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If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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23 Feb 2014, 07:41
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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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23 Feb 2014, 07:42
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SOLUTIONIf \(kmn\neq{0}\), is \(\frac{x}{m}*(m^2+n^2+k^2)=xm+yn+zk\)?Is \(\frac{x}{m}(m^2+n^2+k^2)=xm+yn+zk\)? >multiply both part by \(m\), to get rid of fraction part and open the brackets > \(xm^2+xn^2+xk^2=xm^2+ynm+zkm\) > \(xm^2\) will cancel out and the question becomes is \(xn^2+xk^2=ynm+zkm\) (1) \(\frac{z}{k}=\frac{x}{m}\) > \(zm=kx\) > substitute zm with kx > is \(xn^2+xk^2=ynm+xk^2\) > \(xk^2\) will cancel out and the question becomes is "\(xn^2=ynm\)?" Not sufficient. (2) \(\frac{x}{m}=\frac{y}{n}\) > \(xn=ym\) > substitute ym with xn > is \(xn^2+xk^2=xn^2+zkm\) > \(xn^2\) will cancel out and the question becomes "is \(xk^2=zkm\)?" Not sufficient. (1)+(2) is \(xn^2+xk^2=ynm+zkm\)? > substitute in from (1) and (2) > is \(xn^2+xk^2=xn^2+xk^2\)? Answer is YES. Sufficient. Answer: C.
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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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24 Feb 2014, 03:25
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Option C. From S1: the ques:Is xm+(xn^2)/m + (xk^2)/m=xm+yn+zk ? Replacing x/m by z/k,we get Is xm+(zn^2)/k+zk=xm+yn+zk? Cancelling the common terms on both sides,we still can't tell whether the 2 sides are equal or not.Not suff.
From S2:Is xm+yn+(yk^2)/n=xm+yn+zk ? Cancelling the common terms we still can't tell whether the 2 sides are equal.Not suff.
Combining the 2 terms,we can replace the terms appropriately to find they are equal. LHS:xm +yn +zk=RHS.



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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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24 Feb 2014, 04:44
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Simplifying the question statement given: x/y*(m^2+n^2+k^2)=xm+yn+zk x/y*m^2+x/y*n^2+x/y*k^2 = xm+yn+zk
Thus simplified equation: xm+x/y*n^2+x/y*k^2 = xm+yn+zk
Now Statement 1: z/k = x/m Substituting this in the simplified equation above we get  xm+x/y*n^2+zk = xm+yn+zk Since no other info is provided this statement is insufficient.
Note  Looking at this itself u can figure out that the only info needed to prove both sides of the above equation equal is x/m = y/n. This can be achieved by combining both the statements. So from here itself u can directly conclude that Option (C) is the answer
Statement 2: x/m = y/n Substituting this in the simplified equation above we get  xm+yn+x/y*k^2 = xm+yn+zk Since no other info is provided this statement is insufficient.
Combining both the statements we get x/m = y/n = z/k Substituting this in the simplified equation we get xm+yn+zk = xm+yn+zk
Thus both sides are proved to be equal. Hence Option (C) is the answer.
This question can be solved much faster as explained earlier since both statements are exactly the same. i.e they provide exactly half of the solution.



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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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28 Aug 2014, 13:28
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I use plug in approach and find it easier to understand:
Does x/m(m^2 + n^2 + k^2) = xm + yn + zk ?
1. let z/k = 1/2 = x/m = 2/4, then the above becomes: Does 1/2(16 + n^2 + 4) = 8 + yn + 2 >insuficient
2. let x/m = 2/4 = y/n = 4/8, can guess that we will have something just like (1) > insuficient
1 and 2 together, using the plug in number, we have: Does 1/2 (16 + 84 + 4) = 8 + 32 + 2 > Yes > Answer is C



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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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09 May 2015, 23:58
Bunuel wrote: SOLUTION
If \(kmn\neq{0}\), is \(\frac{x}{m}*(m^2+n^2+k^2)=xm+yn+zk\)?
Is \(\frac{x}{m}(m^2+n^2+k^2)=xm+yn+zk\)? >multiply both part by \(m\), to get rid of fraction part and open the brackets > \(xm^2+xn^2+xk^2=xm^2+ynm+zkm\) > \(xm^2\) will cancel out and the question becomes is \(xn^2+xk^2=ynm+zkm\)
(1) \(\frac{z}{k}=\frac{x}{m}\) > \(zm=kx\) > substitute zm with kx > is \(xn^2+xk^2=ynm+xk^2\) > \(xk^2\) will cancel out and the question becomes is "\(xn^2=ynm\)?" Not sufficient.
(2) \(\frac{x}{m}=\frac{y}{n}\) > \(xn=ym\) > substitute ym with xn > is \(xn^2+xk^2=xn^2+zkm\) > \(xn^2\) will cancel out and the question becomes "is \(xk^2=zkm\)?" Not sufficient.
(1)+(2) is \(xn^2+xk^2=ynm+zkm\)? > substitute in from (1) and (2) > is \(xn^2+xk^2=xn^2+xk^2\)? Answer is YES. Sufficient.
Answer: C. Since this is a DS and not a PS question, couldn't we also solve this without any calculating or simplifying? Statement 1 does not give any information about y, Statement 2 does not give any information about z, Together, Statement 1 + 2 provides information about how every variable relates to one another. Therefore, Answer: C. Could you tell me if my reasoning is flawed?



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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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31 Oct 2015, 17:06
Bunuel wrote: SOLUTION
If \(kmn\neq{0}\), is \(\frac{x}{m}*(m^2+n^2+k^2)=xm+yn+zk\)?
Is \(\frac{x}{m}(m^2+n^2+k^2)=xm+yn+zk\)? >multiply both part by \(m\), to get rid of fraction part and open the brackets > \(xm^2+xn^2+xk^2=xm^2+ynm+zkm\) > \(xm^2\) will cancel out and the question becomes is \(xn^2+xk^2=ynm+zkm\)
(1) \(\frac{z}{k}=\frac{x}{m}\) > \(zm=kx\) > substitute zm with kx > is \(xn^2+xk^2=ynm+xk^2\) > \(xk^2\) will cancel out and the question becomes is "\(xn^2=ynm\)?" Not sufficient.
(2) \(\frac{x}{m}=\frac{y}{n}\) > \(xn=ym\) > substitute ym with xn > is \(xn^2+xk^2=xn^2+zkm\) > \(xn^2\) will cancel out and the question becomes "is \(xk^2=zkm\)?" Not sufficient.
(1)+(2) is \(xn^2+xk^2=ynm+zkm\)? > substitute in from (1) and (2) > is \(xn^2+xk^2=xn^2+xk^2\)? Answer is YES. Sufficient.
Answer: C. managed to do this in 25 seconds by simply seeing through that all 3 fractions are the same before the parenthesis on the left side of the equation from the stem. In visual terms, seeing this saves times aha lot:
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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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29 Jul 2016, 10:09
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Can someone please check my work to see if this is a valid approach:
x/m (m^2 + n^2 + k^2) = xm + yn + zk > n^2x + k^2x = mny + kmz > n^2x  mny = kmz  k^2y > n(nx  my) = k (mz  kx)
1) mz = kx > mz  kx = 0 n(nx  my) = 0 > Not Sufficient
2) nx = my > nx  my = 0 0 = k (mz  kx) > Not Sufficient
T) 0 = 0 > Sufficient



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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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29 Jul 2016, 10:24
charronecrom wrote: Can someone please check my work to see if this is a valid approach:
x/m (m^2 + n^2 + k^2) = xm + yn + zk > n^2x + k^2x = mny + kmz > n^2x  mny = kmz  k^2y > n(nx  my) = k (mz  kx)
1) mz = kx > mz  kx = 0 n(nx  my) = 0 > Not Sufficient
2) nx = my > nx  my = 0 0 = k (mz  kx) > Not Sufficient
T) 0 = 0 > Sufficient Yes, this way also you can solve this question. It's a valid approach.
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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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17 Sep 2017, 01:57
Bunuel wrote: SOLUTION
If \(kmn\neq{0}\), is \(\frac{x}{m}*(m^2+n^2+k^2)=xm+yn+zk\)?
Is \(\frac{x}{m}(m^2+n^2+k^2)=xm+yn+zk\)? >multiply both part by \(m\), to get rid of fraction part and open the brackets > \(xm^2+xn^2+xk^2=xm^2+ynm+zkm\) > \(xm^2\) will cancel out and the question becomes is \(xn^2+xk^2=ynm+zkm\)
(1) \(\frac{z}{k}=\frac{x}{m}\) > \(zm=kx\) > substitute zm with kx > is \(xn^2+xk^2=ynm+xk^2\) > \(xk^2\) will cancel out and the question becomes is "\(xn^2=ynm\)?" Not sufficient.
(2) \(\frac{x}{m}=\frac{y}{n}\) > \(xn=ym\) > substitute ym with xn > is \(xn^2+xk^2=xn^2+zkm\) > \(xn^2\) will cancel out and the question becomes "is \(xk^2=zkm\)?" Not sufficient.
(1)+(2) is \(xn^2+xk^2=ynm+zkm\)? > substitute in from (1) and (2) > is \(xn^2+xk^2=xn^2+xk^2\)? Answer is YES. Sufficient.
Answer: C. Statement 1 does not give any information about y, Statement 2 does not give any information about z, Together, Statement 1 + 2 provides information about how every variable relates to one another. Therefore, Answer: C. Could you tell me if my reasoning is flawed?
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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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24 Sep 2017, 07:10
abhimahna wrote: charronecrom wrote: Can someone please check my work to see if this is a valid approach:
x/m (m^2 + n^2 + k^2) = xm + yn + zk > n^2x + k^2x = mny + kmz > n^2x  mny = kmz  k^2y > n(nx  my) = k (mz  kx)
1) mz = kx > mz  kx = 0 n(nx  my) = 0 > Not Sufficient
2) nx = my > nx  my = 0 0 = k (mz  kx) > Not Sufficient
T) 0 = 0 > Sufficient Yes, this way also you can solve this question. It's a valid approach. abhimahnahi can you please say to me how the manipulation is done to the following ..? "x/m (m^2 + n^2 + k^2) = xm + yn + zk > n^2x + k^2x = mny + kmz > n^2x  mny = kmz  k^2y " thanks in advance...



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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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24 Sep 2017, 08:58
gmatcracker2017 wrote: abhimahnahi can you please say to me how the manipulation is done to the following ..? "x/m (m^2 + n^2 + k^2) = xm + yn + zk > n^2x + k^2x = mny + kmz > n^2x  mny = kmz  k^2y " thanks in advance... Hi gmatcracker2017 , Here I go: x/m (\(m^2\) +\(n^2\) + \(k^2\)) = xm + yn + zk xm + x\(n^2\)/m + x\(k^2\)/m = xm + yn + zk x\(n^2\)/m + x\(k^2\)/m = yn + zk x\(n^2\) + x\(k^2\) = m ( yn + zk) x\(n^2\) + x\(k^2\) = myn + mzk x\(n^2\)  myn = mzk  x\(k^2\) n ( xn  my ) = k(mz kx) Does that make sense?
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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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24 Sep 2017, 10:09
abhimahna wrote: gmatcracker2017 wrote: abhimahnahi can you please say to me how the manipulation is done to the following ..? "x/m (m^2 + n^2 + k^2) = xm + yn + zk > n^2x + k^2x = mny + kmz > n^2x  mny = kmz  k^2y " thanks in advance... Hi gmatcracker2017 , Here I go: x/m (\(m^2\) +\(n^2\) + \(k^2\)) = xm + yn + zk xm + x\(n^2\)/m + x\(k^2\)/m = xm + yn + zk x\(n^2\)/m + x\(k^2\)/m = yn + zk x\(n^2\) + x\(k^2\) = m ( yn + zk) x\(n^2\) + x\(k^2\) = myn + mzk x\(n^2\)  myn = mzk  x\(k^2\) n ( xn  my ) = k(mz kx) Does that make sense? hi man I must thank you for your elaboration.... It can be seen that you have omitted "xm" from both sides, how..? please say to me .. thanks in advance



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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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24 Sep 2017, 11:02
gmatcracker2017 wrote: hi man
I must thank you for your elaboration....
It can be seen that you have omitted "xm" from both sides, how..? please say to me ..
thanks in advance Hi gmatcracker2017 , xm is same on both sides, hence they can be cancelled out. It's similar to if you move xm on the RHS to LHS. you will get xm  xm = 0 Does that make sense?
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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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24 Sep 2017, 11:17
abhimahna wrote: gmatcracker2017 wrote: hi man
I must thank you for your elaboration....
It can be seen that you have omitted "xm" from both sides, how..? please say to me ..
thanks in advance Hi gmatcracker2017 , xm is same on both sides, hence they can be cancelled out. It's similar to if you move xm on the RHS to LHS. you will get xm  xm = 0 Does that make sense? okay that's fine how the below has been done ...? "n^2x + k^2x " please ...



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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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24 Sep 2017, 11:26
gmatcracker2017 wrote: okay that's fine how the below has been done ...?
"n^2x + k^2x "
please ... Do you know this rule? a(b+c) = ab + bc I used the same. Multiplied x/m with the elements inside the parenthesis. Also, do you know this? a/b (c + d) = f can be written as a (c +d) = f * b Try to use these rules, simplify the things and you will understand the concepts.
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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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25 Sep 2017, 00:15
abhimahna wrote: gmatcracker2017 wrote: okay that's fine how the below has been done ...?
"n^2x + k^2x "
please ... Do you know this rule? a(b+c) = ab + bc I used the same. Multiplied x/m with the elements inside the parenthesis. Also, do you know this? a/b (c + d) = f can be written as a (c +d) = f * b Try to use these rules, simplify the things and you will understand the concepts. hi thanks a lot for your generosity .... If you have a close look at it, you will find that the expression says " n is raised to the power (2x), and k is raised to the power (2x)" that is "n^2x + k^2x" charronecrom equated "n^2x + k^2x with mny and kmz" and you have equated "xn^2 + xk^2 with mny and kmz" obviously, "n^ 2x, + k^ 2x", and "xn^2 + xk^2" cannot be the same in magnitude .. yes, now I understand, he meant (n^2) into x and (k^2) into x thanks a lot



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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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22 Oct 2017, 21:46
statement 1 and statement 2 does not give a stable answer.when we combine 1 and 2,if the equality in both the statement is true,then we have yes. What if xn^2 >ynm in statement 1.Then we might get different answer right? Kindly clarify experts.



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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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29 Oct 2017, 11:47
Very useful question, tried to approach it but I had no idea how to solve it. How did you know that you had to simplify the first equation in that specific way? Bunuel: if I have no clue on how to manipulate the equations in order to get favorable combinations of variables, is it always the best way to simplify it in a straight way till the end? Thanks a lot for your support!



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Re: If kmn ≠ 0, is x/m*(m^2 + n^2 + k^2) = xm + yn + zk? [#permalink]
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26 Dec 2017, 10:37
[quote="Bunuel"] The Official Guide For GMAT® Quantitative Review, 2ND EditionIf \(kmn\neq{0}\), is \(\frac{x}{m}*(m^2+n^2+k^2)=xm+yn+zk\)? (1) z/k = x/m (2) x/m = y/n Given \(\frac{x}{m}*(m^2+n^2+k^2)=xm+yn+zk\) where \(kmn\neq{0}\) Find Is LHS=RHS in above equation Now LHS=\(\frac{x}{m}*(m^2+n^2+k^2)\) => Opening the bracket & Multiplying \(\frac{x}{m}\) to all 3 terms AND rearranging the terms we have => \(xm+n*\frac{xn}{m}+k*\frac{xk}{m}\) => Therefore if \(\frac{xn}{m}=y\) AND \(\frac{xk}{m}=z\) then LHS=RHSStatement 1 \(\frac{z}{k} = \frac{x}{m}\) => OR \(z = \frac{xk}{m}\) => Thus 'z' known BUT 'y' NOT known => Therefore NOT SUFFICIENT Statement 2 \(\frac{x}{m} = \frac{y}{n}\) => OR \(y = \frac{xn}{m}\) => Thus 'y' known BUT 'z' NOT known => Therefore NOT SUFFICIENT BOTH Statment 1 & 2 => \(z = \frac{xk}{m}\)  from 1 => \(y = \frac{xn}{m}\)  from 2 => Thus both 'y' and 'z' make LHS=RHS=> Therefore SUFFICIENT Therefore 'C' Thanks Dinesh




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