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

Hi,

i request your help to find out weather statement A is sufficient, when x=8 and Z =9 . Here z is closer to 10 than x, but the average of 10 and X will be equal to z.

Is this correct?

Thanks

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

(1) On the number line, z is closer to 10 than it is to x.
(2) z = 5x

The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

(1) On the number line, z is closer to 10 than it is to x.
(2) z = 5x

If x is a positive number less than 10 is z greater than the average of x and 10?

1) on the number line, z is closer to 10 than it is to x

2) z=5x

Drawing a number line can help you see the answer quickly.


So x is somewhere between 0 and 10 and average of x and 10 is between x and 10.
z can be in any of the 5 regions as shown.

1) on the number line, z is closer to 10 than it is to x

In which regions is z closer to 10 than to x? Only the two rightmost regions. There, z is greater than the average of x and 10. Sufficient.

2) z = 5x
Put x = 1
z = 5 but average of 1 and 10 is 5.5 so z is less than the average. Actually, this is where I stop and mark (A) as the answer without checking to see if there is a value of x for which z is greater than the average. Think, why?

VeritasKarishma
From your explanation:
Statement 1 says: YES
Statement 2 says: NO (before finding YES value)
We should not try for finding YES value in statement 2. Without finding YES value in statement 2, we can confidently say that the correct choice is A, because both statements can't contradict each other!
Is it the right way to choose A?
Thanks__

How can the answer be A? what if x = 1/2 and z = 5? in this case the average between x and 10 is 5.25 and z is 5, thus less than the average. Also, z is closer to 10 than it is to x. Please advice on this question.

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

Given: \(0<x<10\).
Question: is z greater than the average of x and 10? Or: is \(z>\frac{10+x}{2}\)? --> \(2z>10+x\)?

(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.

OR another approach:
Given:
x-----average-----10----- (average of x and 10 halfway between x and 10).

Now, as z is closer to 10 than it (z) is to x, then z is either in the blue area, so more than average OR in the green area, so also more than average. Answer to the question YES.

Sufficient.

(2) z = 5x --> is \(2z>10+x\)? --> is \(10x>10+x\)? --> is \(x>\frac{10}{9}\). We don't now that. Not sufficient. (we've gotten that if \(x>\frac{10}{9}\), then the answer to the question is YES, but if \(0<x\leq{\frac{10}{9}}\), then the answer to the question is NO.)

Answer: A.

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Hi Bunuel

In (1) what if Z= 9 and X = 8, then both are equal as 9.

what if x = 8.99 and z = 10

statement 1 is not sufficient

Bunuel KarishmaB

If z =9
x = 8.99

And average will be ~9.499

Bunuel
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