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

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Senior Manager
Joined: 07 Oct 2017
Posts: 265
Re: In the figure shown, what is the value of x?  [#permalink]

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25 Aug 2018, 23:50
shorteverything wrote:

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:
Triangle.GIF
30 seconds approach
C

This is geometry DS question. I prefer to use technique mentioned in gmat prep now video.

St1) now point S and Q are fixed. So line SQ will have a fixed incline but point U is not fixed. You can move around point U and can get variable values of x

Not sufficient

St2) now point U is fixed but Q is variable.
Not sufficient

Combining
Both points are fixed. X can take only one value. No need to spend time finding the value. We know the figure is restricted now and can only assume bigger sides but the angle will remain same

Consider kudos if that helped

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app
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Thank you =Kudos
The best thing in life lies on the other side of the pain.

Math Expert
Joined: 02 Sep 2009
Posts: 50621
Re: In the figure shown, what is the value of x?  [#permalink]

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26 Aug 2018, 02:54
ravikumarmishra 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 $$\angle{R}+\angle{T}=90$$ --> $$x+\frac{360-90}{2}=180$$ --> $$x=45$$. Sufficient.

Thanks Bunuel, perfect solution. For more practice, can you please provide links for similar problems, thanks.

For other subjects:
ALL YOU NEED FOR QUANT ! ! !
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Math Expert
Joined: 02 Sep 2009
Posts: 50621
Re: In the figure shown, what is the value of x?  [#permalink]

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26 Aug 2018, 02:55
ravikumarmishra 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 $$\angle{R}+\angle{T}=90$$ --> $$x+\frac{360-90}{2}=180$$ --> $$x=45$$. Sufficient.

Thanks Bunuel, perfect solution. For more practice, can you please provide links for similar problems, thanks.

For other subjects:
ALL YOU NEED FOR QUANT ! ! !
_________________
Re: In the figure shown, what is the value of x? &nbs [#permalink] 26 Aug 2018, 02:55

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