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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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22 Jul 2019, 14:50
1
Let angle WYX=x;
angle QSL=angle YST=r; and angle STY=angle RTM=p;
This means that x+r+p=180 and x=180-(r+p) ..........(a)

From statement 1, angle DQS + angle ERT =260 .....(b)
angle DQS+angle LQS=180
And angle LQS=90-r
Hence angle DQS+90-r=180
angle DQS=90+r .....(c)
Also
angle ERT=180-angle MRT
And angle MRT=90-p;
hence angle ERT=90+p ...(d)
From (b), (c), and (d)
angle DQS+angle ERT=260
90+p+90+r=260
p+r=80 ...(e)
From (a), x=180-(p+r)=180-80
Hence x=100. Statement 1 alone is sufficient.

2. angle WQD+angle ERX=100 .....(f)
But angle WQD=90-r and angle ERX=90-p
So 90-r+90-p=100
p+r=80 ....(g)
Now, from (a), x=180-(p+r)
Hence x=100.
Therefore statement 2 alone is sufficient. Answer is therefore D.

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

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22 Jul 2019, 16:46
1
(1) <DQS = <XWY + < WDQ = <XWY + 90
<ERT = <WXY + <XER = <WXY + 90
Since <DQS + <ERT = 260 = <XWY + <WXY + 180, <XWY + <WXY = 80.
Since <XWY + <WXY + <WYX = 180, <WYX = 80. Sufficient

(2) <WDQ + <WQD + <XWY = 90 + <WQD + <XWY = 180.
<XER + <WXY + <ERX = 90 + <WXY + <ERX = 180.
Since <ERX + WQD = 100, <XWY + <WXY = 80,
Since <XWY + <WXY + <WYX = 180, <WYX = 100. Sufficient

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

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22 Jul 2019, 17:15
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.

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

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22 Jul 2019, 19:31
1

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

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22 Jul 2019, 19:37
1
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
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

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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.

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

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22 Jul 2019, 20:34
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 line WX is parallel to line JK, thus angle QDE and DER are both equal to 90
now consider a polygon
Sum of internal angle of polygon = (n-2)*120= 3*180 where n = sides of a polygon (YQDER)

USING equation 1
$$\angle Y$$ + $$\angle Q$$+ $$\angle D$$+ $$\angle E$$+ $$\angle R$$ = 540

The sum of the measures of angles DQS and ERT is 260.
AND$$\angle D$$ + $$\angle E$$ = 180 (EACH ANGLE IS OF $$90 ^{\circ}$$

THUS $$\angle Y$$ + 90 +260 + 90 = 540
$$\angle Y$$= 100
SUFFICIENT
2) The sum of the measures of angles WQD and ERX is 100.
THEREFORE angles DQS and ERT = 360 - ( sum of the measures of angles WQD and ERX ) corresponding linear angles
= 260
Now this is same as equation 1 all other parameter are given ,
Hence it is also sufficient

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

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22 Jul 2019, 20:44
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.

Sol:

Considering Statement (1) alone:
Angle DQS + Angle ERT = 260
We know that Angle WQD + Angle DQS = 180
and also Angle ERX + Angle ERT = 180
=> Angle WQD + Angle DQS + Angle ERX + Angle ERT = 360
=> Angle WQD + Angle ERX = 360 - 260 = 100
From triangle WQD and ERX, we have
Angle W + Angle D + Angle Q + Angle E + Angle X + Angle R = 360 (sum of the angles of two triangles)
=> Angle W + 90 + Angle Q + 90 + Angle X + Angle R = 360
=> Angle W + Angle X = 360 - 180 - 100 = 80
Now, in triangle WYX
Angle W + Angle Y + Angle X = 180
=> Angle Y = 180 - 80
=> Angle Y = 100
SUFFICIENT

Considering statement (2) alone:
Angle Q + Angle R = 100

We also know that
Angle W + Angle D + Angle Q + Angle E + Angle X + Angle R = 360 (sum of the angles of two triangles)
Angle W + 90 + 100 + 90 + Angle X = 360
=> Angle W + Angle X = 360 - 180 - 100 = 80
Now, in triangle WYX
Angle W + Angle Y + Angle X = 180
=> Angle Y = 180 - 80
=> Angle Y = 100
SUFFICIENT

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

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22 Jul 2019, 20:46
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?

In triangle WYX , angle (W + Y +X ) = 180
angle Y = 180 -angle (W+X)

(1) The sum of the measures of angles DQS and ERT is 260
angle DQS = 90 + angle W
angle ERT = 90 + angle X

Given DQS + ERT = 260 => 90+W+90+X = 260
W + X = 80. ----> angle Y = 180 -80 = 100-----> Sufficient

(2) The sum of the measures of angles WQD and ERX is 100.
In triangle WQD , angle D is 90 ==> angle W = 90-angle WQD
in triangle ERX , angle E is 90 ===>angle X = 90-angleERX

from triangle WYX , Y = 180- (W+X)
===> Y = 180-( 90-angle WQD + 90 - angle ERX ) = 180 -( 180-100) = 180-80 = 100---> Sufficient

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

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22 Jul 2019, 22:01
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?
rectangle JKLM ; J=90 K=90 L=90 M=90
Line WX is parallel to Line JK
Based on parallel line properties JDW = DJK = 90 = JKE = KEX ;
if JDW = 90 , WDQ = 180-90 = 90
if KEX = 90 , XER = 180-90 = 90
XWY + WXY + WYX = 180
What is WYX?

(1) The sum of the measures of angles DQS and ERT is 260.
DQS+ERT=260 then DQW+ERX = 360-260 = 100
WDQ=XER=90 (as given)
XWY+WXY = 360 - (WDQ+XER) - (DQW+ERX)
XWY+WXY = 360 - (90+90) - 100 = 360-180-100 = 80
WYX = 180- (XWY+WXY) = 180-80 = 100
Sufficient

(2) The sum of the measures of angles WQD and ERX is 100.
This is exactly what we found in the stmt 1 as step 2 and its directly provided here ,then it follows the same reasoning as above
DQW+ERX = 360-260 = 100
WDQ=XER=90 (as given)
XWY+WXY = 360 - (WDQ+XER) - (DQW+ERX)
XWY+WXY = 360 - (90+90) - 100 = 360-180-100 = 80
WYX = 180- (XWY+WXY) = 180-80 = 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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22 Jul 2019, 22:08
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?

JKLM is a rectangle and WX is parallel to line JK, therefore angle QDE = angle RED = 90 degree.
Sum of all the angles of a pentagon = 540 degrees
QDERY is a pentagon.
If we know the sum of the angles QDE, RED, DQY, ERY then we can find the angle QYR i.e WYX = 540 - (sum of rest of the angles)
QDE+RED = 180. Therefore we only need to know DQY+ERY to find the angle QYR.

(1) The sum of the measures of angles DQS and ERT is 260.
statement 1 gives us angle DQS + angle ERT = 260
i.e. DQY + ERY = 260.
Sufficient.

(2) The sum of the measures of angles WQD and ERX is 100.
WQD+DQY = 180 (straight line)
ERX + ERY = 180 (straight line)
therefore DQY+ERY = 360 - WQD+ERX.
as per the statement WQD+ERX = 100 therefor DQY+ERY=360 -100 = 260
therefore angle WYX can be found.
Sufficient.

Both statements are individually sufficient.

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

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22 Jul 2019, 22:09
1
(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.

Answer is D both are sufficient

This is because you need to find the sum of angles of W and X and subtract it from 180 to find angle Y. Since angles D and E are right angles, both statement above gives information about the sum of angle W and X independently. Hence you can deduce Y

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

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Updated on: 22 Jul 2019, 23:58
1
From Question stem we have DE ‖ JK and ∠QDE = ∠RED = ∠J =∠K =∠M =∠L = 90⁰.
Statement 1) ∠DQS+∠ERT=260⁰.
Since we have sum values of 180⁰ and 260⁰ considering polygon DERTSQ, we may find the ∠WYX. Let’s see:
For ∆YST: ∠SYT + ∠YTS + ∠TSY = 180⁰
 ∠SYT = 180⁰ - ∠YTS - ∠TSY
 ∠SYT = 180⁰ - ∠RTM - ∠LSQ (As ∠YTS = ∠RTM and ∠TSY = ∠LSQ : Opposite ∠’s)
 ∠SYT = 180⁰ - (180⁰ - 90⁰ - ∠TRM) – (180⁰ - 90⁰ - ∠SQL) (As both ∆QLS and ∆TMR are right angled triangles and ∠L and ∠M respectively)
 ∠SYT = ∠TRM + ∠SQL
 ∠SYT = 180⁰ - ∠ERT + 180⁰ - ∠DQS (As ∠ERT + ∠MRT = 180⁰ = ∠DQS + ∠SQL)
 ∠SYT = 180⁰ + 180⁰ - (∠ERT + ∠DQS)
 ∠SYT = 180⁰ + 180⁰ - 260⁰ (From Statement 1)
 ∠SYT = 100⁰
SUFFICIENT
Statement 2) ∠WQD + ∠ERX = 100⁰
METHOD – A:
For ∆WYX: ∠WYX + ∠YWX + ∠WXY = 180⁰ OR
∠Y + ∠W + ∠X = 180⁰ → Eqn. ①
For ∆WQD: ∠WQD + ∠QWD + ∠WDQ = 180⁰
 ∠WQD = 180⁰ - (∠QWD + 90⁰) (As DE ‖ JK, ∠WDQ = ∠QDE =∠DJK = 90⁰)
 ∠WQD = 90⁰ - ∠ QWD → Eqn. ②
For ∆ERX: ∠ERX + ∠RXE + ∠XER = 180⁰
 ∠ERX = 180⁰ - (∠RXE + 90⁰) (As DE ‖ JK, ∠XER = ∠RED =∠EKJ = 90⁰)
 ∠ERX = 90⁰ - ∠RXE → Eqn. ③
Using Eqn. ② and ③ ∠WQD + ∠ERX = 100⁰ becomes
90⁰ - ∠QWD + 90⁰ - ∠RXE = 100⁰
 ∠QWD + ∠RXE = 80⁰ OR ∠W + ∠X = 80⁰ → Eqn. ④
Now from Eqn. ③ and ④
∠Y = 180⁰ - (∠W + ∠X)
 ∠Y = 180⁰ - 80⁰
 ∠Y = 100⁰
METHOD B:
For ∆SQL: ∠SQL + ∠QLS + ∠LSY = 180⁰ → Eqn. ①
For ∆MRT: ∠MRT + ∠RTM + ∠TMR = 180⁰ → Eqn. ②
For ∆SYT: ∠SYT + ∠YTS + ∠TSY = 180⁰
 ∠SYT + ∠RTM + ∠LSQ = 180⁰ (Opposite ∠’s: ∠RTM = ∠YTS & ∠LSQ = ∠TSY)
 ∠SYT + 180⁰ - 90⁰ - ∠MRT + 180⁰ - 90⁰ - ∠LQS = 180⁰ (From Eqn. ① and ②)
 ∠SYT - ∠ERX - ∠DQW = 0 (Opposite ∠’s: ∠ERX = ∠MRT & ∠DQW = ∠LQS)
 ∠SYT = ∠ERX + ∠DQW
 ∠SYT = 100⁰

SUFFICIENT

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Originally posted by lnm87 on 22 Jul 2019, 22:44.
Last edited by lnm87 on 22 Jul 2019, 23:58, 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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22 Jul 2019, 22:48
1
Given,

WX || JK

To find,

mLWYX

Lets consider the options,

Option 1: mLDQS + mLERT = 260

Now sum of all the interior angles of quadrilateral DQSTRE is given by

=> mLEDQ + mLDQS + mLQST + mLSTR + mLTRE + mLRED
=> 90 + (mLDQS + mLTRE) + mLQST + mLSTR + 90
=> 180 + 260 + mLQST + mLSTR
=> 440 + mLQST + mLSTR

But sum of the angles of an n-gon is given by 180 * (n - 2) where n is the number of sides in the n-gon

Hence,

440 + mLQST + mLSTR = 720
mLQST + mLSTR = 280

Now,

mLSYT + mLYST + mLYTS = 180
=> mLSYT + (180 - mLQST) + (180 - mLSTR) = 180
=> mLSYT + 360 - (mLQST + mLSTR) = 180
=> mLSYT + 360 - 280 = 180
=> mLSYT = 100

Hence option 1 is sufficient.

Option 2: mLWQD + mLERX = 100

=> mLLQS + mLTRM = 100

Now,

mLLQS + mLQLS + mLLSQ + mLTRM + mLRMT + mLMTR = 360
=> (mLLQS + mLTRM) + 90 + mLYST + 90 + mLYTS = 360
=> 100 + 180 + mLYST + mLYTS = 360
=> mLYST + mLYTS = 80
=> mLSYT = 100

Hence option 2 is sufficient.

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

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22 Jul 2019, 22:53

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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22 Jul 2019, 23:11
1
Statement1:
<DQS +<ERT = 260.
-->Let's say that <DQS=a and <ERT=b --> a+b=260

Triangle WDQ is right-angled triangle --> <DQW=180-a
------------------------------------<DQW+<W=90 -->180-a+<W=90 --> <W=a-90 (keep it)

Triangle REX is right-angled triangle --> <ERX=180-b
------------------------------------<ERX+<X=90 -->180-b+<X=90 --> <X=b-90 (keep it)

As we know, Triangle WYX is a big triangle and <W+<X+<Y=180
--------------------------------------- a-90+b-90+<Y=180 --> <Y=180+90+90-260=100
Sufficient

Statement2: <WQD+<ERX =100. (<WQD=a, <ERX=b--> a+b=100)
As above mentioned, triangle WDQ is right-angled triangle.
--> <W=90-a

Triangle REX is right-angled triangle.
--> <X=90-b

Triangle WYX is a big triangle and <W+<X+<Y=180
--> 90-a+90-b+<Y=180
<Y=a+b=100
Sufficient

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

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23 Jul 2019, 00:02
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.
Angles DQS and LQS are complementary. LQS = 180 - DQS
In triangle QSL, QSL = 90 - LQS = DQS - 90
QSL = YST = DQS - 90

Similarly, STY = ERT - 90

In triangle STY, anglye WYX = STY = 180 - (YST + TYS) = 180 - (DQS - 90 + ERT - 90)
=360 - (DQS + ERT) = 360 - 260 = 100

Sufficient.

(2) The sum of the measures of angles WQD and ERX is 100.
WQD + ERX = 100
WQD = LQS.
QSL = YST = 90 - LQS = 90 - WQD.
Similarly, STY = 90 - ERX

WYX = STY = 180 - (YST + TYS) = 180 - (90 - WQD + 90 - ERX)
= 180 - 180 +(WQD + ERX)
= 100
Sufficient.

Hence each statement is sufficient alone.

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

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23 Jul 2019, 00:14
1
Let's consider A
The sum of the measures of angles DQS and ERT is 260.
Hence The sum of the measures of angles WQD and ERX is 100.
Which means sum of DWY and EXY 80
Hence WYX = 100
Sufficient

B
The sum of the measures of angles WQD and ERX is 100.
Which means sum of DWY and EXY 80
Hence WYX = 100
Sufficient

Hence answer is D

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

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Updated on: 23 Jul 2019, 00:39
1
Please refer to the diagram below
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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.
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Re: In the diagram above, triangle WXY intersects rectangle JKLM at points  [#permalink]

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23 Jul 2019, 00:37
1
As triangle WXY intersects rectangle JKLM at points D, E, Q, R, S, and T. If line WX is parallel to line JK then:

(1) The sum of the measures of angles DQS and ERT is 260.
DQS + ERT =260
WX || JK : then, DER = EDQ = 90
In Petagon : D + E + R + Y + Q = 540
=> Y = 540 - 90 - 90 - 260
=> Y = 540 - 440
=> Y = 100
Sufficient.

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

WQD + ERX = 100
Let WQD = a , then ERX =100 - a
In Traingle WQD => W = 180 - WDQ - WQD = 180 - 90 - a
In triangle ERX => X = 180 - XER - ERX = 180 - 90 - 100 - a
In triangle WXY: Y = 180 - WQD - ERX = 180 - (90 -a) - (a-10) = 180 - 90 +a - a +10 = 100
Sufficient.

IMO the answer is D.

Please hit kudos if you like the solution.
Re: In the diagram above, triangle WXY intersects rectangle JKLM at points   [#permalink] 23 Jul 2019, 00:37

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

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