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gmatwork
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Re: distance on number line DS

by gmatwork Thu Oct 20, 2011 8:28 pm

I don't agree with RA.......

R and S cannot take the same value 0 since they are placed on different locations on number line. R, S and T need to take different values...none of these can be equal, since they have different positions on the number line.
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Re: distance on number line DS

by tim Wed Nov 23, 2011 1:27 pm

um, RA posted over three years ago and there has been a LOT of discussion since then. did you have a question about some of the intervening discussion?
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Re: distance on number line DS

by jp.jprasanna Wed Aug 15, 2012 10:31 am

tim Wrote: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..


Hi - Even I get confused with number substitution with number line problems such as thiss one, hence wanted to confirm the algebra sol for this problem.

statement 1) s is to the right of zero. - Not sufficient

statement 2) the distance between t and r is the same as the distance between t and -s.

Can be translated to |t-r| = | t - (-s)|
so 2 sol possible

1/ t-r = t+s -> -r=s (0 is in the middle of r and s )
2/ t-r = -t-s -> 2t = r-s (there can be many values for r and s for which 2t = r-s and for which 0 not in the middle of r and s)


Combined - Stat1 and Stat2

Says s is +ve which makes t +ve and bigger than S (because of ---r---s----t---) and also makes t bigger than r

hence forces 1/ t-r = t+s -> -r=s .

Is this reasoning correct?

Cheers
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Re: distance on number line DS

by tim Wed Aug 22, 2012 8:54 am

yes JP, that works!
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Re: distance on number line DS

by bluedot Sun Mar 17, 2013 10:56 am

Another way to prove that statement 2 is not sufficient is to think of a possibility where t=0.
In this case, -s ans s will be equidistant from t.

Am I correct?
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Re: distance on number line DS

by RonPurewal Mon Mar 18, 2013 10:23 am

bluedot Wrote:Another way to prove that statement 2 is not sufficient is to think of a possibility where t=0.
In this case, -s ans s will be equidistant from t.

Am I correct?


sorry, i don't really follow this argument, for two reasons:

1/
i'm not sure where you're going with your last statement (-s & s are equidistant from t). that's not a stated condition or a prompt question anywhere in the problem, so i don't immediately see why that would be a thing here.
if there is a logical progression here, then it's missing a lot of steps; please fill in the essentials (at least enough so that i can follow your logic), thanks.

2/
by definition, this can't be correct anyway, since you're talking about only one case.
remember, to get "not sufficient", you ALWAYS need to get two different things to happen.
i.e., just "think[ing] of a possibility where xxxxx" is never going to prove that a statement is insufficient. instead, you'll always have to come up with at least two possibilities, aiming to get different end results from them.
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Re: distance on number line DS

by tutor_india Mon Mar 18, 2013 11:01 am

You should consider absolute value, not just t-r=t-(-s) -> -r = s. This is partial result.
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Re: distance on number line DS

by tim Wed Mar 20, 2013 12:37 am

let us know if there are any further questions on this one..
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