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25 Nov 2007, 19:39
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Please provide details. Thanks
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25 Nov 2007, 23:38
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C
1. or 2. it is obviously that 1 and 2 along are insufficient. For 1 statement we can move U along PT to change angle x
1&2. QRS and UTS are isosceles triangles
angle(QSR)= (180-angle(QRS))/2
angle(UST)= (180-angle(UTS))/2
x+angle(QSR)+angle(UST)=180 ==>
x=180- (180-angle(QRS))/2-(180-angle(UTS))/2=(angle(QRS)+angle(UTS))/2
but for right triangle: angle(QRS)+angle(UTS)=90
Therefore, x=90/2=45 sufficient
1. or 2. it is obviously that 1 and 2 along are insufficient. For 1 statement we can move U along PT to change angle x
1&2. QRS and UTS are isosceles triangles
angle(QSR)= (180-angle(QRS))/2
angle(UST)= (180-angle(UTS))/2
x+angle(QSR)+angle(UST)=180 ==>
x=180- (180-angle(QRS))/2-(180-angle(UTS))/2=(angle(QRS)+angle(UTS))/2
but for right triangle: angle(QRS)+angle(UTS)=90
Therefore, x=90/2=45 sufficient
26 Nov 2007, 00:23
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alimad
Wow thanks for posting this. After working on this for about 6minutes, i finally solved this problem. I had this on a previous GPrep and guessed C. I wasnt actually able to solve for X though this time around.
Please see drawing.
Ok we have angle X.
S1: for the smaller triangle we have z and y and y. z+2y=180* from here we cannot figure anything else out. Insuff.
S2: we have angle n, and 2 w's. Same as s1, though cannot figure anything else out.
Together we have x+y+w=180*
we also have 2y+z=180*
n+2w=180*.
No look at big triangle we have angles z, 90*, and n. so z+n+90*=180*
z+n=90*.
heres the trick to solve this: add
2y+z=180*
2w+n=180*
we have 2y+2w+z+n=360*---> plug in for z+n--> 2y+2w+90*=360*
y+w=135* No plug into x+y+w --> x= 45*
Answer is C.
Archived Topic
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.
Thank you for understanding, and happy exploring!
