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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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

On st1 i'll draw a number line. 0----------------------x-----------------z---10
We can see that z will always be greater than the average. Sufficient

St2 say that z=5x. we'll choose 2 cases:
Case 1: x=2, so z=10. avarge is 6 and z is greater than the avarage.
Case 2: x=1, so z=5 avarge is 5.5 and z is smaller than the avarage. Insufficient

So ans is A.

Q Is z>(x+10)/2 ; provided that 0<x<10
(1) On the number line, z is closer to 10 than it is to x.

this means distance between z and x is greater than distance between z and 10 on the number line. hence we have
z-x>10-z
z>(x+10)/2
in the above case z is less than 10 and lies between x and 10. If z is greater 10, then it will definitely be greater than average of x+10. hence z is greater than average of x+10. therefore sufficient.

(2) z = 5x

z=5x assume x=1 average of 10 and 1 = 5.5 and z=5(1)
when x=3, average of 10 and 3= 6.5 and z=15

since two different answers are possible hence 2 alone is not sufficient hence correct answer is A

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?
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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

No, the Statement 1 says that - Z is closer to 10 than it is to x and not that z is closer to 10 than x is. Your's is the later case.
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I received a PM about this question.
Here's my solution:

Target question: Is z greater than the mean of x and 10?

Statement 1: On the number line, z is closer to 10 than it is to x.
IMPORTANT: On the number line, the mean of two numbers will lie at the midpoint between those two numbers.
So, the mean of x and 10 will lie halfway between x and 10.
So, if z is closer to 10 than it is to x, then z must lie to the right of the midpoint between x and 10.
This means that z must be greater than the mean of x and 10
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: z = 5x
There are several pairs of numbers that meet this condition. Here are two:
Case a: x=1, z=5, in which case z is less than the mean of x and 10 (mean = 5.5)
Case b: x=4, z=20, in which case z is greater than the mean of x and 10 (mean = 7)
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer:

Cheers,
Brent

RELATED VIDEO

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0 < x < 10
Question : is z greater than the average (arithmetic mean) of x and 10?

Question : is z > (x+10)/2?
OR
Question : is z > Midpoint of x and 10 ?

Statement 1: On the number line, z is closer to 10 than it is to x.

i.e. z is to the right of "midpoint of 10 and x"

i.e. z is GREATER than x

SUFFICIENT

Statement 2: z = 5x

@x = 1, Midpoint = (1+10)/2 = 5.5 and z = 5 i.e. z < AVerage

@x = 2, Midpoint = (2+10)/2 = 6 and z = 10 i.e. z > AVerage

NOT SUFFICIENT

Answer: Option A
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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__
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Yes, absolutely! The two statements won't contradict each other so your answer will be (A). But if I am not running out of time, I will probably get the Yes answer too for stmnt 2, just for my satisfaction!!
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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.

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

If x = 1/2 and z = 5, then the distance between z and 10 is 5, while the distance between z and z is 4.5. So, those numbers violate the statement.
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Hi Bunuel

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

Those numbers do not comply with (1):

9 (z) is NOT closer to 10 than 9 (z) is to 8 (x). The distances are equal.
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