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

 It is currently 06 Dec 2019, 08:57 ### 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

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 59587
In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags 00:00

Difficulty:   65% (hard)

Question Stats: 69% (02:06) correct 31% (02:30) wrong based on 207 sessions

### HideShow timer Statistics 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: VP.png [ 13.56 KiB | Viewed 1929 times ] This question was provided by Veritas Prep for the Game of Timers Competition _________________
Manager  G
Joined: 08 Jan 2018
Posts: 129
In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

1
2
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

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.
SVP  D
Joined: 03 Jun 2019
Posts: 1875
Location: India
In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

2
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
Attachments 2019-07-22_205841.JPG [ 887.45 KiB | Viewed 1678 times ]

Originally posted by Kinshook on 22 Jul 2019, 08:25.
Last edited by Kinshook on 22 Jul 2019, 08:30, edited 1 time in total.
Manager  S
Joined: 30 Aug 2018
Posts: 104
Location: India
Concentration: Finance, Accounting
GMAT 1: 600 Q49 V23 GPA: 3.36
WE: Consulting (Computer Software)
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

3

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 1665 times ]

Senior Manager  G
Joined: 05 Mar 2017
Posts: 261
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]

### Show Tags

2
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  G
Joined: 30 May 2018
Posts: 157
GMAT 1: 710 Q49 V36 GPA: 3.8
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

1
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 - (180-A(DQY)) - 90 = A(DQY)-90 (see A(WDQ) = 90)
Similarly, in triangle EXR, we have A(X) = 180 - (180-A(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.
Manager  S
Joined: 21 Jan 2019
Posts: 100
In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

1
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 - (90-EXR) = 260
90 + DWQ +90+ EXR = 260
Now from 1,
WYX= 260-180
SUFFICIENT

Statement 2,
WQD+ ERX= 100
90 - DWQ +90-EXR = 100
Now from 1,
180-100 = WYX
SUFFICIENT

Originally posted by techloverforever on 22 Jul 2019, 08:42.
Last edited by techloverforever on 22 Jul 2019, 10:46, edited 2 times in total.
Senior Manager  P
Joined: 27 Aug 2014
Posts: 368
Location: Netherlands
Concentration: Finance, Strategy
Schools: LBS '22, ISB '21
GPA: 3.9
WE: Analyst (Energy and Utilities)
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

1

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= 540-180-260 = 100 deg
from 2: we can determine the sum of DQY and ERT as 260, by 360-100, from there we can determine the angle as 100deg.

so in my opinion D
Manager  G
Joined: 28 Jan 2019
Posts: 127
Location: Peru
In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

2
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.
Director  P
Joined: 04 Sep 2015
Posts: 664
Location: India
WE: Information Technology (Computer Software)
In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

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.
(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*(n-2) where n= sides,
so we know that DERYQ=180*3=540
and QDE and DER each is 90 then QYR or WYX =540-180-260 =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 360-180=180

then angle D + angle X= 80 because wqd+erx=100,

now in triangle WYX , we know that total angle in triangle =180, so 180-80 =100

so angle Y=100.

Sufficeint.

D is correct

Originally posted by abhishekdadarwal2009 on 22 Jul 2019, 08:48.
Last edited by abhishekdadarwal2009 on 22 Jul 2019, 23:16, edited 1 time in total.
Manager  S
Joined: 17 Apr 2018
Posts: 107
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

1
WX || JK

DERTSQ --> Hexagon

For Hexagon DERTSQ
D=E =90 degree
Q+R = 260

Let Q be a degrees ==> in triangle DWQ --> W = a-90 degree similarly
Let angle R be "b" degree --> In Triangle EXR Angle X = b-90
a+b=260

For Hexagon Angle S + T = 180-a+180-b
So we can conclude angle SYT

A is sufficient

B is reduced to the same equation as A

Manager  G
Joined: 10 Mar 2019
Posts: 75
Location: Russian Federation
Schools: Booth, Berkeley Haas
GMAT 1: 730 Q50 V39 GPA: 3.95
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

1
(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.
360-260=100.
Now, angles QLS and RMT each equal 90 and we can calculate the sum of angles LSQ and RTM.
360-180-100=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.
180-80=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.
360-180-100=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.
180-80=100
Sufficient.

IMO D
Director  P
Joined: 24 Nov 2016
Posts: 926
Location: United States
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

1
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.

GMAT Club Legend  V
Joined: 18 Aug 2017
Posts: 5428
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

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
Manager  S
Joined: 01 Oct 2018
Posts: 112
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

2
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

Manager  G
Joined: 26 Jan 2016
Posts: 181
In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

1
DQYRE forms a pentagon.

From 1 & 2 each, we get Angle Q & Angle R
And Angle D and Angle E are right-angled.
Hence we can easily get Angle WYX

Hence D

Originally posted by kitipriyanka on 22 Jul 2019, 09:01.
Last edited by kitipriyanka on 22 Jul 2019, 09:51, edited 1 time in total.
Senior Manager  P
Joined: 12 Dec 2015
Posts: 443
In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

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?
Given:
DQYRE is a pentagon:
So sum of all the interior angles of the pentagon DQYRE= (n-2) * 180=3*180
i.e angle-EDQ + angle-DER + angle-DQY + angle-ERY + angel-QYR = 3*180
=> 90 + 90 + angle-DQS + angle-ERT + angel-WYX = 3*180, where angle-EDQ = angle-DER = 90
=> angle-DQS + angle-ERT + angel-WYX = 2*180 --(i)

(1) The sum of the measures of angles DQS and ERT is 260. --> sufficient: angle-DQS + angle-ERT = 260, replacing the value in equation-(i), angel-WYX = 2*180-260 = 100
(2) The sum of the measures of angles WQD and ERX is 100. --> sufficient: angle-WQD + angle-ERX = 100 =>(180-angle-DQS) + (180-angle-ERT) = 100 => angle-DQS + angle-ERT = 260, replacing the value in equation-(i), angel-WYX = 2*180-260 = 100

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.
Director  P
Joined: 16 Jan 2019
Posts: 507
Location: India
Concentration: General Management
WE: Sales (Other)
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

1
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

Attachments Untitled.png [ 26.55 KiB | Viewed 1520 times ]

Intern  B
Joined: 02 Apr 2019
Posts: 7
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

2
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
Manager  S
Joined: 12 Mar 2019
Posts: 158
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

### Show Tags

1
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 Re: In the diagram above, triangle WXY intersects rectangle JKLM at points   [#permalink] 22 Jul 2019, 09:11

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

Display posts from previous: Sort by

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