We have to see if c has any one of 2,3,4,6,12 as factor
From 1
a=3m
so m/2+b/4=12c..we can have m=1,b=44 and c=1 and m=4,b=88 and c=2....not sufficient

From 2
b=3n

a/6+3n/4=12c
or a/72+n/16=c
or (2a+9n)/144=c...for n=2 and a=63 we have c=1 and n=30 and a=9we have c=2..not sufficient

combining

m/2+3n/4=12c
or 2m+3n=48c

again for m=3 and n=14 c will be 1 and n=30 and m=3..c=2..again not sufficient

IMO ans is E

IMO E!!

let us simplfy the given equation as : 2a+3b = 12 *C and it asks whether 12 & C are co- prime or not??

st 1: doesnt speak abt b

Insufficient

St 2:

Doesnt speak abt a :

Insufficient

Combining st 1 & st 2 , we have :

I defactorised a & b as a= 3*X , b= 3 *Y and 144 = 2^4 * 3^2

2 * 3 * X + 3 * 3 * Y= 12 * C = 144 * C

so it becomes:

3( 2 *X + 3*Y) = {2^4 * 3^2} * C

Now depending on value of C , the 3 on both sides will get nullified.

Take few examples :
Let C = 3,

3( 2 *X + 3*Y) = {2^4 * 3^2} * C

=3( 2 *X + 3*Y) = {2^4 * 3^2} * 3

YES 12 & C has one more factor than 1 in common !!

Let C = 7

3( 2 *X + 3*Y) = {2^4 * 3^2} * C

=3( 2 *X + 3*Y) = {2^4 * 3^2} * 7

NO. 12 & C does NOT have any facor in common other than 1.

so Insufficient even after combining 2 sts.

pick E!!

Nice explanation friend.I have just one question,

gmatmachoman wrote: 3( 2 *X + 3*Y) = {2^4 * 3^2} * C

Now depending on value of C , the 3 on both sides will get nullified.

Don't you think that the 3 on both side will nullify even if 3 is not a factor of C..i.e
3( 2 *X + 3*Y) = {2^4 * 3^2} * C
or ( 2 *X + 3*Y) = {2^4 * 3^1} * C
this will be satisfied by any C which does not have a 2,3,4 or 6 as factor,as long as Y is even and X is a multiple of 3.
What say?

yeah u r correct. It will do!!

Actually, i wanted to write more on it, becox of time constraints,I shortened it. yes 3 will get nullified immaterial of value of C.