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M03-15

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M03-15  [#permalink]

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New post 15 Sep 2014, 23:20
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

73% (00:18) correct 27% (00:40) wrong based on 163 sessions

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Re M03-15  [#permalink]

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New post 15 Sep 2014, 23:20
Official Solution:


For one line to be perpendicular to another, the relationship between their slopes has to be negative reciprocal, so if the slope of one line is \(m\) then the line perpendicular to it will have the slope \(-\frac{1}{m}\). In other words, the two lines are perpendicular if and only if the product of their slopes is -1.

(1) \(m*n = -1\). Directly gives an answer YES to the question.

(2) \(m = -n\). Now, if for example \(m=3=-(-3)=-n\) then the lines are not perpendicular but if \(m=1=-(-1)=-n\), so if \(m=1\) and \(n=-1\) (or \(m=-1\) and \(n=1\)) then as \(mn\) would be equal to -1 the lines would be perpendicular. Not sufficient.


Answer: A
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M03-15  [#permalink]

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New post 19 Jul 2017, 05:56
Hi Bunuel,
Could you please elaborate a little on Bold part.
(2) m=−nm=−n. Now, if for example m=3=−(−3)=−nm=3=−(−3)=−n then the lines are not perpendicular but if m=1=−(−1)=−nm=1=−(−1)=−n,

So do you mean to say in the second case when we put the value in the reciprocal equation that is essential for perpendicular condition m*n=3*(-3)=-9 which is not the reciprocal i.e equal to -1. But in the case of 1 and -1 it is.
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Re: M03-15  [#permalink]

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New post 19 Jul 2017, 07:35
2
Prashant10692 wrote:
Hi Bunuel,
Could you please elaborate a little on Bold part.
(2) m=−nm=−n. Now, if for example m=3=−(−3)=−nm=3=−(−3)=−n then the lines are not perpendicular but if m=1=−(−1)=−nm=1=−(−1)=−n,

So do you mean to say in the second case when we put the value in the reciprocal equation that is essential for perpendicular condition m*n=3*(-3)=-9 which is not the reciprocal i.e equal to -1. But in the case of 1 and -1 it is.


The two lines are perpendicular if and only if the product of their slopes is -1.

If \(m=3\) and \(n=-3\), then the product mn is not -1, so the lines are not perpendicular.
If \(m=1\) and \(n=-1\), then the product mn is -1, so the lines are perpendicular.

Two different answers, hence not sufficient.
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Re: M03-15  [#permalink]

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New post 29 Sep 2018, 11:36
Bunuel wrote:
Official Solution:


For one line to be perpendicular to another, the relationship between their slopes has to be negative reciprocal, so if the slope of one line is \(m\) then the line perpendicular to it will have the slope \(-\frac{1}{m}\). In other words, the two lines are perpendicular if and only if the product of their slopes is -1.

(1) \(m*n = -1\). Directly gives an answer YES to the question.

(2) \(m = -n\). Now, if for example \(m=3=-(-3)=-n\) then the lines are not perpendicular but if \(m=1=-(-1)=-n\), so if \(m=1\) and \(n=-1\) (or \(m=-1\) and \(n=1\)) then as \(mn\) would be equal to -1 the lines would be perpendicular. Not sufficient.


Answer: A


Hi Bunuel,

Regarding statement 1, is it not possible for M to be 1 and N -1?

This was the conflict I had in this question. I felt it was a trick answer.
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Re: M03-15  [#permalink]

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New post 30 Sep 2018, 03:30
Mfon wrote:
Bunuel wrote:
Official Solution:


For one line to be perpendicular to another, the relationship between their slopes has to be negative reciprocal, so if the slope of one line is \(m\) then the line perpendicular to it will have the slope \(-\frac{1}{m}\). In other words, the two lines are perpendicular if and only if the product of their slopes is -1.

(1) \(m*n = -1\). Directly gives an answer YES to the question.

(2) \(m = -n\). Now, if for example \(m=3=-(-3)=-n\) then the lines are not perpendicular but if \(m=1=-(-1)=-n\), so if \(m=1\) and \(n=-1\) (or \(m=-1\) and \(n=1\)) then as \(mn\) would be equal to -1 the lines would be perpendicular. Not sufficient.


Answer: A


Hi Bunuel,

Regarding statement 1, is it not possible for M to be 1 and N -1?

This was the conflict I had in this question. I felt it was a trick answer.


Even in this case the slope of one line is a negative reciprocal of the slope of another line.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M03-15 &nbs [#permalink] 30 Sep 2018, 03:30
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