Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

The measures of the interior angles in a polygon are [#permalink]

Show Tags

03 Feb 2011, 15:45

7

This post received KUDOS

9

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

67% (02:56) correct
33% (06:56) wrong based on 165 sessions

HideShow timer Statistics

The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8 B) 9 C) 10 D) 11 E) 13

Sum of Interior Angles of a polygon is \(180(n-2)\) where \(n\) is the number of sides (so is the number of angles);

We are told that the smallest angle is 136 degrees, the next will be 136+1 degrees, ..., and the largest one, \(n_{th}\) angle, will be \(136+(n-1)\) degrees. The sum of the \(n\) consecutive integers (the sum of \(n\) angles) is given by \(\frac{first+last}{2}*# \ of \ terms=\frac{136+(136+n-1)}{2}*n=\frac{271+n}{2}*n\);

So we have that \(180(n-2)=\frac{271+n}{2}*n\) --> \(360(n-2)=(271+n)*n\), now try the answer choices: in order RHS to end with zero (as LHS is because of 360) then \(n\), out of the options listed, could be either 10 or 9, \(n=9\) fits.

The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8 B) 9 C) 10 D) 11 E) 13

When I look at this question, I say, "I know how to find the Interior angle of a regular polygon. But this is not a regular polygon since it has angles 136, 137, 138, 139, 140, 141, 142 ....etc."

Karishma, Can you explain in more details when you say "The average of the given angles can only match 140 (such that effectively, all the angles are 140)". I did not get your point. Thanks _________________

Sum of interior angles for n sides: (n-2)*180 And each interior angle are increasing by 1. This can be written as : 136+137+138+139+... = 136+(136+1)+(136+2)+(136+3)+... For n sides

136n+(1+2+3+4....+n-1) 136n+(n-1)n/2 --- Sum of n-1 natural number

Karishma, Can you explain in more details when you say "The average of the given angles can only match 140 (such that effectively, all the angles are 140)". I did not get your point. Thanks

Sum of interior angles of a polygon = (n-2)*180 (not necessarily regular polygon)

Why? See the figure below:

Attachment:

Ques2.jpg [ 5.14 KiB | Viewed 10610 times ]

A 6 sided polygon can be split into 4 triangles each of which has a sum of interior angles 180 degrees. An n sided polygon can be split into n - 2 triangles.

When the polygon is regular, each angle is same so the sum is divided by the number of sides to get the measure of each angle e.g. in a regular hexagon, each interior angle = 4*180/6 = 120 degrees.

Now if I have a hexagon whose angles are 115, 117, 119, 121, 123 and x, what will be the angle x? We can see it in two ways - 1. The sum of all angles should be 4*180 = 720 So 115 + 117 + 119 + 121 + 123 + x = 720 or x = 125

2. The average of the angles should be 120. (Since the sum of the angles is 720 and there are 6 sides) 119 and 121 average out as 120. (119 is 1 less than 120 and 121 is 1 more than 120) 117 and 123 average out as 120. So 115 and x should average out as 120 too. Therefore, x should be 125.

In the question, the average of the given angles of the polygon can only be 140. So it must have 9 sides. To confirm, 136, 137, 138, 139, 140, 141, 142, 143, 144 - 9 angles with average 140. So the polygon must have 9 sides.

(It cannot be 144 or anything else because 10 angles (136, 137, 138, 139, 140, 141, 142, 143, 144, 145) will not give an average of 144) _________________

tough one. simple solution, but its not an ez one. am i right?

Actually, I wouldn't say it is very tough if we know that the sum of interior angles of a polygon is 180(n -2) (I explained above why it is so). We can also use a very straight forward but long approach.

Interior angles of the given polygon: 136, 137, 138, 139, 140, 141, 142, 143....

Using options:

If the polygon had 8 sides, it would have had 8 interior angles. Their sum: (8-2)180 = 1080 Sum of 8 angles: 136 +137 + 138 + 139 + 140 + 141 + 142 + 143 is more than 1080 hence this polygon does not have 8 sides.

If the polygon had 9 sides, it would have had 9 interior angles. Their sum: (9-2)180 = 1260 Sum of 9 angles: 136 +137 + 138 + 139 + 140 + 141 + 142 + 143 + 144 is 1260. Hence this polygon does have 9 sides. _________________

tough one. simple solution, but its not an ez one. am i right?

I wouldn't say it's very hard either, though I wouldn't recommend trial and error in this cases:

... Calculating the sum of the interior angles of a polygon with 7 sides then checking whether 7 consecutive integers starting from 136 add up to that value; Calculating the sum of the interior angles of a polygon with 8 sides then checking whether 8 consecutive integers starting from 136 add up to that value; Calculating the sum of the interior angles of a polygon with 9 sides then checking whether 9 consecutive integers starting from 136 add up to that value;

Whereas equating two formulas and working on answer choices should give an answer in less time: \(180(n-2)=\frac{271+n}{2}*n\) --> \(360(n-2)=(271+n)*n\) --> \(n=9\). _________________

Hey guys...please let me know whether this method is correct or not. the stem says that the angles are consecutive integers, 136 being the smallest one. Also going by the formula of sum of interior angles, we know atleast this much that sum of angles cant be number like 729, 847, 653, 542 but it can be a number of the form xx0. So going by this knowledge, if the smallest angle is 136 then the sum of angles can only be of the form xx0 when the largest angle is 144. That is 9 sides. Yeah the largest angle could be 154, 164 but in that case the number of sides must be 19 and 29 respectively. Since the largest option is 13, hence the answer is 9. Please correct me if I am wrong. _________________

Re: The measures of the interior angles in a polygon are [#permalink]

Show Tags

11 Feb 2014, 09:52

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: The measures of the interior angles in a polygon are [#permalink]

Show Tags

03 Nov 2014, 08:36

Bunuel wrote:

rxs0005 wrote:

The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8 B) 9 C) 10 D) 11 E) 13

Sum of Interior Angles of a polygon is \(180(n-2)\) where \(n\) is the number of sides (so is the number of angles);

We are told that the smallest angle is 136 degrees, the next will be 136+1 degrees, ..., and the largest one, \(n_{th}\) angle, will be \(136+(n-1)\) degrees. The sum of the \(n\) consecutive integers (the sum of \(n\) angles) is given by \(\frac{first+last}{2}*# \ of \ terms=\frac{136+(136+n-1)}{2}*n=\frac{271+n}{2}*n\);

So we have that \(180(n-2)=\frac{271+n}{2}*n\) --> \(360(n-2)=(271+n)*n\), now try the answer choices: in order RHS to end with zero (as LHS is because of 360) then \(n\), out of the options listed, could be either 10 or 9, \(n=9\) fits.

Answer: B.

Hi Bunnel,

In above post, i could not get " in order RHS to end with zero (as LHS is because of 360) then \(n\), out of the options listed, could be either 10 or 9, \(n=9\) fits'. Why are you saying that LHS is 0.

Re: The measures of the interior angles in a polygon are [#permalink]

Show Tags

03 Nov 2014, 08:39

Expert's post

ammuseeru wrote:

Bunuel wrote:

rxs0005 wrote:

The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have?

A) 8 B) 9 C) 10 D) 11 E) 13

Sum of Interior Angles of a polygon is \(180(n-2)\) where \(n\) is the number of sides (so is the number of angles);

We are told that the smallest angle is 136 degrees, the next will be 136+1 degrees, ..., and the largest one, \(n_{th}\) angle, will be \(136+(n-1)\) degrees. The sum of the \(n\) consecutive integers (the sum of \(n\) angles) is given by \(\frac{first+last}{2}*# \ of \ terms=\frac{136+(136+n-1)}{2}*n=\frac{271+n}{2}*n\);

So we have that \(180(n-2)=\frac{271+n}{2}*n\) --> \(360(n-2)=(271+n)*n\), now try the answer choices: in order RHS to end with zero (as LHS is because of 360) then \(n\), out of the options listed, could be either 10 or 9, \(n=9\) fits.

Answer: B.

Hi Bunnel,

In above post, i could not get " in order RHS to end with zero (as LHS is because of 360) then \(n\), out of the options listed, could be either 10 or 9, \(n=9\) fits'. Why are you saying that LHS is 0.

Regards, Ammu

Because the left hand side (LHS) is 360(n-2) = 10*(36(n-2)) = 10*integer = something with the units digit of 0.

Re: The measures of the interior angles in a polygon are [#permalink]

Show Tags

03 Nov 2014, 09:22

Hi Bunnel,

In above post, i could not get " in order RHS to end with zero (as LHS is because of 360) then \(n\), out of the options listed, could be either 10 or 9, \(n=9\) fits'. Why are you saying that LHS is 0.

Regards, Ammu[/quote]

Because the left hand side (LHS) is 360(n-2) = 10*(36(n-2)) = 10*integer = something with the units digit of 0.

Does this make sense?[/quote]

I understood it now. Thank you Actually I had misinterpreted it . I thought you were saying "RHS to end with zero -->(271+n)*n= 0" . Its my Bad. Thanks for clarification.

Re: The measures of the interior angles in a polygon are [#permalink]

Show Tags

26 Nov 2015, 05:06

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

http://blog.ryandumlao.com/wp-content/uploads/2016/05/IMG_20130807_232118.jpg The GMAT is the biggest point of worry for most aspiring applicants, and with good reason. It’s another standardized test when most of us...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...