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M01-21

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M01-21  [#permalink]

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New post 15 Sep 2014, 23:15
1
4
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

63% (01:08) correct 37% (01:24) wrong based on 211 sessions

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Re M01-21  [#permalink]

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New post 15 Sep 2014, 23:15
Official Solution:


Notice that since ABC is a right triangle then the question basically asks whether angle ACB is equal to angle CAB, because in this case AB would be equal to BC, thus triangle ABC would be an isosceles triangle.

(1) Angle ACB is twice as large as angle ADC. It certainly possible angle ACB to be equal to angle CAB (45 degrees) and angle ADC to be 22.5 degrees but it's also possible angle ACB NOT to be equal to angle CAB, for example if angle ACB is 60 degrees and angle ADC is 30 degrees. Not sufficient.

(2) Angle ACB is twice as large as angle CAB. Directly says that angle ACB is NOT equal to angle CAB, hence triangle ABC is NOT isosceles. Sufficient.


Answer: B
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Re: M01-21  [#permalink]

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New post 31 Oct 2015, 19:57
1 doesn't tell much
2 tells us that angle CAB is 1/2 of angle ACB
therefore we have a 30-60-90 triangle, and not a 45-45-90 triangle. sufficient.
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Re: M01-21  [#permalink]

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New post 11 Mar 2016, 20:59
Hi Bunuel

I was wondering what's the best way to tackle statement 1

by plugging it or by working out the angles ?

Thanks

Michael

Bunuel wrote:
Official Solution:


Notice that since ABC is a right triangle then the question basically asks whether angle ACB is equal to angle CAB, because in this case AB would be equal to BC, thus triangle ABC would be an isosceles triangle.

(1) Angle ACB is twice as large as angle ADC. It certainly possible angle ACB to be equal to angle CAB (45 degrees) and angle ADC to be 22.5 degrees but it's also possible angle ACB NOT to be equal to angle CAB, for example if angle ACB is 60 degrees and angle ADC is 30 degrees. Not sufficient.

(2) Angle ACB is twice as large as angle CAB. Directly says that angle ACB is NOT equal to angle CAB, hence triangle ABC is NOT isosceles. Sufficient.


Answer: B
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Re: M01-21  [#permalink]

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New post 11 Mar 2016, 21:21
Michael KC Chen wrote:
Hi Bunuel

I was wondering what's the best way to tackle statement 1

by plugging it or by working out the angles ?

Thanks

Michael

Bunuel wrote:
Official Solution:


Notice that since ABC is a right triangle then the question basically asks whether angle ACB is equal to angle CAB, because in this case AB would be equal to BC, thus triangle ABC would be an isosceles triangle.

(1) Angle ACB is twice as large as angle ADC. It certainly possible angle ACB to be equal to angle CAB (45 degrees) and angle ADC to be 22.5 degrees but it's also possible angle ACB NOT to be equal to angle CAB, for example if angle ACB is 60 degrees and angle ADC is 30 degrees. Not sufficient.

(2) Angle ACB is twice as large as angle CAB. Directly says that angle ACB is NOT equal to angle CAB, hence triangle ABC is NOT isosceles. Sufficient.


Answer: B


Hi,

In this particular Q, it would be better to PLUG different values..

We are already aware that angle ACB has to be 45 for being ISOSCELES. Now since we are already aware of this, we have to find if plugging 45 and any angle other than 45 works..
Even if you work on variables and find angles, you will have to substitute values to check for condition of ISO..

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Re: M01-21  [#permalink]

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New post 11 May 2016, 15:34
I think this is a high-quality question and I agree with explanation.
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Re: M01-21  [#permalink]

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New post 26 Jun 2016, 04:26
I think this is a high-quality question and I agree with explanation.
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Re: M01-21  [#permalink]

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New post 22 Sep 2016, 22:45
@bunnel
I think statement 1 alone is able to prove the same.
In traingle ACB, applying the exterior angle property angle ACB = angle ADC + angle CAD
Statement 1 : ACB = 2.ADC
if ADC = x -> ACB = 2x and thus CAD = x

Now BAD = 180-(90+x) = 90-x
Also, in Traingle ABC , angle BAC = 180-(90+2x) = 90-2x

So, angle BAC = BAD-CAD = 90-x - 90 + 2x = x

so BAC = x and BCA = 2x
so sufficient to prove ABC is not isoseles

Please correct me if I am wrong in any step
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Re: M01-21  [#permalink]

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New post 28 May 2017, 23:44
I don't agree with the explanation. I think the answer should be D as statement 1 itself is sufficient. You can refer to Gauravategmat's explanation in this thread for that matter
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Re: M01-21  [#permalink]

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New post 29 May 2017, 00:27
gauravategmat wrote:
@bunnel
I think statement 1 alone is able to prove the same.
In traingle ACB, applying the exterior angle property angle ACB = angle ADC + angle CAD
Statement 1 : ACB = 2.ADC
if ADC = x -> ACB = 2x and thus CAD = x

Now BAD = 180-(90+x) = 90-x
Also, in Traingle ABC , angle BAC = 180-(90+2x) = 90-2x

So, angle BAC = BAD-CAD = 90-x - 90 + 2x = x

so BAC = x and BCA = 2x
so sufficient to prove ABC is not isoseles

Please correct me if I am wrong in any step


Hi

you have written that:
angle BAD = 90-x
and in triangle ABC, angle BAC = 90-2x . I completely agree till here. But after that you have written:

angle BAC = angle BAD - angle CAD = ... here you have taken CAD as (90-2x).. I don't understand how?
CAD=x. Its NOT equal to 90-2x.

I think this step is incorrect. Or if I am missing something, please let me know. Thanks
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Re: M01-21  [#permalink]

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New post 29 May 2017, 00:30
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deep14 wrote:
I don't agree with the explanation. I think the answer should be D as statement 1 itself is sufficient. You can refer to Gauravategmat's explanation in this thread for that matter


Hi

I have written a reply to Gauravategmat here: https://gmatclub.com/forum/m01-183532.html#p1860305

you may want to check that and let me know if I am missing something. I think answer should be B only
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Re: M01-21  [#permalink]

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New post 29 May 2017, 00:48
amanvermagmat wrote:
deep14 wrote:
I don't agree with the explanation. I think the answer should be D as statement 1 itself is sufficient. You can refer to Gauravategmat's explanation in this thread for that matter


Hi

I have written a reply to Gauravategmat here: https://gmatclub.com/forum/m01-183532.html#p1860305

you may want to check that and let me know if I am missing something. I think answer should be B only



Yap, it should be B. I have spotted a mistake in my calculation, different from Gaurav's :-D . Anyways thanks for the review.
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Re: M01-21  [#permalink]

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New post 08 Sep 2018, 14:50
Is it safe to say that triangle ACD is isosceles based on statement 1?
It seems so, but what is the universal rule here?
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Re: M01-21 &nbs [#permalink] 08 Sep 2018, 14:50
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