Hi there,
Just want to throw something else in here, even though the explanation already given is exactly right
One other way to think about a problem like this is to focus on the fact that what an HCF (or GCF) is is also the largest factor by which you could divide the numerator and denominator to simplify a fraction. Once we've got the multiples of 15 listed in this problem -- 45, 60, 75, 90 (as manpsingh listed) -- we can think essentially of simplifying (dividing) all of these by that factor of 15 and winding up with 3, 4, 5, 6. There are six possible number-pairs to make here ("4 choose 2" if you're thinking in combinatorics), but two of them -- the 3/6 pair and the 4/6 pair are not in lowest terms, which means each of those two pairs have GREATER common factors than the 15 that we used. So those two number-pairs don't meet our criteria.
On this particular problem, using this strategy doesn't necessarily speed things up, but it might be helpful conceptually, and it's a strategy we can use if the numbers get bigger and more cumbersome on any future GCF problems. To sum up, look at a number pair as a fraction, and then you'll find you've discovered the GCF (whatever is the total factor you're scaled the fraction down by)
when and only when you can't simplify the fraction any further.
Cheers!
Ashley Newman-Owens
GMAT Instructor
Veritas Prep
