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In the diagram above, triangle WXY intersects rectangle JKLM at points

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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

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New post 22 Jul 2019, 09:23
2
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 - 90

i) 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=360-280=80 and So Y = 100

IMO 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  [#permalink]

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New post 22 Jul 2019, 09:29
1
angle EDQ =90=DWQ+DQW
angle DER=90=ERX+EXR(EXTERIOR ANGLE AND CORRESPONDING ANGLE TO RIGHT ANGLE)

1.DQS+ERY=260
=>180-DQW+180-ERX=260
=>180-(90-DWQ)+180-(90-EXR)=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 90-DWQ+90-EXR=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  [#permalink]

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New post 22 Jul 2019, 09:29
1
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 = 100

Triangles 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 sufficient

D
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

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New post 22 Jul 2019, 09:31
1
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 = (5-2)*180 = 540 [Sum = (n-2)*180]
So, angles EDQ+DER+RYQ=540-260=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 = 360-100 = 260
We can consider DERYQ a pentagon having 5 sides and sum of the internal angles = (5-2)*180 = 540 [Sum = (n-2)*180]
So, angles EDQ+DER+RYQ=540-260=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  [#permalink]

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New post 22 Jul 2019, 09:38
1
(1) Let Angle DQS = x
Angle ERT= y

x+y = 100

Angle QSL = 90-x = Angle YST (exterior angle property and vertically opposite angles)
Angle RTM = 90-y = Angle STY (exterior angle property and vertically opposite angles)

Angle YST + Angle STY + Angle SYT = 180
90-x + 90 - y + Angle SYT = 180

Angle SYT can be calculated. Sufficient.

(2) Similarly let angle WQD = a
angle YST = 90-a
Let Angle ERS = b
angle STY = 90-b

a+b = 100
Angle YST + Angle STY + Angle SYT = 180
90-a + 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  [#permalink]

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New post 22 Jul 2019, 09:42
1
IMO-D

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  [#permalink]

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New post 22 Jul 2019, 10:00
1
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  [#permalink]

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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?


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  [#permalink]

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New post 22 Jul 2019, 10:17
1
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  [#permalink]

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New post 22 Jul 2019, 10:22
1
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 - (260-r)
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.

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In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

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New post Updated on: 23 Jul 2019, 03:42
1
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  [#permalink]

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New post 22 Jul 2019, 10:53
1
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=90-angle W; ERX=90-angle X;
WQD+ERX=100, substitute the values 90-W+90-X=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  [#permalink]

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New post 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  [#permalink]

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New post 22 Jul 2019, 11:11
1
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.

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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

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New post 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  [#permalink]

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New post 22 Jul 2019, 12:46
1
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=90-WQD (QDW=90)
In triangle ERX
angles ERX+RXE+EXR=180
angle EXR=90-ERX

now In triangle WYX
angles DWQ+EXR+WYX= 180
90-WQD+90-ERX+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  [#permalink]

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New post 22 Jul 2019, 12:52
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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  [#permalink]

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New post 22 Jul 2019, 13:16
1
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 = 260-90-90=80
angle wyx=180-80=100: Suffiecient

2)angles WQD and ERX is 100.
Angle WQD=180-(angle dwq+90)=90-angle dwg
Angle ERX=90 - angle exr

therefore, 90-angle 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  [#permalink]

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New post 22 Jul 2019, 13:21
1
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 \(\angle{WYX}=180 - (\angle{YST}+\angle{STY})=180-100=100\)

Sufficient

ST2. \(\angle{WQD}+\angle{ERX}= 100\)

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 \(\angle{WYX}\).

Sufficient

Hence D
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

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New post 22 Jul 2019, 13:43
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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.


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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