Ans :only (1) is sufficient to answer
k/6 + m/4 = t/12 => 2k+3m=t
divide throughout by 3 ,
(2k)/3 [ k is a multiple of 3,(2k)/3 is an integer ] + m [m is an integer] = t/3 [Integer + Integer = integer]
Therefore one of the other common factors of t and 12 greater than 1 is 3 .
common factor greater than 1
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sreak1089
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We know, k/6 + m/4 = t/12
=> (2k + 3m)/12 = t/12
=> t = (2k + 3m)
1) says k is multiple of 3
Let's say k = 3p (p is constant)
=> t = (2*3p + 3m)
= 3(2*p + 1)
=> t is a multiple of 3 or t has 3 as its factor.
12 = 3 * 2^2 Hence 3 is a common factor > 1. Hence SUFFICIENT.
2) says m is multipe of 3
Let's say k = 3q (q is constant)
=> t = (2k + 3*3q)
= 2k + 9q
We know that m is multiple of 3 but we don't know anything of k. Hence 2 is NOT SUFFICIENT.
Answer is A.
=> (2k + 3m)/12 = t/12
=> t = (2k + 3m)
1) says k is multiple of 3
Let's say k = 3p (p is constant)
=> t = (2*3p + 3m)
= 3(2*p + 1)
=> t is a multiple of 3 or t has 3 as its factor.
12 = 3 * 2^2 Hence 3 is a common factor > 1. Hence SUFFICIENT.
2) says m is multipe of 3
Let's say k = 3q (q is constant)
=> t = (2k + 3*3q)
= 2k + 9q
We know that m is multiple of 3 but we don't know anything of k. Hence 2 is NOT SUFFICIENT.
Answer is A.
