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
Source: Magoosh
Tough Coordinate Geomtry question
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Line Q:Mo2men wrote: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
Source: Magoosh
5y - 3x = 45
5y = 3x + 45
y = (3/5) + 9.
Draw line Q:
The slopes of perpendicular lines are NEGATIVE RECIPROCALS.
Since line S must be perpendicular to line Q -- and the slope of line Q is 3/5 -- the slope of line S = -5/3.
Thus, the equation for line S is as follows:
y = (-5/3)x + b.
If line S intersects line Q at the x-axis, the following figure is yielded:
Here, line S includes (-15, 0).
Plugging (-15, 0) into y = (-5/3)x + b, we get:
0 = (-5/3)(-15) + b
0 = 25 + b
b = -25.
The result is the following figure:
If line S intersects line Q at the y-axis, we get:
Since lines Q and S may not intersect at either axis, the resulting figure implies that the y-intercept of line S must be an integer value BETWEEN -25 AND 9.
Thus, the y-intercept for line S may be any integer value between -24 and 8, inclusive, for a total of 33 integer options.
The correct answer is B.
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Thanks for you kind supportGMATGuruNY wrote:
Since lines Q and S may not intersect at either axis, the resulting figure implies that the y-intercept of line S must be an integer value BETWEEN -25 AND 9.
Thus, the y-intercept for line S may be any integer value between -24 and 8, inclusive, for a total of 33 integer options.
The correct answer is B.
Hoe come 33? The value between -24 and 8, inclusive should be 32. it is 8 - (-24).
Can you please help
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To count consecutive integers, use the following equation:Mo2men wrote:Hoe come 33? The value between -24 and 8, inclusive should be 32. it is 8 - (-24).GMATGuruNY wrote:
Since lines Q and S may not intersect at either axis, the resulting figure implies that the y-intercept of line S must be an integer value BETWEEN -25 AND 9.
Thus, the y-intercept for line S may be any integer value between -24 and 8, inclusive, for a total of 33 integer options.
The correct answer is B.
Can you please help
biggest - smallest + 1.
Thus, the number of integers between -24 and 8, inclusive, is equal to the following:
8 - (-24) + 1 = 33.
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We could also think of it algebraically.
Since the lines are perpendicular, Line S's slope = -1/(Line Q's slope) = -(5/3).
Line Q is y = 3/5x + 9, so Line S will be y = -(5/3)x + b, where b is an integer. We want to find all possible values of b.
Where the lines intersect, their x and y values are identical, so (3/5)x + 9 = (-5/3)x + b, or (34/15)x + 9 = b.
Now we just need to find a range of values for b. For the intersection to happen in Quadrant II, b must be less than 9, or else the lines will intersect in Quadrant I or on the y-axis. So we've got b < 9.
Similarly, we need Line S's x-intercept to be greater than Line Q's x-intercept, or the lines will intersect in Quadrant III or on the x-axis.
Line Q's x-intercept = -(y-intercept)/slope = -9/(3/5) = -15.
Line S's x-intercept = -(y-intercept)/slope = -b/(-5/3) = (3/5)b.
So we've got -15 < (3/5)b, or -25 < b.
Combining our two inequalities, we have -25 < b < 9, so there are 33 possible integer values of b.
Since the lines are perpendicular, Line S's slope = -1/(Line Q's slope) = -(5/3).
Line Q is y = 3/5x + 9, so Line S will be y = -(5/3)x + b, where b is an integer. We want to find all possible values of b.
Where the lines intersect, their x and y values are identical, so (3/5)x + 9 = (-5/3)x + b, or (34/15)x + 9 = b.
Now we just need to find a range of values for b. For the intersection to happen in Quadrant II, b must be less than 9, or else the lines will intersect in Quadrant I or on the y-axis. So we've got b < 9.
Similarly, we need Line S's x-intercept to be greater than Line Q's x-intercept, or the lines will intersect in Quadrant III or on the x-axis.
Line Q's x-intercept = -(y-intercept)/slope = -9/(3/5) = -15.
Line S's x-intercept = -(y-intercept)/slope = -b/(-5/3) = (3/5)b.
So we've got -15 < (3/5)b, or -25 < b.
Combining our two inequalities, we have -25 < b < 9, so there are 33 possible integer values of b.
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And as an FYI, this question is pretty hard - I wasn't able to do it in under two minutes! It seems more of a conceptual exercise than something you're all that likely to see on the GMAT, but I think I could do it in two minutes next time now that I thought of how to set it up: it was cracking the second condition (Line S's x-int > Line Q's x-int) on top of all the annoying fractional manipulation that put me over time.
That said, it is kindasorta guessable in two minutes. I started with the two slopes, and got to b = (34/15)x + 9. Thinking that the x intercept needs to be an integer, I immediately thought that x is going to be something positive but less than 34, i.e. the numerator of (34/15), so I was gravitating toward 33 from the start. But that ended up being a little murky and hard to justify, forcing me to press on.
That said, it is kindasorta guessable in two minutes. I started with the two slopes, and got to b = (34/15)x + 9. Thinking that the x intercept needs to be an integer, I immediately thought that x is going to be something positive but less than 34, i.e. the numerator of (34/15), so I was gravitating toward 33 from the start. But that ended up being a little murky and hard to justify, forcing me to press on.