Find all School-related info fast with the new School-Specific MBA Forum

It is currently 17 Sep 2014, 06:08

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

DS : TRIANGLE (m09q07)

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
VP
VP
User avatar
Joined: 30 Jun 2008
Posts: 1048
Followers: 11

Kudos [?]: 277 [0], given: 1

GMAT Tests User
DS : TRIANGLE (m09q07) [#permalink] New post 28 Oct 2008, 23:53
3
This post was
BOOKMARKED
On the picture below, is the area of the triangle ABC greater than 1?

1. \angle ABC < 90^\circ
2. Perimeter of the triangle ABC is greater than \frac{4}{a}

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions
Attachments

triangle.jpg
triangle.jpg [ 4.29 KiB | Viewed 9337 times ]


_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Check out my GMAT blog - GMAT Tips and Strategies

Kaplan Promo CodeKnewton GMAT Discount CodesGMAT Pill GMAT Discount Codes
Expert Post
10 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 29664
Followers: 3494

Kudos [?]: 26262 [10] , given: 2708

Re: DS : TRIANGLE (m09q07) [#permalink] New post 22 Feb 2010, 11:50
10
This post received
KUDOS
Expert's post
amitdgr wrote:
On the picture below, is the area of the triangle ABC greater than 1?

1. \angle ABC < 90^\circ
2. Perimeter of the triangle ABC is greater than \frac{4}{a}

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions


Given isosceles triangle ABC with base=AC=\frac{2}{a} and height=a^2.

Question: is area=\frac{1}{2}*base*height=\frac{1}{2}*\frac{2}{a}*a^2=a>1. So we see that basically the question aska: is a>1 true?


(1) \angle ABC < 90^\circ --> assume \angle ABC=90^\circ then hypotenuse is AC=\frac{2}{a}, as ABC becomes isosceles right triangle (45-45-90=1-1-\sqrt{2}) then the leg=BC=AB=\frac{2}{a\sqrt{2}}=\frac{\sqrt{2}}{a}.

But BC=\frac{\sqrt{2}}{a} also equals to \sqrt{(\frac{1}{a})^2+(a^2)^2}, so we have \frac{\sqrt{2}}{a}=\sqrt{(\frac{1}{a})^2+(a^2)^2} --> 2=1+a^6 --> a^6=1 --> a=1 (as a per diagram is positive). Now, if we increase a then the base \frac{2}{a} will decrease and the height a^2 will increase thus making angle ABC smaller than 90 and if we decrease a then the base \frac{2}{a} will increase and the height a^2 will decrease thus making the angle ABC greater than 90. So, as angle ABC is less than 90, a must be more than 1. Sufficient.

(2) Perimeter of the triangle ABC is greater than \frac{4}{a} --> P=BC+AB+AC=2BC+AC=2\sqrt{(\frac{1}{a})^2+(a^2)^2}+\frac{2}{a}>\frac{4}{a} --> \sqrt{\frac{1+a^6}{a^2}}>\frac{1}{a} --> a^6>0 --> a>0. Hence we don't know whether a>1 is true. Not sufficient.

Answer: A.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership


Last edited by Bunuel on 23 Apr 2012, 00:02, edited 2 times in total.
5 KUDOS received
Intern
Intern
avatar
Joined: 10 Aug 2012
Posts: 19
Location: India
Followers: 0

Kudos [?]: 5 [5] , given: 12

Re: DS : TRIANGLE (m09q07) [#permalink] New post 01 Mar 2013, 20:56
5
This post received
KUDOS
Area of the triangle : \frac{1}{2}*AC*OB
= \frac{1}{2}*\frac{2}{a}*a^2
= a

So we to find that a > 1 or not ??

Statement 1:

angle ABC < 90 => angle ABO < 45.
mean angle BAO is greater than angle ABO
that further implies that side OB > side OA
a^2 >\frac{1}{a}
=> a^3> 1 => a > 1 since length of side can not be negative

So A is sufficient

Statement 2:

Perimeter of the triangle is greater than 4/a

So fine perimeter, first we have to find AB

AB^2 = OB^2 + OA ^2

AB^2 = (a^2)^2 + \frac{1}{a^2}

AB^2 = \frac{(a^6 +1)}{a ^2}

AB =\sqrt{(a^6 +1)}/a

Perimeter of triangle = AB + BC +CA
since AB = BC,

Perimeter = 2*AB + CA = 2 * \sqrt{(a^6 +1)} /a + \frac{2}{a}

2 * \sqrt{(a^6 +1)} /a + 2/a > 4/a


\sqrt{(a^6 +1)} + 1 > 2

\sqrt{(a^6 +1)} > 1

a^6 +1 > 1

a^6 > 0

a > 0

So, we can not say anything from statement 2 that a >1 or not.

Hence, correct answer is A.
3 KUDOS received
SVP
SVP
avatar
Joined: 17 Jun 2008
Posts: 1579
Followers: 12

Kudos [?]: 183 [3] , given: 0

GMAT Tests User
Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 00:54
3
This post received
KUDOS
A for me.

Area of triangle = 1/2 * 2/a * a^2 = a. Hence, we need to find out whether a > 1/

From stmt1: AB^2 + BC^2 > AC^2 or 2*AB^2 > AC^2
or 2*(a^4 + 1/a^2) > 4/a^2
or, (a^6 + 1) > 2 or, a^6 > 1 that means a > 1. Hence, sufficient.

Simplifying stmt2 will give a > 0...insufficient.
Expert Post
2 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 29664
Followers: 3494

Kudos [?]: 26262 [2] , given: 2708

Re: DS : TRIANGLE (m09q07) [#permalink] New post 02 Dec 2010, 11:56
2
This post received
KUDOS
Expert's post
MathMind wrote:
Bunuel,

Great explanation - I am not sure about your conclusion "Hence a>1 is not true as when we decrease angle ABC, a will decrease as well and will become less than 1"

I think this may lead to "As the angle ABC decreases, since ht=a^2 when base = 2a (ht/base ratio = a/2), the base will become broader => a will increase

=> This triangle will have area = 1 for ABC = 90*

=> This triangle will have area > 1 for ABC < 90*

=> This triangle will have area < 1 for ABC > 90* "

Thx, JS


First of all base=2/a (and not 2a) and height/base=a^3/2.

Though I did have a typo there. It should be: if we increase a then the base \frac{2}{a} will decrease and the height a^2 will increase thus making angle ABC smaller than 90 and if we decrease a then the base \frac{2}{a} will increase and the height a^2 will decrease thus making the angle ABC greater than 90. So, as angle ABC is less than 90, a must be more than 1.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

1 KUDOS received
SVP
SVP
avatar
Joined: 17 Jun 2008
Posts: 1579
Followers: 12

Kudos [?]: 183 [1] , given: 0

GMAT Tests User
Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 02:24
1
This post received
KUDOS
amitdgr wrote:
scthakur wrote:
A for me.

Simplifying stmt2 will give a > 0...insufficient.


how do we simplify stmt 2 ??

Thanks :)


Perimeter = AB + BC + AC = 2AB + AC = 2*sqrt((1/a^2)+ a^4) + 2/a
= 2*(sqrt(1+a^6))/a + 2/a > 4/a
or, sqrt(1+a^6) + 1 > 2
or, sqrt(1+a^6) > 1 or a > 0.
1 KUDOS received
Intern
Intern
avatar
Joined: 28 Jan 2011
Posts: 7
Followers: 0

Kudos [?]: 4 [1] , given: 0

Re: DS : TRIANGLE (m09q07) [#permalink] New post 18 Feb 2014, 00:49
1
This post received
KUDOS
Bunuel's explanation is excellent, as usual.
Here's a different way that does not involve lots of algebra...

(1) Let's assume that a = 1, the simplest thing to try. Also, let's call the origin (0, 0) point D (got to name it so we can talk about it).

This gives us a triangle with height 1 (BD) and base 2 (AC), so the area is clearly 1.
It should be intuitively obvious, if you look at this triangle in your mind, that it's an isosceles right triangle (angle ABC = 90).
But if you're not sure, consider that if a = 1, BD = DC = 1, so triangle BDC is an isosceles right triangle.
This means that angle BCD is 45 degrees, and since angle BDC is 90 degrees, angle DBC is 45 degrees.
In the same way, we know that angle DBA is 45 degrees, and that therefore angle ABC is 90 degrees.

If we make a bigger, the triangle gets taller and narrower; try a = 2. Height 4, base 1, area 2.
Ah-hah! looks like as a gets bigger, area gets bigger.

One more quick check to validate: try a = \frac{1}{2}. Height \frac{1}{4}, base 4, area \frac{1}{2}.

So it's clear, without doing any algebra: as the triangle gets taller and narrower, its area increases.
As the triangle gets taller and narrower (a > 1), angle ABC gets smaller, i.e. less than 90 degrees. That's what Statement (1) says. Sufficient.

(2) Again, let's try and avoid algebra with some quick numbers. If a = 1, BD = DC = 1, triangle BDC is an isosceles right triangle.
It must have sides in the ratio 1:1:\sqrt{2}; therefore BC = \sqrt{2}.
AB will also = \sqrt{2}, so the perimeter is 2 + 2\sqrt{2}, which is clearly > 4.
(If a = 1, \frac{4}{a}=4).

So when a = 1, the perimeter is > \frac{4}{a}, but the area of triangle ABC is = 1 (see Statement (1) above) and therefore not > 1.
When a = 1, the answer to the overall question, whether the area > 1, is False.

Now if a becomes a larger number (a > 1), the perimeter will increase.

(Try a = 10. Now a^2 = BD = 100. Clearly BC must be still larger.
Therefore the perimeter is much, much larger than \frac{4}{a}, which is \frac{4}{10}, and much larger than 4 as well.)

We already know from Statement (1) above that when a > 1, the answer to the question is True (area ABC > 1).
Therefore, without doing any more work, both True and False answers are possible and Statement (2) is Insufficient.
VP
VP
User avatar
Joined: 30 Jun 2008
Posts: 1048
Followers: 11

Kudos [?]: 277 [0], given: 1

GMAT Tests User
Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 01:12
scthakur wrote:
A for me.

Simplifying stmt2 will give a > 0...insufficient.


how do we simplify stmt 2 ??

Thanks :)
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Check out my GMAT blog - GMAT Tips and Strategies

Senior Manager
Senior Manager
User avatar
Joined: 21 Apr 2008
Posts: 272
Location: Motortown
Followers: 0

Kudos [?]: 82 [0], given: 0

GMAT Tests User
Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 05:32
scthakur wrote:
A for me.

Area of triangle = 1/2 * 2/a * a^2 = a. Hence, we need to find out whether a > 1/

From stmt1: AB^2 + BC^2 > AC^2 or 2*AB^2 > AC^2
or 2*(a^4 + 1/a^2) > 4/a^2
or, (a^6 + 1) > 2 or, a^6 > 1 that means a > 1. Hence, sufficient.

Simplifying stmt2 will give a > 0...insufficient.


scthakur : Can you explain how you simplified stmt1

And from a^6 > 1, how did you come up with a > 1, a could be -2 also right ?, or is it because 1/a is a point on Positive X-axis ?
CEO
CEO
User avatar
Joined: 29 Aug 2007
Posts: 2501
Followers: 53

Kudos [?]: 506 [0], given: 19

GMAT Tests User
Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 07:59
LiveStronger wrote:
scthakur wrote:
A for me.

Area of triangle = 1/2 * 2/a * a^2 = a. Hence, we need to find out whether a > 1/

From stmt1: AB^2 + BC^2 > AC^2 or 2*AB^2 > AC^2
or 2*(a^4 + 1/a^2) > 4/a^2
or, (a^6 + 1) > 2 or, a^6 > 1 that means a > 1. Hence, sufficient.

Simplifying stmt2 will give a > 0...insufficient.


scthakur : Can you explain how you simplified stmt1

And from a^6 > 1, how did you come up with a > 1, a could be -2 also right ?, or is it because 1/a is a point on Positive X-axis ?


nice work by scthakur.

since a is a measurement of length of sides, a cannot be -ve. therefore a has some +ve value.
_________________

Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

CEO
CEO
User avatar
Joined: 29 Aug 2007
Posts: 2501
Followers: 53

Kudos [?]: 506 [0], given: 19

GMAT Tests User
Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 08:10
scthakur wrote:
amitdgr wrote:
scthakur wrote:
A for me.

Simplifying stmt2 will give a > 0...insufficient.


how do we simplify stmt 2 ??

Thanks :)


Perimeter = AB + BC + AC = 2AB + AC = 2*sqrt((1/a^2)+ a^4) + 2/a
= 2*(sqrt(1+a^6))/a + 2/a > 4/a
or, sqrt(1+a^6) + 1 > 2
or, sqrt(1+a^6) > 1 or a > 0.


that should be a^2. if so:
2*(sqrt(1+a^6))/a^2 + 2/a > 4/a
or, sqrt(1+a^6)/a + 1 > 2
or, sqrt(1+a^6) > a
or, (1+a^6) > a^2
or, a^6 - a^2 > -1
or, a^2 (a^4 - 1) > -1
1: a^2 > -1
a > -1

2: a^4 - 1 > -1
a^4 > 0 .

in either case a is only +ve but not sure whether it is >1. so nsf.

A is correct.
_________________

Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

Current Student
avatar
Joined: 28 Dec 2004
Posts: 3404
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 159 [0], given: 2

GMAT Tests User
Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 15:58
Nice work..i almost fell for C..

I agree A is the ans..
SVP
SVP
avatar
Joined: 17 Jun 2008
Posts: 1579
Followers: 12

Kudos [?]: 183 [0], given: 0

GMAT Tests User
Re: DS : TRIANGLE [#permalink] New post 29 Oct 2008, 21:36
GMAT TIGER wrote:
scthakur wrote:
Perimeter = AB + BC + AC = 2AB + AC = 2*sqrt((1/a^2)+ a^4) + 2/a
= 2*(sqrt(1+a^6))/a + 2/a > 4/a
or, sqrt(1+a^6) + 1 > 2
or, sqrt(1+a^6) > 1 or a > 0.


that should be a^2. if so:
2*(sqrt(1+a^6))/a^2 + 2/a > 4/a
or, sqrt(1+a^6)/a + 1 > 2
or, sqrt(1+a^6) > a
or, (1+a^6) > a^2
or, a^6 - a^2 > -1
or, a^2 (a^4 - 1) > -1
1: a^2 > -1
a > -1

2: a^4 - 1 > -1
a^4 > 0 .

in either case a is only +ve but not sure whether it is >1. so nsf.

A is correct.



GT, I think my expression is correct as a is out of sqrt. Within sqrt, it will be a^2. Am I missing something?
Director
Director
User avatar
Joined: 25 Oct 2008
Posts: 611
Location: Kolkata,India
Followers: 9

Kudos [?]: 173 [0], given: 100

GMAT Tests User
Re: DS : TRIANGLE (m09q07) [#permalink] New post 05 Oct 2009, 17:08
Hey guys..not clear..could somebody explain it in more detail?
I understood that area of the triangle=a
so we have to see whther a>1..
Got stuck in stmt 1:How does it imply AB^2 + BC^2 > AC^2 ??
and then 2*(a^4 + 1/a^2) > 4/a^2??????
_________________

countdown-beginshas-ended-85483-40.html#p649902

Director
Director
User avatar
Joined: 25 Oct 2008
Posts: 611
Location: Kolkata,India
Followers: 9

Kudos [?]: 173 [0], given: 100

GMAT Tests User
Re: DS : TRIANGLE (m09q07) [#permalink] New post 06 Oct 2009, 21:54
Manager
Manager
User avatar
Joined: 14 Aug 2009
Posts: 123
Followers: 2

Kudos [?]: 90 [0], given: 13

Re: DS : TRIANGLE (m09q07) [#permalink] New post 22 Feb 2010, 05:24
tejal777 wrote:
Hey guys..not clear..could somebody explain it in more detail?
I understood that area of the triangle=a
so we have to see whther a>1..
Got stuck in stmt 1:How does it imply AB^2 + BC^2 > AC^2 ??
and then 2*(a^4 + 1/a^2) > 4/a^2??????


got the same question:
How does it imply AB^2 + BC^2 > AC^2 ??
_________________

Kudos me if my reply helps!

Intern
Intern
avatar
Joined: 25 Nov 2009
Posts: 46
Location: India
Followers: 0

Kudos [?]: 8 [0], given: 6

Re: DS : TRIANGLE (m09q07) [#permalink] New post 22 Feb 2010, 06:45
flyingbunny wrote:
tejal777 wrote:
Hey guys..not clear..could somebody explain it in more detail?
I understood that area of the triangle=a
so we have to see whther a>1..
Got stuck in stmt 1:How does it imply AB^2 + BC^2 > AC^2 ??
and then 2*(a^4 + 1/a^2) > 4/a^2??????


got the same question:
How does it imply AB^2 + BC^2 > AC^2 ??


Although I'm still confused with the problem (actually the significance of angleABC < 90-degree), still "trying to force" a logic for the above algebric inequality.
Since the AD (y-axis) is perpendicular to BC (x-axis),
AB^2 = AD^2 + BD^2. Thus, AB^2 = (a^6 + 1)/ a^2 = AC^2 ------- (i)
Now for triangle ABC, length of 2-sides must be greater than that of the 3rd side. So,
AB + AC > BC. => 2 AB > BC => sqrt (2(a^6 + 1)/ a^2) > 2/a ------- (ii)
i.e. 2(a^6 + 1)/ a^2 > 4/a^2 => a^6 + 1 > 2 => a^6 > 1.
Since "a" (measure of side of a triangle) cannot be negative, "a^6 > 1" implies a > 1

Can someone help with OE/ OA pl?
Intern
Intern
User avatar
Joined: 29 Dec 2009
Posts: 42
Location: India
Concentration: Finance, Real Estate
Schools: Duke (Fuqua) - Class of 2014
GMAT 1: 770 Q50 V44
GPA: 3.5
WE: General Management (Real Estate)
Followers: 4

Kudos [?]: 10 [0], given: 1

Re: DS : TRIANGLE (m09q07) [#permalink] New post 22 Feb 2010, 07:58
Please explain:

How does angle ABC < 90 deg. lead to the conclusion "AB^2 + BC^2 > AC^2"?
Senior Manager
Senior Manager
User avatar
Joined: 21 Dec 2009
Posts: 268
Location: India
Followers: 9

Kudos [?]: 69 [0], given: 25

GMAT Tests User
Re: DS : TRIANGLE (m09q07) [#permalink] New post 22 Feb 2010, 10:22
St 1:

If Ang (ABC) <90
then Ang (ABO) < 45
So, Ang (BAO) > 45 as Ang (ABO) = 90

OB/OA > 1 as tan (BAO) > 1

a^2/(1/a) > 1

or a^3 > 1 and a>0 as it is a length

so a>1

but

st 2:

a>0 as analysed by others.

So answer is A
_________________

Cheers,
SD

Senior Manager
Senior Manager
User avatar
Joined: 21 Dec 2009
Posts: 268
Location: India
Followers: 9

Kudos [?]: 69 [0], given: 25

GMAT Tests User
Re: DS : TRIANGLE (m09q07) [#permalink] New post 22 Feb 2010, 10:23
deepakdewani wrote:
Please explain:

How does angle ABC < 90 deg. lead to the conclusion "AB^2 + BC^2 > AC^2"?


Its rule. Just memorise it.
_________________

Cheers,
SD

Re: DS : TRIANGLE (m09q07)   [#permalink] 22 Feb 2010, 10:23
    Similar topics Author Replies Last post
Similar
Topics:
22 Experts publish their posts in the topic DS : TRIANGLE (m09q07) amitdgr 41 28 Oct 2008, 23:53
Experts publish their posts in the topic DS: Triangle empanado 5 26 Nov 2007, 19:29
ds: triangle mirhaque 5 20 May 2005, 17:14
DS: Triangle mirhaque 14 07 May 2005, 04:07
DS:Triangle mirhaque 11 23 Mar 2005, 20:23
Display posts from previous: Sort by

DS : TRIANGLE (m09q07)

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page    1   2   3    Next  [ 42 posts ] 

Moderators: Bunuel, WoundedTiger



GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

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®.