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

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12 Oct 2009, 01:20

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Statement 1) Only tells us S is positive but nothing about its distance and nothing about R. Statement 2) This tells us that -S=R but it doesn't tell us anything to either S or R in relation to 0.

Statement 1+2) This tells us S is positive. Hence -S is negative. Since -S=R then the distance between S and 0 is the same as -S and 0 and hence R and 0.

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

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12 Oct 2009, 06:26

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1/ if 0 is to the left of r --> wrong 2/ there are 2 cases case 1: if -s to the right of t then 0 to the right of s,t -> wrong case 2: if -s to the left of t then 0 is between r and s -> right Both 1/ and 2/ then we can eliminate case 1, hence C

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

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08 Nov 2009, 06:15

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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..

ngoctraiden1905 wrote:

1/ if 0 is to the left of r --> wrong 2/ there are 2 cases case 1: if -s to the right of t then 0 to the right of s,t -> wrong case 2: if -s to the left of t then 0 is between r and s -> right Both 1/ and 2/ then we can eliminate case 1, hence C

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

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08 Nov 2009, 23:17

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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.

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

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09 Nov 2009, 01:41

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?

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. "

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

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09 Nov 2009, 06:35

Awesome !!! Three cheers to Bunuel , +1K

One question: For t<-s, meaning t is to left of -s, we have a situation where -s is to the right of s !! That looks strange...are we only talking about magnitude of s ??

One question: For t<-s, meaning t is to left of -s, we have a situation where -s is to the right of s !! That looks strange...are we only talking about magnitude of s ??

There can be 4 cases for t, s, and -s, remember: 1. t and s are fixed, t is to the right of s; 2. Obviously s and -s are always different sides of 0 and |s|=|-s| meaning that they are obviously equidistant from 0.

A. --(-s)---0---s----t--- Means s is positive, t is positive and t+s>0

B. ----s---0---(-s)--t--- Means s is negative, t is positive and t+s>0

C. ---s-----0--t--(-s)--- Means s is negative, t is positive and t+s<0

D. ---s--t--0-----(-s)--- Means s is negative, t is negative and t+s<0

You can see that in every case (C, D) when t is to the left of -s, t+s<0. The cases when -s is to the right of s, just means that s is negative, therefore -s is positive.

Please refer to the discussion above for complete solution.

A mistake you make is that you can not cancel out \(t\) from the absolute values in LHS and RHS as you did: |t-r| = |t-(-s)| |-s| = |r| |-s| = |s| This is not correct.

(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 \(t\) (given on the diagram), 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.

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

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09 Aug 2010, 00:10

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Try to solve this problem in the following approach. We have t>r>s (graph) The question asks is r+s=0 or is r=-s

Statement 1 It is obvious that statement 1 alone is not sufficient since it only provides information regarding point S. s>0, no other information is given thus not sufficient.

Statement 2 t-r = t - (-s) thus -r=s and r=-s hence sufficient.

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

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12 Sep 2010, 06:49

Hi Bunuel,

I am still confused here: "BUT \(t+s\) can be 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). "

If we take relative positioning given in diagram, \(t\) should be right to \(s\). For example, \(s=2 & t=4 or s=-4 & t= -2\). Is my logic right here? If that so, then if we take \(t<-s\), and consider \(s=5 & t=-6 or s=-2 & t=-5\) so that \(t+s\) negative.

Attachment:

illustration.jpg [ 5.75 KiB | Viewed 71027 times ]

Then, in such scenario..doesn't it violate relative positioning given \(s, t\) in question diagram.

I am still confused here: "BUT \(t+s\) can be 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). "

If we take relative positioning given in diagram, \(t\) should be right to \(s\). For example, \(s=2 & t=4 or s=-4 & t= -2\). Is my logic right here? If that so, then if we take \(t<-s\), and consider \(s=5 & t=-6 or s=-2 & t=-5\) so that \(t+s\) negative.

Attachment:

illustration.jpg

Then, in such scenario..doesn't it violate relative positioning given \(s, t\) in question diagram.

Kindly, help!

I'm not sure I understand your question.

Guess you are referring to the statement (2). One of the scenarios is \(t+s<0\) (for example t=-2>s=-4 --> t+s=-2-6=-8<0)

Your examples s=5 & t=-6 (t<s) or s=-2 & t=-5 (t<s) are not correct as relative position of the points implies that t>s so we can not consider them.
_________________

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

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12 Sep 2010, 19:49

Bunuel,

Thanks for taking a look.

What I meant is that..While taking on statement 2:

You have considered \(t-r\) to be positive based on there positioning. "\(t-r\) is always positive as \(r\) is to the left of the \(t\) (given on the diagram), hence \(|t-r|=t-r\)" So, you are believing in the relative position of point \(t, r\) shown in diagram to conclude something. Right?

Now, you have also considered a scenario where \(t+s\) is negative (when \(t<-s\)). In such case, point \(t\) will be the left to \(s\). While in the question, it is shown that point \(t\) is towards right of \(s\). So, it all boils down to my doubt that we should neglect condition \(t<-s\) OR \(|t+s|=-t-s\) as it is not in accordance with the relative positioning of points \(t ,s\) in the question diagram.

Bunuel wrote:

Guess you are referring to the statement (2). One of the scenarios is \(t+s<0\) (for example t=-2>s=-4 --> t+s=-2-6=-8<0)

Your examples s=5 & t=-6 (t<s) or s=-2 & t=-5 (t<s) are not correct as relative position of the points implies that t>s so we can not consider them.

I must confess that this is one of the trickiest DS question, I have come across!
_________________

What I meant is that..While taking on statement 2:

You have considered \(t-r\) to be positive based on there positioning. "\(t-r\) is always positive as \(r\) is to the left of the \(t\) (given on the diagram), hence \(|t-r|=t-r\)" So, you are believing in the relative position of point \(t, r\) shown in diagram to conclude something. Right?

Now, you have also considered a scenario where \(t+s\) is negative (when \(t<-s\)). In such case, point \(t\) will be the left to \(s\). While in the question, it is shown that point \(t\) is towards right of \(s\). So, it all boils down to my doubt that we should neglect condition \(t<-s\) OR \(|t+s|=-t-s\) as it is not in accordance with the relative positioning of points \(t ,s\) in the question diagram.

Bunuel wrote:

Guess you are referring to the statement (2). One of the scenarios is \(t+s<0\) (for example t=-2>s=-4 --> t+s=-2-6=-8<0)

Your examples s=5 & t=-6 (t<s) or s=-2 & t=-5 (t<s) are not correct as relative position of the points implies that t>s so we can not consider them.

I must confess that this is one of the trickiest DS question, I have come across!

Scenario \(t<-s\) means that \(t\) is to the left of \({-s}\) (minus \(s\), not \(s\)), note that even in this case \(s\) could be to the left of \(t\) and relative position of the points shown on the diagram still will be the same.

For example: \(s=-4\), \(t=2\), and \(-s=4\) --> \(s<t<-s\) --> \(--(s)--(t)--(-s)--\).

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

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12 Sep 2010, 20:31

Bunuel,

Thanks a lot. It's all clear now. Manytimes, I get wrong in scenarios when I have to consider a number of conditions dealing with -ve values or with inequalites and absolute values. Any tips for this
_________________

Thanks a lot. It's all clear now. Manytimes, I get wrong in scenarios when I have to consider a number of conditions dealing with -ve values or with inequalites and absolute values. Any tips for this

Q)) On the number line shown, is zero halfway between r and s? ----r---- s---- t--- 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.

1)

Case I: -----r--0--s----t--- 0 is midway between r & s.

Case II: --0--r----s----t--- 0 is not midway between r & s.

Not Sufficient.

2) Case I: Let's say r=-s; r=-2; s=2 t =3 -----r--0--s----t--- |t-r| = |3-(-2)|=5 |t-s| = |3-(-2)|=5 0 is midway between r and s.

Case II: Let's say r=-s; r=-4; s=-2 t =-1; -s=2 -----r--s--t--0----(-s) |t-r| = |-1-(-4)|=3 |t-s| = |-1-(2)|=3 0 is not midway between r and s. Not Sufficient.

Combining both;

r=-2; s=2 t =3 -----r--0--s----t--- |t-r| = |3-(-2)|=5 |t-s| = |3-(-2)|=5 0 is midway between r and s.

@ fluke....got it perfectly.......!!! regards .....ill be posting some other queries too ......and i find ur solutions very helpful and self explanatory thnx so much