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by zachlebo » Wed May 18, 2011 8:22 pm
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
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Source: — Data Sufficiency |

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by Anurag@Gurome » Wed May 18, 2011 8:57 pm
zachlebo wrote: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.
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by manpsingh87 » Wed May 18, 2011 9:04 pm
zachlebo wrote: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
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by zachlebo » Wed May 18, 2011 9:05 pm
Thank you!! now it makes sense!
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by Anurag@Gurome » Wed May 18, 2011 9:06 pm
zachlebo wrote:Thank you!! now it makes sense!
Great!
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by blaster » Wed May 18, 2011 9:46 pm
zachlebo wrote: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.
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by zachlebo » Thu May 19, 2011 7:54 am
so because you can factor out a 3 in statement one and not in statement two makes it sufficient?
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