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Statement 1) if s is to the right of zero then 2 cases arrive Case 1 -----0---r----s-----t----- Case 2 -----r---0----s-----t----- which to choose, hence insufficient

statement 2) the distance b/w t & r is the same as the distance b/w t & -s still 2 cases arrive Case 1 r=-5, s=-3, t=-1, s=3 -----r---------s-------------------t-------------0--------------(+s)----- where +s=3 case 2 -----r------0------s---------------t---------------- r=-s

combining the two statements above, its clear that 0 lies midway to r and s.

Re: On the number line shown, is zero halfway between r and s ? [#permalink]

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31 Mar 2012, 10:14

Graphical approch:-

1. s doesn't lead to answer.

2. 0 could be to the right of t i.e. assuming s to be -ve and -s to be positive or 0 before s meaning r & -s are same point (these are the only two cases as points could be on two sides of t or on same side of t)

Together: Since s is to the right of 0 then -s is to the left of 0... and |r-t| = |t+s| then r must be equal to -s... <=====(r=-s)==(0)===(s)===(t)=====>

Re: On the number line shown, is zero halfway between r and s ? [#permalink]

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04 Apr 2013, 03:05

yangsta8 wrote:

mbaquestionmark wrote:

I have a question guys..

If -s is to the right of t, then wont r be equal to s ? But clearly in the picture, r and s are different points..

So dont u think that option is ruled out ? or is it like we should not go by the pic ? I know we should not go by the scale of the pic.. also this ?

Cos I thought the answer was B.. can someone please explain if I am wrong..

Thanks..

There's a couple of points to remember. Firstly never base your answer on how the diagrams look, they are representative but are by no means accurate. Because a triangle is drawn as equilateral for example, there is no reason to assume it is.

I think you've made a couple of incorrect assumptions in your reasoning: 1) -S is not necessarily to the right of T. Consider the case that 0 is between S and R. Then -S is negative meaning it is to the left of 0 and hence to the left of T. Your assumption is that 0 is on the right of S, but this isn't stated anywhere in Statement 2. 2) No answers state that R and S are the same point. Just that R = negative S.

Hope that clears it up.

hi Yangsta, If 0 is between S and R, then there are two points(r and -s) on the left hand side of T which are distinct yet have the same distance from T?? how this is possible,,can u explain with an e.g if possible?

Re: On the number line shown, is zero halfway between r and s ? [#permalink]

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29 Mar 2014, 09:30

I still don't quite get statement 2

I'm looking at the diagram and the statement and it seems clear that r must be = -s. And therefore zero is in the middle hence answer is sufficient

Also trying when r>s>t and so r=-s, but still 0 is still between s and r and answer is still yes

Is there any case I am missing, could someone please illustrate in number line

Bonus question: I read B's explanation about what IanStewart mentioned regarding graphs in PS, DS and all that stuff mentioning that we could trust the relative position of points in the diagram. Hence in this case r>s>t always?

Thanks! Would throw a lot of Kudos for this Cheers J

I'm looking at the diagram and the statement and it seems clear that r must be = -s. And therefore zero is in the middle hence answer is sufficient

Also trying when r>s>t and so r=-s, but still 0 is still between s and r and answer is still yes

Is there any case I am missing, could someone please illustrate in number line

Bonus question: I read B's explanation about what IanStewart mentioned regarding graphs in PS, DS and all that stuff mentioning that we could trust the relative position of points in the diagram. Hence in this case r>s>t always?

Thanks! Would throw a lot of Kudos for this Cheers J

First of all, the question asks whether 0 is halfway between r and s, not just between r and s.

Below is the case for (2) when 0 is NOT halfway between r and s. (2) The distance between t and r is the same as the distance between t and -s

--r-------s---t-----------(-s)

Here s is negative, -s is positive and 0 is somewhere between t and -s.

As for your second question: from the diagram we can infer that \(t>s>r\), not that \(r>s>t\).

OG13, page 272: A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2). Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight. You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees. All figures lie in a plane unless otherwise indicated.

OG13, page 150: Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

Re: On the number line shown, is zero halfway between r and s ? [#permalink]

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21 Jun 2014, 07:14

DenisSh wrote:

Attachment:

Number line.PNG

On the number line shown, is zero halfway between r and s?

(1) s is to the right of zero (2) The distance between t and r is the same as the distance between t and -s

Hi Bunuel,

Got this question incorrect on GMAT prep test and thus looked at your solution....

Initially, I tried to do the question by myself and here is what I did

The Question basically asks us whether 0 is between r and s or |r-0|=|s-0| or |r|=|s| or r= s or -s (Now, if r=s then answer is no but if r=-s then answer is yes...But I think I did apply the mod statement correctly so in this case do we reject the case of r=s at this stage itself and reduce the question to if r=-s ) St 1 says s>0 not much help here as r can be placed to the right of zero or left or at zero. Not sufficient St 2 says |t-r|=|t+s| This can be interpreted in one of the 2 ways

t-r = t+s or t-r= -(t+s)

So we get either r=-s or 2t=r+s

Not sufficient.

combining we see that r=-s and s>0 therefore r<0 and r=-s which is same
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On the number line shown, is zero halfway between r and s?

(1) s is to the right of zero (2) The distance between t and r is the same as the distance between t and -s

Hi Bunuel,

Got this question incorrect on GMAT prep test and thus looked at your solution....

Initially, I tried to do the question by myself and here is what I did

The Question basically asks us whether 0 is between r and s or |r-0|=|s-0| or |r|=|s| or r= s or -s (Now, if r=s then answer is no but if r=-s then answer is yes...But I think I did apply the mod statement correctly so in this case do we reject the case of r=s at this stage itself and reduce the question to if r=-s ) St 1 says s>0 not much help here as r can be placed to the right of zero or left or at zero. Not sufficient St 2 says |t-r|=|t+s| This can be interpreted in one of the 2 ways

t-r = t+s or t-r= -(t+s)

So we get either r=-s or 2t=r+s

Not sufficient.

combining we see that r=-s and s>0 therefore r<0 and r=-s which is same

On the number line shown, is zero halfway between r and s?

\(k\) is halfway between \(m\) and \(n\) can ALWAYS be expressed as: \(\frac{m+n}{2}=k\).

Is 0 halfway between r and s? --> is \(\frac{r+s}{2}=0\)? --> \(r+s=0\).

The question asks whether we have the following case:

--r---0---s---t--

(1) s is to the right of zero. Clearly insufficient.

(2) The distance between t and r is the same as the distance between t and -s

If s < 0, then we'd have the following case:

--r-------s---t---0-------(-s)

Answer NO.

If s > 0, then we'd have the following case:

--r---0---s---t-------------

Answer YES. Notice that in this case r and -s coincide.

Not sufficient.

(1) + (2) We have the second case from (2). Sufficient.

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The distance between t and r is the same as the distance between t and -s

-s --- t ---r ---- 0 ---------s (taking numbers as below) -10(s) --- -7(t) --- -4(r) ---- 0 -------------- -10 (s) No

-4(r=-s)----- 0(t) ------4 (S)

Go to the original post on this thread. Check out the diagram given with the question. "On the number line shown...." This tells you that r, s and t are on the number line in that order. r to the left of s and s to the left of t. So the cases you have taken are not valid since you have changed the relative positions of r, s and t.
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On the number line shown, is zero halfway between r and s ? [#permalink]

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11 Sep 2015, 14:56

Hello Benuel/Experts,

It is a bit embarrassing for me to post this potential silly question. For whatever reason, I am unable to get this out of my head. I would continue to read the entire thread multiple times while someone kind enough to take some time and answer it.

I am under the impression that, unless stated otherwise, the given figure is to some sort of scale. Regardless of the units of the scale, and also due to the reason that it did not say that the figure is not to scale, why are we even considering the possibility of s = -s = 0 when in fact statement (2) states that the distance between t and r is same as t and -s (assuming that -s needs to be left of s) ?

Thank you so much for your time here.

I answered my own question here. Realizing that both s and -s can take both + and - values helped understand this question.

It is a bit embarrassing for me to post this potential silly question. For whatever reason, I am unable to get this out of my head. I would continue to read the entire thread multiple times while someone kind enough to take some time and answer it.

I am under the impression that, unless stated otherwise, the given figure is to some sort of scale. Regardless of the units of the scale, and also due to the reason that it did not say that the figure is not to scale, why are we even considering the possibility of s = -s = 0 when in fact statement (2) states that the distance between t and r is same as t and -s (assuming that -s needs to be left of s) ?

Thank you so much for your time here.

I answered my own question here. Realizing that both s and -s can take both + and - values helped understand this question.

Be careful with what you have mentioned in red above. Especially in DS questions, be wary of the assumptions!

As per official guide "You may assume that the positions of points, angles, regions, and so forth exist in the order shown. Thus in DS, unless otherwise stated to the contrary, do not assume that the figures are drawn to scale. Only the order is the same. This is a very subtle but important point for this question.
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Re: On the number line shown, is zero halfway between r and s ? [#permalink]

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11 Oct 2016, 04:45

A. s is to the right of zero; in this case r could be both in left and right side of zero If s = 2, r = -3, then No If s = 2, r = 1, then No If s = 1, r = -1, then Yes

Not Sufficient

B. |t-r|=|t-(-s)| =>|t-r|=|t+s| t-r is always positive,since r is to the left of the t,therefore |t-r|=t-r; For example,if t = 3 and r = 2,then t-r=1 Again,if t = 3 and r = -2,then t-r= 3-(-2) = 5 Again,if t = 3 and r = -5,then t-r= 3-(-5)=8

On other hand, |t+s| may be positive or negative For example,if t = 3 and s = 2,then t+s=5 Again,if t =3 and s =-2,then t+s= 3+(-2) =1 Again,if t = 3 and s=-5,then t+s= 3+(-5)=- 8

∴t-r=|t+s| =>t-r=t+s or t-r=-(t+s) =>-r=s or 2t=r-s

Not Sufficient

A+B: From A: Since s>0, then t+s>0 Now applying the condition of A in B we get, t-r=|t+s|=>t-r=t+s=>-r=s

Since r and-r are halfway between 0,then r and s (=-r) will also be halfway between 0.

Re: On the number line shown, is zero halfway between r and s ? [#permalink]

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18 Oct 2016, 12:14

Bunuel wrote:

Let me clear this one:

NOTE: In GMAT we can often see such statement: \(k\) is halfway between \(m\) and \(n\). Remember this statement can ALWAYS be expressed as: \(\frac{m+n}{2}=k\).

Also in GMAT we can often see another statement: The distance between \(p\) and \(m\) is the same as the distance between \(p\) and \(n\). Remember this statement can ALWAYS be expressed as: \(|p-m|=|p-n|\).

Back to original question:

Is 0 halfway between r and s? OR is \(\frac{r+s}{2}=0\)? --> Basically the question asks is \(r+s=0\)?

(1) \(s>0\), clearly not sufficient.

(2) The distance between \(t\) and \(r\) is the same as the distance between \(t\) and -\(s\): \(|t-r|=|t+s|\).

\(t-r\) is always positive as \(r\) is to the left of the \(t\), hence \(|t-r|=t-r\);

BUT \(t+s\) can be positive (when \(t>-s\), meaning \(t\) is to the right of -\(s\)) or negative (when \(t<-s\), meaning \(t\) is to the left of -\(s\), note that even in this case \(s\) would be to the left of \(t\) and relative position of the points shown on the diagram still will be the same). So we get either \(|t+s|=t+s\) OR \(|t+s|=-t-s\).

In another words: \(t+s\) is the sum of two numbers from which one \(t\), is greater than \(s\). Their sum clearly can be positive as well as negative. Knowing that one is greater than another doesn't help to determine the sign of their sum.

Hence: \(t-r=t+s\) --> \(-r=s\); OR \(t-r=-t-s\) --> \(2t=r-s\).

So the only thing we can determine from (2) is: \(t-r=|t+s|\) Not sufficient.

(1)+(2) \(s>0\) and \(t-r=|t+s|\). \(s>0\) --> \(t>0\) (as \(t\) is to the right of \(s\)) hence \(t+s>0\). Hence \(|t+s|=t+s\). --> \(t-r=t+s\) --> \(-r=s\). Sufficient.

Answer: C.

yangsta8 wrote:

Statement 2) This tells us that -S=R but it doesn't tell us anything to either S or R in relation to 0.

This is not correct. If we were able to determine that \(-s=r\), statement (2) would be sufficient. But from (2) we can only say that \(t-r=|t+s|\).

Economist wrote:

This is confusing.. Okay, let me put it this way: for number lines, if we have such points...do we trust the sign of the points? and their relative positioning ? Experts please comment.

eg. here, do we assume that s cannot be 0, as -s and s are supposed to be distinct +ve and -ve values.

also, do we trust the relative positioning ( not distance ) r-s-t as shown in figure?

As for \(s\) to be zero: from statement (1) we can say that \(s\) can not be zero as it states that \(s>0\).

For (2) we don't know whether -s=s=0 or not. If \(-s=s=0\), \(s\) and therefore -\(s\) are to the left of \(t\) and (2) would be sufficient in this case. But we don't know that.

About the relative position of the points on diagram. Do you remember the question about the two circles and point C? (ds-area-between-circles-85958.html) I didn't know at that time if we could trust the diagram about the C being in the circle or not. You said we should, and you were right. I asked this question to Ian Stewart and he gave me the explanation about the "trust" of the diagrams in GMAT:

"In general, you should not trust the scale of GMAT diagrams, either in Problem Solving or Data Sufficiency. It used to be true that Problem Solving diagrams were drawn to scale unless mentioned otherwise, but I've seen recent questions where that is clearly not the case. So I'd only trust a diagram I'd drawn myself. ...

Here I'm referring only to the scale of diagrams; the relative lengths of line segments in a triangle, for example. ... You can accept the relative ordering of points and their relative locations as given (if the vertices of a pentagon are labeled ABCDE clockwise around the shape, then you can take it as given that AB, BC, CD, DE and EA are the edges of the pentagon; if a line is labeled with four points in A, B, C, D in sequence, you can take it as given that AC is longer than both AB and BC; if a point C is drawn inside a circle, unless the question tells you otherwise, you can assume that C is actually within the circle; if what appears to be a straight line is labeled with three points A, B, C, you can assume the line is actually straight, and that B is a point on the line -- the GMAT would never include as a trick the possibility that ABC actually form a 179 degree angle that is imperceptible to the eye, to give a few examples).

So don't trust the lengths of lines, but do trust the sequence of points on a line, or the location of points within or outside figures in a drawing. "

Hope it helps.

I am still confused. If we can not assume that S is right to R, How can we assume that T is right to S and R? Can't they all be in the same poit, 0?

NOTE: In GMAT we can often see such statement: \(k\) is halfway between \(m\) and \(n\). Remember this statement can ALWAYS be expressed as: \(\frac{m+n}{2}=k\).

Also in GMAT we can often see another statement: The distance between \(p\) and \(m\) is the same as the distance between \(p\) and \(n\). Remember this statement can ALWAYS be expressed as: \(|p-m|=|p-n|\).

Back to original question:

Is 0 halfway between r and s? OR is \(\frac{r+s}{2}=0\)? --> Basically the question asks is \(r+s=0\)?

(1) \(s>0\), clearly not sufficient.

(2) The distance between \(t\) and \(r\) is the same as the distance between \(t\) and -\(s\): \(|t-r|=|t+s|\).

\(t-r\) is always positive as \(r\) is to the left of the \(t\), hence \(|t-r|=t-r\);

BUT \(t+s\) can be positive (when \(t>-s\), meaning \(t\) is to the right of -\(s\)) or negative (when \(t<-s\), meaning \(t\) is to the left of -\(s\), note that even in this case \(s\) would be to the left of \(t\) and relative position of the points shown on the diagram still will be the same). So we get either \(|t+s|=t+s\) OR \(|t+s|=-t-s\).

In another words: \(t+s\) is the sum of two numbers from which one \(t\), is greater than \(s\). Their sum clearly can be positive as well as negative. Knowing that one is greater than another doesn't help to determine the sign of their sum.

Hence: \(t-r=t+s\) --> \(-r=s\); OR \(t-r=-t-s\) --> \(2t=r-s\).

So the only thing we can determine from (2) is: \(t-r=|t+s|\) Not sufficient.

(1)+(2) \(s>0\) and \(t-r=|t+s|\). \(s>0\) --> \(t>0\) (as \(t\) is to the right of \(s\)) hence \(t+s>0\). Hence \(|t+s|=t+s\). --> \(t-r=t+s\) --> \(-r=s\). Sufficient.

Answer: C.

yangsta8 wrote:

Statement 2) This tells us that -S=R but it doesn't tell us anything to either S or R in relation to 0.

This is not correct. If we were able to determine that \(-s=r\), statement (2) would be sufficient. But from (2) we can only say that \(t-r=|t+s|\).

Economist wrote:

This is confusing.. Okay, let me put it this way: for number lines, if we have such points...do we trust the sign of the points? and their relative positioning ? Experts please comment.

eg. here, do we assume that s cannot be 0, as -s and s are supposed to be distinct +ve and -ve values.

also, do we trust the relative positioning ( not distance ) r-s-t as shown in figure?

As for \(s\) to be zero: from statement (1) we can say that \(s\) can not be zero as it states that \(s>0\).

For (2) we don't know whether -s=s=0 or not. If \(-s=s=0\), \(s\) and therefore -\(s\) are to the left of \(t\) and (2) would be sufficient in this case. But we don't know that.

About the relative position of the points on diagram. Do you remember the question about the two circles and point C? (ds-area-between-circles-85958.html) I didn't know at that time if we could trust the diagram about the C being in the circle or not. You said we should, and you were right. I asked this question to Ian Stewart and he gave me the explanation about the "trust" of the diagrams in GMAT:

"In general, you should not trust the scale of GMAT diagrams, either in Problem Solving or Data Sufficiency. It used to be true that Problem Solving diagrams were drawn to scale unless mentioned otherwise, but I've seen recent questions where that is clearly not the case. So I'd only trust a diagram I'd drawn myself. ...

Here I'm referring only to the scale of diagrams; the relative lengths of line segments in a triangle, for example. ... You can accept the relative ordering of points and their relative locations as given (if the vertices of a pentagon are labeled ABCDE clockwise around the shape, then you can take it as given that AB, BC, CD, DE and EA are the edges of the pentagon; if a line is labeled with four points in A, B, C, D in sequence, you can take it as given that AC is longer than both AB and BC; if a point C is drawn inside a circle, unless the question tells you otherwise, you can assume that C is actually within the circle; if what appears to be a straight line is labeled with three points A, B, C, you can assume the line is actually straight, and that B is a point on the line -- the GMAT would never include as a trick the possibility that ABC actually form a 179 degree angle that is imperceptible to the eye, to give a few examples).

So don't trust the lengths of lines, but do trust the sequence of points on a line, or the location of points within or outside figures in a drawing. "

Hope it helps.

I am still confused. If we can not assume that S is right to R, How can we assume that T is right to S and R? Can't they all be in the same poit, 0?

You can trust the sequence of points. In the diagram, S is to the right of R so we can take it to be true. Similarly, T is also to the right of S and R. They cannot be the same point.
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