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Senior Manager  D
Joined: 25 Sep 2018
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Location: United States (CA)
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

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1
Solution:
Question stem analysis:

From the given figure, we can determine that angle J, K , L , M are all 90 degrees since JKLM is a rectangle.

We know that segment Jk is parallel to segment WX, and we know that segment Jl is a transversal .By the property of transversal lines, we know that

line one is parallel to line two, and both lines are cut by the transversal, t. A transversal is simply a line that passes through two or more lines at different points. Some important relationships result.
• Vertical angles are equal
• Corresponding angles are equal:
• Supplementary angles sum to 180°
• Any acute angle + any obtuse angle will sum to 180°.

Therefore angle Angle QDE & angle RDE both are 90 degrees each

Statement One analysis:

The sum of the measures of angles DQS and ERT is 260.
We can observe that QDERY is a pentagon. since seg QD,DE,ER,RY &QY form 5 sides

a pentagon is any five-sided polygon or 5-gon. The sum of the internal angles in a simple pentagon is 540°

We determined that angle QDE & angle RED both are 90 degrees each.
& from statement one, we know that angle
The sum of the measures of angles DQS and ERT is 260.

Therefore angle DQS + ERT + QDE + RED + RYQ = 540
260 + 180 + RYQ = 540
Hence angle RYQ = 100 degree
Statement one alone is sufficient we can eliminate C & E

Statement two analysis:

Let us consider two triangles QWD & triangle EXR ,

From question stem we know that QDE & angle RDE both are 90 degrees each, therefore angle WDQ & angle XER are 90 degrees each

From statement two, we know that
The sum of the measures of angles WQD and ERX is 100.
If we sum up the two triangles QWD + triangle EXR we get the total sum as 360.
Therefore from question stem and statement two,
Angle WQD+ angle ERX + angle + angle WDQ + angle REX + angle QWD + angle RXE = 360
100+ 180 + angle QWD + angle RXE = 360
There fore angle QWD + angle RXE = 80 ....(1)

Since the above are the vertices of the triangle, and sum of all the measures of a triangle is 180, and from (1) we can determine that Angle WYX = 100 degrees

Hence both the statements are sufficient

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

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1
As JKLM is a rectangle, all 4 angles of JKLM are $$90^0$$ each.

JK//DE
So, $$\angle$$QDE = $$\angle$$DER = $$90^0$$

Statement 1:
$$\angle$$DQS + $$\angle$$ERT = $$260^0$$ ---> eqn 1

We have inferred from the question stem that $$\angle$$QDE = $$\angle$$DER = $$90^0$$

So, $$\angle$$QDE + $$\angle$$DER = $$180^0$$ ---> eqn 2

We can see that DQYRE is a pentagon.
We know that sum of angles of a pentagon is $$(5-2)*180^0 = 540^0$$
Hence, $$\angle$$QDE + $$\angle$$YQD + $$\angle$$RYQ + $$\angle$$ERY + $$\angle$$DER = $$540^0$$ ---> eqn 3

From eqns 1, 2 and 3, we get,
$$\angle$$RYQ = $$\angle$$WYX = $$540^0 - 180^0 - 260^0$$ = $$100^0$$

Statement 1 is sufficient.

Statement 2:
$$\angle$$WQD + $$\angle$$ERX = $$100^0$$ ---> eqn 4

As WQY and XRY are straight lines, $$\angle$$WQY = $$\angle$$XRY = $$180^0$$
So, $$\angle$$WQY + $$\angle$$XRY = $$360^0$$ ---> eqn 5

$$\angle$$WQY = $$\angle$$WQD + $$\angle$$DQY ---> eqn 6
$$\angle$$XRY = $$\angle$$ERX + $$\angle$$ERY ---> eqn 7

From eqns 5, 6 and 7, we get $$\angle$$WQD + $$\angle$$DQY + $$\angle$$ERX + $$\angle$$ERY = $$360^0$$ ---> eqn 8

From eqn 4 and 8, we get:
$$\angle$$DQY + $$\angle$$ERY = $$260^0$$
This is the same as statement 1 which we have already proved to be sufficient.

So, statement 2 is sufficient.

The answer is (D).
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GMAT 1: 670 Q49 V32 Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

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1
(1) The sum of the measures of angles DQS and ERT is 260.

As shown in the attached picture, we need one of the angles of the highlighted pentagon.
We can find it with 180(n-2) formula.
Sufficient.

(2) The sum of the measures of angles WQD and ERX is 100.

This statement again gives the same data as the first one gives.
Sufficient.

Ans should be (D)
Attachments IMG_0750.PNG [ 142.98 KiB | Viewed 139 times ]

Manager  S
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Posts: 93
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

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1
IMO D.

Given: WXY is a triangle. JKLM is a rectangle. WX || JK.
To find: $$\angle$$ WYX.

Statement 1: The sum of the measures of angles DQS and ERT is 260.
$$\angle$$ DQS = $$\angle$$ ERT. Hence, each angle is 130$$^{\circ}$$ --> As they are alternate angles and equal.
$$\angle$$DQW = $$\angle$$ERX and $$\angle$$WDQ = $$\angle$$XER = 90$$^{\circ}$$
$$\angle$$DQW = $$\angle$$ERX = 180-130 = 50$$^{\circ}$$
$$\angle$$ DWQ = $$\angle$$EXR = 180 -( 90+50) = 40$$^{\circ}$$
And therefore, $$\angle$$ WYX = 180 -(40+40) = 100$$^{\circ}$$
Hence, this statement alone is sufficient.

Statement 2: The sum of the measures of angles WQD and ERX is 100.
$$\angle$$ WQD = $$\angle$$ ERX. Hence, each angle is 50$$^{\circ}$$ --> As they are alternate angles and equal.
$$\angle$$WDQ = $$\angle$$XER = 90$$^{\circ}$$
$$\angle$$ DWQ = $$\angle$$EXR = 180 -( 90+40) = 50$$^{\circ}$$
And therefore, $$\angle$$ WYX = 180 -(50+50) = 80$$^{\circ}$$
Hence, this statement alone is sufficient.

Hence, the answer is D.
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In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

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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?

From the given diagram we can say the following:

DQS + WQD + ERT + ERX = 360 (2 pairs of supplementary angles)

WDQ = XER = 90. (perpendicular lines)

(1) The sum of the measures of angles DQS and ERT is 260.

Here, DQS + ERT = 260.

So, WQD + ERX = 100

DWQ = 180 - WDQ - WQD = 90 - WQD (DWQ is angle W in triangle XYW.)

EXR = 180 - XER - ERX = 90 - ERX (EXR is angle X in triangle XYW.)

Now, in the triangle XYW:

WYX = 180- W - X = 180-90 + WQD - 90 + ERX = (WQD + ERX) = 100.

So, first can answer the question.

(2) The sum of the measures of angles WQD and ERX is 100.

Here, WQD + ERX = 100 ....(Same information as first one.)

Since given information is the same, we can infer that second can also answer the question.

(I hope the concept I am trying to explain is clear. I apologize in advance if I have made any mistake in naming the angles. )

Originally posted by Rohan007 on 23 Jul 2019, 06:28.
Last edited by Rohan007 on 23 Jul 2019, 06:31, edited 1 time in total.
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

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1
(1) The sum of the measures of angles DQS and ERT is 260.
Notice that DQYRE forms a pentagon. Sum of interior angles of a pentagon is 540 degrees. Further, QDE and RED are both right angles.
Angle QDE+RED+ERT+DQS+WYX=540 => WYX=540-260-90-90=100. Sufficient

(2) The sum of the measures of angles WQD and ERX is 100.
WQD=SQL;ERX=TRM ---> Vertically Opposite angles
SQL+DQS=180; TRM+ERT=180 --->Supplementary angles
=> DQS+ERT=260, This is similar to (1). Hence, WYX can be determined to be 100 in the same way. --> Sufficient.

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

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

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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

Posted from my mobile device
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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?

(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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1. Not sufficient.
2. In ∆wqd, angle QWD=90-WQD
Similarly, in ∆erx, angle RXE=90-ERX
In ∆WXY, XWY+WXY+WYX=180
=> WYX=100. Sufficient.

Option B.

Posted from my mobile device
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

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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 = (n-2)*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 = (5-2) * 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.

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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?

(1) The sum of the measures of angles DQS and ERT is 260.
SUFF.
(2) The sum of the measures of angles WQD and ERX is 100.
SUFF.

Edit Note: I now see my error and corrected it.
PSA.... I tried to do "math stuffs" in my head on a red eye flight from the Pacfic region to the East coast. With dimmed cabin lights! .... Don't be like me kids.
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Originally posted by duchessjs on 23 Jul 2019, 03:12.
Last edited by duchessjs on 23 Jul 2019, 13:11, edited 2 times in total.
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

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Got it correct.
Sometimes GMAT tricks you to apply brute force while the ques can be solved without even using the pen.
Nice question.
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