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CAT 2007: Quantitative Questions a Day 1 to 50 - The discussions
Quantitative Questions and Answers Discuss Quantitative and other Math related questions. Post your math doubts and get it solved by the smartest brains this side of the universe !

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Jaskiran Jaskiran is offline
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Re: CAT 2007: Quantitative Questions a Day 1 to 50 - The discussions - 14-05-2007, 10:48 AM

each angle = (p-2)180/p => 180 - 360/p => for angle to b integer p needs 2 b a factor of 360
Total factors of 360(2^3 * 3^2 * 5^1)= 24
p cnt b 1 or 2 so left are 24-22 => ans is 22


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Re: CAT 2007: Quantitative Questions a Day 1 to 50 - The discussions - 14-05-2007, 11:03 AM

Me also getting 22.

Quote:
Originally Posted by Destiny's_Child View Post
For a regular polygon of side p, each angle = (2p - 4)*90/p
= 2p/p - 360/p
= 2 - 360/p
That should be
=2p/p*90-360/p
=2*90-360/p

I presume it was just an error while writing
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Re: CAT 2007: Quantitative Questions a Day 1 to 50 - The discussions - 14-05-2007, 12:11 PM

Hi All ,
I have joined as a membr from 6th may'07 .Pls tell me how to receive the correct answers of the previous posts to check up with my caln.

Cheers,
Aritra
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Re: CAT 2007: Quantitative Questions a Day 1 to 50 - The discussions - 14-05-2007, 12:20 PM

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Originally Posted by aritra13 View Post
Hi All ,
I have joined as a membr from 6th may'07 .Pls tell me how to receive the correct answers of the previous posts to check up with my caln.
I didn't get you what you are exactly trying to ask.??:
Are you asking for solutions for all the previous questions i.e from Q1-Q29.Then you have to go through the whole thread starting from Page1 where you will find members posting there different ways of solving a given problem.But if you are asking for the solutions of questions from 6th May then you have to subscribe the newsletter(I here don't know whether you have subscribed or not) and the solutions will be provided to your mailbox.
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Re: CAT 2007: Quantitative Questions a Day 1 to 50 - The discussions - 14-05-2007, 12:23 PM

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Originally Posted by aritra13 View Post
Hi All ,
I have joined as a membr from 6th may'07 .Pls tell me how to receive the correct answers of the previous posts to check up with my caln.

Cheers,
Aritra
hey the answer comes everyday along with the news letter ... i hope u have subscribed to the QQAD .. if not put ur emailid here http://www.pagalguy.com/index.php?categoryid=65
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Re: CAT 2007: Quantitative Questions a Day 1 to 50 - The discussions - 14-05-2007, 12:34 PM

Quote:
Originally Posted by STALWART View Post
------------------------------------------------------------
Quantitative Question # 29
------------------------------------------------------------
Question:


Consider a regular polygon of p sides .The number of values of p for which the polygon will have angles whose values in degrees can be expressed in integers?

(a) 24 (b) 23 (c ) 22 (d) 20 (e)21


Sorry , Replying late as usual .......

Answer -- 22

For any regular polygon having " p " number of sides , the sum of all the interior angles = ( p-2 ) * 180 = ( 180*p ) - 360 ......... (1)

Now the for finding out the value of each angle = [ (180*p) - 360 ] / p

==>> 180 - ( 360/p ) ......... (2)


For any Number N = (a^p) * ( b^q) * ( c^r ) ........
Where a , b , c , ...... are all prime numbars .
Number of all factors of N ( including 1 and the number itself ) = (p+1)(q+1)(r+1)......

Here 360 = (2^3) * (3^2) * (5^1)
So total number of factors = 4 * 3 * 2 = 24 ....... (3)

But out of them p=1,2 are not acceptable as then in those cases value of (2) will come in negative or 0 .

Or Just imagine , Can we have any polygon which have 1 or 2 number of side/sides ? Impossible !!

So required Number of solutions = 24 - 2 = 22

Answer = 22

..... Sayonara



same slightly different

any angle will be given by (p-2)*180/p

now if 180 is multiple of p then we have integers

180 = 3^2*2^2*5 = 18 factors

but we can have case when we have exhausted all the 2's of 180 but as p-2 and p have a diff of 2 we can cancel there two

so take out 2^2 we are left with 3^2*5
which give 6 factors

so total 24

bhai log ab ye batau where i have counted more


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Re: CAT 2007: Quantitative Questions a Day 1 to 50 - The discussions - 14-05-2007, 12:44 PM

my answer is 22.

explanation same.

each interior angle will be (2p-4)*90/p

boils down to 180-360/p.

number of factors of 360 is 24 but we have to leave p=1,2.

so answer will be 22.
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Re: CAT 2007: Quantitative Questions a Day 1 to 50 - The discussions - 14-05-2007, 02:01 PM

Right answer is indeed 22 . Good show people.

Now this was another typical Cat types .


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Re: CAT 2007: Quantitative Questions a Day 1 to 50 - The discussions - 14-05-2007, 02:18 PM

Quote:
Originally Posted by warrior View Post
same slightly different

any angle will be given by (p-2)*180/p

now if 180 is multiple of p then we have integers

180 = 3^2*2^2*5 = 18 factors

but we can have case when we have exhausted all the 2's of 180 but as p-2 and p have a diff of 2 we can cancel there two

so take out 2^2 we are left with 3^2*5
which give 6 factors

so total 24

bhai log ab ye batau where i have counted more
I dint follow your method....
Can you please explain.....
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Re: CAT 2007: Quantitative Questions a Day 1 to 50 - The discussions - 14-05-2007, 03:01 PM

Quote:
Originally Posted by amit_vce View Post
I am coming up with answer as 75.43 .

Reason , The trajectory of torpedo is a semi circle with radius 20. The distance it has to cover is pi*R/2 = pi*20 /2 = 10*pi.

so time it will take is = 3600 *10*pi /1500 = 75.43 .

this soln is perfect simple and strt
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