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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] New post 27 Jun 2009, 19:40
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In the figure shown, what is the value of x?
Image

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

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-figure-shown-what-is-the-value-of-x-125923.html
[Reveal] Spoiler: OA

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Re: Angle X [#permalink] New post 28 Jun 2009, 18:28
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Let angle RSQ = S1 = RQS
Let angle RTP = T
Let angle TSU = S2 = SUT

S1 + S2 + X = 180.....................1
In triangle TSU we have
S2 + s2 + T = 180
2s2 + T = 180 ........................2

In triangle PRT we have
P + R + T = 180
90 + T + (180 - 2S1) = 180 (angle R is calculated from triangle RQS)
or 2S2 + 2S1 = 270
or S1 + S2 = 135
Put it in eq 1
135 + X = 180
X = 45.
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Re: Angle X [#permalink] New post 30 Jun 2009, 10:03
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Ive noticed that most of the times when information given is very little in gmat questions, the problem requires a conceptual approach, and its usually easier to solve with principles (rules of triangles in this case) rather than taking variables or numbers.

I will just share how i solved it mentally

Sttmt 1) only gives triangle RQS is isoceles, obviously cannot find x
Sttmt 2) Again,... gives that triangle STU is isoceles, not sufficient.

Combining
since the big triangle is right angled, Angles QRS and STU are complementary, ie their sum is also 90......(I)
the two smaller triangles RQS and STU are isoceles, so Angle rqs = rsq and Angles tsu = tus
in each of the smaller triangles, if the sum of two angles qrs + stu = 90 (from I above), then sum of the other two pairs should also be 90
ie. rqs + rsq + tsu + tus = 90

or 2rsq + 2 tsu = 90

As soon as we know this, we know we can solve for x, because we are getting rsq + tsu.
so dont even need to calculate beyond this.


It may look complicated at the first glance (if you look at the wordiness of the explanation) , but really it is very easy if you try make images in your mind and solve.
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Re: Angle X [#permalink] New post 30 Jun 2009, 19:22
[in each of the smaller triangles, if the sum of two angles qrs + stu = 90 (from I above), then sum of the other two pairs should also be 90
ie. rqs + rsq + tsu + tus = 90
.[/quote]

I didn't get this statement. how the sum of other two pairs is equal to 90, it should be 270

(the sum of all the angles of two smaller triangles is 360, if take the complementary angles out then the sum becomes 270). Kindly correct me if I am wrong.
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Re: Angle X [#permalink] New post 01 Jul 2009, 05:31
Hussain15 wrote:

I didn't get this statement. how the sum of other two pairs is equal to 90, it should be 270

(the sum of all the angles of two smaller triangles is 360, if take the complementary angles out then the sum becomes 270). Kindly correct me if I am wrong.


Yes, you are right, should be 270.

I didn't do the calculations, because we were not required to find the value of x, thus, the calculation error
sorry for the inconvenience though. :)
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Re: Angle X [#permalink] New post 28 Aug 2009, 07:29
Thank you for the great explanation. I can't believe that such a question is being asked. I need to spend a lot more time for preparation :oops:

Can somebody please explain the step in red below
...
P + R + T = 180
90 + T + (180 - 2S1) = 180 (angle R is calculated from triangle RQS)

or 2S2 + 2S1 = 270

Thank you
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Re: Angle X [#permalink] New post 28 Aug 2009, 21:46
suppose angle QRS is a
angle STU is b

we can get

(180-a)/2+x+(180-b)/2=180
and a+b=90

x=45

answer C.
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Re: Angle X [#permalink] New post 08 May 2014, 22:23
rashminet84 wrote:
Ive noticed that most of the times when information given is very little in gmat questions, the problem requires a conceptual approach, and its usually easier to solve with principles (rules of triangles in this case) rather than taking variables or numbers.

I will just share how i solved it mentally

Sttmt 1) only gives triangle RQS is isoceles, obviously cannot find x
Sttmt 2) Again,... gives that triangle STU is isoceles, not sufficient.

Combining
since the big triangle is right angled, Angles QRS and STU are complementary, ie their sum is also 90......(I)
the two smaller triangles RQS and STU are isoceles, so Angle rqs = rsq and Angles tsu = tus
in each of the smaller triangles, if the sum of two angles qrs + stu = 90 (from I above), then sum of the other two pairs should also be 90
ie. rqs + rsq + tsu + tus = 90


or 2rsq + 2 tsu = 90

As soon as we know this, we know we can solve for x, because we are getting rsq + tsu.
so dont even need to calculate beyond this.


It may look complicated at the first glance (if you look at the wordiness of the explanation) , but really it is very easy if you try make images in your mind and solve.


I know this is a few years old but could someone help me with the problem? I am trying to wrap my head around the bolded part. What sum of two pairs are we really talking about? Because how I am understanding this shouldn't rqs+rsq+tsu+tus=180? (I know that the calculation was corrected to 270 in another post). I don't understand how that calculation above would equal to 270. I understand the explanation all up until this part where I just get confused and would like to understand this conceptually for any similiar future problems.
Please help. Thanks.
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Re: Angle X [#permalink] New post 09 May 2014, 02:16
Expert's post
amjet12 wrote:
rashminet84 wrote:
Ive noticed that most of the times when information given is very little in gmat questions, the problem requires a conceptual approach, and its usually easier to solve with principles (rules of triangles in this case) rather than taking variables or numbers.

I will just share how i solved it mentally

Sttmt 1) only gives triangle RQS is isoceles, obviously cannot find x
Sttmt 2) Again,... gives that triangle STU is isoceles, not sufficient.

Combining
since the big triangle is right angled, Angles QRS and STU are complementary, ie their sum is also 90......(I)
the two smaller triangles RQS and STU are isoceles, so Angle rqs = rsq and Angles tsu = tus
in each of the smaller triangles, if the sum of two angles qrs + stu = 90 (from I above), then sum of the other two pairs should also be 90
ie. rqs + rsq + tsu + tus = 90


or 2rsq + 2 tsu = 90

As soon as we know this, we know we can solve for x, because we are getting rsq + tsu.
so dont even need to calculate beyond this.


It may look complicated at the first glance (if you look at the wordiness of the explanation) , but really it is very easy if you try make images in your mind and solve.


I know this is a few years old but could someone help me with the problem? I am trying to wrap my head around the bolded part. What sum of two pairs are we really talking about? Because how I am understanding this shouldn't rqs+rsq+tsu+tus=180? (I know that the calculation was corrected to 270 in another post). I don't understand how that calculation above would equal to 270. I understand the explanation all up until this part where I just get confused and would like to understand this conceptually for any similiar future problems.
Please help. Thanks.


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

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

(1) The length of line segment QR is equal to the length of line segment RS --> QRS is isosceles --> <RQS=<QSR=(180-R)/2 (as <RQS+<QSR+<R=180 --> 2<QSR+<R=180 --> <QSR=(180-R)/2). Not sufficient.

(2) The legnth of line segment ST is equal to the length of line segment TU --> UST is isosceles --> <SUT=<UST=(180-T)/2. Not sufficient.

(1)+(2) x+<QSR+<UST=180 --> x+(180-R)/2+(180-T)/2=180 --> x+(360-(R+T))/2=180 --> as R+T=90 --> x+(360-90)/2=180 --> x=45. Sufficient.

Answer: C.

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-figure-shown-what-is-the-value-of-x-125923.html
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Re: Angle X   [#permalink] 09 May 2014, 02:16
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