The material covered on the exam is not inherently complicated, but the combination of subtle wordplay and constant stress about time management creates an environment where test takers often rush through prompts and misinterpret questions. Unfortunately, time management and stress management are two of the major skills being tested on the CAT, so the time limit isn’t going away any time soon. Instead, it’s worth mastering simple techniques to save time and extrapolate patterns based on smaller samples.

As an example, consider a simple question that asks you how many even numbers there are between 1 and 100. Of course, you could write out all 100 terms and identify which ones are even, say by circling them, and then sum up all the circled terms. This strategy would work, but it is completely inefficient and anyone who’s successfully passed the fourth grade would be able to see that you can get the answer faster than this. If every second number is even, then you just have to take the number of terms and divide by 2. The only difficulty you could face would be the endpoints (say 0 to 100 instead), but you can adjust for these easily. The next question might be count from 1 to 1,000, and you definitely don’t want to be doing that manually.

Other questions might not be as straight forward, but can be solved using similar mathematical properties. This means that the goal of the test is not to waste your time executing calculations you would execute on your calculator in real life, but rather to evaluate how you think and whether you can find a logical shortcut that will yield the correct answer quickly.

Let’s look at an example that can waste a lot of time if you’re not careful:

*Brian plays a game in which he rolls two die. For each die, an even number means he wins that amount of money and an odd number means he loses that amount of money. What is the probability that he loses money if he plays the game once?*

*A) 11/12
*

*B) 7/12*

*C) 1/2*

*D) 5/12*

*E) 1/3*

First, it’s important to interpret the question properly. Brian will roll two die, independently of one another. For each even number rolled, he will win that amount of money, so any given die is 50/50. If both end up even, he’s definitely winning some money, but if one ends up even and the other odd, he may win or lose money depending on the values. The probability should thus be close to being 50/50, but a 5 with a 4 will result in a net loss of 1$, whereas a 5 with a 6 will result in a net gain of 1$. Clearly, we need to consider the actual values of each die in some of our calculations.

Let’s start with the brute force approach (similar to writing out 1-100 above). There are 6 sides to a die, and we’re rolling 2 dice, so there are 6^2 or 36 possibilities. We could write them all out, sum up the dollar amounts won or lost, and circle each one that loses money. However, it is essentially impossible to do this in less than 2 minutes (or even 3-4 minutes), so we shouldn’t use this as our base approach. We may have to write out a few possibilities, but ideally not all 36.

If both numbers are even, say 2 and 2, then Brian will definitely win some money. The only variable is how much money, but that is irrelevant in this problem. Similarly, if he rolls two odd numbers, say 3 and 3, then he’s definitely losing money. We don’t need to calculate each value; we simply need to know they will result in net gains or net losses. For two even numbers, in which we definitely win money, this will happen if the first die is a 2, a 4 or a 6, and the second die is a 2, a 4 or a 6. That would leave us with 9 possibilities out of the 36 total outcomes. You can also calculate this by doing the probability of even and even, which is 3/6 * 3/6 or 9/36. Similarly, odd and odd will also yield 9/36 as the possibilities are 1, 3, and 5 with 1, 3, and 5. Beyond this, we don’t need to consider even/even or odd/odd outcomes at all.

The interesting part is when we come to odds and evens together. One die will make Brian win money and the other will make him lose money. The issue is in the amplitude. Since we’ve eliminated 18 possibilities that are all entirely odd or even, we only need to consider the 18 remaining mixed possibilities. There is a logical way to solve this issue, but let’s cover the brute force approach since it’s reasonable at this point. The 18 possibilities are:

Odd then even: Even then odd:

1, 2 3, 2 5, 2 2, 1 4, 1 6, 1

1, 4 3, 4 5, 4 2, 3 4, 3 6, 3

1, 6 3, 6 5,6 2, 5 4, 5 6, 5

Looking at these numbers, it becomes apparent that each combination is there twice ((2,1) or (1,2)). The order may matter when considering 36 possibilities, but it doesn’t matter when considering the sums of the die rolls. (2,1) and (1,2) both yield the same result (net gain of 1), so the order doesn’t change anything to the result. We can simplify our 18 cases into 9 outcomes and recall that each one weighs 1/18 of the total:

(1,2) or (2,1): Net gain of 1$

(1,4) or (4,1): Net gain of 3$

(1,6) or (6,1): Net gain of 5$

Indeed, no matter what even number we roll with a 1, we definitely make money. This is because 1 is the smallest possible number. Next up:

(3,2) or (2,3): Net loss of 1$

(3,4) or (4,3): Net gain of 1$

(3,6) or (6,3): Net gain of 3$

For 3, one of the outcomes is a loss whereas the other two are gains. Since 3 is bigger than 2, it will lead to a loss. Finally:

(5,2) or (2,5): Net loss of 3$

(5,4) or (4,5): Net loss of 1$

(5,6) or (6,5): Net gain of 1$

For 5, we tend to lose money, because 2/3 of the possibilities are smaller than 5. Only a 6 paired with the 5 would result in a net gain. Indeed, all numbers paired with 6 will result in a net gain, which is the same principle as always losing with a 1.

Summing up our 9 possibilities, 3 led to losses while 6 led to gains. The probability is thus not evenly distributed as we might have guessed up front. Indeed, the fact that any 6 rolled with an odd number always leads to a gain whereas any 1 rolled with an even number always leads to a loss helps explain this discrepancy.

To find the total probability of losing money, we need to find the probability of reaching one of these three odd-even outcomes. The chance of the dice being odd and even (in any order) is ½, and within that the chances of losing money are 3/9: (3, 2), (5, 2), and (5, 4). Thus we have 3/9 * ½ = 3/18 or 1/6 chance of losing money if it’s odd/even. Similarly, if it ends up odd/odd, then we always lose money, and that’s 3/9 * 3/9 = 9/36 or ¼. We have to add the two possibilities since any of them is possible, and we get ¼ + 1/6, if we put them on 12 we get 3/12 + 2/12 which equals 5/12. This is answer choice D.

It’s convenient to shortcut this problem somewhat by identifying that it cannot end up at 50/50 (answer choice C) because of the added weight of even numbers. Since 6 will win over anything, you start getting the feeling that your probability of losing will be lower than ½. From there, your choices are D or E, 15/36 or 12/36. Short of taking a guess, you could start writing out a few possibilities without having to consider all 36 outcomes, and determine that all odd/odd combinations will work. After that, you look at the few possibilities that could work ((5,4), (4,5), etc) and determine that there are more than 12 total possibilities, locking you in to answer choice D.

Many students struggle with problems such as these because they appear to be simple if you just write out all the possibilities. Especially when your brain is already feeling fatigued, you may be tempted to try and save mental energy by using brute force to solve problems. Beware, the exam wants you to do this (It’s a trap!) and waste precious time. If you need to write out some possibilities, that’s perfectly fine, but try and avoid writing them all out by using logic and deduction. On test day, if you use logic to save time on possible outcomes, you won’t lose.