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Thinking too much into it

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Thinking too much into it

by vbcannon » Wed Dec 03, 2008 10:14 pm
So maybe I am thinking too much into this but I started thinking about this question and trying to prove each one insufficient and that is why I missed it.

If 4r + 2s =12, what is the value of s?

(1) r + s =5
(2) 2r + s =6

I looked at the first one and said immediately, "Yes it is sufficient because r + s can be (1,4) or (2,3) and it would equal to 12 so it is sufficient"

Then I started remember how these questions try to trick you so I started to think "Wait a minute r + s=5 could also mean r is -3 and s is 8 and then it wouldn't =12 in the stem.

Why would this not be a proper way to assume? Is it because I should take the stem to be TRUE and then figure out a way to find only the right number for r + s =5? I guess I am just confused as to why I couldn't assume that r and s are could have a negative integer and a positive one that will throw off it being completely sufficient.
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by vittalgmat » Wed Dec 03, 2008 10:36 pm
I approached this way.
Pls let me know if I have messed something up.

4r +2s = 12.
simplifying
2r +s = 6

stmt 1:
r +s = 5

the original eqn can be written as
r +r +s = 5
r + (r+s) = 5
r+ 6 = 5
r = -1
s = 6.
sufficient.

stmt 2
2r +s = 6.
This is same as question.
r & s cannot be found.

Insufficient.

So A

have I missed something ??
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by cramya » Wed Dec 03, 2008 10:40 pm
Vittal,
I think u did it right!

The key is to always simplify the question stem which u did(escpecially if its a linear equation in 2 variables) The common trick is to give the simplified version of the equation as one of the statements

For example if statement 2 read

r+s/2 = 3 (its the same as the question stem ; all of us just need to be extra careful of the disguised forms) .

Regards,
Cramya
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by vbcannon » Wed Dec 03, 2008 10:44 pm
vittalgmat wrote:I approached this way.
Pls let me know if I have messed something up.

4r +2s = 12.
simplifying
2r +s = 6

stmt 1:
r +s = 5

the original eqn can be written as
r +r +s = 5
r + (r+s) = 5
r+ 6 = 5
r = -1
s = 6.
sufficient.

stmt 2
2r +s = 6.
This is same as question.
r & s cannot be found.

Insufficient.

So A

have I missed something ??

the original eqn can be written as
r +r +s = 5
r + (r+s) = 5
r+ 6 = 5
r = -1
s = 6.

How can the original equation be rewritten this way from 4r + 2s = 12?

Please excuse my ignorance but I am just trying to make sure I grasp it.

It seems so easy and I am making it difficult. I like the way you simplified the second part of the problem to make it the exact same as the main stem. That is what it said to do in the book. How did you know to do that? My eye didn't even catch it!
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by vbcannon » Wed Dec 03, 2008 10:46 pm
cramya wrote:Vittal,
I think u did it right!

The key is to always simplify the question stem which u did(escpecially if its a linear equation in 2 variables) The common trick is to give the simplified version of the equation as one of the statements

For example if statement 2 read

r+s/2 = 3 (its the same as the question stem ; all of us just need to be extra careful of the disguised forms) .

Regards,
Cramya
Oh ok so it is a common thing for the GMAT makers to do that? Is that why vittal or other people would know to look for that?

The cool thing is, I know to look for it now!!
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by cramya » Wed Dec 03, 2008 10:48 pm
Other way to look at

Question stem:

2r+s=6

Statement I says r+s=5

2 different equations wiht 2 variables . Can be solved for s.

SUFFICIENT
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by cramya » Wed Dec 03, 2008 10:50 pm
Oh ok so it is a common thing for the GMAT makers to do that? Is that why vittal or other people would know to look for that?

The cool thing is, I know to look for it now!!
I am sure u will get to know the common traps as u dive deep in to your prep. This forum is excellent in that way..(U can brwse through problems/solutions posted and u will come across several tips/tricks this being one)

Good luck!
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by vbcannon » Wed Dec 03, 2008 10:50 pm
cramya wrote:Other way to look at

Question stem:

2r+s=6

Statement I says r+s=5

2 different equations wiht 2 variables . Can be solved for s.

SUFFICIENT
OHHH

so the 2r+s=6 is where r+r+s=6?
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by cramya » Wed Dec 03, 2008 10:54 pm
OHHH

so the 2r+s=6 is where r+r+s=6?
I think Vittal had a typo where he said r=-1.Probably he meant to write r=1

For this problem

r will be 1 s will be 4.

2r+s can be written as r+r+s (r+r=2r)
so r+r+s = 6

r+5 = 6 (since its given in stmt I r+s=5)
r=1

r+s=5
1+s=5
s=4

Hope this helps!

Regards,
Cramya
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by vbcannon » Wed Dec 03, 2008 11:00 pm
It does help, but what I was doing was ASSUMING that r+s=5 could have an r value of -4 per se and an s value of 9. In this case the 4r + 2s=12 would not work. That is why I deemed it insufficient. Is that just not the way to think? and why?
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by cramya » Wed Dec 03, 2008 11:04 pm
I think in this problem we are trying to get to a unique vaue for s which we can from the question stem and stmt I
but what I was doing was ASSUMING that r+s=5 could have an r value of -4 per se and an s value of 9. In this case the 4r + 2s=12 would not work.
Yes with r=-4 and s=9 4r + 2s=12 will not work. This tells us that these are not the valeus for r and s. We need to find for what values these will work and hence we use the 2 different equations to get to those values.
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by vbcannon » Wed Dec 03, 2008 11:06 pm
cramya wrote:I think in this problem we are trying to get to a unique vaue for s which we can from the question stem and stmt I
but what I was doing was ASSUMING that r+s=5 could have an r value of -4 per se and an s value of 9. In this case the 4r + 2s=12 would not work.
Yes with r=-4 and s=9 4r + 2s=12 will not work. This tells us that these are not the valeus for r and s. We need to find for what values these will work and hence we use the 2 different equations to get to those values.
Ok so basically instead of me trying to find the unique value, I started trying to find all the ways it could be wrong instead of focusing on seeing if r + s=5 could help me solve the stem.

I see.

Thank you very much!
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by vittalgmat » Thu Dec 04, 2008 12:51 am
Oops, .. I was away from my PC for a while..
That is a delicious dialogue betwn vb and cramya.. nice..

vbcannon,
to answer your question on how did I figure out the simplification, 2 simple
answers: 1) Practise
2) actively reading on this forum. We have some brilliant minds here who have mastered the tricks of the trade...

I keep on the look out and every new trick/trap I encounter, I note in my cheatsheet. (rather I need to call it cheatbook).

Dont worry.. as ur prep progresses u will get better. As a matter of fact, I am seeing some glimmer of hope only in the past 2 weeks...

ht helps
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by vittalgmat » Thu Dec 04, 2008 12:57 am
vbcannon,
I did the following to pick up some of the tricks/traps/juicy discussions.
Paste the following string into the "advanced search" on the top of the BTG page.
ian stewart, stuart kovinsky, stacey koprince, cramya, logitech, parallel_chase

BTW there are others whom I cant just remember.
Browse thru the search output and understand the discussion.
This has helped immensely esp. DS number properties.

HT Helps
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by ivanoski » Thu Dec 04, 2008 2:06 pm
sooo.. whats the officil answer? i get B , statement 2 only

because whith the first statement: r=5-s and s= -4/3

And statement 2: r= (6-s) / 2 and s=0


plugging in each one to the initial question: 4r+2s=12

with statement 1:

4(5-(-4/3)) + s(-4/3) = 22.6667 /= 12

With statement 2:
4((6-0)/2)+2(0) = 12

therefore s MUST be = 0 in order to comply with the initial question , the value of "s" AND (whatever is here) = 12

therefore i think the answere is statement 2 alone, (b) am i right?
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