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Re: If you got the last question right, check this out too. [#permalink]
23 Mar 2011, 04:53

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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 2516 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: If you got the last question right, check this out too. [#permalink]
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]
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]
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]
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]
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]
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]
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.

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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]
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.

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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]
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....

Re: Lines a and b have different y-intercepts. When line a is [#permalink]
21 Aug 2014, 20:49

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