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

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New post 19 Aug 2010, 03:29
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Is \(\frac{x}{m}*(m^2+n^2+k^2)=xm+yn+zk\)?


(1) \(\frac{z}{k}=\frac{x}{m}\)

(2) \(\frac{x}{m}=\frac{y}{n}\)

This is just an awful arithmetic problem. Can't I just pick numbers and PIN them? The problem is: as soon as I choose numbers for all of the variables, I ALWAYS get a definite answer...
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Is x/m*(m^2+n^2+k^2)=xm+yn+zk?  [#permalink]

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New post 19 Aug 2010, 07:57
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heyholetsgo wrote:
Is x/m*(m^2+n^2+k^2)=xm+yn+zk?
(1) z/k=x/m
(2) x/m=y/n


This is just an awful arithmetic problem. Can't I just pick numbers and PIN them? The problem is: as soon as I choose numbers for all of the variables, I ALWAYS get a definite answer...


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.
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Re: Is x/m*(m^2+n^2+k^2)=xm+yn+zk?  [#permalink]

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New post 19 Aug 2010, 06:45
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heyholetsgo wrote:
Is x/m(m^2+n^2+k^2)=xm+yn+zk?
1.)z/k=x/m
2.)x/m=y/n


This is just an awful arithmetic problem. Can't I just pick numbers and PIN them? The problem is: as soon as I choose numbers for all of the variables, I ALWAYS get a definite answer...


This is easier than it looks,

\(\frac{x}{m}(m^2+n^2+k^2)\)

\(xm + \frac{x}{m}*n^2 + \frac{x}{m}*k^2\)

Now if you see the second term can be written as \(\frac{x}{m}*n^2= \frac{x}{m}*n *n = y*n\)

Where \(\frac{x}{m}*n = y\)

Similarly for the third term \(z= \frac{x}{m}*k\)


Now if you see both the statements, you will be able to figure out that Statement 1 is giving you the value for Z and Statement 2 is giving you the value of Y

Hope this helps.
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Re: Is x/m*(m^2+n^2+k^2)=xm+yn+zk?  [#permalink]

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New post 19 Aug 2010, 08:50
Bunuel - Your explanations are excellent. They are clear, articulate and to the point. Thank you for your continued support.
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Re: Is x/m*(m^2+n^2+k^2)=xm+yn+zk?  [#permalink]

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New post 19 Aug 2010, 21:03
What is PIN? Picking numbers? I think for numbers with this many variables, it's much better to go for a conceptual approach. It's too easy to make a careless math error, or some weird math property or cancellation comes up.
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Re: Is x/m*(m^2+n^2+k^2)=xm+yn+zk?  [#permalink]

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New post 20 Aug 2010, 08:55
Thanks guys.
Yes, PIN means plugging in numbers. I just prefer doing that.
However, if I do that I will always get a solution! If both sides of the equation are equal, the solution is YES and if both sides of the equation are unequal the solution would be NO. As it is a DS question, it is solvable no matter if it's YES or NO.
But can I do that? It seems weird that I will always get one solution as soon as the ratio of all variables are given!
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Re: Is x/m*(m^2+n^2+k^2)=xm+yn+zk?  [#permalink]

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New post 22 Dec 2010, 13:02
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A question with too many variables should not be done by picking numbers. Takes too much time, plus prone to errors.
then again, if you are short on time, and have barely half a minute, plug in and run.. Let me take both methods:

Proportion: This question tests your understanding of ratios and proportion.

Concept tested: If \(\frac{x}{m} = \frac{y}{n} = \frac{z}{k}\), then \(\frac{x}{m} = \frac{y}{n} = \frac{z}{k} = \frac{{x+y+z}}{{m+n+k}}\)

Question: Is \(\frac{x}{m}(m^2+n^2+k^2)=xm+yn+zk\)?
or Is \(\frac{x}{m}=\frac{xm+yn+zk}{(m^2+n^2+k^2)}\)?

1.)x/m = z/k
Since we have no information about n and y, this is not sufficient. If you are not convinced, put x = 0, z = 0.
Question becomes: Is \(0=\frac{yn}{(m^2+n^2+k^2)}\)?
We do not know since this depends on the values of y and n. Not sufficient.

2.)x/m=y/n
Similarly statement 2 alone is also not sufficient.

Using both together, we get \(\frac{x}{m} = \frac{y}{n} = \frac{z}{k}\)
which is same as \(\frac{x}{m}=\frac{xm}{m^2} = \frac{yn}{n^2} = \frac{zk}{k^2} = \frac{xm+yn+zk}{(m^2+n^2+k^2)}\)

Sufficient. Answer C.


Plugging in numbers:
1. x/m = z/k
Put x = 0, z = 0.
Question becomes: Is \(0=\frac{yn}{(m^2+n^2+k^2)}\)?
We do not know since this depends on the values of y and n. Not sufficient.

2. x/m=y/n
Put x = 0, y = 0.
Question becomes: Is \(0=\frac{zk}{(m^2+n^2+k^2)}\)?
We do not know since this depends on the values of z and k. Not sufficient.

Using both together, put x = 0, y = 0, z = 0
Question becomes: Is 0=0? Answer 'Yes'
Try putting x = 2, y = 2, z = 2, m = 1, n = 1, k = 1
It satisfies so mark the answer as (C) and move ahead. Remember, this is not foolproof. In some case, you could have taken values which satisfied the equation but not those which did not satisfy it. But if you don't have much time to spare, this gives you a decent probability of getting the answer right.
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Re: Is x/m*(m^2+n^2+k^2)=xm+yn+zk?  [#permalink]

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New post 01 May 2011, 06:39
I do not understand why the answer is not A, as one can deduce that x/m = y/n from the first part. Therefore, we do not learn anything new from the second part.

Please help.
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Re: Is x/m*(m^2+n^2+k^2)=xm+yn+zk?  [#permalink]

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New post 01 Jan 2012, 17:05
Interesting question, I also brute force simplified it. Took me just under 4 minutes, is there a short cut?

I first simplified it into xn²/m + k²x/m = yn + zk by multiplying out the x/m from first part and reducing xm

1. z/k = x/m or zm = kx

Using this, the top equation simplifies down to y/n = z/k simply by plugging z/k instead of x/m and replacing kx by zm (basically removing the X).


2. x/m = y/n

After 1. is used to simplify it's just about y/n = z/k = x/m

The more I think about it, the more I see that this can be simplified very quickly.
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Re: Is x/m*(m^2+n^2+k^2)=xm+yn+zk?  [#permalink]

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Re: Is x/m*(m^2+n^2+k^2)=xm+yn+zk?  [#permalink]

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New post 11 Mar 2014, 11:50
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Are we not considering that K and m are not zero when we are moving that other side.

z/k = x/m => zm=xk
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New post 04 May 2014, 03:42
Is x/m*(m^2+n^2+k^2)=xm+yn+zk?

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

Sol.

Expand the expression,
x/m*(m^2+n^2+k^2)=xm+yn+zk => xm + xn^2/m + xk^2/m

try to match with the R.H.S of the expression in question.
What we want is -
xn^2/m = yn i.e. x/m = y/n
similarly, xk^2/m = zk i.e. x/m = z/k

Now have a look at the options.
Opt.1 gives one part of what we need. Not Sufficient.

Opt.2 gives the other part of what is required. Not Sufficient.

Together, we have the info that we are looking for i.e. . Hence, sufficient

Answer C.
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Re: Is x/m*(m^2+n^2+k^2)=xm+yn+zk?  [#permalink]

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New post 18 Sep 2016, 09:04
I struggled with this one and found the explanations to be difficult to follow...

Here's how I walked away with C being the correct choice..

Together - (1) + (2) gives us (z/k) = (y/m) = (y/n) --> looking at the main eq we can rewrite the following way...
=(x/m)(m^2) + (y/n)(n^2) + (z/k)(k^2) & boom we have what we're looking for!

Hope this is helpful!!
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Re: Is x/m*(m^2+n^2+k^2)=xm+yn+zk?  [#permalink]

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New post 25 Nov 2017, 01:07
what if m is negative, can we still take it to the other side?
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