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In quadrilateral ABCD, angle A measures 20 degrees more than the avera

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In quadrilateral ABCD, angle A measures 20 degrees more than the avera  [#permalink]

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New post Updated on: 24 Jul 2019, 04:58
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

80% (02:10) correct 20% (01:57) wrong based on 25 sessions

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Originally posted by Bunuel on 09 Apr 2019, 23:39.
Last edited by SajjadAhmad on 24 Jul 2019, 04:58, edited 1 time in total.
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Re: In quadrilateral ABCD, angle A measures 20 degrees more than the avera  [#permalink]

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New post 10 Apr 2019, 01:30
Bunuel wrote:
In quadrilateral ABCD, angle A measures 20 degrees more than the average of the other three angles of the quadrilateral. Then angle A =

(A) 70°
(B) 85°
(C) 95°
(D) 105°
(E) 110°


given
a+b+c+d=360
and b+c+d=360-a
and
a=20+(b+c+d/3)
solve for a
we get a = 105
IMO D
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Re: In quadrilateral ABCD, angle A measures 20 degrees more than the avera  [#permalink]

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New post 10 Apr 2019, 14:49
1
We can rewrite angles A,B,C,D as X.
X, in this case, denotes the average of the angles.
The question states angle A is 20 more than the average of the other three angles (B,C,D).

So we can solve this using algebra.

X + X + X + X+20 = 360

All the angles inside a given quadrilateral always equal 360.

So our above equation can be solved as follows:

4x + 20 = 360
-20 -20
4x=340
divide both sides by 4.
x=85

Angle A = 85 + 20 = 105
GMAT Club Bot
Re: In quadrilateral ABCD, angle A measures 20 degrees more than the avera   [#permalink] 10 Apr 2019, 14:49
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In quadrilateral ABCD, angle A measures 20 degrees more than the avera

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