Last visit was: 24 Apr 2024, 09:54 It is currently 24 Apr 2024, 09:54

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
SORT BY:
Date
Tags:
Show Tags
Hide Tags
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3726
Own Kudos [?]: 16832 [34]
Given Kudos: 165
Send PM
Most Helpful Reply
GMAT Tutor
Joined: 24 Jun 2008
Posts: 4128
Own Kudos [?]: 9241 [10]
Given Kudos: 91
 Q51  V47
Send PM
Math Expert
Joined: 02 Sep 2009
Posts: 92902
Own Kudos [?]: 618772 [6]
Given Kudos: 81587
Send PM
General Discussion
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3726
Own Kudos [?]: 16832 [3]
Given Kudos: 165
Send PM
Re: Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
3
Kudos
Expert Reply
Note: This questions is related to the article on Common Errors in Geometry

Kindly go through the article once, before solving the question or going through the solution.


Official Solution


Find: Value of angle CDB


Working:


Analysing Statement 1:


Given: angle CAB = angle ACB = \(30^o\).

In triangle ACD –

angle (ACD + CAD + CDA) = \(180^o\)

We know that –

angle ACD = \(90^o\)
angle CAB = \(30^o\)

and CDA can be written as CDB

Therefore,

\(90^o\) + \(30^o\) + CDB = \(180^o\)

angle CDB = \(60^o\).

Hence Statement 1 is sufficient to the answer the question.


Analysing Statement 2:


Given: AB = BC = BD

We can conclude that triangle ABC and triangle CBD are isosceles triangles.

But that will not help us to find the angle CBD, as we cannot find the value of individuals angles of either triangle ABC or triangle CDB.

Hence statement 2 is not sufficient to answer the question.


Since we are getting our answer from the first statement only.

The correct answer option is A



Thanks,
Saquib
Quant Expert
e- GMAT


Originally posted by EgmatQuantExpert on 22 Nov 2016, 11:12.
Last edited by EgmatQuantExpert on 01 Dec 2016, 01:48, edited 1 time in total.
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3726
Own Kudos [?]: 16832 [0]
Given Kudos: 165
Send PM
Re: Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
Expert Reply
Hey Everyone,

The official solution has been posted. Kindly go through it once and feel free to post your doubts, if any.

Thanks,
Saquib
Quant Expert
e-GMAT
Intern
Intern
Joined: 01 Jun 2016
Posts: 22
Own Kudos [?]: 36 [3]
Given Kudos: 48
Send PM
Re: Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
3
Kudos
EgmatQuantExpert wrote:
Hey Everyone,

The official solution has been posted. Kindly go through it once and feel free to post your doubts, if any.

Thanks,
Saquib
Quant Expert
e-GMAT


EgmatQuantExpert

Can you please explain how statement 2 is insufficient?

Below is my analysis:
Angle CAB + Angle CBA = Angle CBA
Angle BCD + Angle BDC = Angle CBD

Adding both the equations we get:
Angle CAB + Angle CBA + Angle BCD + Angle BDC = 180

as triangle ABC and CBD are isosceles triangle and share a common side so,

4 X Angle BDC = 180
Angle BDC = 45.

Statement 2 is sufficient.
Intern
Intern
Joined: 28 Apr 2016
Posts: 11
Own Kudos [?]: 13 [3]
Given Kudos: 78
Send PM
Re: Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
3
Kudos
EgmatQuantExpert wrote:
Note: This questions is related to the article on Common Errors in Geometry

Kindly go through the article once, before solving the question or going through the solution.


Official Solution


Find: Value of angle CDB


Working:


Analysing Statement 1:


Given: angle CAB = angle ACB = \(30^o\).

In triangle ACD –

angle (ACD + CAD + CDA) = \(180^o\)

We know that –

angle ACD = \(90^o\)
angle CAB = \(30^o\)

and CDA can be written as CDB

Therefore,

\(90^o\) + \(30^o\) + CDB = \(180^o\)

angle CDB = \(60^o\).

Hence Statement 1 is sufficient to the answer the question.


Analysing Statement 2:


Given: AB = BC = BD

We can conclude that triangle ABC and triangle CBD are isosceles triangles.

But that will not help us to find the angle CBD, as we cannot find the value of individuals angles of either triangle ABC or triangle CDB.

Hence statement 2 is not sufficient to answer the question.


Since we are getting our answer from the first statement only.

The correct answer option is A



Thanks,
Saquib
Quant Expert
e- GMAT




As per statement 2:
AB=BD, thus CB is bisecting AD.

As ACD=90 degress, ACB=BCD=45 degrees.

Statement 2 is also sufficient
Manager
Manager
Joined: 07 Jun 2017
Posts: 81
Own Kudos [?]: 19 [0]
Given Kudos: 454
Send PM
Re: Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
Since BC = BD, which makes CBD 1:1:sqrt 2 right triangle, correct?
So we should be able to know the angle CDB. isn't it?
Manager
Manager
Joined: 16 Jan 2013
Posts: 66
Own Kudos [?]: 21 [0]
Given Kudos: 1323
Location: Bangladesh
GMAT 1: 490 Q41 V18
GMAT 2: 610 Q45 V28
GPA: 2.75
Send PM
Re: Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
Not sure why B is insufficient. According to statement 2, BC is a straight line and perpendicular to AD. It can be inferred that Angle BCD = Angle BDC = 45°

Someone please clarify. Thanks.
Manager
Manager
Joined: 02 Nov 2015
Posts: 133
Own Kudos [?]: 57 [0]
Given Kudos: 121
GMAT 1: 640 Q49 V29
Send PM
Re: Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
I am quite confused regarding statement 2. Initially I thought it to be sufficient.

Experts kindly help , how is it insufficient.
My reasoning is same as posted above by two persons.

Sent from my Lenovo TAB S8-50LC using GMAT Club Forum mobile app
Intern
Intern
Joined: 25 Apr 2017
Posts: 14
Own Kudos [?]: 8 [2]
Given Kudos: 26
GMAT 1: 710 Q48 V40
Send PM
Re: Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
2
Kudos
Having a very hard time understanding any of the explanations of why statement 2 is insufficient. The question seems to imply that angle ACD is 90 degrees. If AB = BC = BD, doesn't that mean that angle ACB + BCD = 90?
Intern
Intern
Joined: 25 Apr 2017
Posts: 14
Own Kudos [?]: 8 [1]
Given Kudos: 26
GMAT 1: 710 Q48 V40
Send PM
Re: Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
1
Kudos
Bunuel wrote:
getitdoneright wrote:

Having a very hard time understanding any of the explanations of why statement 2 is insufficient. The question seems to imply that angle ACD is 90 degrees. If AB = BC = BD, doesn't that mean that angle ACB + BCD = 90?


Yes, angle ACD is 90 degrees and ACB + BCD = 90 degrees but the point is that we don't know whether CB is perpendicular to AD. Not knowing that, how are you getting that (2) is sufficient?


It seems that I am assuming wrongly that angle ACB = angle BCD = angle CDB since AB = BD = BC and each of those angles is opposite each respective side. Since ACB + BCD = 90 degrees and ACB = BCD, then ACB and BCD = 45 degree each. Thus CDB = 45 degrees.

Any chance you could help me identify where my thinking is flawed? A few months later and I still can't figure it out. Thanks a ton.
Intern
Intern
Joined: 27 Jun 2015
Posts: 46
Own Kudos [?]: 12 [0]
Given Kudos: 77
Location: India
WE:Information Technology (Computer Software)
Send PM
Re: Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
Bunuel wrote:
getitdoneright wrote:

Having a very hard time understanding any of the explanations of why statement 2 is insufficient. The question seems to imply that angle ACD is 90 degrees. If AB = BC = BD, doesn't that mean that angle ACB + BCD = 90?


Yes, angle ACD is 90 degrees and ACB + BCD = 90 degrees but the point is that we don't know whether CB is perpendicular to AD. Not knowing that, how are you getting that (2) is sufficient?


Hi Bunuel,

Its given,AB=BC=BD
So here, it satisfies this equation: BC^2=AB.BD
This implies, CB is perpendicular to AD.

Can you please tell whether this understanding is flawed?
Math Expert
Joined: 02 Sep 2009
Posts: 92902
Own Kudos [?]: 618772 [5]
Given Kudos: 81587
Send PM
Re: Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
4
Kudos
Expert Reply
kkrrsshh wrote:
Bunuel wrote:
getitdoneright wrote:

Having a very hard time understanding any of the explanations of why statement 2 is insufficient. The question seems to imply that angle ACD is 90 degrees. If AB = BC = BD, doesn't that mean that angle ACB + BCD = 90?


Yes, angle ACD is 90 degrees and ACB + BCD = 90 degrees but the point is that we don't know whether CB is perpendicular to AD. Not knowing that, how are you getting that (2) is sufficient?


Hi Bunuel,

Its given,AB=BC=BD
So here, it satisfies this equation: BC^2=AB.BD
This implies, CB is perpendicular to AD.

Can you please tell whether this understanding is flawed?


Useful property: The median on the hypotenuse of a right triangle always equals to one-half the hypotenuse.

AB = BC = BD means that CB is the median (B is the midpoint of AD). Because of the property above, AB = BC = BD is true for any right angled triangle with median CB. So, CB may or may not be perpendicular to AD.
Tutor
Joined: 12 Oct 2010
Status:GMATH founder
Posts: 893
Own Kudos [?]: 1353 [0]
Given Kudos: 56
Send PM
Re: Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
Expert Reply
EgmatQuantExpert wrote:



In the triangle above, what is the measure of Angle CDB?
(1) Angle CAB = Angle ACB = 30°
(2) AB = BC = BD

\(? = \angle CDB = x\)

(All angles are measured in degrees.)

(1) Using the exterior angle property (to justify 30+30 in red), triangle BCD is equilateral (hence x=60). Sufficient!

(2) The constructions shown below are viable and present two different possible values for x. Insufficient!





This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Director
Director
Joined: 09 Aug 2017
Posts: 689
Own Kudos [?]: 415 [0]
Given Kudos: 778
Send PM
Re: Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
So it is all about notion of 90 degree.
90 degree notion is given between AC & CD but is absent between CB and AD.

If notion is not provided in figure, we should not assume two line perpendicular even if they look exactly perpendicular.

fskilnik wrote:
EgmatQuantExpert wrote:



In the triangle above, what is the measure of Angle CDB?
(1) Angle CAB = Angle ACB = 30°
(2) AB = BC = BD

\(? = \angle CDB = x\)

(All angles are measured in degrees.)

(1) Using the exterior angle property (to justify 30+30 in red), triangle BCD is equilateral (hence x=60). Sufficient!

(2) The constructions shown below are viable and present two different possible values for x. Insufficient!





This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Intern
Intern
Joined: 24 Mar 2018
Posts: 42
Own Kudos [?]: 24 [0]
Given Kudos: 61
Location: India
Concentration: Strategy, Sustainability
GPA: 3.7
WE:Consulting (Consulting)
Send PM
Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
In the two triangles, ABC and CBD , side AB=BC (given), BC=BD (given), hence third side is equal too which is AC=CD (implied). Implying, Angle CAB to be equal to Angle CDB. Hence, by eqaution 2x+2y=180 or 4y=180, y or angle CDB = 45 degrees. How is this reasoning wrong?
Intern
Intern
Joined: 02 Jan 2020
Posts: 26
Own Kudos [?]: 25 [0]
Given Kudos: 45
Send PM
Re: Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
srishti246 wrote:
In the two triangles, ABC and CBD , side AB=BC (given), BC=BD (given), hence third side is equal too which is AC=CD (implied). Implying, Angle CAB to be equal to Angle CDB. Hence, by eqaution 2x+2y=180 or 4y=180, y or angle CDB = 45 degrees. How is this reasoning wrong?


side AC=CD can be true only if angle ABC = angle CBD, which is not given in the statement.. from statement we can only infer that triangles ABC and CBD are isosceles, however we need one angle also to infer that triangles are similar triangles
Director
Director
Joined: 21 Feb 2017
Posts: 521
Own Kudos [?]: 1034 [0]
Given Kudos: 1091
Location: India
GMAT 1: 700 Q47 V39
Send PM
Re: Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
As per st 2:

Angle ACB= x then BAC = x
Angle DCB= y and hence BDC = y
ACB is not equal to DCB here
Hence B isn't sufficient.
BSchool Moderator
Joined: 22 Jul 2018
Posts: 475
Own Kudos [?]: 372 [1]
Given Kudos: 618
GRE 1: Q165 V163

GRE 2: Q165 V163

GRE 3: Q170 V163
WE:Engineering (Computer Software)
Send PM
Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
1
Kudos
Bunuel wrote:
kkrrsshh wrote:
Bunuel wrote:
getitdoneright wrote:

Having a very hard time understanding any of the explanations of why statement 2 is insufficient. The question seems to imply that angle ACD is 90 degrees. If AB = BC = BD, doesn't that mean that angle ACB + BCD = 90?


Yes, angle ACD is 90 degrees and ACB + BCD = 90 degrees but the point is that we don't know whether CB is perpendicular to AD. Not knowing that, how are you getting that (2) is sufficient?


Hi Bunuel,

Its given,AB=BC=BD
So here, it satisfies this equation: BC^2=AB.BD
This implies, CB is perpendicular to AD.

Can you please tell whether this understanding is flawed?


Useful property: The median on the hypotenuse of a right triangle always equals to one-half the hypotenuse.

AB = BC = BD means that CB is the median (B is the midpoint of AD). Because of the property above, AB = BC = BD is true for any right angled triangle with median CB. So, CB may or may not be perpendicular to AD.


Hi Bunuel, chetan2u, VeritasKarishma
In STMT 2,
It is provided that AB=BD=CB,
Which means that Angle CAB = Angle ACB = Angle BCD = Angle CDB = x.

So would it be incorrect to assume that the triangles ACB and CBD are congruent(Angle CAB = Angle CDB and Angle ACB = Angle BCD ) wih BC being the common side. For this to be possible Angle CBA needs to be a right angle and Angle CDB hence becomes 45

Even solving it geometrically,
We would be arriving at Angle CBA = 180 -2x(Angle sum property)
And Angle CBD to be 2x

In Triangle BCD, Angle CDB and Angle BCD are equal to x
so 4x =180
x= 45 and 180-2x =90

Consequently, I feel that the Answer should be D. Could you please point out where I am going wrong?

Thank You
GMAT Club Bot
Common Mistakes in Geometry Questions - Exercise Question #2 [#permalink]
 1   2   
Moderator:
Math Expert
92902 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne