Getting ready to face the unexpected

After studying for the CAT for a few months (or years, in my case), you start to form expectations of exam questions. If you’re doing sentence correction, and you see a pronoun, there’s a good chance that the various answer choices will have different pronouns to ensure that you pick the correct one. If you’re doing math with three or four digit numbers, there’s a good chance that you have to deal with unit digits in order to shortcut the calculations. And if you’re doing geometry, there’s a good chance that the Pythagorean Theorem will show up, directly or indirectly. (My money is on directly.) However, it does sometimes happen that a question shatters your expectations. You see the question, you peruse the answer choices, and you immediately look for the properties that you expect to show up. Then, after reading the question, you still don’t have what you expect, and you’re a little lost as to how you should proceed. After all, if you’ve seen the same type of question ten times in a row, a deviation on the 11^th time can be somewhat discombobulating. And yet, this is a strategy that the CAT frequently employs. At the mid level questions (think 25^th-75^th percentile), the exam tests the same concepts repeatedly, driving home some crucial ideas through repetition. At the higher level questions (above 75^th percentile), the questions tend to get trickier by using your own crutches against you. This throws you out of your comfort zone, and forces you to have to look at a problem through a different vantage point. Let’s look at such a problem: In right triangle ABC, BC is the hypotenuse. If BC = 13 and AB + AC = 15, what is the area of the triangle? A) 2 √7 B) 2 √14 C) 14 D) 28 E) 56 Reading through this problem, we note that it’s a right angle triangle with a hypotenuse of 13. Immediately, my brain jumps to the fairly common 5-12-13 triangle that the CAT likes to use. Apart from the ubiquitous 3-4-5 right angle triangle, the 5-12-13 triangle is the next smallest right triangle with all integer sides. Perusing the rest of the question, I fully expect AB + AC to be 5 + 12 or 17. However, the question states that AB + AC is not equal to 17, which takes me completely by surprise (and almost makes me question my very existence). Now, knowing that this isn’t a 5-12-13 triangle isn’t that big of a deal, but it does shatter my expectations of this problem. Clearly, there’s still a solution because the question is being asked, but it deviates from what I thought I had to do. It’s like going to work and your usual route is blocked off. You won’t head back home and sulk, you just have to find an alternate route. Similarly, I now have to take a different approach to solve this geometry question. Let’s review what we know: it’s a right angle triangle, which means it’s almost guaranteed that we’ll need to use the Pythagorean Theorem. The area is being asked, which is ½ Base * Height, as long as Base and Height are orthogonal to one another. The fact that it’s a right angle triangle and BC is the hypotenuse assures us that AB and AC will be the 90 degree angle we need. All we need to do is multiply AB by AC and divide the product by 2 to get our area. The problem is that we only have one equation given: AB + AC = 15. To solve for two unknowns, we need two (independent) equations. The second equation will have to come from Pythagoras (possibly by text message). We know that the square of the two right angle sides will equal the square of the hypotenuse, meaning here we know AB^2 + AC^2 = BC^2. Since we know BC is 13, we really have AB^2 + AC ^2 = 13^2 or AB^2 + AC ^2 = 169 Combine this with our earlier equation of AB + AC = 15 And we have two equations and two unknowns. This should be solvable, but the fact that one equation is linear and the other is quadratic can be somewhat disconcerting. We can square the second equation and use the elimination method to isolate variables and get to the right answer. AB + AC = 15. We now want to square both sides. (AB + AC)^2 = (15)^2. Remember to square each side, not the individual elements. AB^2 + 2 AB*AC + AC^2 = 225. This is a perfect square on the left hand side. Bringing back in the Pythagorean equation: AB^2 + AC^2 = 169. Using the elimination method to subtract one statement from the other, we can eliminate two variables in one fell swoop, leaving us with: AB^2 + 2 AB*AC + AC^2 = 225 – AB^2 + AC^2 = 169   AB^2 + 2 AB*AC + AC^2 = 225 – AB^2 + AC^2 = 169   2 AB*AC = 56. Meaning that AB*AC = 28. Finally, AB * AC is really just the Base * the Height. Since that is what we’re looking for, we don’t need to manipulate the algebra any further. However, there is one final step. The equation we’re looking for is ½ Base * Height, so we need to divide the result by 2 again, yielding just 14. Answer choice C is thus correct. There are several clues that this solution is on the right track. Firstly, the answer we found is among the answer choices. Moreover, two other answer choices are steps we had to pass through in order to find the final answer, making for perfect trap answer choices for overzealous students. Finally, the area of the triangle is very small, which makes sense because the hypotenuse is 13 and the sum of the other two sides is 15. Even the 5-12-13 triangle, which is a relatively thin triangle with an area of 30 (Pythagoras FTW) is twice as big as this thin triangle. The sides of this triangle won’t be integers, but given their relative sizes, it’s something like 2.5-12.5-13, which is quite thin. If you know the Pythagorean Theorem and can apply the elimination method to two equations, this problem isn’t that difficult once you start solving for variables. The difficulty lies primarily in getting started and not getting caught in trying to backsolve or pick numbers for this problem. When going through it, your mind might automatically think of 5-12-13, or whatever typical information is provided for similar questions. Sometimes you have to think of the problem from a different vantage point in order to solve it.