GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 27 Jun 2019, 03:32 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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  If m>n>0, x=m^2+n^2, and y=2mn,

Author Message
TAGS:

Hide Tags

Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7508
GMAT 1: 760 Q51 V42 GPA: 3.82
If m>n>0, x=m^2+n^2, and y=2mn,  [#permalink]

Show Tags 00:00

Difficulty:   15% (low)

Question Stats: 84% (01:50) correct 16% (02:30) wrong based on 205 sessions

HideShow timer Statistics [GMAT math practice question]

If $$m>n>0, x=m^2+n^2$$, and $$y=2mn$$, what is the value of $$\sqrt{x^2-y^2}$$ in terms of $$m$$ and $$n$$?

A. $$m^2+n^2$$
B. $$2m^2+n^2$$
C.$$m^2+2n^2$$
D. $$n^2-m^2$$
E. $$m^2-n^2$$

_________________
Senior PS Moderator V
Joined: 26 Feb 2016
Posts: 3364
Location: India
GPA: 3.12
If m>n>0, x=m^2+n^2, and y=2mn,  [#permalink]

Show Tags

1
Formula used: $$x^2 - y^2 = (x+y)(x-y)$$

$$\sqrt{x^2-y^2} = \sqrt{(x+y)(x-y)}$$

In terms of m and n, the equation can be written as $$\sqrt{(m^2+n^2+2mn)(m^2+n^2-2mn)}$$

Simplifying, we get $$\sqrt{(m+n)^2(m-n)^2} = (m+n)(m-n) = m^2-n^2$$(Option E)
_________________
You've got what it takes, but it will take everything you've got
Manager  G
Joined: 07 Jun 2017
Posts: 166
Location: India
Concentration: Technology, General Management
GMAT 1: 660 Q46 V38 GPA: 3.6
WE: Information Technology (Computer Software)
Re: If m>n>0, x=m^2+n^2, and y=2mn,  [#permalink]

Show Tags

sqrt (m+n)^2(m−n)^2
m^2 - n^2 = Option E
_________________
Regards,
Naveen
email: nkmungila@gmail.com
Please press kudos if you like this post
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7508
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: If m>n>0, x=m^2+n^2, and y=2mn,  [#permalink]

Show Tags

1
=>

Plugging in the expressions for $$x$$ and $$y$$, and expanding gives:

$$x^2 – y^2 = (m^2+n^2)^2 – (2mn)^2 = m^4 -2m^2n^2 + n^4 = (m^2-n^2)^2.$$
So, since $$m^2 > n^2$$,
$$\sqrt{x^2-y^2}=\sqrt{(m^2-n^2)^2}=|m^2-n^2|=m^2-n^2.$$

_________________
SC Moderator P
Status: GMAT - Pulling Quant and Verbal together
Joined: 04 Sep 2017
Posts: 201
Location: United States (OH)
GPA: 3.6
WE: Sales (Computer Software)
If m>n>0, x=m^2+n^2, and y=2mn,  [#permalink]

Show Tags

pushpitkc wrote:
Formula used: $$x^2 - y^2 = (x+y)(x-y)$$

$$\sqrt{x^2-y^2} = \sqrt{(x+y)(x-y)}$$

In terms of m and n, the equation can be written as $$\sqrt{(m^2+n^2+2mn)(m^2+n^2-2mn)}$$

Simplifying, we get $$\sqrt{(m+n)^2(m-n)^2} = (m+n)(m-n) = m^2-n^2$$(Option E)

I do not understand. I am getting...

$$\sqrt{(m^2+n^2)^2-(2mn)^2}$$

Which then leads me down a long and incorrect path. Thanks in advance for a descriptive solution.
_________________
Would I rather be feared or loved? Easy. Both. I want people to be afraid of how much they love me.

How to sort questions by Topic, Difficulty, and Source:
https://gmatclub.com/forum/search.php?view=search_tags If m>n>0, x=m^2+n^2, and y=2mn,   [#permalink] 27 Feb 2018, 05:38
Display posts from previous: Sort by

If m>n>0, x=m^2+n^2, and y=2mn,  