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abovethehead
 
 

distance on number line DS

by abovethehead Thu Aug 14, 2008 1:57 pm

<------------------R----------S-----T------------>

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

Question, for 2), I got SUFFICIENT
t-r=t-(-s) -> -r = s
so, 0 should be equidistant between r and s right?

Can someone explain where I'm going wrong and also, explain why the answer is C

Thanks
RA
 
 

by RA Sat Aug 16, 2008 7:20 am

Choice (B) does not eliminate the possibility that R & S are zero. Combining the two statements eliminates zero as an answer and gives us a definite "yes" as an answer.

Hope that helps
guest612
 
 

can someone please explain how the answer is C?

by guest612 Mon Aug 25, 2008 7:32 pm

I got E. Clearly statement 1 is insufficient. I thought 2 would prove to be insufficient as well.
RonPurewal
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Re: distance on number line DS

by RonPurewal Sat Sep 20, 2008 12:39 am

abovethehead Wrote:Question, for 2), I got SUFFICIENT
t-r=t-(-s) -> -r = s
so, 0 should be equidistant between r and s right?


watch those assumptions.

the distance between t and (-s) must be a positive number, but the problem is that we don't know which way to subtract to get that positive number. if t > -s, then the distance is t - (-s), as you've written here. however, if -s > t, then the distance is actually (-s - t) instead.
if s is to the left of zero, then -s will be to the right of zero - which could well place -s to the right of t. if that happens, then the distance will become (-s - t), rendering your calculation inaccurate. try drawing out this possibility - put zero WAY to the right of both s and t on the number line, then find -s, and watch what happens).

if s lies to the right of zero, then -s must lie even further to the left than does s itself. since s is already to the left of t, it then follows that -s is also to the left of t. therefore, in that case, you can definitively write the distance as t - (-s), and your calculation is valid. therefore, (c).

--

ironically, the presence of statement (1) should make it easier to see that statement (2) is insufficient. specifically, statement (1) calls your attention to the fact that s could lie to the left of zero, in which case you could get the alternative outcome referenced above. that's something you might not think about if statement (1) weren't there.
victorgsiu
 
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Re: distance on number line DS

by victorgsiu Mon Oct 26, 2009 11:26 pm

Let me see if I can plug in numbers to the number line here:

Statement 1)
if the line reads: r=-1, zero, s=1, t=3, then zero is halfway between r and s.
if the line reads: zero, r=1, s=2, t=3, then zero is not between r and s.
Insufficient.

Statement 2)
by definition zero is halfway between s and -s.

a) if the line reads: -s=r=-2, zero, s=2, and t=4, then (t to r)=(t to -s)=6.
zero is halfway in between r and s.

b) if the line reads: r=-4, s=-2, t=-1, zero, and -s=2, then (t to r) = (t to -s) = 3.
zero is between t and -s.
Insufficient.

Together)
Forces the case 2a). Sufficient.

Hope that helps.



RonPurewal Wrote:
abovethehead Wrote:Question, for 2), I got SUFFICIENT
t-r=t-(-s) -> -r = s
so, 0 should be equidistant between r and s right?


watch those assumptions.

the distance between t and (-s) must be a positive number, but the problem is that we don't know which way to subtract to get that positive number. if t > -s, then the distance is t - (-s), as you've written here. however, if -s > t, then the distance is actually (-s - t) instead.
if s is to the left of zero, then -s will be to the right of zero - which could well place -s to the right of t. if that happens, then the distance will become (-s - t), rendering your calculation inaccurate. try drawing out this possibility - put zero WAY to the right of both s and t on the number line, then find -s, and watch what happens).

if s lies to the right of zero, then -s must lie even further to the left than does s itself. since s is already to the left of t, it then follows that -s is also to the left of t. therefore, in that case, you can definitively write the distance as t - (-s), and your calculation is valid. therefore, (c).
alexei600
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Re: distance on number line DS

by alexei600 Wed Feb 09, 2011 3:48 pm

Hello,
Is it correct that the absolute value of r=s, from the prom.
Can you please provide with full solution.
Thanks
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Re: distance on number line DS

by RonPurewal Fri Feb 11, 2011 6:19 am

alexei600 Wrote:Hello,
Is it correct that the absolute value of r=s, from the prom.


what is "the prom."?
please write your questions in complete sentences, without abbreviated words.

and, no, it is not automatically true that |r| = s; if that statement were automatically true, then we would have an automatic answer of "yes" to the prompt question.

Can you please provide with full solution.
Thanks


there's already quite a bit of material on this page.
what do you already understand? what do you not understand? where are you having difficulty?
alexei600
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Re: distance on number line DS

by alexei600 Fri Feb 11, 2011 4:25 pm

Hello, Ron.
For statement 2. can I put is in algebra way such as
|t-r|=|t-(-s)|, and then since absolute value would provide more than 2 solutions we need statement one , to create a single overlap.
With pluging in number and testing, its very confusing to me. When to choose pos/neg and etc..
Thanks.
tim
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Re: distance on number line DS

by tim Tue Feb 15, 2011 5:20 pm

your absolute value equation will yield only two solutions, but that is enough to know that statement 2 is insufficient..

in general, it is more efficient to use algebra than to pick numbers in DS. you also have the problem that picking numbers will never verify that the statement is sufficient unless you can try all possible numbers. so you should try to pick numbers only as a last resort..
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alexei600
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Re: distance on number line DS

by alexei600 Tue Feb 15, 2011 7:34 pm

So, my mentioned above equation correct? The problem for me is to combine two statements together; the math I do still leans towards 'E"
jnelson0612
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Re: distance on number line DS

by jnelson0612 Thu Feb 17, 2011 6:20 pm

Alexei, please go back and read Ron's explanation above. That should help you see why the answer is C rather than E.
Jamie Nelson
ManhattanGMAT Instructor
shadangi
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Re: distance on number line DS

by shadangi Wed Aug 31, 2011 12:07 am

Ron,

Could you help me understand if s=r=0 has any repercussion in problems like this? Does GMAT consider 0 as half way between s=r=0?

Technically for s=r=0, (r+s)/2=0 but I still want to make sure, as we know that sometimes GMAT world is not the same as real world :). For instance, Sqrt(x) is always +ve and can't be an imaginary number.

Secondly, can we assume that r<s<t STRICTLY based on the figure? If that's the case #2 MUST be sufficient. Reason: r < s given. As per #2, distance between t and r = distance between t and -s, for GIVEN any t > r and s. THE ONLY WAY these 2 conditions are possible is if s > 0.

What kind of assumptions can we make with respect to the figures, such as number lines, that GMAT shows in it's questions?

Thanks.
JohnHarris
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Re: distance on number line DS

by JohnHarris Wed Aug 31, 2011 6:16 pm

shadangi Wrote:....Secondly, can we assume that r<s<t STRICTLY based on the figure? If that's the case #2 MUST be sufficient. ...THE ONLY WAY these 2 conditions are possible is if s > 0....


Hi shadangi,

Your conclusion is incorrect. Look at the previous example by victorgsiu:

r=-4 < s=-2 < t=-1 < 0 < -s=2

The distance from t to -s is 3 is the same as The distance from r to t = 3


The fact that s < t says nothing about the relationship of -s and t. As the previous examples show, given s < t, it is possible for -s to be less than t or for -s to be greater than t. If -s is greater than t, then the distance from t to -s is given by -(s+t), if -s is less than t then the distance is given by t+s. Thus the ambiguity mentioned by RonPurewal previously.
shadangi
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Re: distance on number line DS

by shadangi Wed Aug 31, 2011 7:26 pm

Thanks JohnHarris for the clarification. I missed that! :)
tim
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Re: distance on number line DS

by tim Fri Sep 16, 2011 10:27 pm

Thanks John..
Tim Sanders
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