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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q

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Magoosh GMAT Instructor G
Joined: 28 Dec 2011
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Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q  [#permalink]

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7
38 00:00

Difficulty:   95% (hard)

Question Stats: 44% (03:11) correct 56% (02:53) wrong based on 140 sessions

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Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q, has an integer for its y-intercept, and intersects Q in the second quadrant, then how many possible Line S’s exist? (Note: Intersections on one of the axes do not count.)
(A) 25
(B) 33
(C) 36
(D) 41
(E) 58

For a bank of challenging coordinate geometry problems, as well as the OE to this one, see:
http://magoosh.com/gmat/2014/challengin ... questions/

Mike _________________
Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Manager  Joined: 04 Oct 2013
Posts: 148
Location: India
GMAT Date: 05-23-2015
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Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q  [#permalink]

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3
6
Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q, has an integer for its y-intercept, and intersects Q in the second quadrant, then how many possible Line S’s exist? (Note: Intersections on one of the axes do not count.)
(A) 25
(B) 33
(C) 36
(D) 41
(E) 58

The equation of the given line Q is $$5y-3x=45$$ or $$y = \frac{3}{5}x +9$$

Therefore, slope of the line Q = $$\frac{3}{5}$$

Slope of line S, which is perpendicular to line Q = $$\frac{-5}{3}$$

X-intercept of line Q = $$-15$$

Highest possible value of Y-intercept of the line S that is perpendicular to line Q and that intersects in the second quadrant
= Y-intercept of the line Q =$$9$$

Lowest possible value of Y-intercept of the line that is perpendicular to line Q and that intersects in the second quadrant
= - (slope of line Q) *(X-intercept of line Q or X-intercept of line S that intersects Q at X-axis=-15) = $$-25$$

The number of possible perpendicular line S that intersects line Q in the second quadrant and that its Y-intercept is a integer quantity (excluding intersection at axes)= Integer values between 9 and -25= 9 – (-25)-1 =9+25-1 =$$33$$

Attachments Possible-Perpendicular-Lines.png [ 25.6 KiB | Viewed 4729 times ]

Originally posted by arunspanda on 01 Sep 2014, 02:32.
Last edited by arunspanda on 10 Apr 2016, 04:51, edited 1 time in total.
##### General Discussion
Intern  Joined: 17 Mar 2014
Posts: 34
Location: India
Concentration: Strategy, Marketing
GMAT 1: 760 Q50 V44
WE: Medicine and Health (Health Care)
Re: Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q  [#permalink]

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1
5y - 3x = 45 --> 5y = 3x + 45
at x= 0, y = 9
at y=0, x = -15
Now slope = 3/5
perpendicular 's slope will be -5/3
at ( -15,0) perp line eqn = y-0 = -5/3(x-(-15))
thus y = -5/3 x - 25
thus at x=0, y = - 25
24 + 1 + 8 = 33
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Re: Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q  [#permalink]

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How do we get 24 + 1 + 8 = 33
Magoosh GMAT Instructor G
Joined: 28 Dec 2011
Posts: 4483
Re: Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q  [#permalink]

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2
1
RonnieRon wrote:
How do we get 24 + 1 + 8 = 33

Dear RonnieRon,
Please see a full explanation here:
http://magoosh.com/gmat/2014/challengin ... questions/
It's #7 on that page.
Mike _________________
Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Manager  B
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Re: Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q  [#permalink]

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Finally get the concept but a little too difficult for myself
Intern  Joined: 02 Jun 2014
Posts: 13
Location: India
Concentration: Strategy, Operations
Schools: LBS '16, ISB '15
GMAT 1: 530 Q38 V24
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Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q  [#permalink]

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Hi Mike
thanks - great explanation...+1
Intern  Joined: 28 Dec 2015
Posts: 37
Re: Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q  [#permalink]

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Hello Mike,

Well this was a difficult question for me.

I understood the solution but i have a silly doubt-intersection at the axes means the perpendicular line S that passes through the origin should be excluded?? Magoosh GMAT Instructor G
Joined: 28 Dec 2011
Posts: 4483
Re: Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q  [#permalink]

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1
Ashishsteag wrote:
Hello Mike,

Well this was a difficult question for me.

I understood the solution but i have a silly doubt-intersection at the axes means the perpendicular line S that passes through the origin should be excluded?? Dear Ashishsteag,

I'm happy to respond. Yes, this is an extremely difficult question.

Your question is not silly. Let's look at the prompt:
Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q, has an integer for its y-intercept, and intersects Q in the second quadrant, then how many possible Line S’s exist? (Note: Intersections on one of the axes do not count.)
In this problem, the intersection of concern, the only intersection being discussed, is the intersection of Line Q and Line S. That is the only intersection that matters in the problem. If these two lines intersect on the axis---for example, at the point (0, 9)---then we wouldn't count that as an intersection in QII, precisely because points on either axis are not points in QII. Line S will have its own intersections with the x- and y-axes, but those are irrelevant to the problem. It doesn't matter where Line S intersects the axes, even at the origin. The only thing that matters is the intersection point of Line Q and line S.

Does all this make sense?
Mike _________________
Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Intern  Joined: 28 Dec 2015
Posts: 37
Re: Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q  [#permalink]

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mikemcgarry wrote:
Ashishsteag wrote:
Hello Mike,

Well this was a difficult question for me.

I understood the solution but i have a silly doubt-intersection at the axes means the perpendicular line S that passes through the origin should be excluded?? Dear Ashishsteag,

I'm happy to respond. Yes, this is an extremely difficult question.

Your question is not silly. Let's look at the prompt:
Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q, has an integer for its y-intercept, and intersects Q in the second quadrant, then how many possible Line S’s exist? (Note: Intersections on one of the axes do not count.)
In this problem, the intersection of concern, the only intersection being discussed, is the intersection of Line Q and Line S. That is the only intersection that matters in the problem. If these two lines intersect on the axis---for example, at the point (0, 9)---then we wouldn't count that as an intersection in QII, precisely because points on either axis are not points in QII. Line S will have its own intersections with the x- and y-axes, but those are irrelevant to the problem. It doesn't matter where Line S intersects the axes, even at the origin. The only thing that matters is the intersection point of Line Q and line S.

Does all this make sense?
Mike Actually,I got confused with this part of the question written at the very end:"(Note: Intersections on one of the axes do not count.)",and then the question also speaks about y-intercept in the part:"If Line S is perpendicular to Q, has an integer for its y-intercept",so I thought that since the perpendicular line S does not have any sort of y-intercept at the origin (0,0),so we need to exclude that part?
Even one of the solutions mentioned above with the required figure drawn is subtracting 1 at the very end of the solution.I understand your point that only intersections between line S and line Q need to be considered and they need not be counted on either of the axes.Thanx a lot for your help. Manager  S
Joined: 23 Jan 2016
Posts: 175
Location: India
GPA: 3.2
Re: Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q  [#permalink]

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my question is - we calculated the intersections in the y axis between 9 and -25 to get the answer; why did we not calculate the intersections in the x axis between -15 and 5 instead? Please bear with me if this is a stupid question. Thank you.
Magoosh GMAT Instructor G
Joined: 28 Dec 2011
Posts: 4483
Re: Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q  [#permalink]

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2
TheLordCommander wrote:
my question is - we calculated the intersections in the y axis between 9 and -25 to get the answer; why did we not calculate the intersections in the x axis between -15 and 5 instead? Please bear with me if this is a stupid question. Thank you.

Dear TheLordCommander,

I'm happy to respond. This is a hard question, so questions about it are not "stupid." Part of the requirement of the question is that the y-intercept has to be an integer. If we mark off the boundaries on the y-intercept, then we simply can count integers along the y-axis.

You see, if the x-intercept is an integer, that doesn't guarantee that the y-intercept is an integer (unless the slope is +1 or -1). Certainly for any non-integer slope, the general rule is that for most x-intercepts that are integers, the y-intercept is not an integer, and vice versa. If we start looking at points on the x-axis, we know they have to be -15 and 5, but within that range, we have no idea what spacings of the values of the x-intercept would result in integer values on the y-intercept.

Therefore, it's much easier simply to stick to the y-intercept and count integers.

Does all this make sense?
Mike _________________
Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Manager  S
Joined: 23 Jan 2016
Posts: 175
Location: India
GPA: 3.2
Re: Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q  [#permalink]

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mikemcgarry wrote:
TheLordCommander wrote:
my question is - we calculated the intersections in the y axis between 9 and -25 to get the answer; why did we not calculate the intersections in the x axis between -15 and 5 instead? Please bear with me if this is a stupid question. Thank you.

Dear TheLordCommander,

I'm happy to respond. This is a hard question, so questions about it are not "stupid." Part of the requirement of the question is that the y-intercept has to be an integer. If we mark off the boundaries on the y-intercept, then we simply can count integers along the y-axis.

You see, if the x-intercept is an integer, that doesn't guarantee that the y-intercept is an integer (unless the slope is +1 or -1). Certainly for any non-integer slope, the general rule is that for most x-intercepts that are integers, the y-intercept is not an integer, and vice versa. If we start looking at points on the x-axis, we know they have to be -15 and 5, but within that range, we have no idea what spacings of the values of the x-intercept would result in integer values on the y-intercept.

Therefore, it's much easier simply to stick to the y-intercept and count integers.

Does all this make sense?
Mike Makes a lot of sense Mike. Thanks Non-Human User Joined: 09 Sep 2013
Posts: 15376
Re: Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q  [#permalink]

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_________________ Re: Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q   [#permalink] 17 Apr 2020, 03:45

# Line Q has the equation 5y – 3x = 45. If Line S is perpendicular to Q  