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In a six-sided polygon PQRTV ∠P measures 48° less

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Maksym
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In a six-sided polygon PQRTV ∠P measures 48° less [#permalink] New post   06 Jul 2014, 06:35
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In a six-sided polygon PQRSTV ∠P measures 48° less than the average (arithmetic mean) of the remaining five angles. What does ∠P equal to, in degrees?

(A) 72
(B) 80
(C) 88
(D) 96
(E) 104
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B
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manpreetsingh86
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Re: In a six-sided polygon PQRTV ∠P measures 48° less [#permalink] New post   06 Jul 2014, 09:06
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Maksym wrote:
In a six-sided polygon PQRSTV ∠P measures 48° less than the average (arithmetic mean) of the remaining five angles. What does ∠P equal to, in degrees?

(A) 72
(B) 80
(C) 88
(D) 96
(E) 104



sum of all the internal angles of the six sided polygon=720
which can be written as

p+q+r+s+t+v=720 (p,q,r,s,t,v) are the internal angle of the polygon

also we have

p= ((q+r+s+t+v)/5)-48

5p= 720-p-240
6p=480
p=80
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Re: In a six-sided polygon PQRTV ∠P measures 48° less [#permalink] New post   06 Jul 2014, 10:45
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Maksym wrote:
In a six-sided polygon PQRSTV ∠P measures 48° less than the average (arithmetic mean) of the remaining five angles. What does ∠P equal to, in degrees?

(A) 72
(B) 80
(C) 88
(D) 96
(E) 104


The sum of Interior Angles of a polygon is 180(n-2), where n is the number of sides. Check Polygons chapter of our Math Book

Thus, the sum of the angles in a six-sided polygon is 180(6 - 2) = 720°.

The measure of five angles except ∠P is 720 - P. So, we have that P + 48 = (720 - P)/5 --> P = 80°.

Answer: B.
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Re: In a six-sided polygon PQRTV ∠P measures 48° less [#permalink] New post   06 Jul 2014, 11:41
The sum of interior angle in a polygon is (n-2)*180

so for a six sided figure, (6-2)*180 = 720

let the A.M of the other 5 angles be 'm'

--> m*5 + ∠P = 720, we know ∠P=m-48
--> m*5 + m - 48 = 720, solving we get m=128
hence ∠P= 128-48 = 80

Ans B.
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Re: In a six-sided polygon PQRTV ∠P measures 48° less [#permalink] New post   29 Apr 2015, 10:31
Hi Bunuel,

I did it this way:

I said let the average of the 5 angles be x = (q+r+s+t+v)/5
Then the total of those five angles = 5x

Then angle P = x - 48

Now total of 6 angles = 720 deg

So 5x + x - 48 = 720
6x = 768.

What am I doing wrong here ?
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Re: In a six-sided polygon PQRTV ∠P measures 48° less [#permalink] New post   29 Apr 2015, 14:24
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Hi desiwolverine,

Your approach is fine, but you have not completed your work and you have not answered the question that was ASKED.

I'm going to continue on with the work that you've already done....

6X = 768
X = 128

We're asked for Angle P....

Angle P =

X - 48 =

128 - 48 = 80

Final Answer:
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B


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Re: In a six-sided polygon PQRTV ∠P measures 48° less [#permalink] New post   29 Apr 2015, 19:01
Thanks Rich !
Have to remind myself to read the question carefully and answer what is asked.
I extracted the value of X.
Looked at the answer and did not find it and was confused. :(
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PowerMBACLEVIS
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Re: In a six-sided polygon PQRTV ∠P measures 48° less [#permalink] New post   29 Apr 2015, 19:53
One can solve this using the formulae given below :

Sum of interior angles in an N sided polygon= (N-2) * 180

:D
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Re: In a six-sided polygon PQRTV ∠P measures 48° less [#permalink] New post   12 Sep 2020, 22:59
Given that Angle <P = Avg. of 5 Remaining Angles - 48 deg.

<P = (Q + R + S + T + V) / 5 - 48

Since All the Answer Choices for Angle < P are INTEGERS:

1st) Subtract 720 total degrees for the Polygon by the Answer Choice

2nd) See if the Result is Divisible by 5 ---- because the Remaining 5 Angles must have an Average that is an Integer in order for Angle <P to equal an Integer.

Only when the Angle <P = 80 degrees do you get:

Remaining 5 Angles = 720 - 80 = 640 Degrees, which is a Multiple of 5

Answer has to be -B-
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Re: In a six-sided polygon PQRTV ∠P measures 48° less [#permalink]
12 Sep 2020, 22:59
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