Sep 15 10:00 PM PDT  11:00 PM PDT EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Sep 16 10:00 PM PDT  11:00 PM PDT Join a FREE 1day verbal workshop and learn how to ace the Verbal section with the best tips and strategies. Limited for the first 99 registrants. Register today! Sep 18 12:00 PM EDT  01:00 PM EDT Mindful MBA series Part 1, Fall 2019. Becoming a More Mindful GMAT Taker. Tuesday, September 18th at 12 PM ET Sep 19 12:00 PM PDT  10:00 PM PDT On Demand $79, For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Sep 19 10:00 PM PDT  11:00 PM PDT Join a FREE 1day Data Sufficiency & Critical Reasoning workshop and learn the best strategies to tackle the two trickiest question types in the GMAT! Sep 21 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC Sep 21 08:00 PM PDT  09:00 PM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 18 May 2019
Posts: 243

Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
22 Jul 2019, 14:50
Let angle WYX=x; angle QSL=angle YST=r; and angle STY=angle RTM=p; This means that x+r+p=180 and x=180(r+p) ..........(a)
From statement 1, angle DQS + angle ERT =260 .....(b) angle DQS+angle LQS=180 And angle LQS=90r Hence angle DQS+90r=180 angle DQS=90+r .....(c) Also angle ERT=180angle MRT And angle MRT=90p; hence angle ERT=90+p ...(d) From (b), (c), and (d) angle DQS+angle ERT=260 90+p+90+r=260 p+r=80 ...(e) From (a), x=180(p+r)=18080 Hence x=100. Statement 1 alone is sufficient.
2. angle WQD+angle ERX=100 .....(f) But angle WQD=90r and angle ERX=90p So 90r+90p=100 p+r=80 ....(g) Now, from (a), x=180(p+r) Hence x=100. Therefore statement 2 alone is sufficient. Answer is therefore D.
Posted from my mobile device



Intern
Joined: 08 Jan 2018
Posts: 38

In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
22 Jul 2019, 16:46
(1) <DQS = <XWY + < WDQ = <XWY + 90 <ERT = <WXY + <XER = <WXY + 90 Since <DQS + <ERT = 260 = <XWY + <WXY + 180, <XWY + <WXY = 80. Since <XWY + <WXY + <WYX = 180, <WYX = 80. Sufficient
(2) <WDQ + <WQD + <XWY = 90 + <WQD + <XWY = 180. <XER + <WXY + <ERX = 90 + <WXY + <ERX = 180. Since <ERX + WQD = 100, <XWY + <WXY = 80, Since <XWY + <WXY + <WYX = 180, <WYX = 100. Sufficient
Ans is D



Director
Joined: 19 Oct 2018
Posts: 850
Location: India

Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
22 Jul 2019, 17:15
In the diagram above, triangle WXY intersects rectangle JKLM at points D, E, Q, R, S, and T. If line WX is parallel to line JK, what is the measure of angle WYX?
(1) The sum of the measures of angles DQS and ERT is 260. (2) The sum of the measures of angles WQD and ERX is 100.
∠KJL=∠XDL=∠WDQ =90 ∠JKM= ∠WEM=∠XER=90
Also, ∠XWY+∠WYX+∠WXY=180
∠WYX=180(∠WXY+∠XWY)
Hence if we know the value of (∠WXY+∠XWY), we can easily find ∠WYX.
Statement 1 The sum of the measures of angles DQS and ERT is 260. Hence DQW+ERX=360260=100 (∠WXY+∠XWY)= 180100=80
Sufficient
Statement 2 The sum of the measures of angles WQD and ERX is 100. (∠WXY+∠XWY)= 180100=80
Sufficient
IMO D



Intern
Joined: 10 May 2018
Posts: 17

Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
22 Jul 2019, 19:31
Since JK is parallel to WX, then QDW and REX must be 90 degrees. Since angles in a triangle must add up to 180, we only need DWQ and EXR to solve for WYX. We don't even need to know the individual angles, just the collective sum of DWQ and EXR. 1.The sum of the measures of angles DQS and ERT is 260. You can figure out DQW and ERX with this since both ERT and DQS are on a straight line. Together, they must equal 360  260 =100 (each straight line must equal 180, since there are two, add together to be 360 and then subtract 260 for the sum of the angles). Since you know DQW and XRE equals 100 and you know WDQ and XER are right angles (90 each), you know that DWQ and EXR must equal 360280 = 80 (since you're calculating two triangles, the sum must be 180+180 =360 and 280 comes from 90+90+100 from above). If those two angles equal 80, then WYX must be 100 so that triangle WXY equals 180.Sufficient2.The sum of the measures of angles WQD and ERX is 100. This gives the same information as above but one less stepSufficientD



Senior Manager
Joined: 30 Sep 2017
Posts: 332
Concentration: Technology, Entrepreneurship
GPA: 3.8
WE: Engineering (Real Estate)

In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
22 Jul 2019, 19:37
Question: \(\angle{WYX}\) ? I believe my word explanation is sufficiently clear to understand, but anyway you can find attached sketch for clarity. Line \(WX\) is \(parallel\) to line \(JK\) > Line \(WX\) is \(perpendicular\) to both lines \(JL\) and \(KM\). Statement (1): \(\angle{DQS} + \angle{ERT} = 260.\) Draw line \(YZ\) that intersects line \(JK\) at point \(Z\) and is \(perpendicular\) to line \(JK\) > Line \(YZ\) is \(parallel\) to both lines \(JL\) and \(KM\).  \(YZ\) is \(parallel\) to \(JL\), \(then\) \(\angle{DQS} + \angle{SYZ} = 180\) > \(\angle{SYZ} = 180  \angle{DQS}\)  \(YZ\) is \(parallel\) to \(KM\), \(then\) \(\angle{ERT} + \angle{TYZ} = 180\) > \(\angle{TYZ} = 180  \angle{ERT}\). \(\angle{WYX} = \angle{SYZ} + \angle{TYZ}\) > \(\angle{WYX} = (180  \angle{DQS}) + (180  \angle{ERT})\) > \(\angle{WYX} = 360  (\angle{DQS} + \angle{ERT}) = 360  260 = 100\) Statement (1) is SUFFICIENT Statement (2): \(\angle{WQD} + \angle{ERX}\) = 100. Draw line \(YZ\) that intersects with line \(JK\) at point \(Z\) and is \(perpendicular\) to line \(JK\) > Line \(YZ\) is \(parallel\) to both lines \(JL\) and \(KM\).  \(YZ\) is \(parallel\) to \(JL\), \(then\) \(\angle{WQD} = \angle{SYZ}.\)  \(YZ\) is \(parallel\) to \(KM\), \(then\) \(\angle{ERX} = \angle{TYZ}.\) \(\angle{WYX} = \angle{SYZ} + \angle{TYZ}\) > \(\angle{WYX} = \angle{WQD} + \angle{ERX}\) > \(\angle{WYX} = 100\) Statement (2) is SUFFICIENT ANSWER IS (D)  Each statement alone is sufficient
Attachments
GMAT Quant Question.jpeg [ 69.76 KiB  Viewed 214 times ]



Manager
Joined: 27 Mar 2018
Posts: 79
Location: India

Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
22 Jul 2019, 20:06
1. Not sufficient. 2. In ∆wqd, angle QWD=90WQD Similarly, in ∆erx, angle RXE=90ERX In ∆WXY, XWY+WXY+WYX=180 => WYX=100. Sufficient. Option B. Posted from my mobile device
_________________
Thank you for the kudos. You are awesome!



Director
Joined: 28 Jul 2016
Posts: 569
Location: India
Concentration: Finance, Human Resources
GPA: 3.97
WE: Project Management (Investment Banking)

Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
22 Jul 2019, 20:34
In the diagram above, triangle WXY intersects rectangle JKLM at points D, E, Q, R, S, and T. If line WX is parallel to line JK, what is the measure of angle WYX?
(1) The sum of the measures of angles DQS and ERT is 260. (2) The sum of the measures of angles WQD and ERX is 100.
Given line WX is parallel to line JK, thus angle QDE and DER are both equal to 90 now consider a polygon Sum of internal angle of polygon = (n2)*120= 3*180 where n = sides of a polygon (YQDER)
USING equation 1 \(\angle Y\) + \(\angle Q\)+ \(\angle D\)+ \(\angle E\)+ \(\angle R\) = 540
The sum of the measures of angles DQS and ERT is 260. AND\(\angle D\) + \(\angle E\) = 180 (EACH ANGLE IS OF \(90 ^{\circ}\)
THUS \(\angle Y\) + 90 +260 + 90 = 540 \(\angle Y\)= 100 SUFFICIENT 2) The sum of the measures of angles WQD and ERX is 100. THEREFORE angles DQS and ERT = 360  ( sum of the measures of angles WQD and ERX ) corresponding linear angles = 260 Now this is same as equation 1 all other parameter are given , Hence it is also sufficient
Thus D



Manager
Status: Not Applying
Joined: 27 Apr 2009
Posts: 179
Location: India

Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
22 Jul 2019, 20:44
In the diagram above, triangle WXY intersects rectangle JKLM at points D, E, Q, R, S, and T. If line WX is parallel to line JK, what is the measure of angle WYX? (1) The sum of the measures of angles DQS and ERT is 260. (2) The sum of the measures of angles WQD and ERX is 100. Sol:Considering Statement (1) alone:Angle DQS + Angle ERT = 260 We know that Angle WQD + Angle DQS = 180 and also Angle ERX + Angle ERT = 180 => Angle WQD + Angle DQS + Angle ERX + Angle ERT = 360 => Angle WQD + Angle ERX = 360  260 = 100 From triangle WQD and ERX, we have Angle W + Angle D + Angle Q + Angle E + Angle X + Angle R = 360 (sum of the angles of two triangles) => Angle W + 90 + Angle Q + 90 + Angle X + Angle R = 360 => Angle W + Angle X = 360  180  100 = 80 Now, in triangle WYX Angle W + Angle Y + Angle X = 180 => Angle Y = 180  80 => Angle Y = 100 SUFFICIENTConsidering statement (2) alone:Angle Q + Angle R = 100 We also know that Angle W + Angle D + Angle Q + Angle E + Angle X + Angle R = 360 (sum of the angles of two triangles) Angle W + 90 + 100 + 90 + Angle X = 360 => Angle W + Angle X = 360  180  100 = 80 Now, in triangle WYX Angle W + Angle Y + Angle X = 180 => Angle Y = 180  80 => Angle Y = 100 SUFFICIENTThe answer is (D).
_________________
http://www.wizius.inBetter Prep. Better Scores. Better Schools Guaranteed Admission to Top50 MBA Programs You either getin or get your moneyback.



Senior Manager
Joined: 18 Jan 2018
Posts: 308
Location: India
Concentration: General Management, Healthcare
GPA: 3.87
WE: Design (Manufacturing)

Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
22 Jul 2019, 20:46
In the diagram above, triangle WXY intersects rectangle JKLM at points D, E, Q, R, S, and T. If line WX is parallel to line JK, what is the measure of angle WYX?
In triangle WYX , angle (W + Y +X ) = 180 angle Y = 180 angle (W+X)
(1) The sum of the measures of angles DQS and ERT is 260 angle DQS = 90 + angle W angle ERT = 90 + angle X
Given DQS + ERT = 260 => 90+W+90+X = 260 W + X = 80. > angle Y = 180 80 = 100> Sufficient
(2) The sum of the measures of angles WQD and ERX is 100. In triangle WQD , angle D is 90 ==> angle W = 90angle WQD in triangle ERX , angle E is 90 ===>angle X = 90angleERX
from triangle WYX , Y = 180 (W+X) ===> Y = 180( 90angle WQD + 90  angle ERX ) = 180 ( 180100) = 18080 = 100> Sufficient
Option D is the correct Answer



Manager
Joined: 17 Jul 2014
Posts: 127

Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
22 Jul 2019, 22:01
In the diagram above, triangle WXY intersects rectangle JKLM at points D, E, Q, R, S, and T. If line WX is parallel to line JK, what is the measure of angle WYX? rectangle JKLM ; J=90 K=90 L=90 M=90 Line WX is parallel to Line JK Based on parallel line properties JDW = DJK = 90 = JKE = KEX ; if JDW = 90 , WDQ = 18090 = 90 if KEX = 90 , XER = 18090 = 90 XWY + WXY + WYX = 180 What is WYX?
(1) The sum of the measures of angles DQS and ERT is 260. DQS+ERT=260 then DQW+ERX = 360260 = 100 WDQ=XER=90 (as given) XWY+WXY = 360  (WDQ+XER)  (DQW+ERX) XWY+WXY = 360  (90+90)  100 = 360180100 = 80 WYX = 180 (XWY+WXY) = 18080 = 100 Sufficient
(2) The sum of the measures of angles WQD and ERX is 100. This is exactly what we found in the stmt 1 as step 2 and its directly provided here ,then it follows the same reasoning as above DQW+ERX = 360260 = 100 WDQ=XER=90 (as given) XWY+WXY = 360  (WDQ+XER)  (DQW+ERX) XWY+WXY = 360  (90+90)  100 = 360180100 = 80 WYX = 180 (XWY+WXY) = 18080 = 100 Sufficient
D is the answer!



Manager
Joined: 10 Aug 2016
Posts: 68
Location: India

Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
22 Jul 2019, 22:08
In the diagram above, triangle WXY intersects rectangle JKLM at points D, E, Q, R, S, and T. If line WX is parallel to line JK, what is the measure of angle WYX?
JKLM is a rectangle and WX is parallel to line JK, therefore angle QDE = angle RED = 90 degree. Sum of all the angles of a pentagon = 540 degrees QDERY is a pentagon. If we know the sum of the angles QDE, RED, DQY, ERY then we can find the angle QYR i.e WYX = 540  (sum of rest of the angles) QDE+RED = 180. Therefore we only need to know DQY+ERY to find the angle QYR.
(1) The sum of the measures of angles DQS and ERT is 260. statement 1 gives us angle DQS + angle ERT = 260 i.e. DQY + ERY = 260. Sufficient.
(2) The sum of the measures of angles WQD and ERX is 100. WQD+DQY = 180 (straight line) ERX + ERY = 180 (straight line) therefore DQY+ERY = 360  WQD+ERX. as per the statement WQD+ERX = 100 therefor DQY+ERY=360 100 = 260 therefore angle WYX can be found. Sufficient.
Both statements are individually sufficient.
Answer Choice: D



Intern
Joined: 25 Aug 2015
Posts: 45

Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
22 Jul 2019, 22:09
(1) The sum of the measures of angles DQS and ERT is 260. (2) The sum of the measures of angles WQD and ERX is 100.
Answer is D both are sufficient
This is because you need to find the sum of angles of W and X and subtract it from 180 to find angle Y. Since angles D and E are right angles, both statement above gives information about the sum of angle W and X independently. Hence you can deduce Y
Posted from my mobile device



Manager
Joined: 07 Mar 2019
Posts: 221
Location: India
WE: Sales (Energy and Utilities)

In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
Updated on: 22 Jul 2019, 23:58
From Question stem we have DE ‖ JK and ∠QDE = ∠RED = ∠J =∠K =∠M =∠L = 90⁰. Statement 1) ∠DQS+∠ERT=260⁰. Since we have sum values of 180⁰ and 260⁰ considering polygon DERTSQ, we may find the ∠WYX. Let’s see: For ∆YST: ∠SYT + ∠YTS + ∠TSY = 180⁰ ∠SYT = 180⁰  ∠YTS  ∠TSY ∠SYT = 180⁰  ∠RTM  ∠LSQ (As ∠YTS = ∠RTM and ∠TSY = ∠LSQ : Opposite ∠’s) ∠SYT = 180⁰  (180⁰  90⁰  ∠TRM) – (180⁰  90⁰  ∠SQL) (As both ∆QLS and ∆TMR are right angled triangles and ∠L and ∠M respectively) ∠SYT = ∠TRM + ∠SQL ∠SYT = 180⁰  ∠ERT + 180⁰  ∠DQS (As ∠ERT + ∠MRT = 180⁰ = ∠DQS + ∠SQL) ∠SYT = 180⁰ + 180⁰  (∠ERT + ∠DQS) ∠SYT = 180⁰ + 180⁰  260⁰ (From Statement 1) ∠SYT = 100⁰ SUFFICIENT Statement 2) ∠WQD + ∠ERX = 100⁰ METHOD – A: For ∆WYX: ∠WYX + ∠YWX + ∠WXY = 180⁰ OR ∠Y + ∠W + ∠X = 180⁰ → Eqn. ① For ∆WQD: ∠WQD + ∠QWD + ∠WDQ = 180⁰ ∠WQD = 180⁰  (∠QWD + 90⁰) (As DE ‖ JK, ∠WDQ = ∠QDE =∠DJK = 90⁰) ∠WQD = 90⁰  ∠ QWD → Eqn. ② For ∆ERX: ∠ERX + ∠RXE + ∠XER = 180⁰ ∠ERX = 180⁰  (∠RXE + 90⁰) (As DE ‖ JK, ∠XER = ∠RED =∠EKJ = 90⁰) ∠ERX = 90⁰  ∠RXE → Eqn. ③ Using Eqn. ② and ③ ∠WQD + ∠ERX = 100⁰ becomes 90⁰  ∠QWD + 90⁰  ∠RXE = 100⁰ ∠QWD + ∠RXE = 80⁰ OR ∠W + ∠X = 80⁰ → Eqn. ④ Now from Eqn. ③ and ④ ∠Y = 180⁰  (∠W + ∠X) ∠Y = 180⁰  80⁰ ∠Y = 100⁰ METHOD B: For ∆SQL: ∠SQL + ∠QLS + ∠LSY = 180⁰ → Eqn. ① For ∆MRT: ∠MRT + ∠RTM + ∠TMR = 180⁰ → Eqn. ② For ∆SYT: ∠SYT + ∠YTS + ∠TSY = 180⁰ ∠SYT + ∠RTM + ∠LSQ = 180⁰ (Opposite ∠’s: ∠RTM = ∠YTS & ∠LSQ = ∠TSY) ∠SYT + 180⁰  90⁰  ∠MRT + 180⁰  90⁰  ∠LQS = 180⁰ (From Eqn. ① and ②) ∠SYT  ∠ERX  ∠DQW = 0 (Opposite ∠’s: ∠ERX = ∠MRT & ∠DQW = ∠LQS) ∠SYT = ∠ERX + ∠DQW ∠SYT = 100⁰ SUFFICIENT Answer (D)
_________________
Ephemeral Epiphany..!
GMATPREP1 590(Q48,V23) March 6, 2019 GMATPREP2 610(Q44,V29) June 10, 2019 GMATPREPSoft1 680(Q48,V35) June 26, 2019
Originally posted by lnm87 on 22 Jul 2019, 22:44.
Last edited by lnm87 on 22 Jul 2019, 23:58, edited 1 time in total.



Manager
Joined: 18 Jun 2013
Posts: 136
Location: India
Concentration: Technology, General Management
GPA: 3.2
WE: Information Technology (Consulting)

Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
22 Jul 2019, 22:48
Given,
WX  JK
To find,
mLWYX
Lets consider the options,
Option 1: mLDQS + mLERT = 260
Now sum of all the interior angles of quadrilateral DQSTRE is given by
=> mLEDQ + mLDQS + mLQST + mLSTR + mLTRE + mLRED => 90 + (mLDQS + mLTRE) + mLQST + mLSTR + 90 => 180 + 260 + mLQST + mLSTR => 440 + mLQST + mLSTR
But sum of the angles of an ngon is given by 180 * (n  2) where n is the number of sides in the ngon
Hence,
440 + mLQST + mLSTR = 720 mLQST + mLSTR = 280
Now,
mLSYT + mLYST + mLYTS = 180 => mLSYT + (180  mLQST) + (180  mLSTR) = 180 => mLSYT + 360  (mLQST + mLSTR) = 180 => mLSYT + 360  280 = 180 => mLSYT = 100
Hence option 1 is sufficient.
Option 2: mLWQD + mLERX = 100
=> mLLQS + mLTRM = 100
Now,
mLLQS + mLQLS + mLLSQ + mLTRM + mLRMT + mLMTR = 360 => (mLLQS + mLTRM) + 90 + mLYST + 90 + mLYTS = 360 => 100 + 180 + mLYST + mLYTS = 360 => mLYST + mLYTS = 80 => mLSYT = 100
Hence option 2 is sufficient.
Answer: D



Manager
Joined: 12 Mar 2018
Posts: 83
Location: United States

Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
22 Jul 2019, 22:53
Correct answer A
Stmt 1: angle DQS + angle ERT = 260 Also, DQYRE forms a polygon, with angle D and E being 90 degrees respectively.
Using the formula, Sum of Interior angles of a Polygon = (n2)*180, where n is the no. of sides
for the polygon DQYRE, we get, angle D + angle E + sum of angles DQS and ERT + angle Y = (52) * 180 > 90 + 90 + 260 +angle Y = 3*180 > angle Y = 100
Hence sufficient
Stmt 2: Sum of angles WQD and ERX is 100. There is no way to get to angle Y using the above information. Hence, insufficient.
Correct answer A



Manager
Joined: 25 Jul 2018
Posts: 207

In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
22 Jul 2019, 23:11
Statement1: <DQS +<ERT = 260. >Let's say that <DQS=a and <ERT=b > a+b=260
Triangle WDQ is rightangled triangle > <DQW=180a <DQW+<W=90 >180a+<W=90 > <W=a90 (keep it)
Triangle REX is rightangled triangle > <ERX=180b <ERX+<X=90 >180b+<X=90 > <X=b90 (keep it)
As we know, Triangle WYX is a big triangle and <W+<X+<Y=180  a90+b90+<Y=180 > <Y=180+90+90260=100 Sufficient Statement2: <WQD+<ERX =100. (<WQD=a, <ERX=b> a+b=100) As above mentioned, triangle WDQ is rightangled triangle. > <W=90a
Triangle REX is rightangled triangle. > <X=90b
Triangle WYX is a big triangle and <W+<X+<Y=180 > 90a+90b+<Y=180 <Y=a+b=100 Sufficient
The answer choice is D.



Manager
Joined: 27 May 2010
Posts: 200

Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
23 Jul 2019, 00:02
In the diagram above, triangle WXY intersects rectangle JKLM at points D, E, Q, R, S, and T. If line WX is parallel to line JK, what is the measure of angle WYX? (1) The sum of the measures of angles DQS and ERT is 260. Angles DQS and LQS are complementary. LQS = 180  DQS In triangle QSL, QSL = 90  LQS = DQS  90 QSL = YST = DQS  90 Similarly, STY = ERT  90 In triangle STY, anglye WYX = STY = 180  (YST + TYS) = 180  (DQS  90 + ERT  90) =360  (DQS + ERT) = 360  260 = 100 Sufficient. (2) The sum of the measures of angles WQD and ERX is 100. WQD + ERX = 100 WQD = LQS. QSL = YST = 90  LQS = 90  WQD. Similarly, STY = 90  ERX WYX = STY = 180  (YST + TYS) = 180  (90  WQD + 90  ERX) = 180  180 +(WQD + ERX) = 100 Sufficient. Hence each statement is sufficient alone. Option D.
_________________
Please give Kudos if you like the post



Manager
Joined: 27 Mar 2016
Posts: 110

Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
23 Jul 2019, 00:14
Let's consider A The sum of the measures of angles DQS and ERT is 260. Hence The sum of the measures of angles WQD and ERX is 100. Which means sum of DWY and EXY 80 Hence WYX = 100 Sufficient
B The sum of the measures of angles WQD and ERX is 100. Which means sum of DWY and EXY 80 Hence WYX = 100 Sufficient
Hence answer is D
Posted from my mobile device



Manager
Joined: 30 May 2019
Posts: 108

In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
Updated on: 23 Jul 2019, 00:39
Please refer to the diagram below
Attachments
photo5310010120249388064.jpg [ 140.37 KiB  Viewed 92 times ]
Originally posted by mira93 on 23 Jul 2019, 00:37.
Last edited by mira93 on 23 Jul 2019, 00:39, edited 1 time in total.



Manager
Joined: 08 Jan 2018
Posts: 98
Location: India
GPA: 4
WE: Information Technology (Computer Software)

Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
Show Tags
23 Jul 2019, 00:37
As triangle WXY intersects rectangle JKLM at points D, E, Q, R, S, and T. If line WX is parallel to line JK then:
(1) The sum of the measures of angles DQS and ERT is 260. DQS + ERT =260 WX  JK : then, DER = EDQ = 90 In Petagon : D + E + R + Y + Q = 540 => Y = 540  90  90  260 => Y = 540  440 => Y = 100 Sufficient.
(2) The sum of the measures of angles WQD and ERX is 100.
WQD + ERX = 100 Let WQD = a , then ERX =100  a In Traingle WQD => W = 180  WDQ  WQD = 180  90  a In triangle ERX => X = 180  XER  ERX = 180  90  100  a In triangle WXY: Y = 180  WQD  ERX = 180  (90 a)  (a10) = 180  90 +a  a +10 = 100 Sufficient.
IMO the answer is D.
Please hit kudos if you like the solution.




Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
[#permalink]
23 Jul 2019, 00:37



Go to page
Previous
1 2 3 4
Next
[ 73 posts ]



