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ValleyBall1
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Is x/m(m^2 + n^2 + k^2) = xm + yn + zk? (1) z/k = x/m (2) 27 Oct 2005, 16:35
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

55% (hard)

Question Stats:

60% (04:02) correct 40% (01:58) wrong based on 5 sessions

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Is x/m(m^2 + n^2 + k^2) = xm + yn + zk?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-m-m-2-n-2-k-2-xm-yn-zk-99423.html



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C
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conocieur
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Is x/m(m^2 + n^2 + k^2) = xm + yn + zk? (1) z/k = x/m (2) 27 Oct 2005, 16:49
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I think is C

x/m(m^2 + n^2 + z^2) = xm + yn + zk

that is equal to

xm + xn^2/m + xz^2/n = xm + yn + zk

from I: x/m = y/n then xn = ym and from there (xn*n)/m is equal to
ym*n/m equal to yn

so insuff

from ii, do the same stuff as i and you get zk

so insuff

taking both it works so C
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Is x/m(m^2 + n^2 + k^2) = xm + yn + zk? (1) z/k = x/m (2) 28 Oct 2005, 15:54
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Right again - OA is C.
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Is x/m(m^2 + n^2 + k^2) = xm + yn + zk? (1) z/k = x/m (2) 18 Nov 2011, 03:39
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Is x/m(m^2 + n^2+ k^2) = xm + yn + zk? 

xm^2 + xn^2 + xk^2 = xmm +ynm +zkm?

Is xn^2 + xk^2 = ynm + zkm?

(1) z/k = x/m (xk = mz)

xk^2 = (xk)k = (mzk)

xn^2 + xk^2 = ynm + zkm =

xn^2 = ynm?
xn = my?

(2) x/m = y/n (xn = my)

xn^2 = (xn)n = (my)n

xn^2 + xk^2 = ynm + zkm =

xk^2 = zkm?
xk = mz?

Put (1) and (2) together 

xn^2 + xk^2 = ynm + zkm =

(my)n + (mz)k = ynm + zkm

Both statements used = Answer C

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Is x/m(m^2 + n^2 + k^2) = xm + yn + zk? (1) z/k = x/m (2) 18 Nov 2011, 05:19
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is xm^2+xn^2+xk^2=xm^2+ynm+zkm
i. e. xn^2+xk^2=ynm+zkm
statement 1: xk=zm
xn^2=ynm Not Sufficient
Statement 2: ym=xn
xk^2=zkm Not Sufficient
Together: xn^2+xk^2=xn^2+xk^2
Sufficient
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Is x/m(m^2 + n^2 + k^2) = xm + yn + zk? (1) z/k = x/m (2) 23 Nov 2017, 07:43
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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: is \(xm^2+xn^2+xk^2=xm^2+ynm+zkm\)?

\(xm^2\) 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 \(xn^2+xk^2=ynm+zkm\)? --> substitutein from (1) and (2) --> is \(xn^2+xk^2=xn^2+xk^2\)? Answer is YES. Sufficient.

Answer: C.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-m-m-2-n-2-k-2-xm-yn-zk-99423.html



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Thank you for understanding, and happy exploring!
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