anks17 wrote:Experts,

I had marked the answer of attached question as C as there can be many numbers from A and B that satisfy them but when we combine them, we get the common number satisfying both.Not sure how the answer is something else. Plz help.

We need to find the LCM of x,6 and 9.

With that in mind, consider the statements:

1) LCM of x and 6 is 30. Since we know LCM of two of the numbers, and we know the third number, sufficient.
We can call this Sufficient because:
LCM of 3 numbers can be found by, taking LCM of two numbers, lets call it A, and then taking LCM of A with the third number. In this case, LCM of x,6, and 9 will be equal to LCM of 30 and 9, which equals 90).

2) With the same reasoning as above, statement 2 is Sufficient as well (because LCM of 45 and 6 = 90).

D is correct.


Alternatively, although it would be time consuming, you can find the various values of x and find the LCM in each case.

For statement 1, x can be 5, 10, 15 or 30. All values of x have common factors to 6 and 9 other than 5 as a factor of x. LCM will still be 90 for any value of x, 6 and 9.
Sufficient.

For statement 2, x can be 5,15, or 45. Again, using similar reasoning to above, LCM of any value of x, 6, and 9 = 90. Sufficient again.

D is right.

Thanks Eagleeye. I did found different values X in A and B but thought that since there are many values satisfying A and B, LCM for three numbers in question will be different in each case (depending upon the number we pick from various values satisfying A or B), hence chose E.

this is where the methodology matters..amean I too started with finding different x values without giving a thot to the funda eagleeye pointed out..gets more time consuming and affects overall exam performance :/
temme a way out!!