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

It is currently 17 Jun 2018, 17:07

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

Events & Promotions in June
Open Detailed Calendar

In the diagram above, AB is parallel to EH, and BD is parall

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

Hide Tags

Expert Post
2 KUDOS received
Magoosh GMAT Instructor
User avatar
G
Joined: 28 Dec 2011
Posts: 4668
In the diagram above, AB is parallel to EH, and BD is parall [#permalink]

Show Tags

New post 08 May 2014, 11:57
2
7
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

68% (03:00) correct 32% (03:16) wrong based on 183 sessions

HideShow timer Statistics

Attachment:
Line diagram, two pairs of parallel lines.JPG
Line diagram, two pairs of parallel lines.JPG [ 20.68 KiB | Viewed 3734 times ]

In the diagram above, AB is parallel to EH, and BD is parallel to FH. Also, AB = BC, and EF = FH. If ∠EGC = 70°, then ∠D =
(A) 65°
(B) 70°
(C) 75°
(D) 80°
(E) 85°


For other Geometry practice questions, as well as the OE to this particular question, see:
http://magoosh.com/gmat/2014/gmat-geome ... -problems/

Mike :-)

_________________

Mike McGarry
Magoosh Test Prep

Image

Image

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Magoosh Discount CodesManhattan GMAT Discount CodesJamboree Discount Codes
Intern
Intern
avatar
Joined: 16 Jun 2013
Posts: 9
Schools: Mays '17
Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]

Show Tags

New post 09 May 2014, 00:09
1. AB//EH ==> ^A = ^EGC = 70.
AB = BC ==> ^ACB = 70 = ^DCG
2. ^ABC = 180 - (70+70) = 40
3. AB//EH & BD//FH ==> ^EHF = ^ABC = 40
4. EF = FH ==> ^FEH = ^EHF = 40 ==> ^EFH = 180 - (40 + 40) = 100
5. BD // FH ==> ^D + ^EFH = 180 ==> ^D = 180 - ^EFH = 180 - 100 = 80

OR

4. When ^FEH = 40 ==> ^DEG = 180 - 40 = 140
5. ^D = 360 - (^DEG + ^EGC + ^GCD) = 360 - (140 + 70 + 70) = 360 - 280 = 80

SO D is indeed the correct answer.
1 KUDOS received
Manager
Manager
avatar
Joined: 28 Jun 2012
Posts: 52
Location: Singapore
GMAT 1: 720 Q50 V36
WE: Consulting (Consulting)
GMAT ToolKit User
Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]

Show Tags

New post 11 May 2014, 09:51
1
1
AB||EH and AG is transversal => Angle EGA = Angle GAB = 70 degree [ Alternate Angles]

Angle GAB = Angle CAB = Angle ACB = 70 (AB = CB) => Angle B = 40

again, AB||EH & BD||FH => Angle DBA = Angle FHE = 40 degree

Therefore, Angle EGH = (180 - 2*40) = 100 degrees

Now, BD||FH and DF is transversal, so Angle BDF + Angle HFE= 180
=> BDF + 100 = 180
=> CDE = 80
Answe : D
_________________

Do not hesitate to share appreciation, hit Kudos!!

3 KUDOS received
SVP
SVP
User avatar
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1837
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]

Show Tags

New post 14 Aug 2014, 02:03
3
1
Answer = 80

Refer diagram below
Attachments

Line%20diagram,%20two%20pairs%20of%20parallel%20lines.JPG
Line%20diagram,%20two%20pairs%20of%20parallel%20lines.JPG [ 32.56 KiB | Viewed 3293 times ]


_________________

Kindly press "+1 Kudos" to appreciate :)

Board of Directors
User avatar
P
Joined: 17 Jul 2014
Posts: 2730
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)
GMAT ToolKit User Premium Member Reviews Badge
Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]

Show Tags

New post 03 Apr 2016, 11:33
what a nightmare the question is..tried multiple times to draw on my paper..
Attachments

Line diagram%2C two pairs of parallel lines.JPG
Line diagram%2C two pairs of parallel lines.JPG [ 38.22 KiB | Viewed 2461 times ]

BSchool Forum Moderator
User avatar
D
Joined: 12 Aug 2015
Posts: 2642
GRE 1: 323 Q169 V154
GMAT ToolKit User Premium Member
Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]

Show Tags

New post 03 Apr 2016, 12:28
Expert Post
3 KUDOS received
Magoosh GMAT Instructor
User avatar
G
Joined: 28 Dec 2011
Posts: 4668
Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]

Show Tags

New post 03 Apr 2016, 17:39
3
1
Chiragjordan wrote:
Why is that AB||EH & BD||FH => Angle DBA = Angle FHE
i mean when i look at the figure they sure seem equal
But I couldn't figure that out while answering...

Dear Chiragjordan,
I'm happy to respond. :-) I was the creator of this question.

Think about it this way. Suppose we extended segment B-C-D so that it went off beyond D as a ray. Then, imagine that we extend segment H-G-E off as a ray beyond E. Those two rays would intersect: call that intersection point Q. You may have to draw the figure out on paper to visualize this.

Think about the angle formed by those two rays at Q. Because AB||EH, the long BCD ray is a transversal cutting those two parallel lines, so the angle at Q would equal the angle at B: they would be what are known in Geometry as "Alternate Interior Angles." Because BD||FH, the long HGE ray is a s a transversal cutting those two parallel lines, so the angle at Q would equal the angle at H. The angle at B and the angle at H both equal the angle at Q, so they have to equal each other. That's one way to see it.

Another way is to think about it as follows. Suppose I am standing at B, and my friend is standing at H. When I look from B through point A to the horizon, and my friend looks from point H through point E to the horizon, we both will be facing in the same direction, facing the same point on the horizon. Now, we both turn clockwise. After the turn, I am still at B, looking out through point D to the horizon, and my friend is still at H, looking out through point F to the horizon. We are still facing in the same direction, facing the same point on the horizon. Well, if we started facing the same direction, and then each turned and wound up facing the same direction, we must have turned the same amount. Therefore, the angle of our turning, the angle at B and the angle at H, must be equal.

Yet another way to think about it would be: suppose we dropped the whole diagram into the Coordinate plane. AB||EH, so they have the same slope: call that m1. BD||FH, so they also have the same slope: call that m2. Well clearly, if two lines at two different slopes meet at some angle, then changing the y-intercepts of those lines would move the intersection point around, but as along as the slopes are the same, the angle between the two lines are the same. Thus, if we look at the intersection of any line with a slope of m1 with any line with a slope of m2, they will intersect at the same angle. That's exactly what we have with angle B and angle H.

Does all this make sense?
Mike :-)
_________________

Mike McGarry
Magoosh Test Prep

Image

Image

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

BSchool Forum Moderator
User avatar
D
Joined: 12 Aug 2015
Posts: 2642
GRE 1: 323 Q169 V154
GMAT ToolKit User Premium Member
Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]

Show Tags

New post 03 Apr 2016, 22:41
mikemcgarry wrote:
Chiragjordan wrote:
Why is that AB||EH & BD||FH => Angle DBA = Angle FHE
i mean when i look at the figure they sure seem equal
But I couldn't figure that out while answering...

Dear Chiragjordan,
I'm happy to respond. :-) I was the creator of this question.

Think about it this way. Suppose we extended segment B-C-D so that it went off beyond D as a ray. Then, imagine that we extend segment H-G-E off as a ray beyond E. Those two rays would intersect: call that intersection point Q. You may have to draw the figure out on paper to visualize this.

Think about the angle formed by those two rays at Q. Because AB||EH, the long BCD ray is a transversal cutting those two parallel lines, so the angle at Q would equal the angle at B: they would be what are known in Geometry as "Alternate Interior Angles." Because BD||FH, the long HGE ray is a s a transversal cutting those two parallel lines, so the angle at Q would equal the angle at H. The angle at B and the angle at H both equal the angle at Q, so they have to equal each other. That's one way to see it.

Another way is to think about it as follows. Suppose I am standing at B, and my friend is standing at H. When I look from B through point A to the horizon, and my friend looks from point H through point E to the horizon, we both will be facing in the same direction, facing the same point on the horizon. Now, we both turn clockwise. After the turn, I am still at B, looking out through point D to the horizon, and my friend is still at H, looking out through point F to the horizon. We are still facing in the same direction, facing the same point on the horizon. Well, if we started facing the same direction, and then each turned and wound up facing the same direction, we must have turned the same amount. Therefore, the angle of our turning, the angle at B and the angle at H, must be equal.

Yet another way to think about it would be: suppose we dropped the whole diagram into the Coordinate plane. AB||EH, so they have the same slope: call that m1. BD||FH, so they also have the same slope: call that m2. Well clearly, if two lines at two different slopes meet at some angle, then changing the y-intercepts of those lines would move the intersection point around, but as along as the slopes are the same, the angle between the two lines are the same. Thus, if we look at the intersection of any line with a slope of m1 with any line with a slope of m2, they will intersect at the same angle. That's exactly what we have with angle B and angle H.

Does all this make sense?
Mike :-)



Superb Explanation mikemcgarry ; the only thing that bothers me here is => It still taking me approximately 3 minutes to solve this question..
Wonder what will happen during the Actual Gmat..

Having said that => This is one heck of a question..
Wait ...lemme tap that Kudos..
Thanks
_________________


MBA Financing:- INDIAN PUBLIC BANKS vs PRODIGY FINANCE!

Getting into HOLLYWOOD with an MBA!

The MOST AFFORDABLE MBA programs!

STONECOLD's BRUTAL Mock Tests for GMAT-Quant(700+)

AVERAGE GRE Scores At The Top Business Schools!

Manager
Manager
avatar
Joined: 03 May 2013
Posts: 72
Re: In the diagram above, AB is parallel to EH, and BD is parall [#permalink]

Show Tags

New post 21 Sep 2017, 03:28
missionphd wrote:
AB||EH and AG is transversal => Angle EGA = Angle GAB = 70 degree [ Alternate Angles]

Angle GAB = Angle CAB = Angle ACB = 70 (AB = CB) => Angle B = 40

again, AB||EH & BD||FH => Angle DBA = Angle FHE = 40 degree

Therefore, Angle EGH = (180 - 2*40) = 100 degrees

Now, BD||FH and DF is transversal, so Angle BDF + Angle HFE= 180
=> BDF + 100 = 180
=> CDE = 80
Answe : D




can somebody please let me know .. how come. Angle DBA = Angle FHE = 40 degree
Re: In the diagram above, AB is parallel to EH, and BD is parall   [#permalink] 21 Sep 2017, 03:28
Display posts from previous: Sort by

In the diagram above, AB is parallel to EH, and BD is parall

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


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®.