In the fractions a/b and c/d, where a, b, c and d are positive integer

Using Statement 2, if 2 is the largest divisor of b and d, then clearly 2 is a prime divisor of each. Since b and d share no other divisors, the second prime divisor of b must be different from the second prime divisor of d. So the prime divisors of b are 2 and p, and the prime divisors of d are 2 and q, where p and q are different primes. The product bd will therefore have three prime divisors, 2, p and q, and Statement 2 is sufficient.

Using Statement 1, when fractions are completely reduced, their least common denominator will be the LCM of the two denominators. Since, for two numbers, LCM*GCD = product, then if the LCM of the denominators is half their product, the GCD of the two denominators would be 2. Then we have the situation of Statement 2, discussed above, and we have three prime divisors of bd in total. The problem though is that our fractions might not be reduced. We know a '2' cancels somewhere when we find the least common denominator of the two fractions. It might cancel, as above, when the denominators share a prime divisor of 2 -- the fractions might be 1/6 and 1/10, say. Then the least common denominator is 30, half of the product of 6 and 10, and bd = 60 has three prime divisors. But the 2 might instead cancel because one of the fractions can be reduced. The fractions might instead be 2/14 and 1/15, say. Then the least common denominator is the least common denominator of 1/7 and 1/15, which is 105, and that is (correctly) half the product of 14 and 15. But in this case bd has four prime divisors, 2, 3, 5 and 7, not three. So Statement 1 is not sufficient, and the answer is B.