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All points (x,y) that lie below the line l, shown above [#permalink]
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All points (x,y) that lie below the line l, shown above, satisfy which of the following inequalities? A. y < 2x + 3 B. y < 2x + 3 C. y < x + 3 D. y < 1/2*x + 3 E. y < 1/2*x + 3
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Last edited by Bunuel on 28 Jun 2013, 01:35, edited 1 time in total.
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Re: GMATPrep Lines [#permalink]
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All points (x,y) that lie below the line l, shown above, satisfy which of the following inequalities? A. y < 2x + 3 B. y < 2x + 3 C. y < x + 3 D. y < 1/2*x + 3 E. y < 1/2*x + 3 First of all we should write the equation of the line \(l\): We have two points: A(0,3) and B(6,0). Equation of a line which passes through two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(\frac{yy_1}{xx_1}=\frac{y_1y_2}{x_1x_2}\)So equation of a line which passes the points A(0,3) and B(6,0) would be: \(\frac{y3}{x0}=\frac{30}{06}\) > \(2y+x6=0\) > \(y=\frac{1}{2}x+3\) Points below this line satisfy the inequality: \(y<\frac{1}{2}x+3\) ORThe equation of line which passes through the points \(A(0,3)\) and \(B(6,0)\) can be written in the following way: Equation of a line in point intercept form is \(y=mx+b\), where: \(m\) is the slope of the line and \(b\) is the yintercept of the line (the value of \(y\) for \(x=0\)). The slope of a line, \(m\), is the ratio of the "rise" divided by the "run" between two points on a line, thus \(m=\frac{y_1y_2}{x_1x_2}\) >\(\frac{30}{06}=\frac{1}{2}\) and \(b\) is the value of \(y\) when \(x=0\) > A( 0,3) > \(b=3\). So the equation is \(y=\frac{1}{2}x+3\) Points below this line satisfy the inequality: \(y<\frac{1}{2}x+3\). Actually one could guess that the answer is E at the stage of calculating the slope \(m=\frac{1}{2}\), as only answer choice E has the same slope line in it. Answer: E. For more please check Coordinate Geometry chapter of the Math Book (link in my signature). Hope it's clear.
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Re: GMATPrep Lines [#permalink]
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14 Oct 2009, 04:55
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The line will be in the form of y=mx+b In this case we are looking for a negative m (eliminate option A and D). Finally we are looking for a line with X intercept of 6. So if you make y=0 for all remaining formulas you get
b) x=3/2 wrong c) x=3 wrong e) x=6 RIGHT
ANS = E.



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gmat prep,, PS question [#permalink]
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30 Jun 2010, 01:40
Can someone help me solve this problem please!!!
All points (x,y) that lie below the line L, shown above, satisfy which of the following inequalities..
*I couldnt draw the xyplane but, y is +3 and x is +6 and line L is lined from y to x straight..
1) y<2x+3 2) y<2x+3 3) y<x+3 4) y<1/2x+3 5) y<1/2x+3
the answer is 5...
thanks



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Re: gmat prep,, PS question [#permalink]
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30 Jun 2010, 02:39
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Hi,
compute the slope. The slope is the change in rise over run or (y2y1)/(x2x1). So, slope is (30)/(06) = 1/2. (Or, (03)/(60) = 1/2. And the yintercept of the line (where the line crosses or touches the yaxis) is +3. Thus, using the line equation y = mx + b (in which "y" and "x" is an ordinate pair for any point on the line, "m" is slope, and "b" is yintercept), the equation of the line is y = 1/2x + 3. We're looking for points that lie below this line, so for any given value of x, the y value should be the biggest possible value without going above the line. Thus, the correct answer is choice E.
You can pick numbers to confirm. Let x = 6. We know that when x = 6, according to the line, y = 0. Let's plug x = 6 into the line equation: y = (1/2)*6 + 3 = 0. Yep, that's right. To fall below the line, then, when x = 6, y<0.
If you understand the line equation, then as soon as you computed the slope of "1/2", you know that the answer is E, and you're done (because none of the other choices have "1/2").
We could have also observed that the slope is negative (because the line is "falling" reading from left to right). That observation allows us to cancel choices A and D. The slope is definitely not 1...eliminate C. Choice B is a trap for someone who reversed the given and x and y coordinates or else a trap for someone who computed slope as run/rise rather than rise/run.



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Re: GMATPrep Lines [#permalink]
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30 Jun 2010, 03:15
Your explanation is very clear..
I understood now and you are right I didnt click to think this line going down from left to right which creates negative slope...
thank you so much!!



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Re: GMATPrep Lines [#permalink]
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30 Jun 2010, 13:53
gmatJP wrote: Your explanation is very clear..
I understood now and you are right I didnt click to think this line going down from left to right which creates negative slope...
thank you so much!! You're most welcome!



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Re: GMATPrep Lines [#permalink]
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04 Sep 2010, 23:55
Looks difficult but after solving, becomes eazy.... good.



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Re: All points (x,y) that lie below the line l, satisfy which of [#permalink]
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27 Jun 2013, 23:45



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Re: All points (x,y) that lie below the line l, satisfy which of [#permalink]
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study wrote: All points (x,y) that lie below the line l, satisfy which of the following inequalities? A. y<2x+3 B. y<2x+3 C. y<x+3 D. y<1/2x+3 E. y<1/2x+3 Attachment: new_PS_Lines_E.JPG For any given line L, if the intercepts are given, we can write the equation of the line as \(\frac {x} {xintercept} + \frac {y} {yintercept} 1 = 0\) For the given line, it stands as L = \(\frac {x} {6} + \frac {y} {3} 1 = 0\) Now, notice that when the value of origin is plugged in (0,0), we get L as 0+01 > L<0. Thus, the origin lies on the negative side of the given line. And, as origin lies below the given line, all the points in that region will make L<0 > \(\frac {x} {6} + \frac {y} {3} 1<0\) > \(\frac {y} {3}< 1\frac {x} {6}\) > \(y < 3\frac {x} {2}\) E.
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Re: All points (x,y) that lie below the line l, shown above [#permalink]
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28 Jun 2013, 01:45
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study wrote: Attachment: Line.png All points (x,y) that lie below the line l, shown above, satisfy which of the following inequalities? A. y < 2x + 3 B. y < 2x + 3 C. y < x + 3 D. y < 1/2*x + 3 E. y < 1/2*x + 3 Frankly, I did this question without any calculation. I hope my approach helps you save time. First step: We have equation: \(y = ax + b\) in which a is the slope of the line. I see the line "l" passes through quadrant II and IV ==> The slope of line "l" should be negative==> A, D are out immediately. Second step.We see two points, say A (6, 0) and B (0, 3) on line "l". Let plug in one point, say A (6,0) to B, C, E ==> C is out Let plug in the second point, say B (0,3) to D & E ==> D is out Only E remains and is correct. Hope it helps. PS: You can save a lot of time by using "plug in" method
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Re: All points (x,y) that lie below the line l, satisfy which of [#permalink]
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28 Jun 2013, 01:55
Bunuel wrote: Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HEREWe have X intercept as 6 and Y intercept as 3 The equation of a line having a as X intercept and b as Y intercept is \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 So the equation of the line above would be \(\frac{x}{6}\) + \(\frac{y}{3}\) = 1 > Y = \(\frac{1x}{2}\) + 3 All the points below the line would satisfy the inequality Y < \(\frac{1x}{2}\) + 3 Hence Choice E
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Re: All points (x,y) that lie below the line l, shown above [#permalink]
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30 Jun 2013, 02:59
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study wrote: Attachment: Line.png All points (x,y) that lie below the line l, shown above, satisfy which of the following inequalities? A. y < 2x + 3 B. y < 2x + 3 C. y < x + 3 D. y < 1/2*x + 3 E. y < 1/2*x + 3 As an alternative solution to this question i suggest to plugin 0 for both x and y to find the x and y intercepts. From the graph it is clearly seen that the values of y should be less than 3 and the values of x should be less than 6 so ideally when we find the x and y intercepts should get the y<3 and x<6. a) x=0 then y<3 this part works; y=0, x>1,5 not our target; b) x=0 then y<3 this part works; y=0, x<1,5 not our target; c) x=0 then y<3 this part works; y=0, x<3 not our target; d) x=0 then y<3 this part works; y=0, x>6 not our target; e) x=0 then y<3 this part works; y=0, x<6 BINGO! E is the line in the graph satisfies y< 1/2*x + 3. This method seems timeconsuming but for those who forget the functions of slope and othe formulas this is very basic visual solution. It took just under 2 min, plus just from one glance it is seen that y<3 in all options so no need to spend time for y, just concentrate to find option which satisfies for x. Hope that helps!
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Re: All points (x,y) that lie below the line l, shown above [#permalink]
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23 Aug 2013, 09:59
i just calculated the intercepts for x and y.onle E suffice and slope match



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Re: All points (x,y) that lie below the line l, shown above [#permalink]
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13 Oct 2013, 02:19
All points (x,y) that lie below the line l, shown above, satisfy which of the following inequalities? A. y < 2x + 3 B. y < 2x + 3 C. y < x + 3 D. y < 1/2*x + 3 E. y < 1/2*x + 3 The easiest approach. Firstly, you have to know that a line which slopes downwards from left to right ALWAYS has ve slope =====>options A and D are eliminated WHY? ALL equations are of form y=mx+c, where m is slope. Now, check yintercept i.e., c in the options B,C and E ALL are 3 and in the diagram also yintercept is 3 So, at this point you cannot eliminate any options on basis of yintercept\ BUT note that in the options slopes for all 3 options is DIFFERENTthis is what you should attack. HOW? Observe diagram: two coordinates on the line are (0,3) and (6,0) slope=03/60=1/2 As line is ve sloped(discussed earlier) m=1/2 Only one options exists with this slope and its E.



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Re: All points (x,y) that lie below the line l, shown above [#permalink]
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Took me 5 seconds to figure out the answer: When you look at this graph, you can write it right away in this form: y=mx+b b=point on y coordinate m=negative > decreasing m=positive > increasing m=1/2 = the line goes from b point to (2,2), as the next y,x integer cross, and (4,1) as the next, and (6,0) next (basically 1 in slope means that it goes 1 down, and 2 means it goes 2 down (if slope is negative, in positive one 1 means one up, 2 means 2 right). So just by looking at (6,0) point and b=3 you can firmly say that the line is gonna be y=1/2x+3. So for instance if you draw line with b=2, and slope 5/3, you will get this green line (y=5/3x+2), and if you draw line with b=2 and slope 5/3, you will get orange line (y=5/3x+2)
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Re: All points (x,y) that lie below the line l, shown above [#permalink]
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Re: All points (x,y) that lie below the line l, shown above [#permalink]
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Bunuel wrote: All points (x,y) that lie below the line l, shown above, satisfy which of the following inequalities? A. y < 2x + 3 B. y < 2x + 3 C. y < x + 3 D. y < 1/2*x + 3 E. y < 1/2*x + 3 First of all we should write the equation of the line \(l\): We have two points: A(0,3) and B(6,0). Equation of a line which passes through two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(\frac{yy_1}{xx_1}=\frac{y_1y_2}{x_1x_2}\)So equation of a line which passes the points A(0,3) and B(6,0) would be: \(\frac{y3}{x0}=\frac{30}{06}\) > \(2y+x6=0\) > \(y=\frac{1}{2}x+3\) Points below this line satisfy the inequality: \(y<\frac{1}{2}x+3\) ORThe equation of line which passes through the points \(A(0,3)\) and \(B(6,0)\) can be written in the following way: Equation of a line in point intercept form is \(y=mx+b\), where: \(m\) is the slope of the line and \(b\) is the yintercept of the line (the value of \(y\) for \(x=0\)). The slope of a line, \(m\), is the ratio of the "rise" divided by the "run" between two points on a line, thus \(m=\frac{y_1y_2}{x_1x_2}\) >\(\frac{30}{06}=\frac{1}{2}\) and \(b\) is the value of \(y\) when \(x=0\) > A( 0,3) > \(b=3\). So the equation is \(y=\frac{1}{2}x+3\) Points below this line satisfy the inequality: \(y<\frac{1}{2}x+3\). Actually one could guess that the answer is E at the stage of calculating the slope \(m=\frac{1}{2}\), as only answer choice E has the same slope line in it. Answer: E. For more please check Coordinate Geometry chapter of the Math Book (link in my signature). Hope it's clear. HI Bunuel, Could you please clarify following statement. Points below this line satisfy the inequality: y<\frac{1}{2}x+3



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