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GMATPrep - Practice Test #1 - DS - Please Help


 
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dominate11
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PostPosted: Wed Feb 28, 2007 7:31 am    Post subject: GMATPrep - Practice Test #1 - DS - Please Help Reply with quote

Any insight would be much appreciated...

7. In the triangle ABC above, what is the length of side BC? 1) Line segment AD has length 6 2) x=36 ( I was not sure how to include the example of the triangle)

9. On the sight seeing tour, the ratio of the number of women to the number of children was 5 to 2. What was the number of men on the sight seeing tour? 1) On the sight seeing tour, the ratio of the number of children to the number of men was 5 to 11. 2) The number of women on the sight seeing tour was less than 30.
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gabriel
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PostPosted: Sat Mar 03, 2007 8:54 am    Post subject: Re: GMATPrep - Practice Test #1 - DS - Please Help Reply with quote

dominate11 wrote:
Any insight would be much appreciated...

7. In the triangle ABC above, what is the length of side BC? 1) Line segment AD has length 6 2) x=36 ( I was not sure how to include the example of the triangle)

9. On the sight seeing tour, the ratio of the number of women to the number of children was 5 to 2. What was the number of men on the sight seeing tour? 1) On the sight seeing tour, the ratio of the number of children to the number of men was 5 to 11. 2) The number of women on the sight seeing tour was less than 30.


for the first one i dont see a diagram.....

for the second one .... statement 1 along with the information in the question will give u the ratio of men:children:women... which is 22:10:25.... so this statement alone is not sufficient.... statement again does not give u any worthwhile information so this alone is not sufficient either... but take both the statement together and the only possible value of men that wuld satisfy both the given condition is 22 ... so the answer is C
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Stacey Koprince
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PostPosted: Sun Mar 11, 2007 10:52 pm    Post subject: Reply with quote

Hi, dominate - if you don't know how to post images, you can also just describe the triangle in text.
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Mark Dabral
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PostPosted: Mon Mar 12, 2007 10:01 am    Post subject: Here is the image associated with the problem statement Reply with quote

Here is the problem from GMATPrep. This problem is also in OG11, and I believe there is a typo in the book. The problem as stated here is correct.

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Mark

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PostPosted: Tue Mar 13, 2007 10:54 pm    Post subject: triangle ABC Reply with quote

Hi dominate11 ,

Assuming you figured out the triangle problem. if not,

if you work out the angles, you see that angles BAD=ABD
hence sides AD=BD=BC - hence A is sufficient.
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PostPosted: Mon Sep 08, 2008 11:18 pm    Post subject: Reply with quote

Would someone please post a detailed explanation to this problem? I'm still confused. Thanks!
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PostPosted: Tue Sep 09, 2008 5:59 am    Post subject: Reply with quote

they ask for a length, so statement 2 is not going to be useful - it gives us an angle only, and you can have a triangle of any size, teeny-tiny to huge, with those angles.

So, ACE.

AD is 6. <BAD is x. <BCD is 2x. <BCD is 2x. They want to know side BC.

<BDC is what's called an exterior angle of triangle BAD. You create this by taking any leg and just extending it out in a straight line. That exterior angle is always equal to the sum of the two opposite interior angles - in this case, angles BAD and ABD. So BAD + ABD = BDC. Fill in what you know. x + ABD = 2x. Solve. ABD = x.

So if ABD = x, it's the same angle as BAD. This means the sides opposite these angles are the same also. We know AD is 6, so BD is also 6.

Now look at the triangle on the right. We also have two identical angles there, so the two opposite sides are the same. One of those sides is BD (which we already know is 6) and the other side is BC, which we want to find. BC is also 6.

So statement 1 is sufficient. Answer is A.

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PostPosted: Tue Sep 09, 2008 10:16 am    Post subject: Reply with quote

Thanks Stacey!
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PostPosted: Thu Sep 11, 2008 9:01 pm    Post subject: Reply with quote

For triangle ABD,
x + angleB + (180-2X) = 180
angleB - x = 0
angleB = x

The right triange is isoceles and you can go from there.
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