1-
let consider the first sentence : k = 3 * k'
so we got : t = 12 * (k'/2 + m/4 ) --> t = 3 * (2*k' + m)
then 3 is a common factor of t and 12 -> 1 is sufficient
2-
let's consider the sencond hypothese : m = 3 * m'
so we got : t = 12 * ( k/6 + 3*m'/4) --> t = 2*k + 9*m' --> we can't guess from here any known factor of 12
-> 2 is unsufficient
-> the right answer is : A
Integer
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Combine the fractions over a common denominator using the bowtie method:
k/6+m/4 = (4k+6m)/24 = 2k+3m/12
So
2k+3m/12 = t/12 --> t=2k+3m
Stat. (1): since k is a multiple of 3, then t is the sum of two multiples of 3 (both 2k and 3m are multiples of 3), and will thus be a multiple of 3 itself. whatever t is, it definitely shares at least 3 as a common divisor with 12. Stat. (1)->Yes-> sufficient.
Stat. (2): m is a multiple of 3 tells you nothing definite about t. Plug in a few numbers:
If k=1 and m=3, then t=2*1+9 = 11, which shares no common divisor with 12.
but if k=3 (note, multiple of 3 --> same as stat. (1)) and m=3, then t=3*3+3*3 = 18, which shares 2, 3 and 6 as common divisors with 12. Thus, stat. (2) allows both a yes and a no answer to the question, and is insufficient. Answer is indeed A.
k/6+m/4 = (4k+6m)/24 = 2k+3m/12
So
2k+3m/12 = t/12 --> t=2k+3m
Stat. (1): since k is a multiple of 3, then t is the sum of two multiples of 3 (both 2k and 3m are multiples of 3), and will thus be a multiple of 3 itself. whatever t is, it definitely shares at least 3 as a common divisor with 12. Stat. (1)->Yes-> sufficient.
Stat. (2): m is a multiple of 3 tells you nothing definite about t. Plug in a few numbers:
If k=1 and m=3, then t=2*1+9 = 11, which shares no common divisor with 12.
but if k=3 (note, multiple of 3 --> same as stat. (1)) and m=3, then t=3*3+3*3 = 18, which shares 2, 3 and 6 as common divisors with 12. Thus, stat. (2) allows both a yes and a no answer to the question, and is insufficient. Answer is indeed A.
