Multiple-Choice Strategies

On math contests, and for exams in general, multiple choice tests have benefits for both test-givers and test-takers.  For the test-giver, grading hundreds of papers is as simple as running them through a machine.  But there is a price to that convenience, which you, the test-taker, can and should take advantage of.

For one thing, every question gives the correct answer, hidden in plain sight.  If you’re the math contestant, you only have to find that answer among the choices.  And those answer choices themselves can guide you.

Let’s take a look at two strategies that can be quite effective.  I’ll use the AMC 8 to demonstrate since it is on our students’ minds and that contest features some very clever, challenging problems.  Unfortunately, I can’t repost problems here due to copyright restrictions but I will link to each problem.

So for each problem, follow the link, look at the problem and choices, then return here to see the strategy.


This is a classic multiple-choice strategy that can easily boost your score and speed you up.  Contest writers actually expect students to do this, but surprisingly few seem to actually do it.

The process is simple.  Read the problem.  Then read the answer choices.  Now, if you see any choices that, for a good reason, can’t be right, draw a slash through those letters on the question.

Some important things to note here:  (1) Reading the answer choices is part of reading the problem.  Don’t stop at the question mark.  Read all choices.  (2) You have to eliminate a choice for a good, logical reason, not because it doesn’t look right.  (3) If you can’t eliminate anything for a good, logical reason, that’s OK.  Just start working the problem as usual.  (4) Draw slashes on the question.  Do not draw slashes on the answer sheet!  That is, no slashes in the bubbles!

Here are some examples of how to eliminate choices for a good, logical reason….

2014 Problem 6

All the rectangles have a vase length of 2.  Therefore all the areas we are adding up are even, and the answer must be even.  We can eliminate choices A and B because they are odd.

2012 Problem 8

The shop sells at half price, then discounts again.  So the final price must be more than 50% off.   That eliminates A, B, and C.

More advanced students may be able to eliminate E as well.  Taking 50% off, then 20% off the result, does not result in 70% off; compound discounts don’t work that way.

2011 Problem 6

We can eliminate E immediately.  Why?  If 351 car owners don’t own a motorcycle, that means there are no motorcycle owners in the town of 351.  But that can’t be true: there are 45 motorcycle owners.

We can also eliminate A, B, C because somewhere between 0 and 45 car owners own a motorcycle, so somewhere between 351-0 and 351-45 don’t.  A, B, and C all fall outside this range.

2014 Problem 4

The sun of the primes is odd, so there must be an even and an odd.  That makes their product even.  So eliminate A, B, C, D because they’re all odd!


This takes advantage of the fact that it’s a math contest, after all.  Only one answer choice fits all the requirements.  Why not test answer choices to find that one?  And, even better — what if we knew that we didn’t have to test them all?

Often, you can start at the middle and then go up or down, so you don’t even have to test all choices.

This is a great strategy to use if you don’t see a quick shortcut to the answer right away.  (But if you’re going through this evaluation process, and you see a shortcut, abandon the evaluation and use the shortcut.)


2012 Problem 16

Shortcut:  The digits of each addend should be 9+8, 7+6, 5+4, 3+2, 1+0 to get the maximum sum.  So the correct answer choice should have one digit from each pair in the right place.  Only choice C does:  87431.

Evaluation method:  Suppose you don’t see the shortcut.  Then try this.  Look at the five choices, one at a time.  For each answer choice, figure out the unused digits and arrange them to get the highest possible addend.  Then find the sum.  When you’re done, see which of the five sums is highest.

A:  76531 => unused digits:  0, 2, 4, 8, 9 => 76531 + 98420 = 174951

B:  86724 => unused digits:  0, 1, 3, 5, 9 => 86724 + 95310 = 182034

C:  87431 => unused digits:  0, 2, 5, 6, 9 => 87431 + 96250 = 183951

D:  96240 => unused digits:  1, 3, 5, 7, 8 => 96240 + 87531 = 183771

E:  97403 => unused digits:  1, 2, 5, 6, 8 => 97403 + 86521 = 183924

The highest of the five sums is 183951, which came from choice C.

2013 Problem 17

Shortcut:  With consecutive integers, or any other arithmetic (linear) sequence, the median and mean are equal.  Sum is 2013 => mean is 2013 ÷ 6 = 335.5 => median must also be 333.5 => integer #3 must be 335 and #4 must be 336.  Continuing:  #5 is 337, and #6 is 338.  The answer is B.

Evaluation method:  Start in the middle with choice C and try it:  335 + 336 + 337 + 338 + 339 + 340 = 2025.  Too high.  So go down to B, which should give a lower sum:  333 + 334 + 335 + 336 + 337 + 338 = 2013.  Bingo!  Answer is B.

2014 Problem 24

Shortcut:  Number the customers in order 1 to 100, based on how many cans they bought.  (No need to break ties.)  The median will be the average of the cans for #50 (call it m) and #51 (call it n).  To get the highest possible median, assign the lowest possible number of cans to the other 98 customers:  #1-#49 should have 1 can each, and #52-#100 should have n cans each.  With 252 cans total, #1-#49 get 49, #51-#100 get 50n.  m + 50n = 203.  n can be 1, 2, 3, or 4; to get the maximum median, set n = 4 => m = 203 – 200 = 3.  Median is (m+n)/2 = 3.5.  Answer is C.

Evaluation method:  Again, arrange customers from #1 to #100.  Try each median to see if you can make a sum as small as 252.  If it works, go up.  If it doesn’t, go down. When you have a choice that works and a choice that doesn’t, the lower one will be the answer.

Does C work?  Median of 3.5 means #51 has >= 4.  To make the sum small, give customers #51-100 4 cans each.  To get a median of 3.5, #50 needs to have 3.  Give #1-49 1 each to make the sum even smaller.  #1-49 + #50 + #51-100 = 49(1) + 3 + 50(4) = 252.  Yes, this works.  Go up.

Does D work?  Median of 4.0 means #51 has >=4.  To make the sum small, give customers #51-100 4 cans each.  To get a median of 4.0, #50 needs to have 4.  Give #1-49 1 each to make the sum even smaller.  #1-49 + #50 + #51-100 = 49(1) + 4 + 50(4) = 253.  No, this doesn’t work.

Because C works and D doesn’t, the maximum value is given by C.


What about answer choices like “all of the above,” “none of the above,” or the in-between variants like “II or III only”?  The same strategies apply, but you just have to think carefully through the logic.  For instance:  Only if you eliminate all other choices can you select “none of the above.”  If you evaluate and find that two answer choices fit, you can select “all of the above.”

Sometimes the contest penalizes random guessing by giving more points for a blank than for an incorrect response.  Examples of such contests include the Math Is Cool multiple choice rounds and the AMC 10.  In these cases, evaluation still works and elimination is still a viable strategy.  But the final step of picking a non-eliminated choice is a probability problem gets a little trickier.

Let’s talk about these when the time comes.

There are also other, more questionable “strategies”:  looking for similar answers and picking one; or just picking “C” for everything.  On a contest where there is no penalty for guessing (like AMC 8), I suppose these would be better than leaving something blank.  But let’s try to minimize the need for this.


By correctly applying elimination and evaluation, you improve your chances of a better score on the AMC 8 or other multiple-choice contests.  Try it on your next practice!

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