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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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
stat 2: slope of k < -2 still not sufficient
together also, not sufficient
IMO it should be E
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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.
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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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
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
- MAAJ
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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
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
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