BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Gmatprep3 Problem

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 358
Joined: Thu Apr 18, 2013 9:46 amFollowed by: 7 members
Location: Jeddah, Saudi Arabia
Thanked: 42 times
GMAT Score:730
by faraz_jeddah » Sun Sep 22, 2013 12:48 am
No Idea how to solve this.

Image
A good question also deserves a Thanks.

Messenger Boy: The Thesselonian you're fighting... he's the biggest man i've ever seen. I wouldn't want to fight him.
Achilles: That's why no-one will remember your name.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1556
Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members
Thanked: 448 times
GMAT Score:650

by theCodeToGMAT » Sun Sep 22, 2013 5:25 am
Edited:
Area of the BCE = √2/4

Answer [spoiler][/spoiler]

Sorry, earlier I made a wrong diagram which resulted in answer 1/2
Attachments
area.jpg
Last edited by theCodeToGMAT on Sun Sep 22, 2013 7:24 am, edited 1 time in total.
R A H U L
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 358
Joined: Thu Apr 18, 2013 9:46 amFollowed by: 7 members
Location: Jeddah, Saudi Arabia
Thanked: 42 times
GMAT Score:730

by faraz_jeddah » Sun Sep 22, 2013 5:33 am
I am not sure what the logic you have used but the answer is incorrect.

OA is B
A good question also deserves a Thanks.

Messenger Boy: The Thesselonian you're fighting... he's the biggest man i've ever seen. I wouldn't want to fight him.
Achilles: That's why no-one will remember your name.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1556
Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members
Thanked: 448 times
GMAT Score:650

by theCodeToGMAT » Sun Sep 22, 2013 5:53 am
faraz_jeddah wrote:I am not sure what the logic you have used but the answer is incorrect.

OA is B
Really??????? Can you please share the explanation.. I was very convinced that I was correct :|
R A H U L
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1556
Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members
Thanked: 448 times
GMAT Score:650

by theCodeToGMAT » Sun Sep 22, 2013 7:03 am
faraz_jeddah wrote:I am not sure what the logic you have used but the answer is incorrect.

OA is B
yes, correct.. the answer is indeed .. small mistake :)

I have edited my post which is just below the question.
R A H U L
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 17
Joined: Mon Jul 22, 2013 4:40 pm
Thanked: 1 times

by runningguy » Sun Sep 22, 2013 7:25 am
I'm confused as to why the answer is not 1/2 (A=1/2(b)(h)).

b=1
h=1

A=1/2(1)(1)=1/2
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1556
Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members
Thanked: 448 times
GMAT Score:650

by theCodeToGMAT » Sun Sep 22, 2013 7:32 am
runningguy wrote:I'm confused as to why the answer is not 1/2 (A=1/2(b)(h)).

b=1
h=1

A=1/2(1)(1)=1/2
How did you find that base and height are 1?? that's not possible

if you say that you drew a perpendicular on BE from C and that weighted 1.. then the triangle is not possible as BC & CE are 1.. so perpendicular to BE from C must be less than 1.

if you say that you extended the C and dropped a perpendicular from E on the extended C line.. then also similar rule applies..

Refer my solution just below the question..
R A H U L
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members
Location: Palo Alto, CA
Thanked: 2867 times
GMAT Score:800

by [email protected] » Sun Sep 22, 2013 12:33 pm
Hi All,

You might find this approach to be a bit easier; it's based on spotting a "hidden" 45/45/90 triangle...


Image


GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 17
Joined: Mon Jul 22, 2013 4:40 pm
Thanked: 1 times

by runningguy » Sun Sep 22, 2013 3:27 pm
theCodeToGMAT wrote:
runningguy wrote:I'm confused as to why the answer is not 1/2 (A=1/2(b)(h)).

b=1
h=1

A=1/2(1)(1)=1/2
How did you find that base and height are 1?? that's not possible

if you say that you drew a perpendicular on BE from C and that weighted 1.. then the triangle is not possible as BC & CE are 1.. so perpendicular to BE from C must be less than 1.

if you say that you extended the C and dropped a perpendicular from E on the extended C line.. then also similar rule applies..

Refer my solution just below the question..
I had incorrectly assumed that one of the legs that was equal to 1 was the height. After looking at my flash cards, I see that for an isosceles triangle, you can divide the hypotenuse by sqrt2 to find the length. This was a very tricky question for me.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Mon Sep 23, 2013 2:41 am
Image

In square ABCD, since s=1, BD = √2 and CF = √2/2.
It is given that CE=1.
Thus, EF = √2/2 + 1.

∆BED = (1/2)(BD)(EF) = (1/2)(√2)(√2/2 + 1) = (1/2)(1 + √2) = 1/2 + √2/2.

∆BCD = (1/2)(square ABCD) = 1/2.

Quadrilateral BEDC = ∆BED - ∆BCD = (1/2 + √2/2) - 1/2 = √2/2.

∆BCE = (1/2)(quadrilateral BEDC) = (1/2)(√2/2) = √2/4.

The correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 358
Joined: Thu Apr 18, 2013 9:46 amFollowed by: 7 members
Location: Jeddah, Saudi Arabia
Thanked: 42 times
GMAT Score:730

by faraz_jeddah » Mon Sep 23, 2013 2:51 am
GMATGuruNY wrote:Image

In square ABCD, since s=1, BD = √2 and CF = √2/2.
It is given that CE=1.
Thus, EF = √2/2 + 1.

∆BED = (1/2)(BD)(EF) = (1/2)(√2)(√2/2 + 1) = (1/2)(1 + √2) = 1/2 + √2/2.

∆BCD = (1/2)(square ABCD) = 1/2.

Quadrilateral BEDC = ∆BED - ∆BCD = (1/2 + √2/2) - 1/2 = √2/2.

∆BCE = (1/2)(quadrilateral BEDC) = (1/2)(√2/2) = √2/4.

The correct answer is B.
Thanks Mitch! I must say it would be hard to figure this question out in less than 3 minutes.
A good question also deserves a Thanks.

Messenger Boy: The Thesselonian you're fighting... he's the biggest man i've ever seen. I wouldn't want to fight him.
Achilles: That's why no-one will remember your name.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Newbie | Next Rank: 10 Posts
Posts: 9
Joined: Mon Jul 06, 2009 2:08 am

by shinys » Tue Sep 24, 2013 6:19 pm
Rich,
Can you please help me understand how you figured out that 135 deg angle?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members
Location: Palo Alto, CA
Thanked: 2867 times
GMAT Score:800

by [email protected] » Tue Sep 24, 2013 6:36 pm
Hi shinys,

We're told that the sides BC=CD=CE=1 and we're told that BE=DE. This info tells us that the 2 triangles are IDENTICAL.

Since they're identical, their two "big angles" + the corner of the square = 360 degrees

2(Big angle) + 90 = 360
2(Big angle) = 270
Big angle = 135

Be on the lookout for the "special" right triangles in GMAT Quant questions; sometimes they're obvious, sometimes they're hidden.

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 35
Joined: Sat Aug 31, 2013 1:29 pmFollowed by: 1 member
Thanked: 2 times
GMAT Score:680

by Resp007 » Thu Sep 26, 2013 4:38 am
HI Rich,
I used GMATGuruNY approach and solved it correctly but as pointed its bit time consuming,
So looked into yours and couldn't understand how you found H.
How you were able to conclude its a 45-45-90 triangle please?

[email protected] wrote:Hi All,

You might find this approach to be a bit easier; it's based on spotting a "hidden" 45/45/90 triangle...


Image


GMAT assassins aren't born, they're made,
Rich
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 35
Joined: Sat Aug 31, 2013 1:29 pmFollowed by: 1 member
Thanked: 2 times
GMAT Score:680

by Resp007 » Thu Sep 26, 2013 4:39 am
repeat post...
Last edited by Resp007 on Thu Sep 26, 2013 4:42 am, edited 1 time in total.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply