difficult problems

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.

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.

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

can anyone solve the second question in the attachment..

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.

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.

can some one solve this...

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

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?

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.

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..

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

can u explain how u got 12 when the length of the arc is given as 24?

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