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Strategies and techniques for approaching featured GMAT topics

# M24-08

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Math Expert
Joined: 02 Sep 2009
Posts: 51223

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16 Sep 2014, 00:20
00:00

Difficulty:

65% (hard)

Question Stats:

69% (01:33) correct 31% (01:50) wrong based on 143 sessions

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In quadrilateral $$ABCD$$, $$AB = CD$$ and $$BC = AD$$. If $$\angle CBD = 30$$ degrees and $$\angle BAD = 80$$ degrees, what is the value of $$\angle ADC$$?

A. 30 degrees
B. 50 degrees
C. 70 degrees
D. 100 degrees
E. 120 degrees

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Joined: 02 Sep 2009
Posts: 51223

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16 Sep 2014, 00:20
Official Solution:

In quadrilateral $$ABCD$$, $$AB = CD$$ and $$BC = AD$$. If $$\angle CBD = 30$$ degrees and $$\angle BAD = 80$$ degrees, what is the value of $$\angle ADC$$?

A. 30 degrees
B. 50 degrees
C. 70 degrees
D. 100 degrees
E. 120 degrees

Because $$AB = CD$$ and $$BC = AD$$, $$ABCD$$ is a parallelogram.

$$\angle ADC = \angle BDA + \angle BDC = \angle CBD + \angle ABD = \angle CBD + (180 - \angle BAD - \angle BDA) =$$

$$= 30 + (180 - 80 - 30) = 100$$.

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17 Apr 2016, 11:53
To make the explanation visual, did someone solve this question graphically? thank you.
Intern
Joined: 27 Mar 2014
Posts: 9

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16 Aug 2016, 17:52
Hi Bunuel,

Thanks for all that you do. Your explanations have made the study grind a lot more manageable.

I was under the impression that when you write out an angle expression (ex. ∠ABC), it means the angle is formed by the intersection of line AB with line BC (see attached illustration)

Is this the case? If so, I'm having trouble visualizing how you could form the parallelogram above using the given information. Thanks in advance!
>> !!!

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27 Nov 2016, 18:32
Can someone explain how B can be the vertex of angle CBD if the prompt states AB=CD and AD=BC? I can't figure out how to draw this...
Current Student
Joined: 12 Jul 2013
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07 Dec 2016, 20:45
Hi Bunnel,

Can you please elaborate more on solution .. i got this question wrong.
Manager
Joined: 03 Jan 2016
Posts: 59
Location: India
WE: Engineering (Energy and Utilities)

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05 Apr 2017, 00:55
Question didn't mentioned that Opposite sides are parallel !!!!

can we still assume that given quadrilateral is Parallelogram?

Requesting expert analysis on this !!!

Narayana Raju
Math Expert
Joined: 02 Sep 2009
Posts: 51223

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05 Apr 2017, 01:08
gvvsnraju@1 wrote:
Question didn't mentioned that Opposite sides are parallel !!!!

can we still assume that given quadrilateral is Parallelogram?

Requesting expert analysis on this !!!

Narayana Raju

This is explained in the solution:
Because $$AB = CD$$ and $$BC = AD$$, $$ABCD$$ is a parallelogram.

If two pairs of opposite sides of a quadrilateral are equal in length then the quadrilateral is a parallelogram.
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10 May 2017, 21:53
1
Bunuel wrote:
In quadrilateral $$ABCD$$, $$AB = CD$$ and $$BC = AD$$. If $$\angle CBD = 30$$ degrees and $$\angle BAD = 80$$ degrees, what is the value of $$\angle ADC$$?

A. 30 degrees
B. 50 degrees
C. 70 degrees
D. 100 degrees
E. 120 degrees

Consecutive angles are supplementary => $$\angle BAD + \angle ADC=180$$

$$\angle ADC=180-80=100$$

Am I right here?
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11 May 2017, 01:23
1
Here is the image for visual view
Attachment:
Capture.PNG

>> !!!

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11 May 2017, 01:42
Sirakri wrote:
Bunuel wrote:
In quadrilateral $$ABCD$$, $$AB = CD$$ and $$BC = AD$$. If $$\angle CBD = 30$$ degrees and $$\angle BAD = 80$$ degrees, what is the value of $$\angle ADC$$?

A. 30 degrees
B. 50 degrees
C. 70 degrees
D. 100 degrees
E. 120 degrees

Consecutive angles are supplementary => $$\angle BAD + \angle ADC=180$$

$$\angle ADC=180-80=100$$

Am I right here?

Yes, consecutive angles in a parallelogram are supplementary, add to 180°, so yes you are right.
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17 Apr 2018, 05:00
+1 for option D. The angle is 100.
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17 Apr 2018, 16:16
Does the position of A, B , C and D matter for these types of problems? If we wanted to draw it, what is the correct placement?
Math Expert
Joined: 02 Sep 2009
Posts: 51223

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17 Apr 2018, 21:39
1
bpdulog wrote:
Does the position of A, B , C and D matter for these types of problems? If we wanted to draw it, what is the correct placement?

OFFICIAL GUIDE:

Problem Solving
Figures: All figures accompanying problem solving questions are intended to provide information useful in solving the problems. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

Data Sufficiency:
Figures:
• Figures conform to the information given in the question, but will not necessarily conform to the additional information given in statements (1) and (2).
• Lines shown as straight are straight, and lines that appear jagged are also straight.
• The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero.
• All figures lie in a plane unless otherwise indicated.
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12 Nov 2018, 06:16
Because you know that its a parallelogram, you don't need the information that CBD is 30 degrees. Opposite angles in parallelogram are the same and quadrilateral has 360 degrees in total -> ADC = [360-(2*80)]/2
Intern
Joined: 11 Mar 2017
Posts: 9

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06 Dec 2018, 21:19
Hi,

I solved by simply deducting angle(BAD) from 180 degrees to find the supplementary angel of BAD, which will be equal to angle ADC since AB and DC are parallel.

Is this approach correct?

Bunuel wrote:
Official Solution:

In quadrilateral $$ABCD$$, $$AB = CD$$ and $$BC = AD$$. If $$\angle CBD = 30$$ degrees and $$\angle BAD = 80$$ degrees, what is the value of $$\angle ADC$$?

A. 30 degrees
B. 50 degrees
C. 70 degrees
D. 100 degrees
E. 120 degrees

Because $$AB = CD$$ and $$BC = AD$$, $$ABCD$$ is a parallelogram.

$$\angle ADC = \angle BDA + \angle BDC = \angle CBD + \angle ABD = \angle CBD + (180 - \angle BAD - \angle BDA) =$$

$$= 30 + (180 - 80 - 30) = 100$$.

Re: M24-08 &nbs [#permalink] 06 Dec 2018, 21:19
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# M24-08

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