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MBAlad
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GMATPrep Geometry 07 Dec 2006, 16:45
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Tricky one



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mbagal1
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GMATPrep Geometry 07 Dec 2006, 17:47
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Ans: C

Angle QRS = r
Angle RTP = t
T+r = 90 degrees
Z (Angle RQS) +x+y (Angle TUS) = 180


1) Angle RSQ = Angle RQS = z INSUFF
r = 180 -2z

2) Angle SUT = Angle UST = y INSUFF
T = 180 -2y

Combining and solving we get 3 equations of 3 variables x, y,z. Thus can be solved


X+y = r+z
Z+x+ (180-2y) = 180 =>x+z = 2y
X+y-z = (90-t) = (90-180+2y)
And x+y+z = 180
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mkl_in_2001
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GMATPrep Geometry 07 Dec 2006, 18:03
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from 1. <RQS = <RSQ
from 2. <TUS= <TSU

from diagram
<RSQ+x+<TSU=180
=> <RSQ+<TSU = 180-x

aslo from PQSU
X+<SQP+<P+<PUS =360
=>x+180-<RQS+90 + 180-<TUS=360
=>x+90-(<RQS + <TUS) =0
=>x+90 -(<RSQ+<TSU)=0
=>x+90 -(180-x)=0;
=>x= 45

so c is the ans.
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Hermione
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GMATPrep Geometry 11 Dec 2006, 02:50
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Quote:
Ans: C

Angle QRS = r
Angle RTP = t
T+r = 90 degrees
Z (Angle RQS) +x+y (Angle TUS) = 180


1) Angle RSQ = Angle RQS = z INSUFF
r = 180 -2z

2) Angle SUT = Angle UST = y INSUFF
T = 180 -2y

Combining and solving we get 3 equations of 3 variables x, y,z. Thus can be solved


X+y = r+z
Z+x+ (180-2y) = 180 =>x+z = 2y
X+y-z = (90-t) = (90-180+2y)
And x+y+z = 180
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confused about the part in orange. hsampath, can you explain please?
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yogeshsheth
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GMATPrep Geometry 11 Dec 2006, 13:25
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the answer is C
Each st individually is insuff.

st 1 says qr =rs thus ang qrs= ang qsr =y and ang rqs =180-2y and pq s=2y
st2 says st=tu thus ang ust = ang stu =z and ang sut =180-2z and ang sup= 2z
Also y+z=90 (from the rt triangle)

Now
x+2y+2z+90=360
x=270-2(y+z)
x=270-180=90
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GMATPrep Geometry 26 Dec 2006, 23:56
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hsampath
Ans: C

Angle QRS = r
Angle RTP = t
T+r = 90 degrees
Z (Angle RQS) +x+y (Angle TUS) = 180


1) Angle RSQ = Angle RQS = z INSUFF
r = 180 -2z

2) Angle SUT = Angle UST = y INSUFF
T = 180 -2y

Combining and solving we get 3 equations of 3 variables x, y,z. Thus can be solved


X+y = r+z
Z+x+ (180-2y) = 180 =>x+z = 2y
X+y-z = (90-t) = (90-180+2y)
And x+y+z = 180
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I don't get the red part... can someone explain please? :? why these angles make 180?
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Sumithra
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GMATPrep Geometry 27 Dec 2006, 02:51
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u2lover
hsampath
Ans: C

Angle QRS = r
Angle RTP = t
T+r = 90 degrees
Z (Angle RQS) +x+y (Angle TUS) = 180


1) Angle RSQ = Angle RQS = z INSUFF
r = 180 -2z

2) Angle SUT = Angle UST = y INSUFF
T = 180 -2y

Combining and solving we get 3 equations of 3 variables x, y,z. Thus can be solved


X+y = r+z
Z+x+ (180-2y) = 180 =>x+z = 2y
X+y-z = (90-t) = (90-180+2y)
And x+y+z = 180

I don't get the red part... can someone explain please? :? why these angles make 180?
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From S1 and S2:
angles RQS=RSQ;TUS=TSU
Since angleRSQ+x+angleTSU=180(straight line),angleRQS+x+angleTUS=180
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GMATPrep Geometry 27 Dec 2006, 07:02
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1. Statements 1 or 2 are not sufficent hence A,B and D are ruled out.
2.From Statement 1,
i) Triangle QRS is isoscles, assume, angles RQS,RSQ be A degrees
ii)Let Angle QRS be C
iii)2A+C = 180------------------------I
3) From Statement 2,
i) Triangle STU is isoscles, assume, angles TSU,TUS be B degrees
ii)Let Angle STU be D
iii)2B+D = 180------------------------II
4) From the traingle PRT,
i)C+D =90----------------------------III
ii)A+X+B=180------------------------IV
5)Solving I,II and III,
2A+2B +C+D = 360
A+B = 135 -------------------------------V
6) From IV and V,
x = 45!

Guess it is clear!
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u2lover
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GMATPrep Geometry 27 Dec 2006, 09:42
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thanks guys!!! things are clear today :-D the best part is that you don't have to solve it... just understand how to do it.
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BLISSFUL
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GMATPrep Geometry 29 Dec 2006, 14:07
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MBAlad
Tricky one
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-----------------------
C


1) , 2) aren't suffi..

Together:
PSQ + x+UST = 180 ---a)
90+x+(180-UST) + (180-PSQ)=360 --b)

frm a) & b)

90+x +x =180 -> x=45

C) is the answer
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Andr359
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GMATPrep Geometry 31 Dec 2006, 10:07
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It´s C.

By simple inspection of the figure, you notice that neither (1) nor (2) can be suff for arriving at a single value for x. Thus, right away you use (1) and (2) together and, by working on the inner angles formed by the 3 adjacent shapes, you get x = 45.
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sundar mahadevan
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GMATPrep Geometry 23 Jan 2007, 05:54
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sorry!!! to post this in this section of the forum......

But can anybody help me how I can post a new question in the respective section. I tried to post on geometry problem in the section,, but it says only moderator can initiate a question...please help soon..

Thanks!!



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
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