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diebeatsthegmat: Legendary Member; Posts: 1119; Joined: Fri May 07, 2010 8:50 amFollowed by: 3 members; Thanked: 29 times

by diebeatsthegmat » Fri Apr 15, 2011 11:30 pm

In the xy-plane, if line k has negative slope, is the y-intercept of line k positive?

(1) The x-intercept of line k is less than the y-intercept of line k.

(2) The slope of line k is less than -2.

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Source: Beat The GMAT — Data Sufficiency | Fri Apr 15, 2011 11:30 pm

clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

deleted

by clock60 » Sat Apr 16, 2011 12:09 am

deleted

Last edited by clock60 on Sat Apr 16, 2011 11:44 am, edited 1 time in total.

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srcc25anu: Master | Next Rank: 500 Posts; Posts: 423; Joined: Fri Jun 11, 2010 7:59 amFollowed by: 2 members; Location: Seattle, WA; Thanked: 86 times

by srcc25anu » Sat Apr 16, 2011 3:01 am

stat 1: not sufficient because line with -ve slope and x-intercepy < y-intercept can pass through (1,2) or (-1,-2)

stat 2: slope of k < -2 still not sufficient

together also, not sufficient
IMO it should be E

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force5: Legendary Member; Posts: 582; Joined: Tue Mar 08, 2011 12:48 amFollowed by: 6 members; Thanked: 61 times; GMAT Score:740

by force5 » Sat Apr 16, 2011 3:37 am

Good question can be done in many ways. i find this the best.....

Stmnt1 - x intercept is < y intercept. ( not since slope is negative you can make lines passing positive y axis and negative y axis.

stmnt 2 - slope of line k is < -2 (this statement alone is insufficient)

combining.

First i will do the usual way.

y= mx+c ( m is given less than -2)

y intercept is c
x intercept is c/m

its given that from stmnt 1 that c/m < c
or c/m-c<0
c(1-m/m)<0
now m is -ve (if you want take a value say m= -3)

-c4/3<0 hence c > 0

hence C

Besides this - if we analyse

slope is Y/X
since slope is -ve either of y and x are negative.
since stmnt 1 says x intercept is smaller that means y has to be positive.

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clock60: Legendary Member; Posts: 759; Joined: Mon Apr 26, 2010 10:15 amFollowed by: 3 members; Thanked: 85 times

by clock60 » Sat Apr 16, 2011 11:42 am

hi guys again in this post, i changed my mind, and the answer here is C
k=mx+b, if m<0 is b>0
(1) if y=0 then x intercept, mx+b=0,mx=-b, x=-b/m. and it is less then b
-b/m<b.
-b/m-b<0. b/m+b>0, b(1/m+1)>0
it is possible in two cases
b>0 and (1/m+1)>0 (both are +ve)
1/m+1>0 if m<-1., so if we prove that m<-1, then b>0
the second case
b<0 and 1/m+1<0 (both are -ve)
1/m+1<0 if -1<m<0 so if we find that m>-1 then b<0
but we are not given any restrictions on m
and it comes that 1 st insuff
(2)m<-2, insuff but provides important info
together
if m<-2 then as we proved in the 1 st (if m<-1) then b>0
so both suff
the answer is C

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MAAJ: Master | Next Rank: 500 Posts; Posts: 243; Joined: Sun Jul 12, 2009 7:12 amFollowed by: 2 members; Location: Dominican Republic; Thanked: 31 times; GMAT Score:480

by MAAJ » Mon Apr 18, 2011 8:36 am

IMO [spoiler](C)[/spoiler], what's the OA?

1) The x-intercept of line k is less than the y-intercept of line k.

0 = mx+b
-b = mx
-b/m = x Hence -b/m < b

-b/m - b < 0
(-b-mb)/m < 0
b(-1-m)/m < 0
Insufficient

(2) The slope of line k is less than -2
Insufficient

(3) Combined:
If b(-1-m)/m < 0 and m < -2 Then (-1-m) is always positive, so:
b(positive)/(negative) < 0
Thus, b must be positive

"There's a difference between interest and commitment. When you're interested in doing something, you do it only when circumstance permit. When you're committed to something, you accept no excuses, only results."

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