4 questions from GMAT prep that is on mba.com

ABC = 24 and equilateral
so 3rd arc will be 24/2 = 12

Circumference = 2 PI r = 36

r = 18 / PI
dia = 36 / PI
so final ans is 11





A:
n is not divisible by 2
so n odd
n-1 even and also n+1 even
both n-1 and n+1 will be multiple of 2 and one will be multiple of 4
so (n-1)(n+1) is multiple of 8
but no logic for 24
so Insuff

B:
n is not divisible by 2
n is 3x+1 or 3x-1
hence
n+1 or n-1 one of these has to be divisible by 3
so
(n-1)(n+1) is multiple of 3
but no logic for 24
so Insuff


A+B
multiple of 8*3
so remainder = 0








y+3x+2
(r,s)
s=3r+2

A:
3r+2-s = 0 or 4r+9-s = 0 or both
3r+2-s = 0
s= 3r+2

or
s = 4r+9

not sufficient.



B:
4r-6-s = 0 or 3r+2-s = 0 or both
4r-6-s = 0
4r-6 = s

or
3r+2-s = 0
s= 3r+2

not sufficient.


A+B [one root is common]
s= 3r+2

satisies the equn and sufficient
















pig + cow = 40

A:
cows are more than twice no of pigs

cows, pigs can be (28, 12) (30, 10) and so on

no sufficient



B:
more than 12 pigs
pigs can be 13 14 15..
so cows can be 27 26 25....

no sufficient



A+B
from B minimum pigs = 13
cows = 27 [satisfies A also]

next pigs = 14
cows = 26
A not satisfied

hence A+B has only one answer
so sufficient so ans is c

for question # 1
Length of Arc = 2*3.14*r*(2*Inscribed Angle)/360 as length of Arc ABC= 24
This will lead equation like
24 = 2*3.14*r*(2*60)/360+ 2*3.14*r*(2*60)/360 , solve this will give you ~ 11

# 2 should be solved by picking a number,

#3 Not sure

#4) 2/3of60 = 40
stmt 1 says more then twice as many cows as pigs, in that case # of pigs can be any number from 2,3.... and so on, but if combine stmt 2 you will get pigs as 13 and cow 27.
Hope its clear

Thank you very much! I have understood every one of them

In the first question, when i saw the length of ABC I thought that it was the perimeter of the triangular...