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Re: Co-Ordinate Geometry!! [#permalink]
20 Feb 2010, 23:42
line m and n pass through point (1,2). Is the slope of m greater than the slope of n?
1.) the x-intercept of m is greater than 1 and that of n is less than 1 m has negative slope as x intercept is greater than 1. take a point (2,0) and draw a line n has positive slope. take a point (0,0) and draw a line Sufficient
2.) the y-intercept of m = 4 and that of n = (-2) m has negative slope n has positive slope sufficient
Re: Co-Ordinate Geometry!! [#permalink]
21 Feb 2010, 02:06
4
This post received KUDOS
apoorvasrivastva wrote:
line m and n pass through point (1,2). Is the slope of m greater than the slope of n?
1.) the x-intercept of m is greater than 1 and that of n is less than 1 2.) the y-intercept of m = 4 and that of n = (-2)
D...
Best is to draw graphically. S1: gives u a range for x intercept. But since you have (1,2) fixed, you would see.. Line m has a -ve slope (in the given range) and Line n has a positive slope. Hence SUFF.
S2: Makes it much more clear by giving u two other points. Hence SUFF.
Attachments
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_________________
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|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|
Re: Co-Ordinate Geometry!! [#permalink]
21 May 2010, 05:51
Which each statment you can know that one slope is positive (n) and the other one is negative (m). So one positive number is greater than a negative one = SUF I hope it helps
Re: Co-Ordinate Geometry!! [#permalink]
21 May 2010, 15:49
line m and n pass through point (1,2). Is the slope of m greater than the slope of n?
1.) the x-intercept of m is greater than 1 and that of n is less than 1 2.) the y-intercept of m = 4 and that of n = (-2)
Aren't we missing something ?
To compare the slope of two lines m and n in the context of the probelm, we must not consider the sign of the slope. Question says which slope is greater. By definition a slope indicates how stiff the line is, other way to say, how difficult it is to walk on the line. A slope is measured by (change in Y) / (change is X). A greater change in Y and lesser change in X give stiffer slope. Therefore, even a negative slope can be stiffer than a positive slope. It just happens that a negative slope is in other direction.
To answer the question, I think we need actual value of the slope of two lines, which can be derived from statement 2.
Answer is B.
What is the rule GMAT follow to compare the slope of two lines ? Does the GMAT compare actual value of the slope or absolute value of the slope ? In my understanding, it is absolute value . However , we must follow whatever GMAT follows.
clarification request to Bunuel.... [#permalink]
28 Apr 2012, 14:56
Dear Bunuel, Can you please explain to me that if the question asks to compare the slope, are we suppose to consider the absolute value of slope of the real value? I have seen in the previous examples, you considering the absolute value and in this example, they have compared the real value . It seems inconsistent. Thanks
Re: clarification request to Bunuel.... [#permalink]
03 May 2012, 10:41
1
This post received KUDOS
Expert's post
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Parthomazumdar wrote:
Dear Bunuel, Can you please explain to me that if the question asks to compare the slope, are we suppose to consider the absolute value of slope of the real value? I have seen in the previous examples, you considering the absolute value and in this example, they have compared the real value . It seems inconsistent. Thanks
The question asks: "Is the slope of m greater than the slope of n?" So, we should NOT compare the absolute value of the slopes.
Notice that a higher absolute value of a slope indicates a steeper incline. So:
If the slopes of lines k and l are positive and line k is steeper then it will have the greater slope. If the slopes of lines k and l are negative and line k is steeper then its slope is more negative then the slope of line l (the absolute value of k's slope is greater), which means that the slope of l is greater than the slope of k.
Re: Line m and n pass through point (1,2). Is the slope of m [#permalink]
21 Jun 2012, 07:27
Expert's post
2
This post was BOOKMARKED
pavanpuneet wrote:
With condition 1, when we try to solve algebraic way:
x intercept of line m : -b1/m1>1 and for line b : -b2/m2<1
Given that we do not the sign of slopes m1 and m2, we cant cross multiply, then how do we proceed after this condition?
ALGEBRAIC APPROACH.
Line m and n pass through point (1,2). Is the slope of m greater than the slope of n?
Given: lines \(y_m=mx+b\) and \(y_n=nx+c\) pass through point (1,2). Hence: \(2=m+b\) and \(2=n+c\). Question asks: is \(m>n\)?
(1) The x-intercept of m is greater than 1 and that of n is less than 1. The x-intercept is the value of \(x\) when \(y=0\), so from this statement we have that:
\(-\frac{b}{m}>1\). Now, since from the stem \(b=2-m\), then \(-\frac{2-m}{m}>1\) --> \(\frac{m-2}{m}>1\) --> \(\frac{m}{m}-\frac{2}{m}>1\) --> \(1-\frac{2}{m}>1\) --> \(\frac{2}{m}<0\) --> \(m<0\);
\(-\frac{c}{n}<1\). Now, since from the stem \(c=2-n\), then \(-\frac{2-n}{n}<1\) --> \(\frac{n-2}{n}<1\) --> \(\frac{n}{n}-\frac{2}{n}<1\) --> \(1-\frac{2}{n}<1\) --> \(\frac{2}{m}>0\) --> \(n>0\);
So, we have that \(m<0<n\). Sufficient.
(2) The y-intercept of m = 4 and that of n = (-2). The y-intercept is the value of \(y\) for \(x=0\), so from this statement we have that:
\(b=4\). Now, since from the stem \(b=2-m\), then \(4=2-m\) --> \(m=-2\); \(c=-2\). Now, since from the stem \(c=2-n\), then \(-2=2-n\) --> \(n=4\);
Re: Line m and n pass through point (1,2). Is the slope of m [#permalink]
23 Jan 2014, 11:09
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Re: Line m and n pass through point (1,2). Is the slope of m [#permalink]
23 Jan 2014, 13:56
Bunuel wrote:
pavanpuneet wrote:
With condition 1, when we try to solve algebraic way:
x intercept of line m : -b1/m1>1 and for line b : -b2/m2<1
Given that we do not the sign of slopes m1 and m2, we cant cross multiply, then how do we proceed after this condition?
ALGEBRAIC APPROACH.
Line m and n pass through point (1,2). Is the slope of m greater than the slope of n?
Given: lines \(y_m=mx+b\) and \(y_n=nx+c\) pass through point (1,2). Hence: \(2=m+b\) and \(2=n+c\). Question asks: is \(m>n\)?
(1) The x-intercept of m is greater than 1 and that of n is less than 1. The x-intercept is the value of \(x\) when \(y=0\), so from this statement we have that:
\(-\frac{b}{m}>1\). Now, since from the stem \(b=2-m\), then \(-\frac{2-m}{m}>1\) --> \(\frac{m-2}{m}>1\) --> \(\frac{m}{m}-\frac{2}{m}>1\) --> \(1-\frac{2}{m}>1\) --> \(\frac{2}{m}<0\) --> \(m<0\);
\(-\frac{c}{n}<1\). Now, since from the stem \(c=2-n\), then \(-\frac{2-n}{n}<1\) --> \(\frac{n-2}{n}<1\) --> \(\frac{n}{n}-\frac{2}{n}<1\) --> \(1-\frac{2}{n}<1\) --> \(\frac{2}{m}>0\) --> \(n>0\);
So, we have that \(m<0<n\). Sufficient.
(2) The y-intercept of m = 4 and that of n = (-2). The y-intercept is the value of \(y\) for \(x=0\), so from this statement we have that:
\(b=4\). Now, since from the stem \(b=2-m\), then \(4=2-m\) --> \(m=-2\); \(c=-2\). Now, since from the stem \(c=2-n\), then \(-2=2-n\) --> \(n=4\);
So, we have that \(m=-2<4=n\). Sufficient.
Answer: D.
Hope it's clear.
Hi Bunel I shall try to give a easier approach. If a line's X intercept is ''a'' what information we get from that . Line pasess through point (a,0) similarly if a line's Y intercept is "b" what information we get from that . Line pasess through point (0,b) and if a line pasess through points (x1,y1)and (x2,y2) we compute the slope of the line by (y2-y1)/x2-x1). so let us apply this to the present question Line m and n pass through point (1,2). Is the slope of m greater than the slope of n? 1.The x-intercept of m is greater than 1 and that of n is less than 1 so x intercept of m is greater than 1 so it should pass through points (>1,0) so let us assume its (2,0) x intercept of n is less than 1 so it should pass through points (<1,0) so let us assume its (-1,0) you can actually take any point less than 1 even (0.9,0) also. so now computing slope with slope formula you can find slope of n > slope of m hence sufficient. 2.The y-intercept of m = 4 and that of n = (-2) i.e. m pasess through point (0,4) and n pasess through point (0,-2) and you can straight away compute the slope and determine which is greater IMP: It is important to understand what is meant by intercept hope this helps This would avoid the larger equations. Give me kudos if this helps
Re: Line m and n pass through point (1,2). Is the slope of m [#permalink]
14 Jul 2015, 20:32
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).
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Re: Line m and n pass through point (1,2). Is the slope of m [#permalink]
08 Dec 2015, 17:44
I solved it easily by picturing the lines in the graph. 1 - if x intercept of m is greater then 1, we can automatically deduct that line m has a negative slope. we are also told that x intercept of line n is less than 1, which only gives 1 possibility - slope of line n is positive. thus, this is sufficient to answer the question.
2. we are told that y-intercept of line m is 4, which is above the point of intersection. the only case when this is possible is when slope of m is negative. we are also told the y-intercept of line n, which is -2. since the point of intersection is above, it clearly means that the slope is positive. sufficient to answer the question.
another method is finding the slope of one line, then finding the slope of another line. Both statements, individually, are sufficient to do so, since for each line we have 2 points, which is enough to find the slope.
Re: Line m and n pass through point (1,2). Is the slope of m [#permalink]
17 Dec 2015, 10:31
Slope is defined as rise by run. From the question we know a point on the line. From the intercepts we can detect which has a higher rise and run. Each Statement alone is sufficient
Hence, D. _________________
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Re: Line m and n pass through point (1,2). Is the slope of m
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17 Dec 2015, 10:31
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