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If the sum of two angles of a parallelogram is 100°, what is the avera

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If the sum of two angles of a parallelogram is 100°, what is the average (arithmetic mean) of the measures of the other two angles?

(A) 50
(B) 90
(C) 130
(D) 180
(E) 260
[Reveal] Spoiler: OA

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Re: If the sum of two angles of a parallelogram is 100°, what is the avera [#permalink]

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New post 29 Sep 2017, 02:02
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sum of two angles of a parallelogram is 100°
so sum of the other two angle should be (360-100)/2 = 130
So mean should be (130+130)/2 = 130

Answer should be C

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Bunuel wrote:
If the sum of two angles of a parallelogram is 100°, what is the average (arithmetic mean) of the measures of the other two angles?

(A) 50
(B) 90
(C) 130
(D) 180
(E) 260

The sum of interior angles of a quadrilateral is 360 degrees.

Though we are not sure which two angle measures sum to 100, we are asked only for the average (arithmetic mean) of the other two.

There are (360 - 100) = 260 degrees that remain, and two angles.

\(\frac{260}{2}\) = 130

Answer C

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Re: If the sum of two angles of a parallelogram is 100°, what is the avera [#permalink]

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New post 03 Oct 2017, 16:34
Bunuel wrote:
If the sum of two angles of a parallelogram is 100°, what is the average (arithmetic mean) of the measures of the other two angles?

(A) 50
(B) 90
(C) 130
(D) 180
(E) 260


A parallelogram’s angles always sum to 360 degrees. If the sum of two angles of a parallelogram is 100 degrees, then the remaining 2 angles must sum to 260 degrees, and so the average of those two angles is 260/2 = 130 degrees.

Answer: C
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Re: If the sum of two angles of a parallelogram is 100°, what is the avera   [#permalink] 03 Oct 2017, 16:34
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