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# If kmn#0, is x/m*(m^2+n^2+k^2)=xm+yn+zk?

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23 Feb 2014, 07:41
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If $$kmn\neq{0}$$, is $$\frac{x}{m}*(m^2+n^2+k^2)=xm+yn+zk$$?

(1) z/k = x/m
(2) x/m = y/n

Data Sufficiency
Question: 106
Category: Algebra First- and second- degree equations
Page: 160
Difficulty: 650

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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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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.

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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
1
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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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01 Mar 2014, 05:14
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.

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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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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.

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.

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.

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 ah-a lot:
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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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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.

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.

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] 17 Sep 2017, 01:57
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