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14 Aug 2006, 13:27
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OK, so I really don't think I'm better than Kaplan 
), but am stumped on how they came to this conclusion:
Reference: Kaplan Premier 2006-2007 w CD:Part Two: DS: page 303: Problem # 20. (Figure attached)
20. In the figure above, segments AD and AC divide angle EAB into 3 non overlapping angles that are equal in measure. Are AE and AB equal in length? DS question:
1. AD = AC
2. AC = CB
Answers: A, B, C, D, or E?
Correct answer by Kaplan: A: Statement 1 is sufficient (I agree with this part); statement 2 is not (I disagree). This is why I think answer should be D – both statements are sufficient by themselves:
1. We know: angles EAD = DAC = CAB -> (= x let’s say)
2. We know AC = CB. Therefore angles CEB = CBA = x. -> (isosceles)
3. Therefore angle BCA = (180 – 2x) -> [angle CAB+CBA = x+x = 2x]
4. Therefore angle ACD = 180- angle BCA = 180 – (180 - 2x) = 2x
5. Therefore angle ADC = (180 - DAC + DCA) = 180 – 3x
6. Therefore angle EDA = 180 - angle ADC = 180 – (180 - 3x) = 3x
7. Therefore angle DEA= 180 – (EAD + EDA) = 180 – (x+3x) = 180 – 4x
Now from 8. and 2. above we see that angle AED (180-4x) is not equal to angle CBA (x). Therefore AE and AB are not equal in length. Sufficient to answer.
Hence my answer is D. Where did I do wrong?
), but am stumped on how they came to this conclusion:
Reference: Kaplan Premier 2006-2007 w CD:Part Two: DS: page 303: Problem # 20. (Figure attached)
20. In the figure above, segments AD and AC divide angle EAB into 3 non overlapping angles that are equal in measure. Are AE and AB equal in length? DS question:
1. AD = AC
2. AC = CB
Answers: A, B, C, D, or E?
Correct answer by Kaplan: A: Statement 1 is sufficient (I agree with this part); statement 2 is not (I disagree). This is why I think answer should be D – both statements are sufficient by themselves:
1. We know: angles EAD = DAC = CAB -> (= x let’s say)
2. We know AC = CB. Therefore angles CEB = CBA = x. -> (isosceles)
3. Therefore angle BCA = (180 – 2x) -> [angle CAB+CBA = x+x = 2x]
4. Therefore angle ACD = 180- angle BCA = 180 – (180 - 2x) = 2x
5. Therefore angle ADC = (180 - DAC + DCA) = 180 – 3x
6. Therefore angle EDA = 180 - angle ADC = 180 – (180 - 3x) = 3x
7. Therefore angle DEA= 180 – (EAD + EDA) = 180 – (x+3x) = 180 – 4x
Now from 8. and 2. above we see that angle AED (180-4x) is not equal to angle CBA (x). Therefore AE and AB are not equal in length. Sufficient to answer.
Hence my answer is D. Where did I do wrong?
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14 Aug 2006, 14:02
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Why can't these two angles be equal?
180-4x=x if x=36, they are equal. So the two sides MIGHT be equal in length!
I have to defend Kaplan. I'm a Kaplan instructor
180-4x=x if x=36, they are equal. So the two sides MIGHT be equal in length!
I have to defend Kaplan. I'm a Kaplan instructor
15 Aug 2006, 07:10
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Aah perfect! Thanks this is what I was hoping for, in that case it becomes ambigous.
BTW - since you are a Kaplan instructor, a question for you - a lot of the probability questions, esp. Coin tosses etc. are solved by mapping out "HHH TTT HHT TTH" etc. [heads = H, Tails = T].
Is there a quicker way of doing such problems?
