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

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If x is a positive number less than 10, is z greater than [#permalink]

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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 23 Mar 2012, 01:57, edited 1 time in total.
Edited the question and added the OA

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Re: x is a positive number less than 10 [#permalink]

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amitgovin wrote:
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

*********************************
please explain. thanks.


IMO A

1) The distance between Z and 10 is lesser than distance between Z and X. So Z is greater than the center of the line segment X to 10. That means Z is greater than the average of X and 10. Sufficient.
2) Z = 5X
If X=1, then Z = 5. Avg of X and 10 = 5.5. So Z<Avg.
If X=2, then Z = 10. Avg of X and 10 = 6. So Z>Avg.
Statement is Insufficient.

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Re: x is a positive number less than 10 [#permalink]

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Yup A.

stmt1:
|10-z| < |z-x| , here x and z are +ve ( as Z is closer to 10 than to x, and x is +ve )
10-z < z-x
=> z > Av of 10 and x.

stmt2:
x=1, z=5, 5 < 5.5
x=2, z=10, 10 > 6

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Re: DS Problem [#permalink]

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amp0201 wrote:
Hi,

Can anyone please suggest me the solution. I arrived at B, but unfortunately it was wrong. Thanks.


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. \(z\leq{10}\) --> \(|10-z|=10-z\) --> \(|10-z|<|z-x|\) 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.

Similar question: 600-level-question-95138.html?hilit=number%20line%20closer
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Re: DS Problem [#permalink]

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New post 19 Apr 2011, 14:07
hello Bunuel,

how did u get Z>10 ? above in red. from your inequality i got 10>x

thx

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Re: x is a positive number less than 10 [#permalink]

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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.

Attachment:
Ques2.jpg
Ques2.jpg [ 4.71 KiB | Viewed 22180 times ]

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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Re: x is a positive number less than 10 [#permalink]

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@Karishma
We stop in (2) because we've found something contrary to the answer in (1), while in earlier choice we had a definite answer. So the answer can't be B,C,D oe E.

The goal in DS is to get No as answer first for the choices, or to find the inconsistencies.

(1)
---------------0--------x-----Av--------10----z------------

---------------0--------x-----Av------z--10----------------

From the diagram above, it's evident that the answer is Yes

(2) is insufficient

if x is very close to 0, then z may not be closer to 10

e.g, x = 0.1, then z = 0.5

if x = 2, then z = 10 (greater than average)

Answer - A
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Re: x is a positive number less than 10 [#permalink]

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subhashghosh wrote:
@Karishma
We stop in (2) because we've found something contrary to the answer in (1), while in earlier choice we had a definite answer. So the answer can't be B,C,D oe E.

The goal in DS is to get No as answer first for the choices, or to find the inconsistencies.

(1)
---------------0--------x-----Av--------10----z------------

---------------0--------x-----Av------z--10----------------

From the diagram above, it's evident that the answer is Yes

(2) is insufficient

if x is very close to 0, then z may not be closer to 10

e.g, x = 0.1, then z = 0.5

if x = 2, then z = 10 (greater than average)

Answer - A


Yes, it is not possible to get inconsistent answers from the two statements i.e. we cannot have 'Yes. z is greater. Sufficient Alone.' from statement 1 and 'No. z is smaller. Sufficient Alone.' from statement 2.
From statement 2, we will either get 'Yes. z is greater. Sufficient Alone.' or 'z can be smaller or greater so not sufficient alone'

From statement 1, we got that z is greater than the average.
From statement 2, we tried a value and we got that z is less than the average. Then there must be a value for which z is greater than the average too in statement 2. We don't even need to try it. Definitely statement 2 alone is not sufficient and answer is (A).
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Re: If x is a positive number less than 10, is z greater than [#permalink]

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New post 25 Apr 2012, 17:22
just wondering about statement 1, am i misunderstanding the question?

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

z = 11
x = 8
so z is closer to 10 than x is

is 11 > (8+10)/2
11 > (18/2) becomes 11 > 9 so the answer is YES

z = 9
x = 8
so z is closer to 10 than x is

is 9 > (8+10)/2
9 > (18/2) becomes 9 > 9 so the answer is NO

or is the question stating z is closer to 10 than z is to x? i interpreted it as z is closer to 10 than 10 is to x.

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Re: If x is a positive number less than 10, is z greater than [#permalink]

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chamisool wrote:
just wondering about statement 1, am i misunderstanding the question?

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

z = 11
x = 8
so z is closer to 10 than x is

is 11 > (8+10)/2
11 > (18/2) becomes 11 > 9 so the answer is YES

z = 9
x = 8
so z is closer to 10 than x is

is 9 > (8+10)/2
9 > (18/2) becomes 9 > 9 so the answer is NO

or is the question stating z is closer to 10 than z is to x? i interpreted it as z is closer to 10 than 10 is to x.


(1) says: "On the number line, z is closer to 10 than it is to x". Here "it" refers to z.

So, it's: "On the number line, z is closer to 10 than z is to x".

Similar question to practice: if-x-is-a-positive-number-less-than-10-is-z-less-than-the-131322.html

Hope it helps.
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Re: If x is a positive number less than 10, is z greater than [#permalink]

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New post 17 Dec 2014, 16:50
Hmm...I don't understand why the answer is A.

for Statement A it states that "On the number line, z is closer to 10 than it is to x". Let's say Z is 9 and X is 8...then the average of X and 10 is 9. This equals Z... OR, let's say Z is 2 and X is 1...then Z < avg (x & 10). How can statement A be sufficient?

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Re: If x is a positive number less than 10, is z greater than [#permalink]

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New post 17 Dec 2014, 21:01
GMATAnnihilator wrote:
Hmm...I don't understand why the answer is A.

for Statement A it states that "On the number line, z is closer to 10 than it is to x". Let's say Z is 9 and X is 8...then the average of X and 10 is 9. This equals Z... OR, let's say Z is 2 and X is 1...then Z < avg (x & 10). How can statement A be sufficient?



The case you are considering with z=9,x=8 then z is equidistant from x and 10 both....So this is not valid.

Case 2 where z=2 ie. average of x+10 ----> (x+10)/2=2 or x+10=4 or x =-6...But we are told x is positive number less than 10...so x=-6 is not possible...

Hope it is clear
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Re: If x is a positive number less than 10, is z greater than [#permalink]

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WoundedTiger wrote:
GMATAnnihilator wrote:
Hmm...I don't understand why the answer is A.

for Statement A it states that "On the number line, z is closer to 10 than it is to x". Let's say Z is 9 and X is 8...then the average of X and 10 is 9. This equals Z... OR, let's say Z is 2 and X is 1...then Z < avg (x & 10). How can statement A be sufficient?



The case you are considering with z=9,x=8 then z is equidistant from x and 10 both....So this is not valid.

Case 2 where z=2 ie. average of x+10 ----> (x+10)/2=2 or x+10=4 or x =-6...But we are told x is positive number less than 10...so x=-6 is not possible...

Hope it is clear


Thanks Wounded Tiger...now where do I find that facepalm emoticon. I clearly misread the question... one of more deficiencies with timed exams... This will have to do: http://vignette3.wikia.nocookie.net/car ... 1217081420

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Re: If x is a positive number less than 10, is z greater than [#permalink]

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New post 04 Jan 2015, 10:26
Quote:
Hmm...I don't understand why the answer is A.

for Statement A it states that "On the number line, z is closer to 10 than it is to x". Let's say Z is 9 and X is 8...then the average of X and 10 is 9. This equals Z... OR, let's say Z is 2 and X is 1...then Z < avg (x & 10). How can statement A be sufficient?


For semplicity sake: rephrase the question into is 2z > 10 + x?

1) We know a few interesting things about x: it has to be smaller than 10; it can be a rational number; it is positive.
According to this statement --> if z=9 then x<8, say x=7,9 so that if 7,9 works any smaller value of x will work too. 2z=18 > 10+x. Thus z is always going to be greater than the average of 10 and x.

hope it helps :P
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Re: If x is a positive number less than 10, is z greater than [#permalink]

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New post 18 May 2015, 06:36
Hi Karishma,

This is regarding the number line approach that you have explained.Please share some problems where i could apply this technique.

Regards,

Kirti

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Re: If x is a positive number less than 10, is z greater than [#permalink]

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New post 18 May 2015, 06:49
Hi Karishma,

How can we be sure that the average is not more than z.The pictorial representation suits your explanation but somewhere i am not able to convince myself with it.

Would be glad if we could discuss

Regards

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Re: If x is a positive number less than 10, is z greater than [#permalink]

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New post 19 May 2015, 02:46
kirtivardhan wrote:
Hi Karishma,

How can we be sure that the average is not more than z.The pictorial representation suits your explanation but somewhere i am not able to convince myself with it.

Would be glad if we could discuss

Regards


Dear kirtivardhan

Before even beginning to think about z and whether the average is more or less than z, let's first simply think about what the average would be. This is the average of only 2 numbers: x and 10. So, the average will simply be the mid-point of x and 10, isn't it? So, let's depict this first of all (note that we're given that x lies between 0 and 10).

Image

With this out of the way, let's now process what the question is asking: Is z greater than the average of x and 10?

That is, does z lie in the yellow region shown below?

Image

Let's now analyze the statements:

St. 1 says that z is closer to 10 than to x on the number line.

Please note that information given in St. 1 is a FACT. So, now, ask yourself: in how many ways can z be closer to 10 than to x on the number line?

You'll find that this happens either in:

Case A: where z lies between the mean and 10, or in
Case B: where z lies on the Right hand side of 10

Image

Both these cases lie in the yellow region.

So, we can say for sure that all the possible values of z that satisfy the fact convey in St. 1 will definitely be greater than the mean. So, St. 1 is sufficient.

St. 2 says that z = 5x

Now, if x = 1, then mean = (1+10)/2 = 5.5 And z = 5*1 = 5.
In this case, z IS NOT greater than the mean

If x = 2, then mean = (2+10)/2 = 6. And z = 5*2 = 12
In this case, z IS greater than the mean

Since from St. 2 alone, we are not able to say for sure if z is greater than the mean or not, it is insufficient.

A follow-up query if something still remains unclear will be most welcome! :-D

Best Regards
Japinder

P.S.: I took the liberty of answering this query because Karishma may not have noticed it (since you only mentioned her name, didn't tag her in your post). In the future, to make sure that an expert gets a notification when you request help from them, please make use of the 'Mention this user' tab (highlighted in the image below) in the Post Editor. You know that you've used this tab successfully if the expert's name has now become hyperlinked. Hope this helped :)

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Re: If x is a positive number less than 10, is z greater than [#permalink]

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New post 19 May 2015, 02:55
Sure I will make a note of it.

Thank you so much for the explanation

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Re: If x is a positive number less than 10, is z greater than [#permalink]

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kirtivardhan wrote:
Hi Karishma,

This is regarding the number line approach that you have explained.Please share some problems where i could apply this technique.

Regards,

Kirti



Here is a post discussing the use of number line: http://www.veritasprep.com/blog/2013/08 ... mber-line/
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Re: If x is a positive number less than 10, is z greater than [#permalink]

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kirtivardhan wrote:
Hi Karishma,

How can we be sure that the average is not more than z.The pictorial representation suits your explanation but somewhere i am not able to convince myself with it.

Would be glad if we could discuss

Regards


I am guessing I don't need to give the explanation of the solution anymore.

A Tip:

When you make a diagram to get the answer - either number line in number properties or geometry - ensure that you keep the diagram as generic as possible. So I know that x is positive, that brings me to the right of 0. I also know that x is less than 10 so I put x anywhere between 0 and 10 - not in the centre. I don't know where z is so I consider all "different" regions where it can be. So what we deduce from such a diagram will hold in all cases.
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