DS Question

[zachlebo]

If k, m and t are positive integers and k/6 + m/4 = t/12, do t and 12 have a common factor greater than 1?

1) k is a multiple of 3
2) m is a multiple of 3








OA says A

i do not understand why 2 is not sufficient

[Anurag@Gurome]

If k, m and t are positive integers and k/6 + m/4 = t/12, do t and 12 have a common factor greater than 1?

1) k is a multiple of 3
2) m is a multiple of 3

OA says A
i do not understand why 2 is not sufficient

k/6 + m/4 = t/12 or 2k + 3m = t

(1) Let k = 3x. Then t = 2(3x) + 3m = 3(2x) + 3m, which implies t is a multiple of 3, and hence t and 12 have also a common factor greater than 1.
So, (1) is SUFFICIENT.

(2) Let m = 3y. Then t = 2k + 3(2y), clearly this does not help in answering the question.
So, (2) is NOT SUFFICIENT.

The correct answer is A.

[manpsingh87]

If k, m and t are positive integers and k/6 + m/4 = t/12, do t and 12 have a common factor greater than 1?

1) k is a multiple of 3
2) m is a multiple of 3








OA says A

i do not understand why 2 is not sufficient

k/6+m/4=t/12;
2k+3m=t;

1) k=3p; here p=1,2,3,4,,,

2(3p)+3m=t;
3(2p+3m)=t;
now since t is a multiple of 3 thus t and 12 have 3 as a common factor which is greater than 1 hence 1 alone is sufficient to answer the question..!!

2)m=3p;
2k+3(3p)=t;

now since this doesn't give any information about the nature of t; hence 2 alone is not sufficient to answer the question hence it should be A

[zachlebo]

Thank you!! now it makes sense!

[Anurag@Gurome]

Thank you!! now it makes sense!

Great!

[blaster]

If k, m and t are positive integers and k/6 + m/4 = t/12, do t and 12 have a common factor greater than 1?

1) k is a multiple of 3
2) m is a multiple of 3

when we simplify equation it looks like this
2k+3m = t

statement 1: if k is a multiple of 3 then we can do something like this 2*3*x+3m=t . so we can make equation like this 3* ( 2*x+m) = t . from here we can see that 3 is factor of t and 12.
statement 2: here as we see, m being a multiple of 3 is no make any differcen, it's already is a multiple of 3.

[zachlebo]

so because you can factor out a 3 in statement one and not in statement two makes it sufficient?