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:

Re: If you got the last question right, check this out too. [#permalink]

Show Tags

20 Mar 2011, 21:57

S1 insufficient Y axis has no reflection about y axis. Assume line a= y axis. The answer is NO

If line a has slope -1. Slop of b is 1. Reflection of line a about y axis is parallel to line b. The answer is YES

S2 insufficient. Slope of line a is unknown

1) + 2) sufficient Since b>0 the line b cannot be x axis. Line a cannot be y axis - a is perpendicular to b. That means when a has reflection about y axis it is parallel to b. The answer to the question is YES

Re: If you got the last question right, check this out too. [#permalink]

Show Tags

20 Mar 2011, 22:09

But my initial guess was A alone sine we know that line a and line b have known and "different y intercepts". This precludes a from being the y axis . So I have two answers a or c. My bet 50/50 on both tough one!

Re: If you got the last question right, check this out too. [#permalink]

Show Tags

21 Mar 2011, 15:34

But according to the 'bagrettin' explanation to my previous question: a --- y = mx + b b --- y = (-1/m)x + b Reflection of line a --- x = ym + c ( shouldn't we flip x and y values to find reflection?) Y = (1/m) x - ( c/ m ) So now slope of line b is (-1/m) & slope of reflected line is (1/m) It looks like they will never be parallel because if one is + ve another is - ve. Help me ? I don't know whether we should proceed this way or not.
_________________

Consider me giving KUDOS, if you find my post helpful. If at first you don't succeed, you're running about average. ~Anonymous

Re: If you got the last question right, check this out too. [#permalink]

Show Tags

21 Mar 2011, 19:55

Hello bhandariavi x coordinate flips sign due to reflection about y axis. Draw two perpendicular lines in any quadrant and you will see that reflection of line a (whose slope -1) Reflected line will have slope = 1 is parallel to line b (slope 1) - ( knowing that line a is perpendicular to b) Visualize the problem that is easier than algebra or you can request bagrettin for algebraic solution. Cheers

Re: If you got the last question right, check this out too. [#permalink]

Show Tags

21 Mar 2011, 20:22

bhandariavi wrote:

But according to the 'bagrettin' explanation to my previous question: a --- y = mx + b b --- y = (-1/m)x + b Reflection of line a --- x = ym + c ( shouldn't we flip x and y values to find reflection?) Y = (1/m) x - ( c/ m ) So now slope of line b is (-1/m) & slope of reflected line is (1/m) It looks like they will never be parallel because if one is + ve another is - ve. Help me ? I don't know whether we should proceed this way or not.

bhandariavi, For reflection on y- axis you don't have to flip x and y values .Just negate x-coordinate The reflection of the point (x, y) across the y-axis is the point (-x, y). The reflection of the point (x, y) across the line y = x is the point (y, x). The reflection of the point (x, y) across the line y = -x is the point (-y, -x). The reflection of the point (x, y) across the x-axis is the point (x, -y).

Re: If you got the last question right, check this out too. [#permalink]

Show Tags

21 Mar 2011, 21:07

Hello onell Due to reflection about y axis Not only the x flips sign but the slope of the line "also" flips sign. I believe your equation did not account for slope change. Pls correct me if this not true.

Re: If you got the last question right, check this out too. [#permalink]

Show Tags

21 Mar 2011, 21:32

gmat1220 wrote:

Hello onell Due to reflection about y axis Not only the x flips sign but the slope of the line "also" flips sign. I believe your equation did not account for slope change. Pls correct me if this not true.

Posted from my mobile device

If it flips the sign of x coordinate and changes the sign of a slope : You will get a original line Consider a line passes through (x ,y) y = mx + b

Upon reflection it passes through (-x ,y) and changes slope to -m (As you have written) y = -m (-x) + c y=mx +c (Equation of a original line) Am I missing sth ?

Re: If you got the last question right, check this out too. [#permalink]

Show Tags

21 Mar 2011, 21:44

Hello onell Strange! Let me try will coordinates. Let's say line a (slope 1) passes through (2,3) and upon reflection passes through (-2,3) slope=-1 Equation of line a (slope 1) is (y-3)/(x-2) =1 y-3 =x-2 y=x+1 -----------(1)

The equation of the reflected line (slope -1) is (y-3)/[x-(-2)]=-1 y-3=-1(x+2) y-3 = -x - 2 y = -x + 1 ----------(2)

Two different equations. The slope of the line b which is perpendicular to a is -1. The slope of the reflected line is also -1. Hence E cannot be the answer.

Posted from my mobile device

Last edited by gmat1220 on 22 Mar 2011, 03:40, edited 1 time in total.

Re: If you got the last question right, check this out too. [#permalink]

Show Tags

21 Mar 2011, 22:01

gmat1220 wrote:

Hello onell Strange! Let me try will coordinates. Let's say line a (slope 1) passes through (2,3) and upon reflection passes through (-2,3) slope=-1 Equation of line a is Y-2/(x-3) =1 Y-2 =x-3 Y=x-1

The equation of the reflected line is Y-(-2)/(x-3)=-1 y+2=-x+3 y=-x+1=1-x

Posted from my mobile device

Strange Indeed. However if you substitute (2,3) and slope 1 and(-2,3) and slope -1 for reflected line in equation y=mx+c . You get the same equation for both the line.. Bunuel, Please help....

Upon request, I am providing the solution of the question above: I will provide a graphical approach since that is what I favor always but later will give an algebraic approach too...

The two diagrams below illustrate the case where we take both statements together. In one case, reflection of a is not parallel to b and in the other reflection of a is parallel to b. Hence even with both statements we cannot say whether reflection of a is parallel to b. Answer (E).

Attachment:

Ques2.jpg [ 23.2 KiB | Viewed 3714 times ]

If a and b make 45 degrees angle with the y axis (as shown, technically I will not say that they are both making 45 degrees angle with y axis but let's not worry about it here), when a is reflected along y axis, its angle with y axis is still 45. In this case a and b are parallel.

Algebraic approach: A line is defined by 2 things - its slope and y intercept. When we reflect a line along the y axis, its slope flips sign but y intercept remains unchanged. For more on this: http://www.veritasprep.com/blog/2010/12 ... he-graphs/ Now, a -> y = mx + c b -> y = nx + d Reflected a -> y = -mx + c Ques: Is -m = n? (Parallel lines have the same slope.)

Stmnt 1: mn = -1 m = -1/n. If m = 1 and n = -1, -m is equal to n If m = -1 and n = 1, -m is equal to n If m = -1/2 and n = 2, -m is not equal to n Not sufficient.

Stmnt 2: n>0 If m = -1 and n = 1, -m is equal to n If m = -1/2 and n = 2, -m is not equal to n Not sufficient.

Taking both together, If m = -1 and n = 1, -m is equal to n If m = -1/2 and n = 2, -m is not equal to n Not sufficient. Answer (E).
_________________

Re: Lines a and b have different y-intercepts. When line a is [#permalink]

Show Tags

21 Aug 2014, 20:49

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: Lines a and b have different y-intercepts. When line a is [#permalink]

Show Tags

17 Aug 2016, 01:58

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...