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In the figure shown, what is the value of x?

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In the figure shown, what is the value of x? [#permalink]

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New post 31 Aug 2009, 04:29
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In the figure shown, what is the value of x?

(1) The length of line segment QR is equal to the length of line segment RS

(2) The length of line segment ST is equal to the length of line segment TU


OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/in-the-figur ... 25923.html
[Reveal] Spoiler: OA
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Re: In the figure shown, what is the value of x? [#permalink]

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New post 31 Aug 2009, 06:07
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C it is
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Re: In the figure shown, what is the value of x? [#permalink]

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New post 31 Aug 2009, 06:11
Thanks and kudos!
I've seen a similar problem before, but forgot... :oops:
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Re: In the figure shown, what is the value of x? [#permalink]

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New post 02 Sep 2009, 06:18
Intially I couldn't see how you got angle 1 + angle 6 to be 90.....but I got to open my eyes to see the bigger picture. Good explanation.
Definitely agree with C.
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Re: In the figure shown, what is the value of x? [#permalink]

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New post 04 Sep 2009, 10:08
IMO C..too...gr8 effort ...clicking the image on web cam...
n thx...i m taking angles as RQS.....RSQ....taking 1,2,3, makes problem more easy..
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Re: In the figure shown, what is the value of x? [#permalink]

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New post 19 Mar 2011, 03:16
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it is C:
(1) <R + <T = 90 in which:

(2) <R = 180 - <RQS - <RSQ = 180 - 2<RSQ (as RQ = RS)
(3) <T = 180 - <UST - <SUT = 180 - 2<UST ( as ST = TU)

(1) + (2) + (3):
180 - 2<RSQ + 180 - 2<UST = 90
=>360 - 2(<RSQ + <UST) = 90
=> 360 - 2 (180 -x) = 90
=> x= 45
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Re: In the figure shown, what is the value of x? [#permalink]

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New post 13 Jun 2017, 23:12
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In the figure shown, what is the value of x?

\(x+\angle{QSR}+\angle{UST}=180\) (straight line =180) and \(\angle{R}+\angle{T}=90\) (as PRT is right angle)


(1) The length of line segment QR is equal to the length of line segment RS --> triangle QRS is isosceles --> \(\angle{RQS}=\angle{QSR}=\frac{180-\angle{R}}{2}\) (as \(\angle{RQS}+\angle{QSR}+\angle{R}=180\) --> \(2*\angle{QSR}+\angle{R}=180\) --> \(\angle{QSR}=\frac{180-\angle{R}}{2}\)). Not sufficient.


(2) The length of line segment ST is equal to the length of line segment TU --> triangle UST is isosceles --> \(\angle{SUT}=\angle{UST}=\frac{180-\angle{T}}{2}\). Not sufficient.


(1)+(2) \(x+\angle{QSR}+\angle{UST}=180\) --> \(x+\frac{180-\angle{R}}{2}+\frac{180-\angle{T}}{2}=180\) --> \(x+\frac{360-(\angle{R}+\angle{T})}{2}=180\) --> since \(\angle{R}+\angle{T}=90\) --> \(x+\frac{360-90}{2}=180\) --> \(x=45\). Sufficient.

Answer: C.

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Re: In the figure shown, what is the value of x?   [#permalink] 13 Jun 2017, 23:12
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