can someone explain this one also?

I couldn't think of a more scientific method, but here's how I see it:

First look at what we've got and break it down into prime factors:
6 = 2*3
9 = 3*3

1. tells you that the least common multiple of x and 6 is 30.
30 = 2*3*5. We're lucky, since you can't make too many numbers out of just 2, 3 and 5, with the restriction that 5 MUST be one of x's divisors (there must be a reason for the 5 in 30 - it can't just pop out of nowhere). You can easily tell that there are only 4 possibilities for x:
a. x = 5, case when the lcm of x, 6 and 9 will be 90
b. x = 10, case when the lcm of x, 6 and 9 will be 90
c. x = 15, case when the lcm of x, 6 and 9 will be 90
d. x = 30, case when the lcm of x, 6 and 9 will be 90.
Since you get 90 all over, 1 is sufficient.

2. 45 = 3*3*5. Again, there aren't a lot of possibilities, even fewer in this case:
a. x = 15, with lcm 90
b. x = 45, with lcm 90.
Answer's 90 all over, meaning that 2 is sufficient as well.

according to 1 - x has to be 5 since LCM is 30 so that is suff

according to 2 - X has to be 5 since LCM is 45 so that is suff

now since we know that x is 5, we can find the LCM of x, 6, 9 to be 90 since x is 5.

I hope I explained correctly. it takes some time to undestand this quesiton, but I am starting to learn.......

Thank you

ST

Well, while your reasoning provided the correct answer, it's not enough, because you missed some of the possible values of x.

Stmt 1 : LCM of x and 6 is 30.
lcm of x,6,9 will be the LCM of 30,9

Stmt1 is sufficient.

similarly

Stmt2 is sufficient.

Thanks
raama

krisraam wrote:Stmt 1 : LCM of x and 6 is 30.
lcm of x,6,9 will be the LCM of 30,9

Stmt1 is sufficient.

similarly

Stmt2 is sufficient.

Thanks
raama

Raama's soln is the simplest and elegent.

His soln is based on the following property of LCM:

LCM(a, b, c) = LCM(a, LCM(b,c) ) = LCM( LCM(a,b) , c )

This is your major takeaway.