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Re: OG 12 DS Question - Line concept Very good one (key wrong ?) [#permalink]
17 Jan 2010, 12:14

5

This post received KUDOS

Expert's post

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.

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]
17 Jan 2010, 15:50

Thanks a lot, as usual this is solid solution. Question: My problem was visualising the line I guess, I see with negative slope and r (y coordinate) being posiive, only second quardant is possible .I miss some point I guess.. COuld you give me idea of two angles of two lines from x_axis to visualise better ? sorry, but I miss some basic points I guess.

Re: OG 12 DS Question - Line concept Very good one (key wrong ?) [#permalink]
17 Jan 2010, 16:08

Expert's post

GMATMadeeasy wrote:

Thanks a lot, as usual this is solid solution. Question: My problem was visualising the line I guess, I see with negative slope and r (y coordinate) being posiive, only second quardant is possible .I miss some point I guess.. COuld you give me idea of two angles of two lines from x_axis to visualise better ? sorry, but I miss some basic points I guess.

I'm not sure I understand your question... Can you please specify? _________________

Re: OG 12 DS Question - Line concept Very good one (key wrong ?) [#permalink]
18 Jan 2010, 05:22

Bunuel wrote:

GMATMadeeasy wrote:

Thanks a lot, as usual this is solid solution. Question: My problem was visualising the line I guess, I see with negative slope and r (y coordinate) being posiive, only second quardant is possible .I miss some point I guess.. COuld you give me idea of two angles of two lines from x_axis to visualise better ? sorry, but I miss some basic points I guess.

I'm not sure I understand your question... Can you please specify?

Thanks Bunuel. You bring me luck, you speak and problem is solved.

I meant that i was not able to visualise two cases where x intercepts could be positive or negative , but I can not. I did dig a little in lines and got the answer.

Thanks again for iron proof answer by you.

P.S. I am going to review/attempt all Qs and concept on number properties by you as this is an area I need more confidence. wil hear more from me. My exam is roughly in four weeks.

Re: OG 12 DS Question - Line concept Very good one (key wrong ?) [#permalink]
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]
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]
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]
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]
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]
20 Feb 2010, 03:14

9

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 7921 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]
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|

Re: In the xy-plane, if line k has negative slope and passes [#permalink]
16 Nov 2011, 21:36

Expert's post

MariaBez wrote:

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]
21 Apr 2012, 03:16

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]
21 Feb 2013, 10:21

algebra and slope. do not use diagram.

this is hard and dose not appear in og many times.

I wan to follow this post not to forget this question. _________________

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Re: In the xy-plane, if line k has negative slope and passes
[#permalink]
21 Feb 2013, 10:21

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