• Philip Martin

Backsolving - The Math Multiple Choice Advantage

Updated: Nov 9

When I took Calculus I in college, I encountered something I had never seen: a multiple choice math test. The teacher gave us four answer options per question, and I immediately realized the advantage in the students’ favor. First, if I got an answer and it wasn’t listed, obviously I was wrong. But more importantly I realized this: I could “try out” answers sometimes, and if it “worked,” I had found the correct answer.


This didn’t work all the time, or as often as I’d like. But, in ACT Math (which is the only of the four tests on the ACT that requires that you choose the 1 correct answer for every question as opposed to some that look for the best answer), you may have this advantage on a number of questions. Knowing this is possible, and how it works, is one of many strategies that a student needs in his or her pocket on test day to maximize success.


In fact, because the ACT knows this strategy exists, the test itself is different. What I mean is this: the English, Reading, and Science tests have questions that feature 4 answer options (A/B/C/D, etc.), but the ACT Math test gives students FIVE answers to choose from (A/B/C/D/E, etc.). This isn't arbitrary, and it's not a way for the ACT to willy-nilly make the test more difficult. Instead, knowing that there are many questions in which having the answers is valuable, they increase the number of answers to make the process take longer.


Now, for the strategy!

This strategy is called Backsolving, and it is simple: instead of always solving the problem, try plugging in the answer choices to find the correct one.


Before we get to a question from a recent ACT, here’s a very simple example to illustrate the point:


Solve the following equation for x: 3x = 9


A. -3

B. -1

C. 0

D. 3

E. 9


I think you know the way to solve this, and could do it in your sleep, but let’s pretend you didn’t. What should you do? Try to figure it out for 2 minutes before guessing? Of course not. You should backsolve, plugging in the answer choices for x until you find which fits, like the right puzzle piece.


Given that the answer choices are in numerical order, we’ll start in the middle. Why? If I determine I need a higher number, I’ll simply move from trying C to D. If I determine I need a smaller number, I’ll simply move from trying C to B.


Starting with C: 3(0) = 9…...0=9


Because 0 is too low, we need a higher number for x, so let’s move “up” to D.


Moving to D: 3(3) = 9…...9=9. That’s it!


Thus, we know the answer to be D. Let’s look at a real question from a recent ACT.


You might know right away to set the exponents equal to each other and solve for x, since the the bases (2) are equal. But, let's assume you didn't know that.


If you solve the right side of this equation on , you will find that 2^15 =32,768. So, you need a value for x that also results in this same number on the left side. You start in the middle (the answers are in ascending order) and choose 11.


2(11) + 7 = 22, and 2^22 = 4,194,304. Too big; go lower. Try B.


2(4) + 7 = 15, and 2^15 = 32,768...That's it! The answer is B!


Let's look at one more, a little different this time, but from the same recent ACT:

Maybe you might notice that you can write an equation for this problem, that would look like this: 2x + 2(x+5) = 40. But, if not, there's backsolving.



Start with the middle choice H and assume it is the width of the rectangle.


Finding the perimeter: 15 + 15 + (15+5) + (15+5) = 70...too big, go lower, try choice G.


Finding the perimeter: 8 + 8 + (8+5) + (8+5) = 42...again too big! By default, we know the choice if F, 7.5


Take advantage of the multiple choice options; put the backsolving arrow in your quiver and make the most of your potential on ACT test day!