Test Cases on Theory Problems

CAT Exam
Try this  problem: * ” If xy + z = x(y + z), which of the following must be true? “(A) x = 0 and z = 0 “(B) x = 1 and y = 1 “(C) y = 1 and z = 0 “(D) x = 1 or z = 0 How did it go? This question is called a “theory” question: there are just variables, no real numbers, and the answer depends on some characteristic of a category of numbers, not a specific number or set of numbers. Problem solving theory questions also usually ask what must or could be true (or what must not be true). When we have these kinds of questions, we can use theory to solve—but that can get very confusing very quickly. Testing real numbers to “prove” the theory to yourself will make the work easier. The question stem contains a given equation: xy + z = x(y + z) Whenever the problem gives you a complicated equation, make your life easier: try to simplify the equation before you do any more work. xy + z = x(y + z) xy + z = xy + xz z = xz Very interesting! The y term subtracts completely out of the equation. What is the significance of that piece of info? Nothing absolutely has to be true about the variable y. Glance at your answers. You can cross off (B) and (C) right now! Next, notice something. We stopped at z = xz. and didn’t divide both sides by z. Why? In general, never divide by a variable unless you know that the variable does not equal zero. Dividing by zero is an “illegal” move in algebra—and it will cause you to lose a possible solution to the equation, increasing your chances of answering the problem incorrectly. The best way to finish off this problem is to test possible cases. Notice a couple of things about the answers. First, they give you very specific possibilities to test; you don’t even have to come up with your own numbers to try. Second, answer (A) says that both pieces must be true (“and”) while answer (D) says “or.” Keep that in mind while working through the rest of the problem. z = xz Let’s see. z = 0 would make this equation true, so that is one possibility. This shows up in both remaining answers. If x = 0, then the right-hand side would become 0. In that case, z would also have to be 0 in order for the equation to be true. That matches answer (A). If x = 1, then it doesn’t matter what z is; the equation will still be true. That matches answer (D). Wait a second— Both answers can’t be correct. Be careful about how you test cases. The question asks what MUST be true. Go back to the starting point that worked for both answers: z = 0. It’s true that, for example, 0 = (3)(0). Does z always have to equal 0? Can you come up with a case where z does not equal 0 but the equation is still true? Try 2 = (1)(2). In this case, z = 2 and x = 1, and the equation is true. Here’s the key to the “and” vs. “or” language. If z = 0, then the equation is always 0 = 0, but if not, then x must be 1; in that case, the equation is z = z. In other words, either x = 1 OR z = 0. The correct answer is (D). The above reasoning also proves why answer (A) could be true but doesn’t always have to be true. If both variables are 0, then the equation works, but other combinations are also possible, such as z = 2 and x = 1. Key Takeaways: Test Cases on Theory Problems (1) If you didn’t simplify the original equation, and so didn’t know that y didn’t matter, then you still could’ve tested real numbers to narrow down the answers, but it would’ve taken longer. Whenever possible, simplify the given information to make your work easier. (2) Must Be True problems are usually theory problems. Test some real numbers to help yourself understand the theory and knock out answers. Where possible, use the answer choices to help you decide what to test. (3) Be careful about how you test those cases! On a must be true question, some or all of the wrong answers could be true some of the time; you’ll need to figure out how to test the cases in such a way that you figure out what must be true all the time, not just what could be true.

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