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# The equation of line k is y=mx+b, where m and b are constant

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Manager
Joined: 21 Sep 2010
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The equation of line k is y=mx+b, where m and b are constant [#permalink]

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29 Apr 2014, 12:21
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Difficulty:

95% (hard)

Question Stats:

37% (01:01) correct 63% (01:01) wrong based on 191 sessions

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The equation of line k is y=mx+b, where m and b are constants. What is the value of m?

1) (b,2b) belongs to k.
2) (2,2) belongs to k.

Had a difficult time understanding the explanation. Anyone want to take a stab?

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Joined: 25 Apr 2012
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Re: The equation of line k is y=mx+b, where m and b are constant [#permalink]

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29 Apr 2014, 22:05
1
TwoThrones wrote:
The equation of line k is y=mx+b, where m and b are constants. What is the value of m?

1) (b,2b) belongs to k.
2) (2,2) belongs to k.

Had a difficult time understanding the explanation. Anyone want to take a stab?

I will give it a try

St1 says (b,2b) belong to k so, we have
2b=mb+b or b=mb or (m-1)*b=0 ----->This implies either m=1 or b=0. St1 is not sufficient. So A and D ruled out

St 2: 2=2m+b or $$m= (2-b)/2$$ or m =1-(b/2).. But we don't know anything about b. So not sufficient. Option B ruled out

Combining we get that b=0 then m=1. Alternatively if m=1 then b=0.

Therefore ans is C
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Re: The equation of line k is y=mx+b, where m and b are constant [#permalink]

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29 Apr 2014, 22:06
3
2
TwoThrones wrote:
The equation of line k is y=mx+b, where m and b are constants. What is the value of m?

1) (b,2b) belongs to k.
2) (2,2) belongs to k.

Had a difficult time understanding the explanation. Anyone want to take a stab?

Stmnt 1:

If by statement 1 you mean that the point (b, 2b) lies on the line k, you can plug in the point in the equation of the line y=mx+b to get 2b = m*b + b.
2b = b(m + 1)
b = mb
(m-1)b = 0
Either m is 1 or b = 0 or both. In case we prove that b is not 0, then m must be 1. If b is 0, m could be anything including 1. Not sufficient.

Stmnt 2:
If line passes through (2, 2), we get 2 = m*2 + b
We have 2 unknowns and only one equation. Not enough to get the value of m.

Using both together, we have
2m + b = 2 .... (I)
such that either b is 0 or m is 1 or both.

If b is 0, put b = 0 in (I) which gives us m = 1
If b is not 0, m must be 1.

So in any case, m is always 1.

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 25 Jan 2014 Posts: 45 Concentration: Strategy, International Business GMAT 1: 600 Q44 V29 GMAT 2: 710 Q48 V38 GPA: 3.35 WE: Analyst (Computer Software) Re: The equation of line k is y=mx+b, where m and b are constant [#permalink] ### Show Tags 15 May 2014, 07:08 Can anybody explain why m cant be 0 ? Intern Joined: 25 Jan 2014 Posts: 45 Concentration: Strategy, International Business GMAT 1: 600 Q44 V29 GMAT 2: 710 Q48 V38 GPA: 3.35 WE: Analyst (Computer Software) Re: The equation of line k is y=mx+b, where m and b are constant [#permalink] ### Show Tags 15 May 2014, 07:11 Also when 2b = mb + b, why can i not cancel b to get m+1 = 2 and m =1? Math Expert Joined: 02 Sep 2009 Posts: 46280 Re: The equation of line k is y=mx+b, where m and b are constant [#permalink] ### Show Tags 15 May 2014, 07:23 gaurav1418z wrote: The equation of line k is y=mx+b, where m and b are constants. What is the value of m? 1) (b,2b) belongs to k. 2) (2,2) belongs to k. Can anybody explain why m cant be 0 ? From (1): m = 1 or b = 0 or both. From (2): 2m + b = 2. If m is not 1, then b must be 0. But if b = 0, then from 2m + b = 2, we get that m = 1. So, in any case, m = 1. Hope it's clear. _________________ Math Expert Joined: 02 Sep 2009 Posts: 46280 Re: The equation of line k is y=mx+b, where m and b are constant [#permalink] ### Show Tags 15 May 2014, 07:26 3 1 gaurav1418z wrote: Also when 2b = mb + b, why can i not cancel b to get m+1 = 2 and m =1? If you divide (reduce) $$2b = mb + b$$ by b you assume, with no ground for it, that b does not equal to zero thus exclude a possible solution (notice that both b = 0 AND m = 1 satisfy the equation). Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. Hope it's clear. _________________ Intern Joined: 25 Jan 2014 Posts: 45 Concentration: Strategy, International Business GMAT 1: 600 Q44 V29 GMAT 2: 710 Q48 V38 GPA: 3.35 WE: Analyst (Computer Software) Re: The equation of line k is y=mx+b, where m and b are constant [#permalink] ### Show Tags 15 May 2014, 23:56 Thank you Bunuel, much appreciated Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 5600 GMAT 1: 800 Q59 V59 GPA: 3.82 Re: The equation of line k is y=mx+b, where m and b are constant [#permalink] ### Show Tags 21 Nov 2015, 09:50 1 Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. The equation of line k is y=mx+b, where m and b are constants. What is the value of m? 1) (b,2b) belongs to k. 2) (2,2) belongs to k. In the original condition, we need to know the gradient and the y-intercept, so we need 2 equations There are 2 equations given by the 2 conditions, so there is high chance (C) will be our answer. Looking at the conditions together, from 2b=mb+b, b=mb, b(m-1)=0, we get b=0 or m=1, and from 2=2m+b, if b=0, m=1. This is sufficient, and the answer becomes (C). For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Re: The equation of line k is y=mx+b, where m and b are constant [#permalink]

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26 Feb 2017, 03:42
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Re: The equation of line k is y=mx+b, where m and b are constant   [#permalink] 26 Feb 2017, 03:42
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