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In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 08:00
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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. Attachment:
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In the diagram above, triangle WXY intersects rectangle JKLM at points
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Updated on: 23 Jul 2019, 01:10
The data that can be deduced from the given figure: Angle EDQ = Angle RED = 90 degrees. > (a) DQYRE is a polygon with five sides. So, sum of internal angles of the polygon = 180 (5 – 2) = 540 > (b)
(1) The sum of the measures of angles DQS and ERT is 260. From (b) we have Angle DQS + Angle QYR + Angle YRE + Angle RED + Angle EDQ = 540 >(f)
From (a) we have 260 + Angle QYR + Angle RED + Angle EDQ = 540 260 + Angle QYR + 90 + 90 = 540 Angle QYR = 100
From the given figure, we can see that Angle QYR = Angle WYX Hence, Angle WYX = 100 degrees
Sufficient
(2) The sum of the measures of angles WQD and ERX is 100. If Angle WQD = a, Then Angle DQS = 180 – a > (c)
If Angle ERX = b, Then Angle YRE = 180 – b > (d)
From (c) and (d) we have Angle DQS + Angle YRE = 180 – a + 180 – b 360 – (a + b) 360 – 100 Angle DQS + Angle YRE = 260 > (e)
From (a), (b), (e) and (f) we have: Angle DQS + Angle QYR + Angle YRE + Angle RED + Angle EDQ = 540 260 + Angle QYR + 90 + 90 = 540 Angle QYR = 100
From the given figure, we can see that Angle QYR = Angle WYX Hence, Angle WYX = 100 degrees
Sufficient
Answer D
Originally posted by Sayon on 22 Jul 2019, 08:23.
Last edited by Sayon on 23 Jul 2019, 01:10, edited 2 times in total.



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In the diagram above, triangle WXY intersects rectangle JKLM at points
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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. Please see solution in the attached file IMO D
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Originally posted by Kinshook on 22 Jul 2019, 08:25.
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 08:29
https://gmatclub.com/forum/download/fil ... c11c353308Clearly A is sufficient. IF a is sufficient B is also sufficient as B is a subset of A
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 08:32
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 a medium level difficulty problem. It tests your ability to infer.
Let's get to the statements.
(1) The sum of the measures of angles DQS and ERT is 260. This statement is insufficient, as multiple possible values are possible for the break up 260.
(2) The sum of the measures of angles WQD and ERX is 100. This again is insufficient as multiple possible values are possible for break up of 100.
Combining the two we know that 100 will be broken up in 60, 40. Combing it all the perpendicular angles of 90 we can infer the other triangles and get the value of Y.
IMO C.



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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 08:40
D
First note that both statements are basically saying the same thing, since A(WQD) (angle WQD) and A(DQS) make 180. Similarly, A(KRX) and A(KRT) make 180. So, sum all 4 angles, we should have 360. So, statement two is simply 360  260 = 100.
So, if we can get A(WYX) from statement 1, we will get in from St 2 as well.
St: A(DQY) + A(KRT) = 260
Look in triangle WDQ, we have A(W) = 180  (180A(DQY))  90 = A(DQY)90 (see A(WDQ) = 90) Similarly, in triangle EXR, we have A(X) = 180  (180A(KRT))90 = A(KRT)90
So, in triangle WYX, A(W) + A(X) + A(Y) = 180 or A(Y) = 180  (A(W)+ A(Y)) = 180  A(DQY) + 90 A(KRT) + 90 = 360  A(DQY)  A(KRT) = 360  260 = 100.



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In the diagram above, triangle WXY intersects rectangle JKLM at points
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Updated on: 22 Jul 2019, 10:46
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? 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. As JKLM is a rectangle and line WX is parallel to line JK, triangle WDQ, XER are right angled triangle therefore, angle W + angle Q= 90. and angle R + angle X = 90. ALso in triangle WXL, angle W + angle X + angle L = 180> (1) Statement 1, angle DQS + ERT = 260 180 DQW + 180 ERX = 260 180(90  DWQ) + 180  (90EXR) = 260 90 + DWQ +90+ EXR = 260 Now from 1, WYX= 260180 SUFFICIENT Statement 2, WQD+ ERX= 100 90  DWQ +90EXR = 100 Now from 1, 180100 = WYX SUFFICIENT Option 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, 08:45
IMO answer is D:
from 1, we know from pentagon DQYRE, sum of interior angles = 540, we know other two angles, are right anlge so sum is 180, therefore angle WYX= 540180260 = 100 deg from 2: we can determine the sum of DQY and ERT as 260, by 360100, from there we can determine the angle as 100deg.
so in my opinion D



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In the diagram above, triangle WXY intersects rectangle JKLM at points
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Updated on: 22 Jul 2019, 20:08
From (1) The sum of the measures of angles DQS and ERT is 260 we have that:
DQS =x ERT = y
So, LQS= 180  x, and LSQ= x90 and then since LSQ and YST are opposite angles, they are equal.
Also, TRM = 180 y, and MRT= y90 and then since MRT and STY are opposite angles, they are equal.
Finally, since all angles of triangle SYT must sum 180, we have that:
x90+ y90 + WYX = 180
WYX = 360  x  y
WYX = 360  (x+y)= 100, so sufficient.
From (2) The sum of the measures of angles WQD and ERX is 100, we have that:
In this one you have the complementary angles of solution in (1), and each complementary angle is 180  x and 180 y, and if you sum up you arrive to the same formula in (1) 360  x y = 100, this one is also sufficient.
(D) is our answer.
Originally posted by Mizar18 on 22 Jul 2019, 08:45.
Last edited by Mizar18 on 22 Jul 2019, 20:08, edited 1 time in total.



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In the diagram above, triangle WXY intersects rectangle JKLM at points
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Updated on: 22 Jul 2019, 23:16
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. (2) The sum of the measures of angles WQD and ERX is 100.
Sol:
1)DQS +ERT=260 then we know that DQSTRE is a six sided figure and the angle of 6 side fig is 180*(n2) where n= sides, so we know that DERYQ=180*3=540 and QDE and DER each is 90 then QYR or WYX =540180260 =100 ,we get solution of angle y
sufficient.
2)WQD+ERX=100 then that means triangle WQD and triangle ERX =180*2=360 and both are right triangles, then
we get XER and WDQ=180 both are 90 thus 360180=180
then angle D + angle X= 80 because wqd+erx=100,
now in triangle WYX , we know that total angle in triangle =180, so 18080 =100
so angle Y=100.
Sufficeint.
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, 08:50
WX  JK DERTSQ > Hexagon
For Hexagon DERTSQ D=E =90 degree Q+R = 260
Let Q be a degrees ==> in triangle DWQ > W = a90 degree similarly Let angle R be "b" degree > In Triangle EXR Angle X = b90 a+b=260
For Hexagon Angle S + T = 180a+180b So we can conclude angle SYT
A is sufficient
B is reduced to the same equation as A
Hence, 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, 08:55
(1) The sum of the measures of angles DQS and ERT is 260. Using this info we can calculate the sum of the external angles LQS and MRT. 360260=100. Now, angles QLS and RMT each equal 90 and we can calculate the sum of angles LSQ and RTM. 360180100=80 Since the sum of LSQ and RTM equals to the sum of angles YST and YTS we can find the measure of angle SYT, which is the same angle as WYX. 18080=100 Sufficient. (2) The sum of the measures of angles WQD and ERX is 100.
This is the same as the first statement, but a little bit easier. The sum of the measures of angles WQD and ERX equals to the sum of the measures of angles LQS and MRT. Now, angles QLS and RMT each equal 90 and we can calculate the sum of angles LSQ and RTM. 360180100=80 Since the sum of LSQ and RTM equals to the sum of angles YST and YTS we can find the measure of angle SYT, which is the same angle as WYX. 18080=100 Sufficient.
IMO D



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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 08:55
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?
(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: the top side of the rectangle and triangle are parallel. rule: straight line angle has 180 degrees rule: triangle interior angles sum to 360 degrees (1) The sum of the measures of angles DQS and ERT is 260: sufficient; with this we can find the sum of the adjacent angles forming 180 degrees, then we can find the sum of the interior angles of the small adjacent triangles, then we can the sum of the angles of the adjacent straight line and the sum of the following straight line which is the interior angles of the triangle formed by angle WXY. (2) The sum of the measures of angles WQD and ERX is 100: sufficient; this gives the same information as above. 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, 08:57
Image 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. #1 individual angles not know insufficient #2 individual angles not know insufficient from 1& 2 nothing can be determined in common IMO E
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 09:00
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?
In order to solve this question, you need to see 2 things: 1. DQYER  pentagon, => TotalSUMofvertex = 180*(52) = 540 2. We know measures of vertex D and E = 90
In order to know vertex Y, we need to know the measures of vertex Q and R or their sum.
(1) The sum of the measures of angles DQS and ERT is 260. Q+R = 260 SUFFICIENT (2) The sum of the measures of angles WQD and ERX is 100. if WQD + ERX = 100: DQS + WQD = 180 ERT+ERX = 180 ERT+ERX + DQS + WQD = 360 ERT + DQS +100 = 360 ERT + DQS = 260
SUFFICIENT
ANSWER D



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In the diagram above, triangle WXY intersects rectangle JKLM at points
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Updated on: 22 Jul 2019, 09:51
DQYRE forms a pentagon. From 1 & 2 each, we get Angle Q & Angle R And Angle D and Angle E are rightangled. Hence we can easily get Angle WYX Hence D
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Originally posted by kitipriyanka on 22 Jul 2019, 09:01.
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In the diagram above, triangle WXY intersects rectangle JKLM at points
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Updated on: 22 Jul 2019, 09:15
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? Given: DQYRE is a pentagon: So sum of all the interior angles of the pentagon DQYRE= (n2) * 180=3*180 i.e angleEDQ + angleDER + angleDQY + angleERY + angelQYR = 3*180 => 90 + 90 + angleDQS + angleERT + angelWYX = 3*180, where angleEDQ = angleDER = 90 => angleDQS + angleERT + angelWYX = 2*180 (i)
(1) The sum of the measures of angles DQS and ERT is 260. > sufficient: angleDQS + angleERT = 260, replacing the value in equation(i), angelWYX = 2*180260 = 100 (2) The sum of the measures of angles WQD and ERX is 100. > sufficient: angleWQD + angleERX = 100 =>(180angleDQS) + (180angleERT) = 100 => angleDQS + angleERT = 260, replacing the value in equation(i), angelWYX = 2*180260 = 100
So the answer is D
Originally posted by hiranmay on 22 Jul 2019, 09:03.
Last edited by hiranmay on 22 Jul 2019, 09:15, edited 1 time in total.



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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 09:03
Since WX is parallel to JK, \(\angle JDW=\angle KEX=90^{\circ}\) By property of intersecting angles, \(\angle DQW = \angle SQL, \angle QSL=\angle TSY, \angle STY=\angle RTM, \angle TRM=\angle ERX\) From the figure, we can see that we need to find \(a+b\) (1) The sum of the measures of angles DQS and ERT is 260. \(x+y=260\) Note that \(a+x=180, b+y=180\) \(a+x+b+y=360\) \(a+b+260=360\) \(a+b=100\) 1 is Sufficient (2) The sum of the measures of angles WQD and ERX is 100. We are directly given that \(a+b=100\) 2 is sufficient 1 and 2 are independently sufficient Answer is (D)
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points
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22 Jul 2019, 09:10
JDEK forms a rectangle ad the lines are parallel. So JDE=KDE=90" SUBSEQUENTLY WDQ=REX=90" St. 1 DQS+ERT= 260" DQW+DQS= 180" and ERX+ERT=180" Add the equations, we get DQW+ERX= 100 Taking both triangles, we should get sum of 360" from 6 angles We have found two angles and sum of two angles, Adding up we get DQW+WDQ+DWQ+ERX+EXR+REX= 360 Insert the values, DWQ+EXR= 80 Taking the triangle WXY, we get 80+WYX= 180 So, angle Y=100  Suff St 2 provides the second step were we calculated sum of angles as 100. Hence, suff 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, 09:11
Its a fairly simple problem disguised by the way of complicated image. Make a pentagon described including angle we need to find : Starting from Point D : DQYER : Sum of Pentagon all angles is : 540 As line wx is parallen it will make angle 90 at angle D and E : So we have now 2 angles : D =e= 90 ,
Statement 1: GIves sum of other 2 angles of pentagon : So 90+ 90+ 260 + X = 540 X =100 So we can have , No need to solve, but solution is rather simple. Sufficient
Statement2: Sufficient : This also gives 2 angles making 180 with angles given in statement 1 : So we can find sum of angles given in statement 1 : and we will have same solution, in other way, So both statements are sufficient




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