Assume that 0<a<b<c

For the 1st statement:
If a >= 2, we have a+b+c >= 2+3+4=9 > 7 => incorrect. So a must equal 1.
To continue with b. b>a means b>1.
If b >= 3, we have a+b+c >= 1+3+4=8 > 7 => incorrect. So b must equal 2.
Then, we find c=7-1-2=4. Statement 1 is sufficient.

Similarly, we apply this way for 2nd statement and find (a,b,c) = (1,2,4)

Therefore, each one is sufficient.

a, b, and c are three distinct positive integers. What is the product abc?

1) a + b + c = 7
2) ab + bc + ca = 14

For Statement 1, if we make all of the integers greater than 1, the smallest that they can be is 2, 3 and 4.

2 + 3 + 4 = 9. That's already too big. So one of them has to be 1.

1 + 3 + 4 = 8 Too big.

Must be 1 + 2 + 4. So there is only one set that works, and Statement 1 is sufficient.

For Statement 2, if we make all of the integers greater than 1, the smallest that they can be is 2, 3, and 4.

Plugging in 2, 3 and 4, we get 6 + 12 + 8 = 26 Too big. So one must be 1.

If a is 1 then the equation becomes b + bc + c = 14

Hmm. I'm thinking 8 + 6. That works with b = 2 and c = 4. 2 + 8 + 4 = 14.

Can other numbers work? If b = 3 and c = 4, the sum is going to be greater than 14.

So the integers must be 1, 2 and 4. So Statement 2 is sufficient.

The correct answer is D.