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Re: In the xy-plane, if line k has negative slope and passes [#permalink]

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21 Jun 2012, 04:46

2

This post received KUDOS

Expert's post

In the xy-plane, if line k has negative slope and passes through the point (-5,r), is the x-intercept of line k positive?

(1) The slope of line k is -5 (2) r>0

Graphic approach:

If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and Y intersects of the line with negative slope have the same sign. Therefore if X and Y intersects are positive, the line intersects quadrant I; if negative, quadrant III.

When we take both statement together all we know is that slope is negative and that it crosses some point in II quadrant (-5, r>0) (this info is redundant as we know that if the slope of the line is negative, the line WILL intersect quadrants II). Basically we just know that the slope is negative - that's all. We can not say whether x-intercept is positive or negative from this info.

Below are two graphs with positive and negative x-intercepts. Statements that the slope=-5 and that the line crosses (-5, r>0) are satisfied.

Statement I first gives x = r/5 -5
Statement II tells us that r is positive which gives us y = -mx + h when y = 0 we get mx = h so x = h/m. we need to know h.

Isn't this wrong y = mx+c and the line passes through (-5,r) -5=mr+c == > r = (-5 -c)/m

shouldn't this be r=-5m+c

Is so can you now explain the answer

Yes, it should be r=-5m+c.

In the xy-plane, if line k has negative slope and passes through the point (-5,r), is the x-intercept of line k positive?

This question can be done with graphic approach (just by drawing the lines) or with algebraic approach.

Algebraic approach:

Equation of a line in point intercept form is \(y=mx+b\), where: \(m\) is the slope of the line, \(b\) is the y-intercept of the line (the value of \(y\) for \(x=0\)), and \(x\) is the independent variable of the function \(y\).

We are told that slope of line \(k\) is negative (\(m<0\)) and it passes through the point (-5,r): \(y=mx+b\) --> \(r=-5m+b\).

Question: is x-intercept of line \(k\) positive? x-intercep is the value of \(x\) for \(y=0\) --> \(0=mx+b\) --> is \(x=-\frac{b}{m}>0\)? As we know that \(m<0\), then the question basically becomes: is \(b>0\)?.

(1) The slope of line \(k\) is -5 --> \(m=-5<0\). We've already known that slope was negative and there is no info about \(b\), hence this statement is insufficient.

(2) \(r>0\) --> \(r=-5m+b>0\) --> \(b>5m=some \ negative \ number\), as \(m<0\) we have that \(b\) is more than some negative number (\(5m\)), hence insufficient, to say whether \(b>0\).

(1)+(2) From (1) \(m=-5\) and from (2) \(r=-5m+b>0\) --> \(r=-5m+b=25+b>0\) --> \(b>-25\). Not sufficient to say whether \(b>0\).

Re: In the xy-plane, if line k has negative slope and passes [#permalink]

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22 Jun 2012, 01:55

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Expert's post

sanjoo wrote:

Bunuel wrote:

riteshgupta wrote:

Isn't this wrong y = mx+c and the line passes through (-5,r) -5=mr+c == > r = (-5 -c)/m

shouldn't this be r=-5m+c

Is so can you now explain the answer

Yes, it should be r=-5m+c.

In the xy-plane, if line k has negative slope and passes through the point (-5,r), is the x-intercept of line k positive?

This question can be done with graphic approach (just by drawing the lines) or with algebraic approach.

Algebraic approach:

Equation of a line in point intercept form is \(y=mx+b\), where: \(m\) is the slope of the line, \(b\) is the y-intercept of the line (the value of \(y\) for \(x=0\)), and \(x\) is the independent variable of the function \(y\).

We are told that slope of line \(k\) is negative (\(m<0\)) and it passes through the point (-5,r): \(y=mx+b\) --> \(r=-5m+b\).

Question: is x-intercept of line \(k\) positive? x-intercep is the value of \(x\) for \(y=0\) --> \(0=mx+b\) --> is \(x=-\frac{b}{m}>0\)? As we know that \(m<0\), then the question basically becomes: is \(b>0\)?.

(1) The slope of line \(k\) is -5 --> \(m=-5<0\). We've already known that slope was negative and there is no info about \(b\), hence this statement is insufficient.

(2) \(r>0\) --> \(r=-5m+b>0\) --> \(b>5m=some \ negative \ number\), as \(m<0\) we have that \(b\) is more than some negative number (\(5m\)), hence insufficient, to say whether \(b>0\).

(1)+(2) From (1) \(m=-5\) and from (2) \(r=-5m+b>0\) --> \(r=-5m+b=25+b>0\) --> \(b>-25\). Not sufficient to say whether \(b>0\).

Answer: E.

Bunuel .. i chose A bacause , u wrote x=-B/M>0?

from A we know slope is -5 .. so we have x=-b/5 and from the question we knw X is -5 so -5=B/-5= SO b = 25.. WE GOT B.. now we will put B value (y intercept ) , and x value in equation and puting value 0 in the y we wil get the x intercept.. where i m wrong?? i cant understand.

First of all if you substitute \(m=-5\) in \(x=-\frac{b}{m}\), you'll get \(x=\frac{b}{5}\). The question asks whether this value is positive. so whether \(x=\frac{b}{5}>0\), or which is the same whether \(b>0\).

Next, what does it mean "from the question we knw X is -5"? We know that the line passes through the point (-5, r), but it does not mean that all values of x of the line are negative (if it were so it would mean that the line is parallel to y-axis).

-5 - c/m > 0 i.e -5 -c > 0 i.e c< -5 X intercept -c/ m but we don't know whether m is + or - hence Insuff

cool_jonny009, Can you please simplify on this.

y = mx+c and the line passes through (-5,r)
-5=mr+c == > r = (-5 -c)/m

as the stmt says r > 0 so (-5 -c)/m > 0
which means c < -5

now X intercept is when y = 0 which will be -c/m
as we know c is negative but we don't know any thing about m ...so can't conclude whether the X intercept is - ve or + ve

-5 - c/m > 0 i.e -5 -c > 0 i.e c< -5 X intercept -c/ m but we don't know whether m is + or - hence Insuff

cool_jonny009, Can you please simplify on this.

y = mx+c and the line passes through (-5,r) -5=mr+c == > r = (-5 -c)/m

as the stmt says r > 0 so (-5 -c)/m > 0 which means c < -5

now X intercept is when y = 0 which will be -c/m as we know c is negative but we don't know any thing about m ...so can't conclude whether the X intercept is - ve or + ve

OE: Using (1) and (2) together does not help in the determination of the x-intercept, since the point (-5, r) could have any positive y coordinate and thus line k could cross the x-axis at many different places. _________________

Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...

-5 - c/m > 0 i.e -5 -c > 0 i.e c< -5 X intercept -c/ m but we don't know whether m is + or - hence Insuff

cool_jonny009, Can you please simplify on this.

y = mx+c and the line passes through (-5,r) -5=mr+c == > r = (-5 -c)/m

as the stmt says r > 0 so (-5 -c)/m > 0 which means c < -5

now X intercept is when y = 0 which will be -c/m as we know c is negative but we don't know any thing about m ...so can't conclude whether the X intercept is - ve or + ve

Isn't this wrong y = mx+c and the line passes through (-5,r) -5=mr+c == > r = (-5 -c)/m

shouldn't this be r=-5m+c

Is so can you now explain the answer _________________

_______________________________________________________________________________________________________________________________ If you like my solution kindly reward me with Kudos.

Re: In the xy-plane, if line k has negative slope and passes [#permalink]

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21 Jun 2012, 05:42

Bunuel wrote:

In the xy-plane, if line k has negative slope and passes through the point (-5,r), is the x-intercept of line k positive?

(1) The slope of line k is -5 (2) r>0

Graphic approach:

If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and Y intersects of the line with negative slope have the same sign. Therefore if X and Y intersects are positive, the line intersects quadrant I; if negative, quadrant III.

When we take both statement together all we know is that slope is negative and that it crosses some point in II quadrant (-5, r>0) (this info is redundant as we know that if the slope of the line is negative, the line WILL intersect quadrants II). Basically we just know that the slope is negative - that's all. We can not say whether x-intercept is positive or negative from this info.

Below are two graphs with positive and negative x-intercepts. Statements that the slope=-5 and that the line crosses (-5, r>0) are satisfied.

_______________________________________________________________________________________________________________________________ If you like my solution kindly reward me with Kudos.

Re: In the xy-plane, if line k has negative slope and passes [#permalink]

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21 Jun 2012, 23:10

Bunuel wrote:

riteshgupta wrote:

Isn't this wrong y = mx+c and the line passes through (-5,r) -5=mr+c == > r = (-5 -c)/m

shouldn't this be r=-5m+c

Is so can you now explain the answer

Yes, it should be r=-5m+c.

In the xy-plane, if line k has negative slope and passes through the point (-5,r), is the x-intercept of line k positive?

This question can be done with graphic approach (just by drawing the lines) or with algebraic approach.

Algebraic approach:

Equation of a line in point intercept form is \(y=mx+b\), where: \(m\) is the slope of the line, \(b\) is the y-intercept of the line (the value of \(y\) for \(x=0\)), and \(x\) is the independent variable of the function \(y\).

We are told that slope of line \(k\) is negative (\(m<0\)) and it passes through the point (-5,r): \(y=mx+b\) --> \(r=-5m+b\).

Question: is x-intercept of line \(k\) positive? x-intercep is the value of \(x\) for \(y=0\) --> \(0=mx+b\) --> is \(x=-\frac{b}{m}>0\)? As we know that \(m<0\), then the question basically becomes: is \(b>0\)?.

(1) The slope of line \(k\) is -5 --> \(m=-5<0\). We've already known that slope was negative and there is no info about \(b\), hence this statement is insufficient.

(2) \(r>0\) --> \(r=-5m+b>0\) --> \(b>5m=some \ negative \ number\), as \(m<0\) we have that \(b\) is more than some negative number (\(5m\)), hence insufficient, to say whether \(b>0\).

(1)+(2) From (1) \(m=-5\) and from (2) \(r=-5m+b>0\) --> \(r=-5m+b=25+b>0\) --> \(b>-25\). Not sufficient to say whether \(b>0\).

Answer: E.

Bunuel .. i chose A bacause , u wrote x=-B/M>0?

from A we know slope is -5 .. so we have x=-b/5 and from the question we knw X is -5 so -5=B/-5= SO b = 25.. WE GOT B.. now we will put B value (y intercept ) , and x value in equation and puting value 0 in the y we wil get the x intercept.. where i m wrong?? i cant understand. _________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

Re: In the xy-plane, if line k has negative slope and passes [#permalink]

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22 Jun 2012, 13:19

Thank you bunuel.. now i got that..if it were in question that all points contain (-5,r) then we wud have taken x -5 rite?? but in question its saying line intersect at (-5,r) that means its not neccesary all points contains -5 ...m i rite bunuel?? _________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

Re: In the xy-plane, if line k has negative slope and passes [#permalink]

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30 May 2013, 10:58

Defining line k as y=ax+b(a<0) it can be also said r=-5a+b ∴b=r+5a 'cause like k passes (-5,r)

The question asks whether b<0 or not.

(1) If applying a=-5 to the equation, b=r-25 Plus or minus of b depends on the value of r. So, it's insufficient.

(2) If applying r>0 to the equation, r=-5a+b>0 ∴b>5a It can't be determined whether b is over 0 or not since a is less than 0. So, (2) is insufficient.

(1) and (2) shows b>5a=-25 This allows b to be both over and below 0. So, this is insufficient.

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