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

(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

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