problem here.
But for the second case Lets suppose x=8 then average of 8 and 10 is 9. z can be 8.1 which is closer to 10 but less than 9 and also z can be 9.1 whihch is closer to 10 and greater than 9 so not sufficent is not it?


If x=8 and z=8.1, then z is closer to x not to 10, which violates the second statement.
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Re: If x is a positive number less than 10, is z less than the  [#permalink]

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New post 29 Apr 2014, 21:03
Bunuel wrote:
agavaqif wrote:
Bunuel wrote:
If x is a positive number less than 10, is z less than the average (arithmetic mean) of x and 10 ?

Given: \(0<x<10\). Question: is \(z<\frac{x+10}{2}\)? --> is \(2z<x+10\)?

(1) z = 6x --> the question becomes: is \(2*6x<x+10\)? --> is \(x<\frac{10}{11}\)? We don't know that. Not sufficient.

(2) On the number line, z is closer to 10 than it is to x.

-----x-----average-----10----- (average of \(x\) and 10 is halfway between x and 10). So the question asks whether \(z\) is in the GREEN area.

Since, \(z\) is closer to 10 than it (\(z\)) is to \(x\) then this statement directly tells us that \(z\) is in the RED area. So, the answer to the question is NO. Sufficient.

Answer: B.

First case is not sufficent no problem here.
But for the second case Lets suppose x=8 then average of 8 and 10 is 9. z can be 8.1 which is closer to 10 but less than 9 and also z can be 9.1 whihch is closer to 10 and greater than 9 so not sufficent is not it?


If x=8 and z=8.1, then z is closer to x not to 10, which violates the second statement.

ouchhh I got the question wrong .. Ok thanks
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Re: If x is a positive number less than 10, is z less than the  [#permalink]

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New post 02 Aug 2015, 19:05
Shouldn't option A be sufficient too?
If we take the highest (9) and lowest (1) possible value of x and take the average of x and 10, then we can see that 6x (or z) is always greater than the average.
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Re: If x is a positive number less than 10, is z less than the  [#permalink]

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New post 02 Aug 2015, 19:27
arunmoitra wrote:
Shouldn't option A be sufficient too?
If we take the highest (9) and lowest (1) possible value of x and take the average of x and 10, then we can see that 6x (or z) is always greater than the average.


The mistake you are making is that x can be any POSITIVE NUMBER <10 and not POSITIVE INTEGER < 10

Try x = 0.5 (a positive number), will give you z (=6x=3) < (x+10)/2 (= 5.25)

Thus you do not get a definite yes or now and hence A is not sufficient.

Another way you can look at what values will negate this condition is :

For limits on values of x for which z < average of x and 10 ----> z < (x+10)/2 ----> 2z<x+10 ----> 2*6x < x+10 ----> x<10/11 . Thus all positive numbers less than 10/11 will give a "no" for z less than the average. Any value > 10/11 will give you a "yes" for it.
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Re: If x is a positive number less than 10, is z less than the  [#permalink]

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New post 02 Aug 2015, 22:13
the main trick inthe question is " x is a positive number" not an intiger

so when we plug the number from 0 to 1 we can get the answer clear to eliminate A
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