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

 It is currently 11 Dec 2018, 16:49

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 December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• Free GMAT Prep Hour

December 11, 2018

December 11, 2018

09:00 PM EST

10:00 PM EST

Strategies and techniques for approaching featured GMAT topics. December 11 at 9 PM EST.
• The winning strategy for 700+ on the GMAT

December 13, 2018

December 13, 2018

08:00 AM PST

09:00 AM PST

What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.

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

Author Message
TAGS:

Hide Tags

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6619
GMAT 1: 760 Q51 V42
GPA: 3.82
If m>n>0, x=m^2+n^2, and y=2mn,  [#permalink]

Show Tags

30 Nov 2017, 00:31
00:00

Difficulty:

15% (low)

Question Stats:

86% (01:51) correct 14% (02:37) wrong based on 192 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. $$m^2-n^2$$

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Senior PS Moderator Joined: 26 Feb 2016 Posts: 3325 Location: India GPA: 3.12 If m>n>0, x=m^2+n^2, and y=2mn, [#permalink] Show Tags 30 Nov 2017, 00:51 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 Joined: 07 Jun 2017 Posts: 174 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 30 Nov 2017, 04:11 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 Joined: 16 Aug 2015 Posts: 6619 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If m>n>0, x=m^2+n^2, and y=2mn, [#permalink] Show Tags 03 Dec 2017, 17:38 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 $$m^2 > n^2$$, $$\sqrt{x^2-y^2}=\sqrt{(m^2-n^2)^2}=|m^2-n^2|=m^2-n^2.$$ Therefore, the answer is E. Answer: E _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Manager
Status: Studying SC
Joined: 04 Sep 2017
Posts: 115
GPA: 3.6
WE: Sales (Computer Software)
If m>n>0, x=m^2+n^2, and y=2mn,  [#permalink]

Show Tags

27 Feb 2018, 04:38
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, &nbs [#permalink] 27 Feb 2018, 04:38
Display posts from previous: Sort by