GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 Sep 2019, 06:22

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 25 Sep 2018
Posts: 411
Location: United States (CA)
Concentration: Finance, Strategy
GMAT 1: 640 Q47 V30
GPA: 3.97
WE: Investment Banking (Investment Banking)
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

23 Jul 2019, 05:23
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
_________________
Why do we fall?...So we can learn to pick ourselves up again
Intern
Joined: 24 Mar 2018
Posts: 48
Location: India
Concentration: Operations, Strategy
Schools: ISB '21
WE: Project Management (Energy and Utilities)
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

23 Jul 2019, 05:35
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).
Manager
Joined: 07 Dec 2018
Posts: 112
Location: India
Concentration: Technology, Finance
GMAT 1: 670 Q49 V32
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

23 Jul 2019, 05:37
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
Joined: 18 Feb 2017
Posts: 93
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

23 Jul 2019, 05:47
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.
Manager
Joined: 24 Jan 2019
Posts: 106
Location: India
Concentration: Strategy, Finance
GMAT 1: 730 Q51 V38
GPA: 4
In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

Updated on: 23 Jul 2019, 06:31
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. )

ANSWER : D

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.
Manager
Joined: 12 Jan 2018
Posts: 112
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

23 Jul 2019, 06:30
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
_________________
"Remember that guy that gave up?
Neither does anybody else"
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 4768
Location: India
Concentration: Sustainability, Marketing
Schools: INSEAD, HEC '22, IIM
GPA: 4
WE: Marketing (Energy and Utilities)
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

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
_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.
Manager
Joined: 06 Aug 2018
Posts: 98
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

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

Posted from my mobile device
Intern
Joined: 26 May 2018
Posts: 45
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

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
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=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
_________________
Thank you for the kudos. You are awesome!
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 = (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.

Correct answer A
Intern
Joined: 14 Mar 2017
Posts: 40
Location: United States (VA)
GPA: 2.9
WE: Science (Pharmaceuticals and Biotech)
In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

Updated on: 23 Jul 2019, 13:11
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.
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.
*** The more you know ***

Correct Answer: D
_________________
"What is my purpose?"
"You pass butter."
"Oh my god..."
"Yeah, welcome to the club, pal."

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.
Senior Manager
Joined: 10 Aug 2018
Posts: 316
Location: India
Concentration: Strategy, Operations
WE: Operations (Energy and Utilities)
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

23 Jul 2019, 08:44
Got it correct.
Sometimes GMAT tricks you to apply brute force while the ques can be solved without even using the pen.
Nice question.
_________________
Don't forget to give KUDOS.
On the way to conquer the GMAT and I will not leave it until I win. WHATEVER IT TAKES.
Target 720+

" I CAN AND I WILL"

Your suggestions will be appreciated here---> https://gmatclub.com/forum/your-one-advice-could-help-me-poor-gmat-scores-299072.html

1) Gmat prep: [620 Q48 V27], 2) Gmat prep: [610 Q47 V28], 3) Gmat prep: [620 Q47 V28],
4) Gmat prep: [660 Q47 V34], 5) Gmat prep: [560 Q37 V29], 6) Gmat prep: [540 Q39 V26],
7) Veritas Cat: [620 Q46 V30], 8) Veritas Cat: [630 Q45 V32]
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points   [#permalink] 23 Jul 2019, 08:44

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

Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne