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Data Sufficiency Questions

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Data Sufficiency Questions [#permalink] New post 01 Nov 2003, 13:22
Can someone help me with an explanation for the following data sufficiency questions? Please refer to the attachment for the diagrams.


Data Sufficiency Question 1 (refer Figure 1)

In the given circle with center O, how many degrees is angle ACB?
(1) Angle OAB = 45
(2) Angle AOB = 100

The answer is D.



Data Sufficiency Question 2 (refer Figure 2)

In the figure 2, what is angle x equal to?
(1) ABCD is a square
(2) PDC is an equilateral triangle

The answer is C.
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 [#permalink] New post 01 Nov 2003, 16:00
Can you please not puplish the answer when you post the question first time? This way it will give others chance to actually WORK on the problem and that can add to the learning? Of course it would be great if you could publish the answers after 2-3 people have tried their hand at the problem. That way every body can have a confirmation also.

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 [#permalink] New post 15 Jan 2004, 07:27
The answer to first should be E.
A) is in sufficient
B) is in sufficient

D would be the answer if line OC made 90 degrees with the horizontal line.
Also you cannot combine A and B because OA = OB so OAB = OBA
if OAB = 45 then OBA = 45 and AOB = 90
if AOB = 100 then OAB = OBA = 40

The answer to second question is C. Angle x = 60
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 [#permalink] New post 03 Feb 2004, 14:38
anandnk wrote:
The answer to first should be E.
A) is in sufficient
B) is in sufficient

D would be the answer if line OC made 90 degrees with the horizontal line.
Also you cannot combine A and B because OA = OB so OAB = OBA
if OAB = 45 then OBA = 45 and AOB = 90
if AOB = 100 then OAB = OBA = 40

The answer to second question is C. Angle x = 60


Anand,

I think the answer D is ok for question 1

OA = OB so OAB = OBA

Statement 1: OAB = 45 then OBA = 45 and AOB = 90

So if we consider the angle AOB on larger arc, it would be 360 - 90 = 270.

Angle ACB is then = 270 /2 = 135

So statement 1 suffecient

Statement 2:

Angle AOB = 100 (I undersatnd this is wrong. This statement should have been "Angle AOB = 90" because that is the value we derived in the first statement and on GMAT if both the statements are given values for a variable it is usually same)

Anyhow, AOB = 100 so the angle on the larger arc is = 260

So angle ACB = 260 / 2 = 130

Statement 2 is SUFFECIENT

Answer D.
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 [#permalink] New post 03 Feb 2004, 17:42
INteresting property. I didnt know that. Are you sure about the larger arc being 270 degrees. Where can I find info about this ?
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 [#permalink] New post 04 Feb 2004, 06:45
anandnk wrote:
INteresting property. I didnt know that. Are you sure about the larger arc being 270 degrees. Where can I find info about this ?


I do not know where to find out about this property. It should be part of high school geometry.

The property is: the angle formed by any arc at the center of the circle is twice the angle formed by the same arc at any point on the circumference of the circle.
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 [#permalink] New post 04 Feb 2004, 09:37
I know the answer to uestion 2 is C.

But how did you derive the degree measure of x as 60. I think it should be 45.

In Traingle PBC all we know is PC=BC. So this is should be an isosceles. how can it be an equilateral.

Please explain
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 [#permalink] New post 04 Feb 2004, 16:10
For Q2
if PDC is equilateral triangle then
pd = dc = pc
Since ABCD is a square pc = bc
since pc = bc we have angle cpb = cbp
also angle pcb = 90-angle pcd = 90-60 = 30
So angle cpb+angle cbp = 180-30 = 150
so each angle is 75 degree.

My answer 60 is wrong.
  [#permalink] 04 Feb 2004, 16:10
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