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655-705 Level|   Arithmetic|      
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If x is a positive number less than 10, is z less than the 25 Apr 2012, 11:22
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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55% (01:55) correct 45% (02:22) wrong based on 607 sessions

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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.

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Actually, it's a 700+ 8-)

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If x is a positive number less than 10, is z less than the 25 Apr 2012, 12:24
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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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https://gmatclub.com/forum/if-x-is-a-po ... 28086.html
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If x is a positive number less than 10, is z less than the 26 Apr 2012, 13:06
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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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If x is a positive number less than 10, is z less than the 26 Apr 2012, 13:32
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metallicafan
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
Show more

(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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If x is a positive number less than 10, is z less than the 29 Apr 2014, 03:48
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Bunuel
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.
Show more

Hi
I did not understand in statement 1..why x<10/11 ..not sufficient???Can you pls explain ??
Thanks
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If x is a positive number less than 10, is z less than the 29 Apr 2014, 04:03
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Bunuel
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
Show more

(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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If x is a positive number less than 10, is z less than the 29 Apr 2014, 04:16
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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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If x is a positive number less than 10, is z less than the 29 Apr 2014, 07:18
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maaadhu
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.
Show more

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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If x is a positive number less than 10, is z less than the 29 Apr 2014, 07:27
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Bunuel
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.
Show more
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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If x is a positive number less than 10, is z less than the 29 Apr 2014, 07:43
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agavaqif
Bunuel
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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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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If x is a positive number less than 10, is z less than the 29 Apr 2014, 22:03
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Bunuel
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Bunuel
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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ouchhh I got the question wrong .. Ok thanks
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If x is a positive number less than 10, is z less than the 02 Aug 2015, 20:05
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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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If x is a positive number less than 10, is z less than the 02 Aug 2015, 20:27
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arunmoitra
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.
Show more

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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If x is a positive number less than 10, is z less than the 02 Aug 2015, 23:13
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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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If x is a positive number less than 10, is z less than the 23 Mar 2019, 08:53
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question: 2z<x+10 ?
I> x can be 1 or 9
thus from the statement, 12<11 or 18<19... nor A neither D
II> this means. 10-z>z-x and eventually, 10+z>2z
answer B
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If x is a positive number less than 10, is z less than the 15 Dec 2022, 22:51
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Question -
If x is a positive number less than 10, is z less than the average (arithmetic mean) of x and 10 ?

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

Solution --
Transforming words to numbers - we need to find relationship between z and (x+10)/2, given that x belongs to any of 1,2,3,4,5,6,7,8,9

From statement (1)
For 6x to be less than (x+10)/2 would mean 11x < 10 or x < 10/11, which is not possible x must be an integer

From statement (2)
z is closer to 10 than it is to x
Which literally says that z is greater than the average of x and 10, so we can say with surety that z is NOT less than average

Hence statement 2 alone is enough to answer the question
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If x is a positive number less than 10, is z less than the 05 May 2025, 19:20
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why can't z be on the left side of both x and 10 on number line and then closer to x than 10, that makes z smaller than both z--x--------10. That probably would make the statement 2 also insufficient.

Bunuel
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.

Similar questions to practice:
https://gmatclub.com/forum/if-x-is-a-po ... 67734.html (OG)
https://gmatclub.com/forum/if-x-is-a-po ... 28086.html
https://gmatclub.com/forum/if-y-is-a-ne ... 95138.html
https://gmatclub.com/forum/if-a-is-a-po ... 05650.html
https://gmatclub.com/forum/if-x-and-z-a ... 55651.html
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If x is a positive number less than 10, is z less than the 06 May 2025, 00:06
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mm92325
why can't z be on the left side of both x and 10 on number line and then closer to x than 10, that makes z smaller than both z--x--------10. That probably would make the statement 2 also insufficient.

Bunuel
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.

Similar questions to practice:
https://gmatclub.com/forum/if-x-is-a-po ... 67734.html (OG)
https://gmatclub.com/forum/if-x-is-a-po ... 28086.html
https://gmatclub.com/forum/if-y-is-a-ne ... 95138.html
https://gmatclub.com/forum/if-a-is-a-po ... 05650.html
https://gmatclub.com/forum/if-x-and-z-a ... 55651.html
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Z cannot be to the left of both x and 10 because in that case, the distance between z and x would be smaller than the distance between z and 10, which contradicts the condition that z is closer to 10 than to x.
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