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

It is currently 21 Jul 2018, 12:34

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

In the figure above, x =

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

Hide Tags

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47169
In the figure above, x =  [#permalink]

Show Tags

New post 11 Oct 2017, 00:44
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

84% (00:48) correct 16% (01:49) wrong based on 45 sessions

HideShow timer Statistics

Senior Manager
Senior Manager
User avatar
S
Joined: 05 Dec 2016
Posts: 259
Concentration: Strategy, Finance
GMAT 1: 620 Q46 V29
GMAT ToolKit User
Re: In the figure above, x =  [#permalink]

Show Tags

New post 11 Oct 2017, 00:59
x = 45 + (bottom leg of side triangle * 2)
Using Pythagorean formula we can find:
bottom leg of side triangle= (50^2-40^2)^1/2=30
x=45+30+30=105

Answer D
Manager
Manager
User avatar
B
Joined: 22 May 2017
Posts: 92
GMAT 1: 580 Q41 V29
GMAT 2: 580 Q43 V27
Re: In the figure above, x =  [#permalink]

Show Tags

New post 12 Oct 2017, 04:39
Alexey1989x wrote:
x = 45 + (bottom leg of side triangle * 2)
Using Pythagorean formula we can find:
bottom leg of side triangle= (50^2-40^2)^1/2=30
x=45+30+30=105

Answer D



side of two parallelogram should be 45 plus plus 45 .
So answer can be 90
option B
Senior Manager
Senior Manager
User avatar
S
Joined: 05 Dec 2016
Posts: 259
Concentration: Strategy, Finance
GMAT 1: 620 Q46 V29
GMAT ToolKit User
Re: In the figure above, x =  [#permalink]

Show Tags

New post 12 Oct 2017, 05:01
Adityagmatclub wrote:
Alexey1989x wrote:
x = 45 + (bottom leg of side triangle * 2)
Using Pythagorean formula we can find:
bottom leg of side triangle= (50^2-40^2)^1/2=30
x=45+30+30=105

Answer D



side of two parallelogram should be 45 plus plus 45 .
So answer can be 90
option B


Sorry, I do not understand your logic.
Could you please specify your line of reasoning.
Manager
Manager
User avatar
B
Joined: 22 May 2017
Posts: 92
GMAT 1: 580 Q41 V29
GMAT 2: 580 Q43 V27
Re: In the figure above, x =  [#permalink]

Show Tags

New post 12 Oct 2017, 05:19
Yes, i was wrong but i can't understand which triangle you are considering for Pythagoras theorem.

Sent from my Redmi Note 3 using GMAT Club Forum mobile app
1 KUDOS received
Senior Manager
Senior Manager
User avatar
S
Joined: 05 Dec 2016
Posts: 259
Concentration: Strategy, Finance
GMAT 1: 620 Q46 V29
GMAT ToolKit User
Re: In the figure above, x =  [#permalink]

Show Tags

New post 12 Oct 2017, 05:28
1
Adityagmatclub wrote:
Yes, i was wrong but i can't understand which triangle you are considering for Pythagoras theorem.

Sent from my Redmi Note 3 using GMAT Club Forum mobile app


Look at the picture in attachment, I've applied Pyth formula to find AF and ED.
Attachments

IMG_3315.jpg
IMG_3315.jpg [ 309.61 KiB | Viewed 583 times ]

Manager
Manager
User avatar
B
Joined: 22 May 2017
Posts: 92
GMAT 1: 580 Q41 V29
GMAT 2: 580 Q43 V27
Re: In the figure above, x =  [#permalink]

Show Tags

New post 12 Oct 2017, 06:22
Yes, easy problem.

Sent from my Redmi Note 3 using GMAT Club Forum mobile app
Re: In the figure above, x = &nbs [#permalink] 12 Oct 2017, 06:22
Display posts from previous: Sort by

In the figure above, x =

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

Events & Promotions

PREV
NEXT


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®.