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If k, m, and t are positive integers and k/6 + m/4 = t/12 , [#permalink]

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07 Oct 2008, 23:07

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If k, m, and t are positive integers and k/6 + m/4 = t/12 , does t and 12 have a common factor greater than 1 ? (1) k is a multiple of 3. (2) m is a multiple of 3. _________________

To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." -Edward Bulwer Lytton

If k, m, and t are positive integers and k/6 + m/4 = t/12 , does t and 12 have a common factor greater than 1 ? (1) k is a multiple of 3. (2) m is a multiple of 3.

k/6 + m/4 = t/12 => (2k + 3m)/12 = t/12

(1) Sufficient. If k=3*n, the equation becomes (6n + 3m)/12 => 3 is a common factor of t and 12 (2) Insufficient. If m=3*n, the equation becomes (2k + 9m)/12 => no common factor visible

If k, m, and t are positive integers and k/6 + m/4 = t/12 , does t and 12 have a common factor greater than 1 ? (1) k is a multiple of 3. (2) m is a multiple of 3.

k/6 + m/4 = t/12 => (2k + 3m)/12 = t/12

(1) Sufficient. If k=3*n, the equation becomes (6n + 3m)/12 => 3 is a common factor of t and 12 (2) Insufficient. If m=3*n, the equation becomes (2k + 9m)/12 => no common factor visible

Hence (A)

If there is no common factor visible, shouldn't (B) also be sufficient? as this is a YES or NO type question. _________________

To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." -Edward Bulwer Lytton

If k, m, and t are positive integers and k/6 + m/4 = t/12 , does t and 12 have a common factor greater than 1 ? (1) k is a multiple of 3. (2) m is a multiple of 3.

k/6 + m/4 = t/12 => (2k + 3m)/12 = t/12

(1) Sufficient. If k=3*n, the equation becomes (6n + 3m)/12 => 3 is a common factor of t and 12 (2) Insufficient. If m=3*n, the equation becomes (2k + 9m)/12 => no common factor visible

Hence (A)

If there is no common factor visible, shouldn't (B) also be sufficient? as this is a YES or NO type question.

(B) does not guarantee of a common factor (A, on the other hand, does). It's possible (in B) for t to have something in common with 12 even when no common factor is obvious.

For instance, if k=3, m=4, then common factor = 2 However, k=5, m=3 does not give any common factor _________________

This is a good question. Especially when it comes to attacking the second statement

2K + 3M / 12 = T/12

Statement 1-----------SUFFICIENT

K=3N Try N =1, (M CAN TAKE ANY VALUES AND STILL BE A MUTILPLE OF 3 AS IT HAS 3 AS ITS COEFFECIENT) 6 + 3M /12 =T/12 3(2+M)/12 =T/12 Therefore 3 is a factor of T and 12 other than 1 Hence A

Statement 2------------INSUFFICIENT

M=3N Try N=1 2K + 9/ 12 = T/12 No common factor other than 1 But if we put K =6, AND N=1 We will get 12 + 9 / 12 =T/12 =3 (4+3)/12 = T/12 Common factor 3

Two different values. Hence Insufficient

Hence A

gmatclubot

Re: Highest Common Factor
[#permalink]
17 Dec 2009, 02:19