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The figure shows that AB=AC=DE=DC, ∠BAC = 30o, ∠PED = 50o and △ABC an

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The figure shows that AB=AC=DE=DC, ∠BAC = 30o, ∠PED = 50o and △ABC an  [#permalink]

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New post Updated on: 12 Jul 2020, 08:54
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[GMAT math practice question]

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The figure shows that \(AB=AC=DE=DC, ∠BAC = 30^o, ∠PED = 50^o\) and \(△ABC\) and \(△CDE\) are congruent. What is the measure of angle \(∠ABP = x\)?

A. \(10^o \)

B. \(15^o\)

C. \(20^o\)

D. \(25^o\)

E. \(30^o\)

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Originally posted by MathRevolution on 06 Jul 2020, 00:40.
Last edited by Bunuel on 12 Jul 2020, 08:54, edited 2 times in total.
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Re: The figure shows that AB=AC=DE=DC, ∠BAC = 30o, ∠PED = 50o and △ABC an  [#permalink]

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New post 06 Jul 2020, 02:55
MathRevolution , Figure is missing.

Please edit the question with figure.
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Re: The figure shows that AB=AC=DE=DC, ∠BAC = 30o, ∠PED = 50o and △ABC an  [#permalink]

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New post 06 Jul 2020, 02:56
Math Revolution GMAT Instructor
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Re: The figure shows that AB=AC=DE=DC, ∠BAC = 30o, ∠PED = 50o and △ABC an  [#permalink]

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New post 07 Jul 2020, 23:22
=>

\(∠ABC = ∠DEC = \frac{(180° - 30°) }{ 2} = 75°.\)

Then we have \(∠BEC = 180° – ( 50° + 75° ) = 55°.\)

Since exterior \(∠BEC\) of triangle \(ABE\) is the sum of non-adjacent interior angles, \(30°\), and \(x\), we have \(x + 30° = 55°\) or \(x = 25°.\)

Therefore, D is the correct answer.
Answer: D
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Re: The figure shows that AB=AC=DE=DC, ∠BAC = 30o, ∠PED = 50o and △ABC an   [#permalink] 07 Jul 2020, 23:22

The figure shows that AB=AC=DE=DC, ∠BAC = 30o, ∠PED = 50o and △ABC an

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