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If x is a positive number less than 10, is z greater than th [#permalink]
08 Nov 2005, 03:40

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

40% (02:55) correct
60% (01:50) wrong based on 86 sessions

This one is from GMATPREP as well. DS:

If x<10 and n is a positive integer, is z>(x+10)/2? (average of x and 10)

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

I am getting (D) as answer because (1) seems to be good enough info to conclude that z>(x+10)/2. Ofcourse, it is clear that (2) is ok too. But the answer is B.

Thanks!

MODERATOR: EDITED THE QUESTION.

ORIGINAL QUESTION:

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.

Re: number line problem [#permalink]
08 Nov 2005, 05:12

rianah100 wrote:

This one is from GMATPREP as well. DS:

If x<10 and n is a positive integer, is z>(x+10)/2? (average of x and 10)

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

I am getting (D) as answer because (1) seems to be good enough info to conclude that z>(x+10)/2. Ofcourse, it is clear that (2) is ok too. But the answer is B.

Thanks!

uhm, the bold part is easily misunderstood ....some can understand that z is closer to 10 than to x ...others may think: z is closer to 10 than x is . IMO, it is the latter one . They are totally different and thus lead to completely different outcomes.

Re: number line problem [#permalink]
08 Nov 2005, 10:37

rianah100 wrote:

This one is from GMATPREP as well. DS:

If x<10 and n is a positive integer, is z>(x+10)/2? (average of x and 10)

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

I am getting (D) as answer because (1) seems to be good enough info to conclude that z>(x+10)/2. Ofcourse, it is clear that (2) is ok too. But the answer is B.

Thanks!

I don't know why there is an 'n' in the stem. Anyway.. I got B.

(1) assume x = 8, z = 8.1.
(x+10)/2 = 9
Hence z < (x+10)/2

If x = 1 and z is 8.
z > (x+10)/2
since (x +10)/2 = 5.5.

so insuff.

(2) If z = 5x

stem can be is 5x > (x+10)/2
which is 5x > x/2 + 5 which is always true.

Re: number line problem [#permalink]
30 Apr 2010, 09:01

1

This post received KUDOS

Expert's post

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This post was BOOKMARKED

vscid wrote:

If x = 1, z = 5, answer to the question is NO. If x= 4, z = 20, answer to the question is YES.

Why is the answer B?

Edited the question. the original question is below:

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. Q: 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<10 --> |10-z|=10-z --> |10-z|<|z-x| becomes: 10-z<z-x --> 2z>10+x. Answer to the question YES.

B. z\geq{10} --> |10-z|=z-10 --> |10-z|<|z-x| becomes: z-10<z-x --> z>10 --> now, as z>10, then 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.)

Re: This one is from GMATPREP as well. DS: If x<10 and n is a [#permalink]
15 Oct 2013, 13:05

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