If abc 0, then (ab + bc + ac)/(1/a + 1/b + 1/c) = ?

I'd do this exactly how divyadna did above, but there are other ways to get to an answer that are also fast:

- if the expression in the question is equal to one of the answer choices, that means they're equal for every set of values of a, b and c that we could plug in (provided we observe the restriction that abc is nonzero). If you just plug in a = b = c = 1, the expression in the question is equal to 1, but the only two answer choices that are equal to 1 are answers B and D. You can then plug in, say, a = b = c = 2 to decide between those two choices. EDIT: Regor60 was kind enough to point out that I made a silly arithmetic mistake plugging a = b = c = 1 into answer D, and it's actually equal to 3, not to 1, so we don't even need to plug in any other numbers. In a way I'm happy I made that error, because it illustrates one of the reasons I prefer divyadna's approach, and would always use it; if you plug in numbers, you're doing five calculations instead of one, and you thus give yourself five times as many opportunities to make a careless error. But it can be a good fallback strategy when you don't see an alternative.

- if we imagine plugging in very large numbers (e.g. 1000 for each) for a, b and c, the numerator ab + bc + ac of the fraction in the question will be huge, while the denominator 1/a + 1/b + 1/c will be tiny. If you divide a very big number by a very small one (when everything is positive), you get a big answer, and the only answers that are plausible are B and E. But answer E can't be right, because it's identical to the numerator of the fraction in question, and so it will only be equal to the fraction in question when the denominator is precisely equal to 1.

But if you can solve questions like this algebraically, as divyadna did above, that will give you the most flexibility, and most of the time that will be the fastest approach on the real test.