The test gurus at Testmasters are at it again with more tricks and tips for conquering the GRE. Today’s problem involves questions with multiple answers. Both paper exams and the computerized test will have math questions that could have more than one correct answer, and to get credit for answering the question correctly, you will have to give all of the answers for that question. Many of these problems are algebra based, such as the one below.

If the numbers 31, 24, and x are averaged, the value will be somewhere between 20 and 30, inclusive. Which of the values below could be x?

1. 4
2. 5
3. 20
4. 35
5. 40

To solve this problem as quickly as possible, without plugging in all five answers, find the smallest and largest possible values for x based on the largest and smallest possible averages for the set of numbers. Recall that an average is the sum of terms divided by the number of terms. Plug in the information from the problem into this, so the average equals the lowest value of 20. Solve for x by first multiplying both sides by 3.    The lowest possible value for x is 5, which eliminates answer (A) as a possible choice, and it verifies that (B) is one of the correct answers.

Now, repeat the process with the largest value for the average, 30. Solve for x by first multiplying both sides by 3.    This means that the largest value for x is 35, so any answer choices between 5 and 35, inclusive will be answers. For this problem, you would select answers (B), (C), and (D).

This entry was posted in Example Problems and tagged , , . Bookmark the permalink.