Last visit was: 25 Jul 2024, 09:11 It is currently 25 Jul 2024, 09:11
Toolkit
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

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

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

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94616
Own Kudos [?]: 643822 [13]
Given Kudos: 86753
Manager
Joined: 08 Jan 2018
Posts: 73
Own Kudos [?]: 239 [3]
Given Kudos: 374
GMAT Club Legend
Joined: 03 Jun 2019
Posts: 5313
Own Kudos [?]: 4251 [2]
Given Kudos: 161
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Intern
Joined: 30 Aug 2018
Posts: 45
Own Kudos [?]: 60 [3]
Given Kudos: 111
Location: India
Concentration: Finance, Accounting
GMAT 1: 600 Q49 V23
GMAT 2: 650 Q49 V29
GPA: 3.36
WE:Consulting (Computer Software)
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
3
Kudos

Clearly A is sufficient.
IF a is sufficient B is also sufficient as B is a subset of A
Attachments

IMG_20190722_205731.jpg [ 114.41 KiB | Viewed 6257 times ]

Manager
Joined: 05 Mar 2017
Posts: 179
Own Kudos [?]: 176 [2]
Given Kudos: 687
Location: India
Concentration: General Management, Marketing
GPA: 3.6
WE:Marketing (Hospitality and Tourism)
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
2
Kudos
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.
Manager
Joined: 28 Jan 2019
Posts: 178
Own Kudos [?]: 257 [2]
Given Kudos: 130
Location: Peru
In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
2
Kudos
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= x-90 and then since LSQ and YST are opposite angles, they are equal.

Also, TRM = 180 -y, and MRT= y-90 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:

x-90+ y-90 + 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.

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.
Intern
Joined: 01 Oct 2018
Posts: 26
Own Kudos [?]: 73 [3]
Given Kudos: 104
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
3
Kudos
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*(5-2) = 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

Intern
Joined: 02 Apr 2019
Posts: 3
Own Kudos [?]: 4 [2]
Given Kudos: 87
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
2
Kudos
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"
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,
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
Senior Manager
Joined: 22 Nov 2018
Posts: 430
Own Kudos [?]: 496 [2]
Given Kudos: 292
Location: India
GMAT 1: 640 Q45 V35
GMAT 2: 740 Q49 V41
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
2
Kudos
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.
Director
Joined: 27 Oct 2018
Status:Manager
Posts: 673
Own Kudos [?]: 1901 [2]
Given Kudos: 200
Location: Egypt
GPA: 3.67
WE:Pharmaceuticals (Health Care)
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
2
Kudos
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
Attachments

VP.png [ 14.33 KiB | Viewed 4489 times ]

Manager
Joined: 26 Mar 2019
Posts: 63
Own Kudos [?]: 104 [1]
Given Kudos: 142
Location: Azerbaijan
Concentration: Finance, Strategy
GMAT 1: 730 Q50 V38
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
1
Kudos
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.

Retired Moderator
Joined: 19 Oct 2018
Posts: 1868
Own Kudos [?]: 6668 [1]
Given Kudos: 705
Location: India
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
1
Kudos
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.

∠KJL=∠XDL=∠WDQ =90
∠JKM= ∠WEM=∠XER=90

Also, ∠XWY+∠WYX+∠WXY=180

∠WYX=180-(∠WXY+∠XWY)

Hence if we know the value of (∠WXY+∠XWY), we can easily find ∠WYX.

Statement 1- The sum of the measures of angles DQS and ERT is 260.
Hence DQW+ERX=360-260=100
(∠WXY+∠XWY)= 180-100=80

Sufficient

Statement 2- The sum of the measures of angles WQD and ERX is 100.
(∠WXY+∠XWY)= 180-100=80

Sufficient

IMO D
Intern
Joined: 10 May 2018
Posts: 6
Own Kudos [?]: 15 [1]
Given Kudos: 14
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
1
Kudos

Since JK is parallel to WX, then QDW and REX must be 90 degrees. Since angles in a triangle must add up to 180, we only need DWQ and EXR to solve for WYX. We don't even need to know the individual angles, just the collective sum of DWQ and EXR.
1.The sum of the measures of angles DQS and ERT is 260.
You can figure out DQW and ERX with this since both ERT and DQS are on a straight line. Together, they must equal 360 - 260 =100 (each straight line must equal 180, since there are two, add together to be 360 and then subtract 260 for the sum of the angles). Since you know DQW and XRE equals 100 and you know WDQ and XER are right angles (90 each), you know that DWQ and EXR must equal 360-280 = 80 (since you're calculating two triangles, the sum must be 180+180 =360 and 280 comes from 90+90+100 from above). If those two angles equal 80, then WYX must be 100 so that triangle WXY equals 180.
Sufficient
2.The sum of the measures of angles WQD and ERX is 100.
This gives the same information as above but one less step
Sufficient

D
Director
Joined: 30 Sep 2017
Posts: 943
Own Kudos [?]: 1285 [1]
Given Kudos: 402
GMAT 1: 720 Q49 V40
GPA: 3.8
In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
1
Kudos
Question: $$\angle{WYX}$$ ?

I believe my word explanation is sufficiently clear to understand, but anyway you can find attached sketch for clarity.

Line $$WX$$ is $$parallel$$ to line $$JK$$ --> Line $$WX$$ is $$perpendicular$$ to both lines $$JL$$ and $$KM$$.

Statement (1): $$\angle{DQS} + \angle{ERT} = 260.$$
Draw line $$YZ$$ that intersects line $$JK$$ at point $$Z$$ and is $$perpendicular$$ to line $$JK$$ --> Line $$YZ$$ is $$parallel$$ to both lines $$JL$$ and $$KM$$.

- $$YZ$$ is $$parallel$$ to $$JL$$, $$then$$ $$\angle{DQS} + \angle{SYZ} = 180$$ --> $$\angle{SYZ} = 180 - \angle{DQS}$$
- $$YZ$$ is $$parallel$$ to $$KM$$, $$then$$ $$\angle{ERT} + \angle{TYZ} = 180$$ --> $$\angle{TYZ} = 180 - \angle{ERT}$$.

$$\angle{WYX} = \angle{SYZ} + \angle{TYZ}$$ --> $$\angle{WYX} = (180 - \angle{DQS}) + (180 - \angle{ERT})$$ --> $$\angle{WYX} = 360 - (\angle{DQS} + \angle{ERT}) = 360 - 260 = 100$$
Statement (1) is SUFFICIENT

Statement (2): $$\angle{WQD} + \angle{ERX}$$ = 100.
Draw line $$YZ$$ that intersects with line $$JK$$ at point $$Z$$ and is $$perpendicular$$ to line $$JK$$ --> Line $$YZ$$ is $$parallel$$ to both lines $$JL$$ and $$KM$$.

- $$YZ$$ is $$parallel$$ to $$JL$$, $$then$$ $$\angle{WQD} = \angle{SYZ}.$$
- $$YZ$$ is $$parallel$$ to $$KM$$, $$then$$ $$\angle{ERX} = \angle{TYZ}.$$

$$\angle{WYX} = \angle{SYZ} + \angle{TYZ}$$ --> $$\angle{WYX} = \angle{WQD} + \angle{ERX}$$ --> $$\angle{WYX} = 100$$
Statement (2) is SUFFICIENT

ANSWER IS (D) - Each statement alone is sufficient
Attachments

GMAT Quant Question.jpeg [ 69.76 KiB | Viewed 4410 times ]

Current Student
Joined: 30 May 2019
Posts: 118
Own Kudos [?]: 254 [1]
Given Kudos: 1696
Location: Tajikistan
Concentration: Finance, General Management
Schools: Simon '24 (A)
GMAT 1: 610 Q46 V28
GMAT 2: 730 Q49 V40 (Online)
GPA: 3.37
WE:Analyst (Consulting)
In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
1
Kudos
Please refer to the diagram below
Attachments

photo5310010120249388064.jpg [ 140.37 KiB | Viewed 4273 times ]

Originally posted by mira93 on 23 Jul 2019, 00:37.
Last edited by mira93 on 23 Jul 2019, 00:39, edited 1 time in total.
SC Moderator
Joined: 25 Sep 2018
Posts: 1100
Own Kudos [?]: 2247 [1]
Given Kudos: 1665
Location: United States (CA)
Concentration: Finance, Strategy
GPA: 3.97
WE:Investment Banking (Investment Banking)
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
1
Kudos
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

Ross School Moderator
Joined: 07 Dec 2018
Posts: 101
Own Kudos [?]: 108 [1]
Given Kudos: 111
Location: India
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
1
Kudos
(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 4327 times ]

Non-Human User
Joined: 09 Sep 2013
Posts: 34093
Own Kudos [?]: 853 [0]
Given Kudos: 0
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points [#permalink]
Moderator:
Math Expert
94614 posts