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

It is currently 22 Feb 2019, 06:26

Close

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
Your Progress

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.

Close

Request Expert Reply

Confirm Cancel
Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
  • Free GMAT RC Webinar

     February 23, 2019

     February 23, 2019

     07:00 AM PST

     09:00 AM PST

    Learn reading strategies that can help even non-voracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT
  • FREE Quant Workshop by e-GMAT!

     February 24, 2019

     February 24, 2019

     07:00 AM PST

     09:00 AM PST

    Get personalized insights on how to achieve your Target Quant Score.

If a = 1 and b = -2, then (2a^2 + b)(x + y) + (a + b)(x - y) =

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

 
Manager
Manager
avatar
Joined: 18 Feb 2015
Posts: 84
If a = 1 and b = -2, then (2a^2 + b)(x + y) + (a + b)(x - y) =  [#permalink]

Show Tags

New post 23 Jul 2016, 13:19
6
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

87% (00:57) correct 13% (01:32) wrong based on 106 sessions

HideShow timer Statistics

If a = 1 and b = -2, then (2a^2 + b)(x + y) + (a + b)(x - y) =

(A) 0
(B) 2x
(C) y - x
(D) x - y
(E) x + y
Manager
Manager
User avatar
Status: 2 months to go
Joined: 11 Oct 2015
Posts: 111
GMAT 1: 730 Q49 V40
GPA: 3.8
If a = 1 and b = -2, then (2a^2 + b)(x + y) + (a + b)(x - y) =  [#permalink]

Show Tags

New post 23 Jul 2016, 15:12
\(\left( 2a^{2\; }+\; b \right)\left( x+y \right)\; +\; \left( a\; +\; b \right)\left( x\; -\; y \right)\;\)

\(=\; \left( 2-2 \right)\left( x+y \right)\; -1\left( x-y \right)\;\)

\(=\; o\left( x+y \right)\; -\; x+y\;\)

\(=\; y-x\)

\(Answer\; \mbox{C}\)

Is it really from Gmatprep? It's really strange that they didn't put the x - 3y choice.
That choice would have probably been the most common mistake on this question.
Current Student
User avatar
D
Joined: 12 Aug 2015
Posts: 2621
Schools: Boston U '20 (M)
GRE 1: Q169 V154
GMAT ToolKit User Premium Member
Re: If a = 1 and b = -2, then (2a^2 + b)(x + y) + (a + b)(x - y) =  [#permalink]

Show Tags

New post 29 Jul 2016, 14:14
1
Intern
Intern
avatar
B
Joined: 13 Aug 2018
Posts: 2
Re: If a = 1 and b = -2, then (2a^2 + b)(x + y) + (a + b)(x - y) =  [#permalink]

Show Tags

New post 05 Sep 2018, 07:04
DensetsuNo wrote:
\(\left( 2a^{2\; }+\; b \right)\left( x+y \right)\; +\; \left( a\; +\; b \right)\left( x\; -\; y \right)\;\)

\(=\; \left( 2-2 \right)\left( x+y \right)\; -1\left( x-y \right)\;\)

\(=\; o\left( x+y \right)\; -\; x+y\;\)

\(=\; y-x\)

\(Answer\; \mbox{C}\)

Is it really from Gmatprep? It's really strange that they didn't put the x - 3y choice.
That choice would have probably been the most common mistake on this question.



Is there a PEMDAS application?
GMAT Club Bot
Re: If a = 1 and b = -2, then (2a^2 + b)(x + y) + (a + b)(x - y) =   [#permalink] 05 Sep 2018, 07:04
Display posts from previous: Sort by

If a = 1 and b = -2, then (2a^2 + b)(x + y) + (a + b)(x - y) =

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.