Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

What is the answer and why ?

I am for B but it is not OG answer. Based on second condition r > 0, x-intercept for this line is negative, so we can determine definitively sign of x-intercept.

Let the \(x\) intercept be the point \((x,0)\). Slope \(m\) is rise over run and for two points \((-5,r)\) and \((x,0)\) would be \(m=\frac{r-0}{-5-x}=\frac{r}{-5-x}\) --> \(x=\frac{-r-5m}{m}\).

Question: is \(x>0\)? --> is \(x=\frac{-r-5m}{m}>0\)?

(1) \(m=-5\) --> \(x=\frac{-r-5m}{m}=\frac{-r+25}{-5}>0\)? \(x=\frac{-r+25}{-5}=\frac{r}{5}-5>0\)? We can not determine whether \(\frac{r}{5}-5>0\) or not. Not sufficient.

(2) \(r>0\) and \(m<0\) --> \(x=\frac{-r-5m}{m}=\frac{-r}{m}-5>0\)? \(\frac{-r}{m}\) is some positive value (as \(m<0\)) but we don't know whether it's more than \(5\) or not. Not sufficient.

(1)+(2) \(x=\frac{r}{5}-5\) and \(r>0\) --> \(r=5x+25>0\) --> \(x>-5\). \(x\) can be positive as well as negative. Not sufficient.

Answer: E.

This can be done by visualizing the question. Statement (2) tells us that the point \((-5,r)\), as \(r>0\), is in the II quadrant. Line with negative slope through the point in the II quadrant can have \(x\) intercept positive as well as negative.

Taken together: as we don't know the exact location of the point \((-5,r)\) in II quadrant we can not say even knowing the slope whether the \(x\) intercept would be positive or negative.
_________________

Re: OG 12 DS Question - Line concept Very good one (key wrong ?) [#permalink]

Show Tags

27 Jan 2010, 02:41

yes, good way too Gurpreet. lets play with it further to strenghtne the concepts .

If I modify the second statement to

2. r < 0 ,

let's try to solve it in two ways i.e. First by visualising it and second, by using the basic equation y = mx +c

Using the second method,

The equation is r = 25+c and we know r < 0, so only possibility is c < -25 , so this can tell definitely the answer. Am I correct ? But if we want to visualise it, is there any possibility it can pass through First quadrant ? Second quadrant i can imagine easy .

Also, How a line can make more than 180° angle ? I am from good maths background but do not remember exacly how a line makes more than 180° . So little help will break the code.

Re: OG 12 DS Question - Line concept Very good one (key wrong ?) [#permalink]

Show Tags

27 Jan 2010, 04:17

GMATMadeeasy wrote:

yes, good way too Gurpreet. lets play with it further to strenghtne the concepts .

If I modify the second statement to

2. r < 0 ,

let's try to solve it in two ways i.e. First by visualising it and second, by using the basic equation y = mx +c

Using the second method,

The equation is r = 25+c and we know r < 0, so only possibility is c < -25 , so this can tell definitely the answer. Am I correct ? But if we want to visualise it, is there any possibility it can pass through First quadrant ? Second quadrant i can imagine easy .

Also, How a line can make more than 180° angle ? I am from good maths background but do not remember exacly how a line makes more than 180° . So little help will break the code.

r = 25+c and r>0 is given that means 25+c >0 => 25> -c

=> -25 <c thus we cannot determine whether c is -ve or +ve

r-25-c should have same or different sign with -C to know whether that number is on same or diff side of origin.
_________________

Re: OG 12 DS Question - Line concept Very good one (key wrong ?) [#permalink]

Show Tags

27 Jan 2010, 05:17

Let's take it a step further. Adding one more question that will take discussion further . (Source: GMATprep)

Lnes n and p lie in the xy plane . Is the slope of line n less than the slope of line p?

1. Lines n and p intersect at the point (5,1) 2. The y intercept of line n is greater than the y intercept of line p.

1. Clearly insufficient : For equation n y=m1.x+c1 and for equation p y=m2.x+c2 => First equation of line :1 = 5.m1+c1 Second equation of line: 1= 5.m2+c2

Clearly, m1 and m2 Can take any vset of values that m1>m2 and m1<m2, both are possible

2. Insufficent again : y=m1.x+c1 and y=m2.x+c2 => given is c1 > c2 => m1 and m2 can take any set of values again

1 and 2 statement togather :

First line : 1 = 5.m1+c1 Second Line: 1= 5.m2+c2

From this we can conclude that 5.m1+c1 = 5.m2+c2 or 5 (m1 - m2) = c2 - c1 now we are given c1 > c2 , I beleive that Statement refers to absolute values so c1 could be 4 and c2 as 3 => c2-c1 gives -1 or c1 could be -4 and c2 as 3 => c2-c1 gives 7 or c1 could be 4 and c2 as -3 => c2-c1 gives -7 or c1 could be -4 and c2 -3 => c2-c1 gives 1

that means m1 and m2 can be greater or lesser than each other.So both together not sufficinet.

If I ignore my assumption that the values are absolute : In that case, c2-c1 has to be negative if c1 > c2 . so m1 - m2 has to be negative too .

m1 - m2 < 0 or m1 < m2 , Sufficient

Question 1 : Which of the case one should assume ? answer is C that both together are sufficient that means we can't assume absolute values.

Question 2 : Sloe -1 and slope -2 , generally speaking -1 is greater than -2 but slope if we look at -2 is greater than -1.

Re: OG 12 DS Question - Line concept Very good one (key wrong ?) [#permalink]

Show Tags

11 Feb 2010, 21:46

jshah wrote:

I still cant visualize.

In stmt 2: we know that like passes through second quadrant and it has negative slope. How can it intercept at positive x?

Negative slope always goes bottom to top from left to right.

Please help me visualizing this, I am sure i am missing something.

take a line that is passing through the origin and bisects the II quadrant and IV quadrant. Now draw parallel lines above and below that line .. one will have positive x intercept and other will have negative x intercept and the original line will have 0 x-intecept.

Re: OG 12 DS Question - Line concept Very good one (key wrong ?) [#permalink]

Show Tags

15 Feb 2010, 09:14

@testprep2010 : Your response is E. I do not understand what values such posts bring.

1. This is OG question so answer is known. 2. The post does not just ask to answer just an option 3. How do I know your answer is wrong or right ? 4. Even if everyone votes A for this, it will not change the answer.

I feel sad why people put thier time answering like this on these forums.

Re: OG 12 DS Question - Line concept Very good one (key wrong ?) [#permalink]

Show Tags

20 Feb 2010, 03:14

11

This post received KUDOS

GMATMadeeasy 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

What is the answer and why ?

I am for B but it is not OG answer. Based on second condition r > 0, x-intercept for this line is negative, so we can determine definitively sign of x-intercept.

Best Approach to use a graph!

Attachments

OG 12 DS Question.png [ 27.25 KiB | Viewed 13194 times ]

_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: OG 12 DS Question - Line concept Very good one (key wrong ?) [#permalink]

Show Tags

20 Feb 2010, 09:15

Thanks Bunuel.. for the Kudos! A kudos from u.. means a lot!
_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Bunuel, how do you know from statement 2 that m<0? Statement 2 only says that r>0. Could you please explain it to me? Thank you!

The question stem tells you that line k has negative slope so m < 0.

Also, you may want to look at the diagram made by jeeteshsingh in the first reply to this question. It uses both the statements together and shows you two possible cases (through two different dotted lines). In one case, line k has positive x intercept; in the second case it has negative x intercept. Hence both statements together are not sufficient.
_________________

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

Show Tags

21 Apr 2012, 03:16

1

This post received KUDOS

GMATMadeeasy 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

What is the answer and why ?

I am for B but it is not OG answer. Based on second condition r > 0, x-intercept for this line is negative, so we can determine definitively sign of x-intercept.

y=mx+c substitute (-5,r)

r=-5m+c

C= r+5m

1) m=-5

y=mx+c

put y=0 to find x intercept

and m=-5

0= -5x+r+5(-5) 5x=r-25

Dont know r so insufficient

2) r> 0, certainly not sufficient . . . 0= mx+r+5m , we do not know m and r

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

Show Tags

25 Sep 2016, 03:29

GMATMadeeasy 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

Hi, Can you please tell where I am wrong. I assumed eq of line as y=mx+c. Let x intercept be A(x,0) At point A, 0=mx+c =>x=-c/m Hence, sign of x will depend on sign of c/m Since m is negative, i. if c<0, -c/m <0 =>x<0 ii. if c>0, -c/m >0 =>x>0

Statement 1 is not sufficient. As per statement 2, r>0 => -5m+c>0 =>-5m>-c =>5m<c =>5<c/m or c/m >5 => both c and m have same sign i.e. m,c<0 => x<0 => Statement 2 is sufficient

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

Show Tags

23 Feb 2017, 03:56

PROMPT ANALYSIS The equation of the will be y = r + m(x+5) where m is the slope. X intercept is -r/m -5

Superset The answer will be either yes or no.

Translation In order to find the answer we need: 1# Exact value of m and r 2# Equations to solve m and r

STATEMENT ANALYSIS

St 1: m = -5. For this x intercept is r/5 -5. Cannot say about the sign of the expression because r can be any real number.INSUFFICIENT. Hence option a and d eliminated.

St 2: r>0. The value of -r/m -5 will depend on the value of m, which can be any real number. INSUFFICIENT. Hence option b eliminated.

St 1&st 2: r/5 -5 will be positive for natural number r>25, these kind of constraints are not given. SO INSUFFICIENT. HEnce option c eliminated. Option E.

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

We need to determine whether the x-intercept of line k is positive, given that line k has a negative slope and passes through the point (-5, r).

We can let the slope of line k be m and the x-intercept of line k be a; that is, line k passes through the point (a, 0). Using the slope formula m = (y_2 - y_1)/(x_2 - x_1), we have:

m = (0 - r)/(a -(-5))= -r/(a + 5)

Statement One Alone:

The slope of line k is -5.

Using the information in statement one, we can say the following:

-r/(a + 5) = -5

a + 5 = r/5

a = r/5 - 5

Since we don't know the value of r, we cannot determine the value of a.

For example, if r = 5, a = -4, which is negative. However, if r = 50, a = 5, which is positive. Statement one alone is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

r > 0

Knowing r is positive does not give us enough information to determine whether a, the x-intercept of line k, is positive. Statement two alone is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Looking at our work from statement one, and keeping in mind that r > 0 from statement two, we see that a can be positive or negative.

For example, if r = 5 then a = -4, which is negative. However, if r = 50 then a = 5, which is positive. The two statements together are still not sufficient to answer the question.

Answer: E
_________________

Jeffery Miller Head of GMAT Instruction

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions