Data Sufficiency

This is a Data sufficiency question. I know I shouldn't solve those but I really hoped someone would help me solve it so I can understand how to deal with these kinds of questions

In a rectangular coordinate system shown above (I didn't attach since it is just a regular coordinate system), does the line k (not shown) intersect quadrant 2?

Statement 1: the slope of the line is -1/6

Statement 2: the y intercept of k is -6

(the answer btw is A)

Thank's a lot...Maya

maya2008 wrote:This is a Data sufficiency question. I know I shouldn't solve those but I really hoped someone would help me solve it so I can understand how to deal with these kinds of questions

In a rectangular coordinate system shown above (I didn't attach since it is just a regular coordinate system), does the line k (not shown) intersect quadrant 2?

Statement 1: the slope of the line is -1/6

Statement 2: the y intercept of k is -6

(the answer btw is A)

Thank's a lot...Maya

First thing to note: quadrant II is the top left quadrant.

So, the question is, does the line pass through the top left section of the co-ordinate system?

Let's start with the easier statement:

(2) the y-int is -6

This tells us that the line passes through (0,-6). We could draw a horizontal line through that point which never touches quadrant 2. We can also draw a diagonal line that does go through quadrant 2. Since the answer is "maybe", (2) is insufficient.

(1) The slope of the line is -(1/6)

You could use trial and error to see that a line with this slope will, eventually, go through sector 2.

Alternatively, you could know this rule:

If a line has a positive slope, it definitely goes through sectors 1 and 3.

If a line has a negative slope, it definitely goes through sectors 2 and 4.

As an aside:

If a line is parallel to the x or y axis, it will go through exactly 2 different sectors.

(Except, of course, for the lines x=0 and y=0, which are the axes.)

If a line is not parallel to the x or y axis it will:

go through exactly 2 sectors if it passes through the origin; or

go through exactly 3 sectors if it does not pass through the origin.

Hey, thanx a million...you really helped me!!!

I also stumbled at the same question. Sort of familiar with Stuart's explanation but the following thought/example got me confused:

what if we think of following scenario:

A: (-3,-1) and B: (4,-6)
Slope = m = -3/5

As we can see, slope is negative but the line k does not necessarily passes through quadtant II ? -

How should we deal with this? - Am I still missing some concept here?

Here u should know what meaning of "slope"
if slope is positive, line goes from left-bottom corner to right-top corner
if negative, line goes from right-bottom corner to left-top corner

You have negative slope, so line will obviously pass though quadrant II.
Answer is A


More serious question is will it pass though the quadrant II and quadrant III? Here statement (2) will help-but here it is not needed. But the way, it will pass through only II, III and IV quadrants if we take (2)

No, I agree with -ve and +ve slope and the direction of line it will go to but I wasnt so sure whether the line will extended through Quadrant II in all cases -- as in the example I provided above -- the line doesnt go through quad II.

Since in DS questions we usually think of examples that goes against given statement and if there arent any examples only then we select that statement. That is why I was thinking why cant we consider my given example above to disprove statement I , but later I understood that we cant use that example since in the given question it says a line k with no given co-ordinates, so therefore, a line with no specific co-ordinates can essentially be extended to infinite and in that case the question asks will it 'ever' intersent quadrant II ?? - and for that statement I provides explanation that yes, it can be extended to quadrant II.

California4jx wrote:I also stumbled at the same question. Sort of familiar with Stuart's explanation but the following thought/example got me confused:

what if we think of following scenario:

A: (-3,-1) and B: (4,-6)
Slope = m = -3/5

As we can see, slope is negative but the line k does not necessarily passes through quadtant II ? -

How should we deal with this? - Am I still missing some concept here?

The line you described will pass through quadrant II.

The line SEGMENT connecting your two points might not, but that's not the same thing.

In this question, I am being mis-led by the term "intersect"... the explanation provided here use "pass through" and not intersect... are they same?[/quote]

What about this line ? This has a negative slope and passes only through Sector 4 and not through Sector 2? What am i missing here ?

batbond007 wrote:


What about this line ? This has a negative slope and passes only through Sector 4 and not through Sector 2? What am i missing here ?

You've mislabeled the sectors: (+,+) is sector I.

Oh Thanks Stuart. It makes sense now..... Forgot the regular sectors

Stuart Kovinsky wrote:

maya2008 wrote:This is a Data sufficiency question. I know I shouldn't solve those but I really hoped someone would help me solve it so I can understand how to deal with these kinds of questions

In a rectangular coordinate system shown above (I didn't attach since it is just a regular coordinate system), does the line k (not shown) intersect quadrant 2?

Statement 1: the slope of the line is -1/6

Statement 2: the y intercept of k is -6

(the answer btw is A)

Thank's a lot...Maya

First thing to note: quadrant II is the top left quadrant.

So, the question is, does the line pass through the top left section of the co-ordinate system?

Let's start with the easier statement:

(2) the y-int is -6

This tells us that the line passes through (0,-6). We could draw a horizontal line through that point which never touches quadrant 2. We can also draw a diagonal line that does go through quadrant 2. Since the answer is "maybe", (2) is insufficient.

(1) The slope of the line is -(1/6)

You could use trial and error to see that a line with this slope will, eventually, go through sector 2.

Alternatively, you could know this rule:

If a line has a positive slope, it definitely goes through sectors 1 and 3.

If a line has a negative slope, it definitely goes through sectors 2 and 4.

As an aside:

If a line is parallel to the x or y axis, it will go through exactly 2 different sectors.

(Except, of course, for the lines x=0 and y=0, which are the axes.)

If a line is not parallel to the x or y axis it will:

go through exactly 2 sectors if it passes through the origin; or

go through exactly 3 sectors if it does not pass through the origin.

Great explaination stuart.

You make thing so simple and easy to follow..

For this question, is it okay to approach it like this?

Since slope = y intercept/x intercept, it means line k passes through (-1,6), which belongs to 2nd Quadrant - Sufficient
Statement 2 doesn't give x- intercept's value, hence NOT sufficient.

We don't need to know the rules -

If a line has a positive slope, it definitely goes through sectors 1 and 3.

If a line has a negative slope, it definitely goes through sectors 2 and 4.


Right? Can someone please clarify?

Stuart Kovinsky wrote:

maya2008 wrote:This is a Data sufficiency question. I know I shouldn't solve those but I really hoped someone would help me solve it so I can understand how to deal with these kinds of questions

In a rectangular coordinate system shown above (I didn't attach since it is just a regular coordinate system), does the line k (not shown) intersect quadrant 2?

Statement 1: the slope of the line is -1/6

Statement 2: the y intercept of k is -6

(the answer btw is A)

Thank's a lot...Maya

First thing to note: quadrant II is the top left quadrant.

So, the question is, does the line pass through the top left section of the co-ordinate system?

Let's start with the easier statement:

(2) the y-int is -6

This tells us that the line passes through (0,-6). We could draw a horizontal line through that point which never touches quadrant 2. We can also draw a diagonal line that does go through quadrant 2. Since the answer is "maybe", (2) is insufficient.

(1) The slope of the line is -(1/6)

You could use trial and error to see that a line with this slope will, eventually, go through sector 2.

Alternatively, you could know this rule:

If a line has a positive slope, it definitely goes through sectors 1 and 3.

If a line has a negative slope, it definitely goes through sectors 2 and 4.

As an aside:

If a line is parallel to the x or y axis, it will go through exactly 2 different sectors.

(Except, of course, for the lines x=0 and y=0, which are the axes.)

If a line is not parallel to the x or y axis it will:

go through exactly 2 sectors if it passes through the origin; or

go through exactly 3 sectors if it does not pass through the origin.

crimson2283 wrote:For this question, is it okay to approach it like this?

Since slope = y intercept/x intercept, it means line k passes through (-1,6), which belongs to 2nd Quadrant - Sufficient
Statement 2 doesn't give x- intercept's value, hence NOT sufficient.

Just because the slope is in the ratio of the y-int/x-int doesn't mean that the slope passes through those points.

For example, a line with a slope of 1 (i.e. 1/1) could pass through (0,5) and (5,0) instead of (0,1) and (1,0).