Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 0410:00 AM PDT
-11:00 AM PDT
What does test anxiety really look like on the GMAT? For many test takers, it does not look like a panic attack. Instead, it shows up as brain fog, rereading, negative comparisons, mental noise....
Sep 0201:00 PM EDT
-11:59 PM EDT
Take 30% off all study plans with code COURSEWORK - Expires on Monday 9/7
Sep 0609:30 AM EDT
-12:30 PM EST
Top scores are possible when you enroll in a powerful EA course, taught live online + 6 months access to TTP OnDemand video courses included! Perfect class schedule and easy course access for working professionals. Class starts Sundays Sept. 6, 2026.- Sep 07
12:01 AM PDT
-11:59 PM PDT
On the 7th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PDT - Sep 07
10:00 AM PDT
-11:00 AM PDT
A complete walkthrough of GMAT Club’s free 12-week GMAT study plan: what you study each week, how progress and goals are tracked, how the error log works, and how your practice unlocks paid tools at no cost. - Sep 09
07:00 AM PDT
-10:00 AM PDT
Meet AdComs and explore top Master’s programs - MiM, MiF, MSc, MSBA and more. - Application Fee Waivers - Free 1-Week of GMAT Club Tests: - Master's Application Toolkit - Grand Prize Giveaway - Sep 09
10:00 AM PDT
-11:00 AM PDT
Three MBA applications. Three rejections. No interview invites. A year later, Aman was admitted to Cambridge Judge, SMU, and IMD. In this episode of MBA Admit Stories, Aman shares how he rebuilt his MBA application... 
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2108:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
17 Jul 2017, 23:31
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
59% (02:37) correct
41%
(02:54)
wrong
based on 102
sessions
History
Date
Time
Result
Not Attempted Yet
In the diagram below, triangle ABC is right-angled at B. What is the measure of ∠BAC?
(1) BD = BC and AE = AD.
(2) BE = DE.
2017-07-18_1030.png [ 6.99 KiB | Viewed 4874 times ]
(1) BD = BC and AE = AD.
(2) BE = DE.
Attachment:
2017-07-18_1030.png [ 6.99 KiB | Viewed 4874 times ]
18 Jul 2017, 05:03
Kudos
Bookmarks
i have tried every possible thing to get the answer but knowing that two sides are same, i cant get the angle of the other side. I cant find the relationship between the two small triangles even...so E for me
18 Jul 2017, 05:36
Kudos
Bookmarks
Bunuel
It should be C
Statement 1: Insufficient
(1) BD = BC and AE = AD.
This tells us both top and bottom triangles are isosceles but it does not provide us any information about the central triangle.
Statement 2: Insufficient
This tells us central triangle is an isosceles but it does not give out information about the angles.
Lets combined 1 and 2.
From 1, Since \(BD = BC\), we know \(∠BDC = ∠BCD\). Let us call this angle x. In the triangle \(BDC\), third angle \(∠CBD\) would be equal to \(180 - 2x\).
Since triangle ABC is right angled at B, \(∠EBD = 90 - (180 - 2x) = -90 + 2x\)
From 2, we know \(BE = DE\), hence \(∠EBD = ∠EDB = -90 + 2x\)
The third angle in the central triangle would be \(180 - 2 (-90 + 2x) = 360 - 4x\)
Lets name the two equal angles from triangle \(AED = y\)
At point D, we have
\(x + (-90 + 2x) + y = 180\)
\(3x + y = 270\) --- (1)
At point E, we have
\(360 - 4x + y = 180\)
\(4x - y = 180\) --- (2)
Solving the two equations, we get the value of y and since two angles in triangle AED are y, the third \(∠BAC\) can be calculated using \(180 - 2y\).
Edit: resizing diagram as it was too large.
Attachments
diagram.jpg [ 28.2 KiB | Viewed 4314 times ]
