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

Lets approach the easier statement (2) first: -

z = 5x and x < 10 lets pick nos

x = 4; Arithematic Mean = (4+10)/2 = 7 and z = 5x4 = 20, So z>arithematic mean.

No lets assume x = 1/4, Arithematic mean = (1/2+10)2 = 5.25, and z = 5x1/2 = 2.5, so z< arithematic mean.

So (2) is insufficient, the answer choices boils down to A & D

Now look at (1)

On number line Z is closer to 10 than x.

Assume x =2, then A.M = 6. If Z has to be closer to 10 then the Z > 6 so Z > A.M
Assume x = 6 then A.M = 8, then Z >9 so Z > A.M

for all +ve values of x<10, Z > A.M Hence (1) is sufficient.

Answer is (A)

Great explanation!! Thanks a lot. (A) is the right answer.

Hello! thanks so much for going through the problem.

MGMAT staff! I have a question re: rephrasing the original statement. How would you rephrase this problem?
Thanks so much :-)

Rephrase for 1:

AM = Equidistant from both 10 and X
If Z is closer to 10 and all 3 quantities are positive then obviously z > AM


Rephrasing for 2:

z > (x+10)/2
z> x/2 + 5

By 2:

5x > x/2 +5

9/2 x >5
x could be anything. within 0 and 10.

there are a couple of ways.

(1) VISUAL
this is the best way. in general, if you have visual 'number line' rephrases, those are pretty much the best way to go, pretty much all the time.
the average of x and 10 is the midpoint on the number line between x and 10.
therefore, greater than that average means to the right of that midpoint.
if you're to the right of the midpoint, then you're closer to the greater number, which, in this case, is 10.
therefore, you can rephrase the question, using this visual approach, as:
is z closer to 10 than to x?

this is dynamite, because it makes statement #1 absolutely trivial.

(2) ALGEBRAIC
question:
is z > (x + 10)/2 ?
is 2z > x + 10 ?

statement 1:
either z > 10, or 10 - z < z - x (i.e., the distance between z and 10 is smaller than the distance between x and z -- notice that "closer" can be rephrased to "smaller distance")
if z > 10, then 2z is more than 20, and x + 10 is less than 20 (because x is less than 10). therefore, YES in that case.
if 10 - z > z - x, you can rearrange that to get x + 10 < 2z, which means YES to the prompt question (that's the exact form of the prompt question).
therefore, unilaterally YES, so, sufficient.

--

statement 2:
no indication of the size of x and z is given, so just consider the extreme possibilities.
if they're tiny (like x = 0.0001 and z = 0.0005), then the average of x and 10 is about 5, which is colossally huge compared to z. so, NO.
if they're huge (like x = 1000 and z = 5000), then the average of x and 10 is less than x (and therefore WAY less than z). so, YES.
insufficient.

answer = a

Thanks

glad it helped.

A can't actually be correct.

If we picked Z to be 7, and X to be 6, then we end up with an AM of 8, smaller.

but if we picked Z to be 9 and X to 7, then we end up with an AM of 8.5, larger.

Unless I'm missing something significant, this can't be right.

Me? I'm tempted to go with E

ell_wu,
I am not sure you got the answer. Anyway, read the question carefully.

Statement 1 says z is closer to 10 than z is to x.

This means that, if you draw the number line, we can place z near to 10 than to x.

0-----x------...-z----10 { '...' sign for AM}

if z is more closer to x, then AM (Arithmetic mean) will be after z and AM will be more near to 10 than z. Since the statement 1 states that z is closer to 10, we can say that z is > AM

TIP: If you happen to see similar problems, draw a number line and place the numbers/values on it. This method is better than taking numbers and doing the calculations.

Hope this helps.

this is a nice explanation.

@ ell_wu, the short version: your first example doesn't count, because it doesn't actually satisfy the conditions of statement 1.
statement 1 says that z must be closer to 10 than to x; i.e., the distance between z and 10 must be less than the distance between z and x.
this is not true if z = 7 and x = 6; in that case, the distance between z and 10 is 3, but the distance between z and x is only 1.

Hi Staff,

in the algebraic approach, how do we get z>10 for statement 1?

thanks,
Kumar

i'm going to assume you're referring to the algebraic approach given in this post:
post18085.html#p18085
if this is not the case, please tell what algebraic approach you are referring to.

in the problem it is given that x < 10, so x is to the left of 10. therefore, if z > 10, then x, z, and 10 are arranged on the number line like this:
---------x----------10-----------z---------------
in this arrangement, z is always closer to 10 than it is to x, so all cases of z > 10 satisfy the statement.

Yes, i was referring to the same post.

Thanks for the explanation, it helped.

regards,
Kumar

ok.