5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Gmatprep3 Problem
-
faraz_jeddah
- Master | Next Rank: 500 Posts
- Posts: 358
- Joined: Thu Apr 18, 2013 9:46 am
- Location: Jeddah, Saudi Arabia
- Thanked: 42 times
- Followed by:7 members
- GMAT Score:730
No Idea how to solve this.
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.
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.
-
theCodeToGMAT
- Legendary Member
- Posts: 1556
- Joined: Tue Aug 14, 2012 11:18 pm
- Thanked: 448 times
- Followed by:34 members
- GMAT Score:650
Edited:
Area of the BCE = √2/4
Answer [spoiler][/spoiler]
Sorry, earlier I made a wrong diagram which resulted in answer 1/2
Area of the BCE = √2/4
Answer [spoiler][/spoiler]
Sorry, earlier I made a wrong diagram which resulted in answer 1/2
- Attachments
-
Last edited by theCodeToGMAT on Sun Sep 22, 2013 7:24 am, edited 1 time in total.
R A H U L
-
faraz_jeddah
- Master | Next Rank: 500 Posts
- Posts: 358
- Joined: Thu Apr 18, 2013 9:46 am
- Location: Jeddah, Saudi Arabia
- Thanked: 42 times
- Followed by:7 members
- GMAT Score:730
I am not sure what the logic you have used but the answer is incorrect.
OA is B
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.
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.
-
theCodeToGMAT
- Legendary Member
- Posts: 1556
- Joined: Tue Aug 14, 2012 11:18 pm
- Thanked: 448 times
- Followed by:34 members
- GMAT Score:650
Really??????? Can you please share the explanation.. I was very convinced that I was correctfaraz_jeddah wrote:I am not sure what the logic you have used but the answer is incorrect.
OA is B
R A H U L
-
theCodeToGMAT
- Legendary Member
- Posts: 1556
- Joined: Tue Aug 14, 2012 11:18 pm
- Thanked: 448 times
- Followed by:34 members
- GMAT Score:650
yes, correct.. the answer is indeed .. small mistakefaraz_jeddah wrote:I am not sure what the logic you have used but the answer is incorrect.
OA is B
I have edited my post which is just below the question.
R A H U L
-
runningguy
- Junior | Next Rank: 30 Posts
- Posts: 17
- Joined: Mon Jul 22, 2013 4:40 pm
- Thanked: 1 times
-
theCodeToGMAT
- Legendary Member
- Posts: 1556
- Joined: Tue Aug 14, 2012 11:18 pm
- Thanked: 448 times
- Followed by:34 members
- GMAT Score:650
How did you find that base and height are 1?? that's not possiblerunningguy 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
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
GMAT/MBA Expert
-
[email protected]
- Elite Legendary Member
- Posts: 10392
- Joined: Sun Jun 23, 2013 6:38 pm
- Location: Palo Alto, CA
- Thanked: 2867 times
- Followed by:511 members
- GMAT Score:800
Hi All,
You might find this approach to be a bit easier; it's based on spotting a "hidden" 45/45/90 triangle...
GMAT assassins aren't born, they're made,
Rich
You might find this approach to be a bit easier; it's based on spotting a "hidden" 45/45/90 triangle...
GMAT assassins aren't born, they're made,
Rich
-
runningguy
- Junior | Next Rank: 30 Posts
- Posts: 17
- Joined: Mon Jul 22, 2013 4:40 pm
- Thanked: 1 times
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.theCodeToGMAT wrote:How did you find that base and height are 1?? that's not possiblerunningguy 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
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..
-
GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
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
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
-
faraz_jeddah
- Master | Next Rank: 500 Posts
- Posts: 358
- Joined: Thu Apr 18, 2013 9:46 am
- Location: Jeddah, Saudi Arabia
- Thanked: 42 times
- Followed by:7 members
- GMAT Score:730
Thanks Mitch! I must say it would be hard to figure this question out in less than 3 minutes.GMATGuruNY wrote:
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.
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.
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.
GMAT/MBA Expert
-
[email protected]
- Elite Legendary Member
- Posts: 10392
- Joined: Sun Jun 23, 2013 6:38 pm
- Location: Palo Alto, CA
- Thanked: 2867 times
- Followed by:511 members
- GMAT Score:800
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
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
-
Resp007
- Senior | Next Rank: 100 Posts
- Posts: 35
- Joined: Sat Aug 31, 2013 1:29 pm
- Thanked: 2 times
- Followed by:1 members
- GMAT Score:680
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?
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...
GMAT assassins aren't born, they're made,
Rich
