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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 09:23
We require the Sum of angles W and X to arrive at the angle of Y as YXW is a triangle whose sum of angles must be equal to 180 degrees Angle J and K are 90 degrees (Rectangle). Since WK is parallel using traversal lines property  Angle WDQ and REX  90i) Sufficient, Can derive Sum of DQW and ERX as 100 degrees. (Sum of Angles in a straight line at Q and R = 360 and Sum of DQS and ERT is 260. Which means that sum of angles at SQL and MRT is 100 and being vertically opposite angles DQW and ERX also has 100). So Sum of all angles in two triangles WQD and RXE other than Sum of angles W and X is 90*2+100. So W+X=360280=80 and So Y = 100IMO D ii) Sufficient, Directly given the first inference of (i) that is Sum of DQW and ERX as 100 degrees.
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 09:29
angle EDQ =90=DWQ+DQW angle DER=90=ERX+EXR(EXTERIOR ANGLE AND CORRESPONDING ANGLE TO RIGHT ANGLE)
1.DQS+ERY=260 =>180DQW+180ERX=260 =>180(90DWQ)+180(90EXR)=260 SOLVING we get value to the sum of DWQ and EXR so SYT is 180 sum Hence 1 is sufficient 2.WQD+ERX=100 So 90DWQ+90EXR=100 Again we get the sum of other 2 angled of the triangle so 3 Rd angle can be found Hence 2 is also sufficient Hence D
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 09:29
from statement (1), angle (1) + (2) = 260, as (1) a complement of (3), and (2) a complement of (4), then angle (3) + (4) = 360  260 = 100Triangles WDQ and XER are right angled triangles at D and E respectively, so (3) + (5) = 90, and (4) + (6) = 90 (5) + (6) = 90  (3) + 90  (4) = 180  100 = 80 so angle (7) = 180  [(5) + (6)] = 18  80 = 100 > sufficient from statement (2), angle (3) + (4) = 100 > which is typically the same conclusion of statement (1), so Also sufficientD
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 09:31
IMO D
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.
From St.1, angles DQS + ERT = 260, We can consider DERYQ a pentagon having 5 sides and sum of the internal angles = (52)*180 = 540 [Sum = (n2)*180] So, angles EDQ+DER+RYQ=540260=280 Given that line WX is parallel to line JK, which means angles EDQ+DER=90+90=180 So, angle RYQ = angle WYX = 100 Sufficient
(2) The sum of the measures of angles WQD and ERX is 100.
From St.2, angles WQD + ERX = 100, We know angles WQD + DQS = 180 and angles ERX + ERT = 180, That means angles DQS + ERT = 360100 = 260 We can consider DERYQ a pentagon having 5 sides and sum of the internal angles = (52)*180 = 540 [Sum = (n2)*180] So, angles EDQ+DER+RYQ=540260=280 Given that line WX is parallel to line JK, which means angles EDQ+DER=90+90=180 So, angle RYQ = angle WYX = 100 Sufficient



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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 09:38
(1) Let Angle DQS = x Angle ERT= y
x+y = 100
Angle QSL = 90x = Angle YST (exterior angle property and vertically opposite angles) Angle RTM = 90y = Angle STY (exterior angle property and vertically opposite angles)
Angle YST + Angle STY + Angle SYT = 180 90x + 90  y + Angle SYT = 180
Angle SYT can be calculated. Sufficient.
(2) Similarly let angle WQD = a angle YST = 90a Let Angle ERS = b angle STY = 90b
a+b = 100 Angle YST + Angle STY + Angle SYT = 180 90a + 90  b + Angle SYT = 180 Angle SYT can be calculated. Sufficient.
D is correct.



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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 09:42
IMOD
Statement (1) The sum of the measures of angles DQS and ERT is 260.
Angle (DQS + ERT) = 260 Angle DQS & ERT are exterior angles of Triangle QSL & RTM resp.
Angle DQS = 90 + QSL Angle ERT = 90 + RTM
Angle (DQS + ERT) = 260 = 180 + QSA + RTM QSL + RTM = 80
Angle QSL = Angle YST & RTM = YTS (Vertically opposite angles)
Angle YST + YTS = 80 So, SYT= 180 80 = 100
Sufficient.
Statement (2) The sum of the measures of angles WQD and ERX is 100.
angles WQD + ERX= 100 or, LQS + TRM = 100 (Vertically opposite angles)
LSQ= 180  (90 + LQS) & MTR= 180  (90 + TRM) LSQ + MTR = 180 (LQS+TRM) = 80 Angle LSQ = Angle YST & MTR = YTS (Vertically opposite angles) Angle YST + YTS = 80 So, SYT= 180 80 = 100
Sufficient



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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 10:00
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?
I AM USING VERTEX NAMES TO REPRESENT ANGLES FOR BETTER READABILITY. SO Y MEANS WYX etc.
We want to find Y
W+X+Y = 180
So we can find Y if we know W + X
(1) The sum of the measures of angles DQS and ERT is 260. DQS+ERT = 260
DQS = W + D ... because exterior angle theorem... extrior angle = sum of remote interior angles DQS = W + 90.... D is 90 because WX is parallel to JK
ERT = X + E ... because exterior angle theorem... extrior angle = sum of remote interior angles ERT = X + 90.... E is 90 because WX is parallel to JK
Substituting both above
W+X+180 = 260
W+X = 80
This is sufficient because W+X+Y = 180, so 80+Y = 180, Y = 100
(1) IS SUFFICIENT
(2) The sum of the measures of angles WQD and ERX is 100.
WQD + W + 90 = 180... because triangle.. and D is 90 because WX is parallel to JK ERX + X + 90 = 180.... because triangle.. and E is 90 because WX is parallel to JK
Adding both
WQD + ERX + W + X + 180 = 360
100 + W + X = 180.... because given that WQD + ERX = 100
W+X = 80 .... This is same as (1) and sufficient to find Y = 100
(2) IS SUFFICIENT
ANSWER: D  EACH IS SUFFICIENT



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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 10:09
Quote: 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? This is the Data Sufficiency (DS) question where were are asked to learn the measure of an angle WYX. Let us analyze the data from each statement first and try to identify if the given data is enough: Statement 1: (1) The sum of the measures of angles DQS and ERT is 260. Let us divide the solution in several steps: a) We know that angles DQS and ERT are 260 degrees. Angle DQS is external angle for the angle WQD. Angle ERT is external angle to ERX. We know that \(internal angle = 180 degrees  external angle\) In this case, angles \(WQD + ERX = (180  DQS) + (180  ERT) = 360  DQS  ERT = 360  (DQS + ERT)\) As \(DQS + ERT = 260 degrees\), => \(WQD + ERX = 360  260 = 100 degrees\) b) Also, one may note that angles WQD and SQL are vertical and vertical angles are equal. The same applies to the angles ERX and TRM. In this case \(WQD + ERX = SQL + TRM = 100\) c) Now, as triangles SQL and TRM are right triangles, we have 2 angles QLS and RMT each of 90 degrees. In this case, sum of angles QSL and RTM are equal to \((180  90  SQL) + (180  90  TRM) = 90  SQL + 90  TRM = 180  (SQL +TRM) = 180  100 = 80\) d) Angles QSL and YST are vertical and because of it they are equal. The same applies to angles RTM and STY. Thus, \(QSL + RTM = YST + STY = 80\). e) Since all angles of a triangle are equal to 180 degrees, angle \(SYT = 180  80 = 100\) We found an angle SYT what was required by the task. Please note that calculations in Data Sufficiency questions are sometimes unnecessary waste of time, and it is required to understand the concept without solving the task itself. Sufficient. Statement 2:(2) The sum of the measures of angles WQD and ERX is 100. This is what we found at the end of step a during analysis of 1st statement. Thus statement 2 is enough. Sufficient. Both statements are sufficient. Answer: D



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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 10:17
Answer D Each statement gives us enough information to get <WYX?
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 10:22
Given the diagram above: Let angle Y = y, angle at W = w, angle at R = r, angle at X = x, angle at Q = q, and angle WQD = u and ERX = b w + x + y = 180 w + u = 90 x + b = 90 Statement 1: q + r = 260 Therefore, q = 260  r u + q = 180 u = 180  q u = 180  (260r) u = r  80 Remember; w + u = 90 w + r  80 = 90 Thus, w = 170  r Similarly, r + b = 180 b = 180  r and r = 260  q Thus, b = 180  260 + q Thus, b = q  80 Remember , x = 90  b Thus, x = 90  q + 80 x = 170  q Combining the values for w, x and y We have, y = 180  x  w y = 180  170 + q  170 + r y = 180  340 + q + r y = 260  160 y = 100 Sufficient AD Statement 2: given that WQD + ERX = 100 u + b = 100 Remember that, u + w + b + x = 90 + 90 Thus, 100 + u + w = 180 Therefore, u + w = 80 Remember that, y + u + w = 180 Thus, y = 100 Sufficient  D Hence answer choice D. Thank you for the kudos. Posted from my mobile device
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In the diagram above, triangle WXY intersects rectangle JKLM at points
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Updated on: 23 Jul 2019, 03:42
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 :from the figure angle DWQ = angle TSY ..............1 angle EXR = angle STY ................2 adding 1 and 2 gives angle DWQ + angle EXR = angle TSY + angle STY ............. 3 subtracting 3 from 180 on both sides of equation 180  (angle DWQ + angle EXR) = 180  (angle TSY + angle STY) .............4 angle WYX = angle SYT ...................................5 THIS IS THE ANGLE WE ARE ASKED TO FIND (1) The sum of the measures of angles DQS and ERT is 260.as angle DQS = angle DWQ + angle WDQ ( exterior opposite angles) and angle ERT = angle REX + angle EXR ( exterior opposite angles) this gives 260 = angle DWQ + angle WDQ + angle REX + angle EXR where ( angle WDQ and angle REX are 90 degress each due to right angle) 260 = 180 + angle DWQ + angle EXR angle DWQ + angle EXR = 80 hence from equation 4, we can find the angle of wyx = 100. hence sufficient (2) The sum of the measures of angles WQD and ERX is 100.so 180  angle WQD + 180  angle ERX = angle DQS + angle ERT = 360  ( angles WQD and ERX) = 260 hence we framed the condition 1, which is proved above. so this is sufficient to find the angle wyx so the ans is D
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Originally posted by ccheryn on 22 Jul 2019, 10:32.
Last edited by ccheryn on 23 Jul 2019, 03:42, edited 1 time in total.



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In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 10:53
We are asked angle Y, SO if we find the sun of angles W and X we will get angle Y=180(W+X)
1 stm > DQS=angle W+90 and ERT=X+90 (we could deduce that by using concepts: 1) outside angle=sum of opposite inner angles; 2) sum of the angles of triangle=180 and sum of adjascent angles equal 180) Now we know that DQS+ERT=260, substitute the values W+90+X+90=260 We can find sum of W+X;
Sufficient
2 stm > WQD=90angle W; ERX=90angle X; WQD+ERX=100, substitute the values 90W+90X=100, We can find X+W=80
Sufficient
IMO Ans: D



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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 11:02
For me it's C
As we know if we take statememt one
Then we are given sum of DQS and ERT
From these two we can't find the required angles
Similary for statement 2
But when we combine these two
Be it any values , the required angles always comes as 100
Beacause combination of.QWD and RXE always comes as 80
So hence it is C
C it is
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 11:11
One of the easiest ways to tackle this question is to realize that statement 2 is same as statement 1: See the following: Statement 2: WQD + ERX = 100 And we know WQD + DQS = 180 and ERX + ERT = 180 Let DQS = a and ERT = b So WQD = 180  a and ERX = 180  b So, (180  a) + (180  b) = 100 => a+b = 260 same as Statement 1. Now, if we prove that statement 1 in itself is correct then we are good and our answer is D or it is E. Please find the solution for statement 1. So as per the diagram explained below, answer is D. Regards, Rishav
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 12:11
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. Not sufficient.
(2) The sum of the measures of angles WQD and ERX is 100. Sufficient. here we can determine the angle



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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 12:46
STATEMENT (1) The sum of the measures of angles DQS and ERT is 260. Since DQYRE is a pentagon so the sum of an interior angle of pentagon = 540
angles QDE + DER + ERY + RYQ + YQD = 540 (QDE = 90 DER = 90 since DE is parallel to JK) 90 + 90 + ERT + WYX + DQS = 540 (ERY is same as ERT and YQD is same as DQS RYQ is same as WYX) 180+260+WYX=540 WYX=100 SUFFICIENT
STATEMENT (2) The sum of the measures of angles WQD and ERX is 100.
In triangle WQD angles WQD+QDW+DWQ=180 angle DWQ=90WQD (QDW=90) In triangle ERX angles ERX+RXE+EXR=180 angle EXR=90ERX
now In triangle WYX angles DWQ+EXR+WYX= 180 90WQD+90ERX+WYX=180 WYX=WQD+ERX=100 SUFFICIENT
D is the answer



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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 12:52
IMO correct answer is D  Each statement alone is sufficient. Explanation is provided as attachment
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 13:16
1) angle DQS + angle ERT=260 angle DQS= angle DWQ + 90 angle ERT=angle EXR +90
Angle WYX=180(angle DWQ +angle EXR)
therefore, angle dwq+ angle exr = 2609090=80 angle wyx=18080=100: Suffiecient
2)angles WQD and ERX is 100. Angle WQD=180(angle dwq+90)=90angle dwg Angle ERX=90  angle exr
therefore, 90angle dwg +90  angle exr =100 or angle dwg+angle exr=80 or angle wyx=100: suffiencient
Ans: D



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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 13:21
What is \(\angle{WYX} ?\) ST1. \(\angle{DQS}+\angle{ERT}= 260\) If \(\angle{LQS}+\angle{DQS}= 180\) and \(\angle{ERT}+\angle{MRT}= 180\), then \(\angle{LQS}+\angle{DQS}+\angle{ERT}+\angle{MRT}= 360\). If \(\angle{DQS}+\angle{ERT}= 260\), then \(\angle{LQS}+\angle{MRT}= 100\). Since both \(\triangle{LQS}\) and \(\triangle{MRT}\) are right triangles with right angles at \(\angle{QLS}\) and \(\angle{RMT}\), the sum of their 4 sharp angles must be \(\angle{LQS}+\angle{QSL}+\angle{RTM}+\angle{MRT}= 180\). If \(\angle{LQS}+\angle{MRT}= 100\), then \(\angle{QSL}+\angle{RTM}= 80\). Since \(\angle{QSL}=\angle{YST}\) and \(\angle{RTM}=\angle{STY}\), then \(\angle{YST}+\angle{STY}=80\). Hence \(\angle{WYX}=180  (\angle{YST}+\angle{STY})=180100=100\) SufficientST2. \(\angle{WQD}+\angle{ERX}= 100\) Since \(\angle{WQD}=\angle{LQS}\) and \(\angle{ERX}=\angle{MRT}\), we can safely conclude that \(\angle{LQS}+\angle{MRT}= 100\). While solving ST1 we have already seen how this information is enough to figure out \(\angle{WYX}\). SufficientHence D
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 13:43
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.
First, we should calm down and figure out the angles mentioned in the question. Our decision will be based on triangles WDQ and XER. We know from the text that angles WDQ and XER are both equal 90 (because JKLM is a rectangle and WX is parallel to line JK).
1)The sum of the measures of angles DQS and ERT is 260, so the sum of angles WQD and ERX is (180+180)  260 = 100. So the sum of angles W and X will be (180+180)  100  180 = 80. Thus the angle WYX = 100 Sufficient
2)The sum of the measures of angles WQD and ERX is 100  this information brings us even closer to the same decision. The sum of angles W and X will be (180+180)  100  180 = 80. Thus the angle WYX = 100 Sufficient
The answer is D




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