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In ∆ PQR above, w =

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Expert Post
Math Expert
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V
Joined: 02 Sep 2009
Posts: 41884

Kudos [?]: 128932 [0], given: 12183

In ∆ PQR above, w = [#permalink]

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New post 17 Aug 2017, 23:47
Expert's post
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Difficulty:

  25% (medium)

Question Stats:

100% (00:31) correct 0% (00:00) wrong based on 36 sessions

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Kudos [?]: 128932 [0], given: 12183

Director
Director
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G
Joined: 18 Aug 2016
Posts: 511

Kudos [?]: 140 [0], given: 123

GMAT 1: 630 Q47 V29
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Re: In ∆ PQR above, w = [#permalink]

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New post 18 Aug 2017, 01:07
Bunuel wrote:
Image
In ∆ PQR above, w =

(A) 50
(B) 55
(C) 60
(D) 65
(E) 75

[Reveal] Spoiler:
Attachment:
2017-08-18_1028.png

Small Triangle
Sum of Two unknown angles = 180-140= 40
In bigger triangle PQR
Sum of all angles = 180 = w+35 + 40 + sum of two unknown angles of smaller triangle
180 = w+35+40+40
w=65
D
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Kudos [?]: 140 [0], given: 123

Intern
Intern
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B
Joined: 13 Jul 2015
Posts: 21

Kudos [?]: 1 [0], given: 1

Re: In ∆ PQR above, w = [#permalink]

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New post 18 Aug 2017, 02:12
Ans- D -65

Let the unknwon angles be x and y ,

Considering the smaller traingle , x+y+ 140 = 180.
x+y = 40

Considering the bigger triangle , 35+x+40+y+w = 180 .

Substituting hte value of x+y , w = 65.

Kudos [?]: 1 [0], given: 1

Intern
Intern
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B
Joined: 17 Apr 2017
Posts: 9

Kudos [?]: 1 [0], given: 11

Location: India
WE: Analyst (Consulting)
Re: In ∆ PQR above, w = [#permalink]

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New post 19 Aug 2017, 03:24
Let the Inner triangles unknown angles be x and y,

1. For inner triangle,
x+y+140=180 (sum of all angles in a triangle)

x+y=40 --(1)

2. For triangle PQR,
Sum of all angles = 35+x+y+40+w=180
using (1)
35+40+40+w=180
w=65

Kudos [?]: 1 [0], given: 11

Re: In ∆ PQR above, w =   [#permalink] 19 Aug 2017, 03:24
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In ∆ PQR above, w =

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