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by envyramana » Thu Jul 10, 2008 5:46 am
Each voting district can be divided with 12000 people. Plug in the values and see the results.
Ex: if a district is reduced to 10900, the next min highest value is 12110 (if we distribute this 1100 among other 10 districts), but 11990 is the limit for 10%.

similarly, if a district is reduced to 11000, the next min highest value is 12100 (if we distribute this 1000 among other 10 districts), hence this is the least possible value for 10%.

Hence Ans 11000

Let me know if my understanding is not correct.
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Any more solutions?

by evansbd » Fri Jul 11, 2008 12:57 pm
This problem 'seems' simple but are there any other insights into solving it?

I got a different answer which is undoubtedly wrong, but I wanted to get opinions the insight needed to solve this problem.
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easy solution

by anksbhandari » Tue Jul 15, 2008 7:17 am
Firstly, think when the population of a town can be least ?
Ans. when the population of all other towns is maximum.
i.e. X (population of the least populated town) + 10% of X= 1.1X

therefore the population of all the 10 towns will be 10*1.1X= 11X

now total population is 11X + x = 12X = 13200

So, X = 13200/12 = 11000 Ans
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by sudhir3127 » Tue Jul 15, 2008 9:07 am
can anyone solve the second question in the attachment..
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by parallel_chase » Tue Jul 15, 2008 9:20 am
sudhir3127 wrote:can anyone solve the second question in the attachment..
Ques 2.


n is between 10 - 99 is n<80 ?


Statement I - sum of the 2 digits of the N is prime,
N could be 11, 21, 41, 47 or 83. NOT SUFFICIENT.


Statement II - each of the two digits of the N is prime.

23, 57, 72, anything less than 77, because over 77 both digits of the number cannot be prime.

Hence B.

Let me know if you have any problem in understanding the above reasoning.
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by Delph » Tue Jul 15, 2008 9:21 am
sudhir3127 wrote:can anyone solve the second question in the attachment..
Each of the 2 digits of n is a prime number. For all integers from 80 to 99 there will be either 8 or 9 as one of the digits. Neither 8 not 9 is a prime number. Thus the answer is yes - n<80.
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by sudhir3127 » Tue Jul 15, 2008 9:33 am
can some one solve this...
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by parallel_chase » Tue Jul 15, 2008 9:58 am
sudhir3127 wrote:can some one solve this...
Arc abc = 24

The triangle is an equilateral triangle, angle A=B=C = 60

Therefore, arc AC = 60/360 * 2 pi r = 1/3 pi r

and, AC + 24 = 2 pi r

substitute the value of AC here

1/3 pi r + 24 = 2 pi r => 72 = 5 pi r

pi = 3.14~ 3 (the question says we have to calculate approx. value of the diameter.)


72 = 15 r

r = 4.8

2r = 9.6 which is greater than 8 but closer to 11.

Let me if you could find a better way to do this.

Thanks
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by sudhir3127 » Tue Jul 15, 2008 10:12 am
i also did the same way but i dint approximate pi . hence got the diameter as 9.17 which is closer to 8 than 11. any take on this?
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by parallel_chase » Tue Jul 15, 2008 10:42 am
sudhir3127 wrote:i also did the same way but i dint approximate pi . hence got the diameter as 9.17 which is closer to 8 than 11. any take on this?
You are absolutely right. I found another way which is simpler and easier.

We know that arc ABC = 24, since the triangle inscribed is an equilateral triangle, all the sides will be equal.

Therefore,
arc AB = arc BC = arc AC

2 AB = 24 => AB = 12

12+12+12 = 2 pi r => 18 = 22/7 r

r= 5.7

2r = 11.4

I hope this is the correct method of solving this question, if this is then i guess GMAT test makers just did not think another way of solving this question LOL.
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by sudhir3127 » Tue Jul 15, 2008 10:04 pm
Thanks dude.. but i am not very sure about the correctness of the above method. can someone please justify it.. i think the whole purpose of giving the length of the arc is to use length of the formula ( 2*pi*r * theta)/360,

require expert advice on it . please..
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by jslavi01 » Wed Jul 16, 2008 7:49 am
Re: the triangle / circle problem

DO NOT OVERCOMPLICATE IT!!

The arc A to B and B to C is 24. Arc ABC is only 2/3 of the perimeter.

The perimeter is 36.

Perimeter = 12+12+12 = 36

Perimeter = pi*diameter

Do not calculate the radius!!

Perimeter = 36 = ~3*(x)

x = (approx) 11

3* 8 = 24
3* 11 = 33
3* 15 = 45
Last edited by jslavi01 on Wed Jul 16, 2008 9:35 am, edited 1 time in total.
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by sudhir3127 » Wed Jul 16, 2008 8:19 am
can u explain how u got 12 when the length of the arc is given as 24?
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my answers

by anksanks » Wed Jul 16, 2008 8:48 am
1.)

districts = 11
to find minimum population in one, we assume other 10 to be maximum.
maximum = 10% more then minimum
let minimum in 1 = x.
then each of other ten will be x + 10% x = 1.1 x
now total = 132,000

hence minimum + ten maximum = 132000
x+10(1.1)x=132,000
=> x= 11,000
Shape me with your bounds,
Wrap me all around,
Define this adorable thought,
And break the silence.
(c) Anks
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