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Hello. I am doing the PR1012, and I came across this question. I got it wrong to begin with, but I also do not understand their brief explanation. Please help me. Just for the record I am in the college class of 2015, but I need to take the GMAT before the end of my sophomore year for a combined degree program (shooting for 700; I know it is really hard!). Eventually I may want an MBA, though. I am studying now to ease into it. I am pretty smart for the most part, but I do badly on tests. I got a 1380 SAT (670M, 710CR) with a little studying. Anyway, here is the question.

After one day of studying I already hate the data questions.

In the figure above, if angle ABC = 110 degrees, and angle AED = 100 degrees, then angle ADE is how many degrees smaller than angle BCD?

Re: In the figure above, if angle ABC = 110 degrees, and angle AED = 100 d [#permalink]

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24 Jun 2011, 13:28

St 2 is very simple, as discussed in the above post. But I solved the first stmt like below, I hope it'll help you! Let <BCD = y; <ADE = x; therefore <CDA = 180 - x (since, <CDA and <ADE are supplimentary)

Now sum of angles of quadrilateral ABCD is 360; <DAB + <ABC + <BCD + <CDA = 360 => 45 + 110 + Y + 180 - X = 360

You can get the difference b/t y and X which is the answer for the question. Therefore the answer is D

Hello. I am doing the PR1012, and I came across this question. I got it wrong to begin with, but I also do not understand their brief explanation. Please help me. Just for the record I am in the college class of 2015, but I need to take the GMAT before the end of my sophomore year for a combined degree program (shooting for 700; I know it is really hard!). Eventually I may want an MBA, though. I am studying now to ease into it. I am pretty smart for the most part, but I do badly on tests. I got a 1380 SAT (670M, 710CR) with a little studying. Anyway, here is the question.

After one day of studying I already hate the data questions.

In the figure above, if angle ABC = 110 degrees, and angle AED = 100 degrees, then angle ADE is how many degrees smaller than angle BCD?

1. Angle DAB = 45 degrees

2. Angle ADC + angle BCD = 205 degrees

Important property of polygons: Sum of Interior Angles of a polygon is \(180(n-2)\) where \(n\) is the number of sides.

So for quadrilateral sum of the angles equals to \(180(4-2)=360\) degrees.

Given: \(\angle{ABC}=110\) and \(\angle{AED}=100\). Question: \(\angle{BCD}-\angle{ADE}=?\).

Attachment:

DS.jpg

Now as \(\angle{ABC}=110\) and \(\angle{AED}=100\) and the sum of all angles is 360 degrees then the sum of other two angles in quadrilateral must be \(\angle{BCD}+\angle{BAE}=360-(110+100)=150\).

(1) \(\angle{DAB}=45\) --> \(\angle{BAE}=\angle{DAB}+\angle{DAE}=45+\angle{DAE}\), but \(\angle{DAE}=180-100-\angle{ADE}=80-\angle{ADE}\) (the sum of interior angles in triangle ADE equals to 180 degrees) --> \(\angle{BAE}=45+80-\angle{ADE}=125-\angle{ADE}\). So \(\angle{BCD}+\angle{BAE}=150\) becomes \(\angle{BCD}+125-\angle{ADE}=150\) --> \(\angle{BCD}-\angle{ADE}=25\). Sufficient.