It is currently 21 Oct 2017, 05:50

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 06 Jul 2010
Posts: 76

Kudos [?]: 184 [5], given: 12

### Show Tags

19 Aug 2010, 03:29
5
KUDOS
23
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

76% (01:43) correct 24% (05:27) wrong based on 793 sessions

### HideShow timer Statistics

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...
[Reveal] Spoiler: OA

Kudos [?]: 184 [5], given: 12

Senior Manager
Joined: 25 Jun 2009
Posts: 298

Kudos [?]: 150 [3], given: 6

### Show Tags

19 Aug 2010, 06:45
3
KUDOS
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.

Kudos [?]: 150 [3], given: 6

Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 129094 [10], given: 12194

### Show Tags

19 Aug 2010, 07:57
10
KUDOS
Expert's post
6
This post was
BOOKMARKED
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 --> $$xm^2+xn^2+xk^2=xm^2+ynm+zkm$$ --> $$xm^2$$ 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.

_________________

Kudos [?]: 129094 [10], given: 12194

Intern
Joined: 22 Dec 2009
Posts: 22

Kudos [?]: 71 [0], given: 1

### Show Tags

19 Aug 2010, 08:50
Bunuel - Your explanations are excellent. They are clear, articulate and to the point. Thank you for your continued support.

Kudos [?]: 71 [0], given: 1

Manager
Joined: 06 Apr 2010
Posts: 58

Kudos [?]: 30 [0], given: 13

### Show Tags

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.
_________________

If you liked my post, please consider thanking me with Kudos! I really appreciate it!

Kudos [?]: 30 [0], given: 13

Manager
Joined: 06 Jul 2010
Posts: 76

Kudos [?]: 184 [0], given: 12

### Show Tags

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!

Kudos [?]: 184 [0], given: 12

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7676

Kudos [?]: 17375 [8], given: 232

Location: Pune, India

### Show Tags

22 Dec 2010, 13:02
8
KUDOS
Expert's post
3
This post was
BOOKMARKED
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 $$\frac{x}{m}=\frac{xm}{m^2} = \frac{yn}{n^2} = \frac{zk}{k^2} = \frac{xm+yn+zk}{(m^2+n^2+k^2)}$$

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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17375 [8], given: 232

Intern
Joined: 01 May 2011
Posts: 1

Kudos [?]: [0], given: 0

### Show Tags

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.

Kudos [?]: [0], given: 0

Intern
Joined: 19 Dec 2009
Posts: 36

Kudos [?]: 10 [0], given: 8

### Show Tags

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.

Kudos [?]: 10 [0], given: 8

Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 129094 [1], given: 12194

### Show Tags

03 Jul 2013, 01:18
1
KUDOS
Expert's post
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Algebra: algebra-101576.html

DS Algebra Questions to practice: search.php?search_id=tag&tag_id=29
PS Algebra Questions to practice: search.php?search_id=tag&tag_id=50

Special algebra set: new-algebra-set-149349.html

_________________

Kudos [?]: 129094 [1], given: 12194

Manager
Joined: 17 Oct 2013
Posts: 56

Kudos [?]: 29 [1], given: 0

### Show Tags

11 Mar 2014, 11:50
1
KUDOS
Are we not considering that K and m are not zero when we are moving that other side.

z/k = x/m => zm=xk

Kudos [?]: 29 [1], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 41891

Kudos [?]: 129094 [1], given: 12194

### Show Tags

11 Mar 2014, 11:54
1
KUDOS
Expert's post
viksingh15 wrote:
Are we not considering that K and m are not zero when we are moving that other side.

z/k = x/m => zm=xk

No, because both of them are in the denominator and thus cannot be 0. If either of them is zero, then the ratio will be undefined and z/k won't be equal to x/m as given in the statement.

Hope it's clear.
_________________

Kudos [?]: 129094 [1], given: 12194

Current Student
Joined: 14 Jul 2013
Posts: 32

Kudos [?]: 8 [0], given: 39

### Show Tags

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

Kudos [?]: 8 [0], given: 39

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16605

Kudos [?]: 273 [0], given: 0

### Show Tags

30 Jun 2015, 21:04
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 273 [0], given: 0

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16605

Kudos [?]: 273 [0], given: 0

### Show Tags

09 Jul 2016, 06:46
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 273 [0], given: 0

Senior Manager
Status: DONE!
Joined: 05 Sep 2016
Posts: 408

Kudos [?]: 23 [0], given: 283

### Show Tags

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!

Kudos [?]: 23 [0], given: 283

Re: Is x/m*(m^2+n^2+k^2)=xm+yn+zk?   [#permalink] 18 Sep 2016, 09:04
Display posts from previous: Sort by