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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? (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 Attachment:
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Re: In the figure shown, what is the value of x?
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26 Jul 2010, 14:56
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 \(\angle{R}+\angle{T}=90\) > \(x+\frac{36090}{2}=180\) > \(x=45\). Sufficient. Answer: C.
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Re: In the figure shown, what is the value of x?
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16 Mar 2011, 21:19
From (1) RSQ = RQS From (2) SUT = TSU Because these 3 angles are on a straight line: x + RSQ + TSU = 180 As all the angles of a quadrilateral sum up to 360: x+90 + 180  RSQ + 180  TSU = 360 => x + 360  (RSQ + TSU) + 90 = 360 => x + 90 (180x) = 0 => 2x  90 = 0 => x = 45 So the answer is C.
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Re: In the figure shown, what is the value of x?
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05 Nov 2010, 18:32
Well this is how I would solve this.
Original statement:
What is x?
Without jumping into statements, we can clearly see that x = 180  ARSQ  ATSU.
Statement 1:
Statement 1 says that triangle QRS is isosceles. with ARSQ = ARQS
therefore we know that ARSQ = (180  ASRQ)/2.
However, without information about triangle TSU cannot solve the above equation about x.
Insufficient
Statement 2:
Statement 2 says that triangle TSU is isosceles with ATSU = (180  ASTU)/2. Similar to statement 1, not enough information about triangle RSQ to find x.
Insufficient.
Statement 1 + 2:
From statement 1: we know that ARSQ = (180ASRQ)/2 From statement 2: we know that ATSU = (180ASTU)/2
Plugging these information into the x = 180  ARSQ  ATSU we see that
x = 180  (180ASRQ)/2  (180ASTU)/2 = 180  (360/2) + (ASRQ + ASTU)/2 = (ASRQ + ASTU)/2
Since we know points RTP makes a right triangle, we know that ASRQ + ASTU = 90.
x = 45.
Sufficient.
My answer would be C.



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Re: In the figure shown, what is the value of x?
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08 Jan 2012, 18:45
C it is. Two equations, x = 180  (a + b), where a and b are RQS and SUT. x = 360 [(180a) + (180b) + 90] = (a+b)90. Solve both, you get x.
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Re: In the figure shown, what is the value of x?
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24 Nov 2014, 09:55
The triangle in the figure has 2 triangles and a quadrilateral in it. S1 and S2 insufficient S1+S2 Considering the toptriangle: assign 'y' to equal angles Considering the bottom triangle: assign 'z' to the equal angles x+y+z= 180.............I Considering the quadrilateral now: 90+(180z)+x+(180y)=360 xyz=90................II Adding I and II gives me a value for x (in the spirit of seeing things through 2x=90 > x=90/2=45)



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Re: In the figure shown, what is the value of x?
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16 Mar 2015, 04:40
How do I know it is not actually equilateral?



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Re: In the figure shown, what is the value of x?
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Re: In the figure shown, what is the value of x?
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16 Mar 2015, 06:59
Ok. Here is what I don't get. To solve this question it is assumed that the two triangles RQS and SUT are isosceles triangles. This assumption is based on the condition QS=QR. My problem however is that an equilateral triangle QS=QR=SR would also satisfy the condition QS=QR. In that way the angles could be either 454590 or 606060. So what am I missing that gives proof that these triangles are Isosceles? As far as I know one should never make assumptions based on the graphs in the GMAT.
thanks for quick response



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Re: In the figure shown, what is the value of x?
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16 Mar 2015, 07:07
Petermeterdeter wrote: Ok. Here is what I don't get. To solve this question it is assumed that the two triangles RQS and SUT are isosceles triangles. This assumption is based on the condition QS=QR. My problem however is that an equilateral triangle QS=QR=SR would also satisfy the condition QS=QR. In that way the angles could be either 454590 or 606060. So what am I missing that gives proof that these triangles are Isosceles? As far as I know one should never make assumptions based on the graphs in the GMAT.
thanks for quick response Two notes on this: 1. An isosceles triangle has at least two sides of the same length, which means that if a triangle is equilateral it could still be considered as isosceles. 2. Triangle QSR is not necessarily 454590 or 606060. Why cannot it be say, 1515150?
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Re: In the figure shown, what is the value of x?
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Updated on: 22 Oct 2015, 08:12
Hi for geom it is often more convenient to name angles your own way to avoid excessive abstraction: a, b, c, d (please the attached file)  because many of the angles get cancelled out during calculations and overall the less astractive the more comprehensible. We are given that angle RPT=90. We derive that QSTU is a rectangle where one of the angles = x = 360  90  c  d. We also understand that x = 180  a  b. ANSWER C: ST 1: QR=RS hence a=a=angle RQS=angle RSQ  insufficient, because we do not know how other angles relate to each other and to x. At this point we understand that knowing b would suffice the question. ST 2: ST=UT hence b=b  insufficient for the same reasons stated above. Clearly this statement gives just the same piece of information as the statement 1 does. We also understand it is a good thing to combine both statements. ST 2 + ST 1: at this point we can stop and choose C because the stimulus gives you a bold hint that one of the angles of the triangle equals 90, x is within the rectangle where one of the angles is 90, 2 of the smaller triangles are isosceles, those base angles in the triangles are adjacent to x  having taken a short glimpse of all those factors you could figure out halfway that this is pretty much enough for a geom question like that to avoid answer E and choose C. However, this is the solution for C: 1. x=180(a+b) and x=360(90+c+d) 2. a=180c and b=180d 3. combine (1) and (2): 36090cd = 180  (180c)  (180d) 4. c+d=225 5. x=360(90+225)=45
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Originally posted by shasadou on 01 Sep 2015, 03:13.
Last edited by shasadou on 22 Oct 2015, 08:12, edited 1 time in total.



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Re: In the figure shown, what is the value of x?
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05 Sep 2015, 11:57
Bunuel wrote: 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 \(\angle{R}+\angle{T}=90\) > \(x+\frac{36090}{2}=180\) > \(x=45\). Sufficient. Answer: C. Wow, praise the lord (bunuel) for this!
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Re: In the figure shown, what is the value of x?
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24 Aug 2016, 11:07
subhashghosh wrote: From (1) RSQ = RQS
From (2) SUT = TSU
Because these 3 angles are on a straight line:
x + RSQ + TSU = 180
As all the angles of a quadrilateral sum up to 360:
x+90 + 180  RSQ + 180  TSU = 360
=> x + 360  (RSQ + TSU) + 90 = 360
=> x + 90 (180x) = 0
=> 2x  90 = 0
=> x = 45
So the answer is C. Hi subhashghosh, Let's break your statement into two parts. x + 90 + 180  RSQ + 180  TSU = 360 1) Xº + 90º + 180º = 360º > They make up the sum of all the angles of the quadrilateral PQSU. 2) 180  RSQ  TSU = Xº Hence, you are saying that 360 º + X = 360º, which is incorrect.



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Re: In the figure shown, what is the value of x?
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26 Nov 2016, 10:04
blazov wrote: In the figure shown , what is the value of x?
(1) The length of segment QR is equal to the length of line segment RS
(2) The length of segment ST is Equal to the length of line segment TU Statement 1: Gives us nothing , since we don't know any other angles or property to compute other angles Statement 2: Gives us nothing , since we don't know any other angles or property to compute other angles Combined: We know that \(\angle\)RQS and \(\angle\)RSQ are same, since have also equal sides RS=RQ. Also angles \(\angle\)TSU and \(\angle\)TUS are same since ST=TU We will name angles as below for simplicity; \(\angle\) RQS=Z, \(\angle\)RSQ=Z, \(\angle\)QRS=B \(\angle\)TSU=W, \(\angle\)TUS=W, \(\angle\)STU=C So, we know that: \(90^{\circ}\)+B+C=180 meaning, B+C=\(90^{\circ}\) I we add up angles of both triangles of QRS and STU: \(\angle\)B+2 *\(\angle\)Z+\(\angle\)C+2 *\(\angle\)W=\(360^{\circ}\) Since we know \(\angle\)B+\(\angle\)C=\(90^{\circ}\), we subtract it: 2 *\(\angle\)Z+2 *\(\angle\)W=\(270^{\circ}\) We can divide both sides by 2 and get : \(\angle\)Z+\(\angle\)W=\(135^{\circ}\) To find \(\angle\)SQP and \(\angle\)SUP 360\(\angle\)(Z+W)=\(225^{\circ}\) \(\angle\)SQPU=\(360^{\circ}\) 90+225+x=\(360^{\circ}\) X=\(45^{\circ}\) Both statements together are enough. Answer C +1 if this helped, I put a lot effort
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Re: In the figure shown, what is the value of x?
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23 Feb 2017, 05:02
Prompt analysis Let angle PRS be a, angle PTR be b. Super set The angle x will be in the range of 0 to 180. Translation In order to find x, we need 1# the exact value of all the angles 2# the exact value of all the sides 3# some information/ property that will lead us to find the value of x Statement analysis St 1: Since the triangle QRS is an isosceles triangle we can say that angle RSQ = (180  a)/2. But we don't have any other information about the rest of the angles. INSUFFICIENT. Hence option A. D eliminated. St 2: Since the triangle STU is an isosceles triangle we can say that angle TSU = (180  b)/2. But we don't have any other information about the rest of the angles. INSUFFICIENT. HEnce option A,B, D eliminated. St 1 & St 2: angle RSQ + angle QSU + angle TSU = 180. Or (180  a)/2 + x+ (180  b)/2 = 180; 180 (a+b)/2 +x =180; x = (a+b)/2 as it is a straight line. Also if we look at the triangle PRT we can say that a +b + 90 = 180 or a +b =90. Putting the value the equation x = (a+b)/2, we get x =45. ANSWER Option C
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Re: In the figure shown, what is the value of x?
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06 May 2017, 13:02
Bunuel wrote: 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 \(\angle{R}+\angle{T}=90\) > \(x+\frac{36090}{2}=180\) > \(x=45\). Sufficient. Answer: C. Hi Bunuel, I'm trying to understand the working after \(x+\angle{QSR}+\angle{UST}=180\) but I can't see how you reach the answer. Can you help clarify for me?



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Re: In the figure shown, what is the value of x?
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06 May 2017, 13:15
Cez005 wrote: Bunuel wrote: 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 \(\angle{R}+\angle{T}=90\) > \(x+\frac{36090}{2}=180\) > \(x=45\). Sufficient. Answer: C. Hi Bunuel, I'm trying to understand the working after \(x+\angle{QSR}+\angle{UST}=180\) but I can't see how you reach the answer. Can you help clarify for me? For (1)+(2) we have \(x+\angle{QSR}+\angle{UST}=180\). From (1): \(\angle{QSR}=\frac{180\angle{R}}{2}\). From (2): \(\angle{UST}=\frac{180\angle{T}}{2}\). Substitute these into \(x+\angle{QSR}+\angle{UST}=180\) to get \(x+\frac{180\angle{R}}{2}+\frac{180\angle{T}}{2}=180\). This leads to \(x+\frac{360(\angle{R}+\angle{T})}{2}=180\). Since \(\angle{R}+\angle{T}=90\), then \(x+\frac{36090}{2}=180\) > \(x=45\). Hope it helps.
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In the figure shown, what is the value of x?
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18 Jun 2017, 14:53
Here is a picture which should help. I used fake number "7's" to show it was isosceles. Also the 7's could have been changed to any number and the 2 triangles didn't have to both have 7's. It was just an easier way to remind myself that the triangle was isosceles.
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Re: In the figure shown, what is the value of x?
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29 Jul 2017, 16:46
I'd like to get a suggestion as to whether my approach is valid or coincidental. Unlike an algebra combination problem (eg. solve for x+y), this is a geometry problem solving for x. Hence the x variables and x equations rule should be valid.
One can notice that there are 9 angles  including x and hence 9 variables. One can also see that there are 7 equations that can be formed that are unique: Triangle QRS, Triangle SUT, Big Triangle PRT, Quadrilateral PQSU, Line RST, Line RQP, and Line PUT. Hence in order to solve for x I need to reduce it down to 7 variables.
S1 reduces it from 9 variables/angles to 8 since one is the same. S2 essentially does the same thing. S1+S2 reduces it from 9 now to 7, which makes this solvable.
The key thing I believe  is to ensure that we are not solving for combinations as everyone may know from algebra problems that although something may not look solvable it can be because something cancels out.
Feedback would be much appreciated whether this approach is valid or not. If it is, then this monstrosity of a problem can be solved simply in less than a minute.



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In the figure shown, what is the value of x?
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25 Aug 2018, 14:51
Bunuel wrote: 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 \(\angle{R}+\angle{T}=90\) > \(x+\frac{36090}{2}=180\) > \(x=45\). Sufficient. Answer: C. Thanks Bunuel, perfect solution. For more practice, can you please provide links for similar problems, thanks.
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