Sometimes the correct answer to a multiple-choice math question is easier to figure out by determining which choices are not correct. If you can show that a choice isn’t correct, you can narrow down the list until you find the right answer.
With some multiple-choice questions, you can prove that an answer choice is incorrect by substituting numbers into the variable expression or equation from either the question or answer choice in order to prove it is not correct. This approach is known as finding counterexamples. Solving by finding counterexamples often involves less complicated math formulas and calculations, and therefore leaves room for fewer errors in computations. This approach is useful not only for multiple-choice questions with a single answer, but also for multiple-choice questions with more than one answer.
Finding counterexamples is a good strategy to use when variables are involved, whether they need to be evaluated or compared. The variables may appear in the question itself, in the answer choices, or in both places. Finding counterexamples is also a good strategy to use when a question asks which must be true. If you can find a case where an answer choice is not always true, then you know that choice is not correct.
So, how do you go about solving using the counterexamples strategy? For each answer choice, pick a number or numbers to substitute for the given situation and show that it is not true. If you can do this for all but one answer choice—in the case of multiple-choice questions with a single answer—you know that answer choice must be correct.
Here’s an example of a question where using substitution to find counterexamples is useful:
If 0 < x < y, which of these expressions must have a value greater than 1?
E x + y
This question involves both variables and the phrase “must have,” which indicate that solving by finding counterexamples would be a good strategy. You can pick appropriate numbers to represent x and y, substitute them into the expressions for each answer choice, and try to find cases in which the expressions are not greater than 1. You do not always need to, nor should you, pick the same numbers to substitute every time. But do try and pick numbers that are easy to work with: it makes the math a lot easier! For this question, you know that x and y are both positive. Keep in mind that x and y can represent both integers and non-integers.
In Choice (A), if x = ½ and y = 1, then xy = (½)(1) = ½. Since ½ is not greater than 1, this represents a counterexample, and Choice (A) cannot be correct.
In Choice (B), if x = ½ and y = 1, then (xy)2 = (½ · 1)2 = ¼. Since ¼ is not greater than 1, Choice (B) cannot be correct.
In Choice (C), if x = 1 and y = 2, then x/y = ½. Since ½ is not greater than 1, Choice (C) cannot be correct.
In Choice (D), if x = ½ and y = 1, then y/x = 2. This is greater than 1. But finding one case where the situation is true doesn’t always mean the answer choice is correct. You can try several other numbers and see whether they always result in a number greater than 1 as well, but it is still a good idea to try to find a counterexample for Choice (E). If you can do that, you know Choice (D) must be correct.
In Choice (E), if x = ¼ and y = ½, then x + y = ¼ + ½ = ¾. Since ¾ is not greater than 1, Choice (E) cannot be correct.
You are able to show that all of the choices except Choice (D) are incorrect. Therefore, the correct answer must be Choice (D).
Use counterexamples as a strategy to boost your math score on the ACT or SAT!