Data Sufficiceny

For two lines to be perpendicular they must have opposite slopes.
We are given in the question coordinates for one point of line k and one point of line m - we are unable to derive the line equations with only one point. We need the statements now ... after-wards we will examine the slopes of line equations.

st(1) intersection point of lines k and m is the point (1,-1) this can be counted as the other, second missing point for line k Only, BUT line m is still having one point -hence Not Sufficient;

st(2) line k has one more point (the second point) given here, BUT where's point for line m? Line m can go in any direction now -hence Not Sufficient;

Combined st(1&2): Still not sufficient, as we are having already three points for line k but only one points for line m

IOM E

Tagne wrote:1. In the xy-plane, line k pass through the points (1,1) and line m passes through the points (1,-1). Are the lines k and m perpendicular to each other?
a. Lines k and m intersect at the point (1,-1)
b. Line k intersects the x axis at the point (1,0)

zt< -3 means Either z<0, t>0 Or z>0, t<0. Find if z<4?

st(1) z<9, z can be -ve or +ve; Not Sufficient;
st(2) t<-4, t is always -ve Hence z>0. Now z multiplied by t should give <-3 OR (t<-4)*z <-3. If t=-90, z=1/30. If t=<-4 (assumption), z=<3/4. Z cannot be greater than 4, Sufficient.

IOM B

Tagne wrote: 2. If zt < -3, is Z<4?
a. Z<9
b. T<-4

2. If zt < -3, is Z<4?
a. Z<9
b. T<-4

zt<-3

a) z<9; lets assume 8 ; depends on t ; NS
b) T<-4; lets assume -5; again depends on Z; zt<-3 will be satisfied for various values of z. NS.

Together : Strategy: Both Algebra & number picking seems to work fine here.

zt<-3 is possible only when z & t have opp. signs. So if T<-4 means T has to be negative & thus z has to be positive. NS

Another Possible way: pick T=-5 & Z= 5 & -4 & respectively . A yes & no . So NS

E

1. In the xy-plane, line k pass through the points (1,1) and line m passes through the points (1,-1). Are the lines k and m perpendicular to each other?
a. Lines k and m intersect at the point (1,-1)
b. Line k intersects the x axis at the point (1,0)

Hi Night Reager,

K & M are perpendicular when product of their slopes are =-1. So we have to find their slopes in order to answer the question.

1) We have two points for each line. The intersection point will satisfy the Eqn of both lines. So we have two points for each line. Sufficient to find slope. S



2) Only Slope of line k can be found. NS

A

Can u pls clarify as to why the intersection point will not be considered a point for both the lines.?

st(1) intersection point of lines k and m is the point (1,-1) this can be counted as the other, second missing point for line k Only, BUT line m is still having one point -hence Not Sufficient;

we are given the same one point for line m (1,-1)...

ankur.agrawal wrote:

1. In the xy-plane, line k pass through the points (1,1) and line m passes through the points (1,-1). Are the lines k and m perpendicular to each other?
a. Lines k and m intersect at the point (1,-1)
b. Line k intersects the x axis at the point (1,0)

Hi Night Reager,

K & M are perpendicular when product of their slopes are =-1. So we have to find their slopes in order to answer the question.

1) We have two points for each line. The intersection point will satisfy the Eqn of both lines. So we have two points for each line. Sufficient to find slope. S



2) Only Slope of line k can be found. NS

A

Can u pls clarify as to why the intersection point will not be considered a point for both the lines.?

the poster has posted this question again, because no previous solution was correct

here's the link with my today's attempt https://www.beatthegmat.com/data-suffici ... tml#348942

Tagne wrote: 2. If zt < -3, is Z<4?
a. Z<9
b. T<-4