The sum of two numbers is 56. Their difference is 1/7 of their sum. Their LCM is:

The correct option is 96.

Let the two numbers be $x$ and $y$ (where $x>y$).

According to the given problem statement:
1. The sum of the two numbers is 56:

$x+y=56$     --- (Equation 1)

2. Their difference is 1/7 of their sum:

$x-y=\frac{1}{7}\times 56$

$x-y=8$     --- (Equation 2)

Adding Equation 1 and Equation 2 gives:

$(x+y)+(x-y)=56+8$

$2x=64$

$x=32$

Substituting $x=32$ into Equation 1:

$32+y=56$

$y=24$

Thus, the two numbers are 32 and 24.

Now, we find the Least Common Multiple (LCM) of 32 and 24 by prime factorization:

$32=2^5$

$24=2^3\times 3$

Taking the highest powers of all prime factors:

$\text{LCM}(32,24)=2^5\times 3=32\times 3=96$

Therefore, the LCM of the two numbers is 96.