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Detailed solution will be appreciated. Thanks

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VP
Joined: 18 May 2008
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Detailed solution will be appreciated. Thanks [#permalink]

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10 Jun 2008, 04:39
1
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Detailed solution will be appreciated.
Thanks

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Manager
Joined: 24 Apr 2008
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10 Jun 2008, 05:46
2
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Let the equations be:
y=mx+c and y=ax+b

St1: we get 2 eqs - 5m+c=1 and 5a+b=1 or 5m+c=5a+b nothing can be said about values of m and a from this relation hence insuffic.

st2: c>b doent provide any relation btw m and a.

now combining them since 5m+c=5a+b and c>b then mhas to be less than a thus sufficient.

Intern
Joined: 10 Jun 2008
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10 Jun 2008, 06:12
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Statement 1 tells us the lines do not have the same slope *EDIT* (this is true only given statement 2)
Statement 2 tells us that N is "above" P at somewhere at X=0

There are three possibilities:
Both lines are pointing up (positive slope) => SLOPE(N) > SLOPE(P)
Both lines are pointing down (negative slope) => SLOPE(N) < SLOPE(P)
One line is pointing up, one line is pointing down
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Joined: 07 Nov 2007
Posts: 1760
Location: New York

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04 Aug 2008, 21:24
1
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iamcartic wrote:
Let the equations be:
y=mx+c and y=ax+b

St1: we get 2 eqs - 5m+c=1 and 5a+b=1 or 5m+c=5a+b nothing can be said about values of m and a from this relation hence insuffic.

st2: c>b doent provide any relation btw m and a.

now combining them since 5m+c=5a+b and c>b then mhas to be less than a thus sufficient.

Good explanation.

Agreed.
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Senior Manager
Joined: 01 Jan 2008
Posts: 499

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04 Aug 2008, 21:33
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Excellent explanation "iamcartic"
Director
Joined: 27 May 2008
Posts: 539

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05 Aug 2008, 08:50
1
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line n crosses y axis at (0,a)
line p crosses y axis at (0,b)

as per statement 1 both lines pass through (5,1)

slope of line n = (1-a)/5 ------ using slope = (y2 - y1) / (x2 - x1)
slope of line p = (1-b)/5

we have to determine if 1-a < 1-b

as per statement 2 : a > b
multiply by -1 on both sides, the inequality sign will change
-a < -b
1-a < 1-b

Option C
Current Student
Joined: 11 May 2008
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05 Aug 2008, 18:11
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durgesh... is this an ok way of analysing??? or its not advisable?

its like this .. keep one line(LINE n) fixed at 5,1 and a certain y intercept. then take the other line(line P) and fix one point at 5,1 and y intercept < 1st y intercept.
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Last edited by arjtryarjtry on 05 Aug 2008, 18:45, edited 1 time in total.
VP
Joined: 17 Jun 2008
Posts: 1322

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05 Aug 2008, 18:31
ritula wrote:
Detailed solution will be appreciated.
Thanks

take line n : y=nx+c
line p : y=px+c1

hence now (1) says (5,1) satisfies above equations =>
1=5n+c
1=5p+c1
5(n-p)=c1-c
not solvable since c>c1 or c<c1
INSUFFI

(2) c>c1 does not help since both are indepedent eqns
(n-p)x=c1-c => x is a variable
INSUFFI

COMBINING (1) and (2):

5(n-p)=c1-c and c>c1 => n<p=> SUFFI
IMO C
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Director
Joined: 27 May 2008
Posts: 539

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05 Aug 2008, 21:32
arjtryarjtry wrote:
durgesh... is this an ok way of analysing??? or its not advisable?

its like this .. keep one line(LINE n) fixed at 5,1 and a certain y intercept. then take the other line(line P) and fix one point at 5,1 and y intercept < 1st y intercept.

i think its ok.. make sure you use both conditions on xy plane only after eliminating that each statement alone is not suff...

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Re: GMATPREP- lines   [#permalink] 05 Aug 2008, 21:32
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