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

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
_________________

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.
_________________

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.
_________________

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.
_________________

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

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.
_________________

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

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

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