amitansu wrote:Ans 'D'.
From stem 1:
x could be 5 or 30
and using both for each cases we can find it is 90.
So sufficient.
From stem 2:
x could be 5or 45.
using both for each cases we can find it is 90.
So sufficient.
Amit
You've got the right answer, but you missed a few possible values for x. From statement 1, x could be 5, 10, 15 or 30; from statement 2, x could be 5, 15 or 45.
This approach (finding values for x) works fine for this question, but could be time consuming if the numbers in the question were large. We actually don't need to find values for x at all here. From statement 1, the LCM of x and 6 is 30. The LCM of x, 6 and 9 must then be equal to the LCM of 30 and 9. If you think about how you calculate LCMs, it should be clear why this is true. Or you can think about what the LCM means: if the LCM of 6 and x is 30, then 30 is the smallest positive integer which is divisible by both 6 and x. Thus the LCM of 6, x and
anything must be a multiple of 30, and the LCM of 6, x and 9 must be the smallest multiple of 30 that is also a multiple of 9. So 1) is sufficient (and by the same reasoning, so is 2)).
