Few things about GRE math questions

The math tested by the GRE mainly consists of junior high school level arithmetic, algebra and geometry. A number of the GRE problem solving questions will come in sets associated with a graph or graphs. The intention of the math section is to test how expert you are at ruling the trouble-free solutions. The idea is that if you use a lot of time working out long solutions you will not come to an end as much of the test as students who spot the short, effortless solutions. So if you find yourself performing long calculations or applying advanced mathematics then don’t do like this. Do the practice on GRE Math practice test. Let see one example here: Column A: Number of square tiles needed to cover a space on a wall of 72” x 108”, using tiles that each has a perimeter of 36 inches. Column B: Number of square tiles needed to make a rectangular border one tile wide around the outside of the same 72” x 108” space, using tiles that each has an area of 9 square inches. A The quantity in Column A is greater.rnB The quantity in Column B is greater.rnC The two quantities are equal.rnD The relationship cannot be determined from the information given. Substitution is a very useful technique for solving GRE math problems. It often reduces hard problems to routine ones. In the substitution method, we choose numbers that have the properties given in the problem and plug them into the answer-choices. During the course of a GRE, you will probably find at least a few multiple-choice questions that you have no idea how to solve. You may practice such questions on GRE quantitative practice questions. Take a moment to look answer choices. Often two or three of them are absurd. Eliminate those and guess one of the others. Sometimes four of the choices are absurd. When this occurs, your answer is no longer a guess. Read carefully – did you notice that the tiles specified in Column A had a perimeter of 36 inches, while the tiles in Column B were specified using their area? This is especially easy to miss since 36 is a perfect square. For Column A, the tiles have side length of 36/4 = 9”. You will need 72/9 = 8 tiles along the width, and 108/9 = 12 tiles along the length; (12)(8) = 96 tiles. Now, since the square border tiles have area of 9 square inches, they must have side length of 3”. The width can be bordered by 72/3 = 24 tiles, while the length can be bordered by 108/3 = 36 tiles. Don’t forget there are two lengths and two widths, for a total of 24 + 24 + 36 + 36 = 120 tiles. (You might also want to add another 4 tiles, for the corners, but it doesn’t make a difference for this problem). So, more 3” small tiles are needed for the border (120) than large 9” tiles for the area (96). If you had used 6” tiles for the area as the problem attempted to mislead you, you would have needed 216 of them and mistakenly chosen A.