GMAT Math

The Beat the GMAT flashcards suggest that we should backsolve starting with answer choice E, but Ive read from other sources that one should start with answer choice C. Which is better and why?

thank you.

Good question, Feyfey - there are good arguments to be made for a few different backsolving strategies, so I'd argue that the key is to pick the one that you feel most comfortable with, but also to remain flexible. Of the two that you mentioned, the advantages are:

Picking E: I think the thought process behind this one is that the GMAT is less likely to reward backsolvers by coincidentally having the first answer choice, A, be correct (that would save you a ton of time, as you wouldn't have to try the other four answers.


Picking C: In many cases, the number you plug in, if it's not the correct answer, will tip you off as to whether you need a smaller number or a larger one. By picking C first, you can limit the number of choices you need to plug in - if C is too big, the answer must be A or B. If C is too small, and you need a bigger number, it has to be D or E.

Then, because after picking C you'll have it narrowed down to two (A/B, or D/E), if you choose one of those two to plug in, it's either that one or the other. Here, you can limit the number of choices you have to plug in to two.

Picking B or D: Similar to the "Picking C" philosophy, on many questions you'll have a hunch as to whether you need a larger or smaller number. Say, for example, that the question asks you for the volume of the larger of two cylinders, or for the length of the longer piece of rope; there's a good chance that the GMAT authors will have included the other (smaller volume, longer rope) value as a potential trap answer. Because you know that, you may have an edge: say you know that you want the smaller of two potential values. Pick choice B, because if B is too big, then the answer is definitely A. You have a better than 40% chance of getting the answer with just one step, because you know that the answer is likely to come from choices A, B, or C.

Then, if B isn't correct, you can pick D next. If D is too big, the answer must be C (because A, B were too small), and if D is too small, it's E.

This strategy means that you'll get the answer in two steps, and also that you have a pretty good shot of getting it in one.


Given these three, I'd rate the effectiveness as:

1) Picking B or D (often you solve it in one step, and you'll usually get it in two or fewer)

2) Picking C (you'll get the answer in one or two steps, and it may feel a little cleaner than the first option, if slightly less efficient)

3) Picking E (it's fairly good insight, but if the answer is B, you're doing a lot of work).


However, I'll give my favorite, hybrid backsolving strategy:

Be lazy: Pick from the above selections, using whichever numbers will be the easiest to backsolve. If you have answer choices like 0, 1, 10, or other easy-to-calculate numbers, you might as well plug those in, as you can probably do it in your head without much thought.

From there, let the results dictate what you do next using the theory behind the above strategies. If 0 turns out to be too much, try smaller numbers; if 1 is too small, try larger ones. Minimize the amount of work you have to do...and to do so, you may not want to be 100% beholden to any one strategy.

One of my favorite tutoring stories from my beginning foray into teaching involves this same philosophy. My student had worked previously with a tutor he loved and raved about...a guy with a Stanford MBA and all kinds of other credentials (it was like having a new girlfriend who had previously dated Brad Pitt, and mentioned him all the time). There was one question, though, that the old tutor couldn't solve and said was probably a "bad question". To his defense, the algebra led to a messy quadratic, so it was pretty ugly. But the answer choices were something like:

200/120; 191/120; 171/120; 180/120; 160/120;

I just grabbed 180/120, because it boiled down pretty easily to 3/2, and the quadratic made it look like you'd be multiplying by 2 in some form which would cancel the denominator. Sure enough, that was it.

Thank you for your reply.