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

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Question Stats: 53% (02:09) correct 47% (02:18) wrong based on 214 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.

Actually, it's a 700+ Source: http://www.gmathacks.com

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

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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.

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

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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) $$- 10 + z < z - x$$
$$x > 10$$

2) $$10 - z < z - x$$
$$10 + x < 2z$$

$$\frac{(10 + x)}{2} < z$$

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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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) $$- 10 + z < z - x$$
$$x > 10$$

2) $$10 - z < z - x$$
$$10 + x < 2z$$

$$\frac{(10 + x)}{2} < z$$

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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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.

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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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.

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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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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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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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.

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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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.

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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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.

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

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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 Re: If x is a positive number less than 10, is z less than the   [#permalink] 23 Mar 2019, 08:53
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