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Set7 Q5

Expert replies
Source: — Data Sufficiency |

by josephcho77 » Thu Oct 04, 2007 7:27 am
I will go with "A"

k/6+m/4=t/12 is equal to 2K+3m=t

1) if k is a multiple of 3, no matter what m is, t must have factor of 3.
Therefore, since 12 has a factor of 3, t and 12 have a common factor
greater than 1

2) if m is a multiple of 3, t can be prime numbers or non prime number.
When t is a prime number, t and 12 can't have a common factor
greater than 1 but when t is a non prime number, t and 12 have a
common factor greater than 1.
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by arocks » Thu Oct 04, 2007 7:42 am
Thanks. Very clear.
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