Many people struggle with math anxiety or outright math phobia, but the reality is that we’re all born with an innate “number sense.” On some level, we know how to count things naturally, almost from birth. For instance, when you’re choosing a grocery store checkout lane, you can tell instantly which lanes are more or less crowded, without counting. And humans aren’t alone—many animals have evolved with the ability to distinguish between larger and smaller groups (called ‘numerosity’), a useful ability for survival and reproduction. Some GMAT problems can be approached by utilizing this natural number sense and working backwards from the answer choices. For instance, here’s a simple one—take a minute or two to try this out: If x and y are positive integers and x + y = 400 and x – y = 50, what is the value of x? a) 250 b) 225 c) 200 d) 150 e) 125 Got it? You could have solved for x with algebra using the substitution or elimination methods. But another option is just to ask yourself, “Do I know which is bigger, x or y?” We know x + y = 400, and if x – y = 50, then x must be 50 bigger than y. If x and y were equal, then they’d each be 200 (half of 400), but we know that’s not the case. Look at the answers—x must be greater than 200, so that eliminates answer choices C, D, and E. If x were 250 (answer A), then y would have to be 150 in order for x + y to be 400, but 250 is not 50 bigger than 150, so C’s out. X has to be 225 (and y is 175), so the answer is B. If you got that answer via algebra, then nicely done—but I encourage you to experiment with solving problems like this via working backwards. It’s always better to have more tools in your toolkit. Using your natural number sense in this way can be a great alternative to slogging through the algebra on every problem—it exercises a different mental muscle and can help prevent you from making careless mistakes because you’re thinking about the numbers in a more concrete, less abstract way. But there are other types of situations in which our natural number sense can be dangerous. When percent change gets involved, our brains aren’t quite as good at intuiting what a reasonable answer should look like. For instance: if you invest $1,000 in a stock, and the stock goes up by 10% on the first day, then goes back down by 10% on the second day, did you break even? It might feel like you should break even, but in this scenario you actually lost money. On the first day, your stock went up 10% from $1,000 to $1,100, for a gain of $100. On the second day, your stock lost 10% of $1,100, for a loss of $110, leaving you with only $990—$10 less than what you started with on day one. The GMAT writers know that that sort of math is counter-intuitive for a lot of people, and they like to write questions to exploit this weakness. Here’s a problem from GMATPrep. Take 2 minutes to attempt the problem (guess if you need to!), and if you finish before the 2 minutes is up, I’d like you to see if you can spot the trap (wrong) answer, and what mistake someone would have to make to get that answer. Okay, here goes: “A used-car dealer sold one car at a profit of 25 percent of the dealer’s purchase price for that car and sold another car at a loss of 20 percent of the dealer’s purchase price for that car. If the dealer sold each car for $20,000, what was the dealer’s total profit or loss, in dollars, for the two transactions combined? “(A) $1,000 profit “(B) $2,000 profit “(C) $1,000 loss “(D) $2,000 loss “(E) $3,334 loss” Did you spot the trap? It’s A. Here’s the mistake they want you to make: “The dealer made a 25% profit on the first car, and a 20% loss on the second car. 25 is 5 more than 20, so the dealer must have made a profit of 5%. The price of each car was $20,000, and 5% of 20,000 is $1,000, so the answer is A.” That’s wrong, but it’s an easy mistake to make, because our natural number sense is screaming at us that 25% is more than 20%. The problem is that the 25% profit and the 20% loss are percents of two different numbers, and neither one is out of $20,000. The $20,000 is what the dealer sold each car for, but the 25% and 20% are based on the two different prices the dealer paid for the cars (i.e., the wholesale cost). My colleague Stacey Koprince has a great walkthrough of this problem here using the “make it real” approach, which I wholeheartedly endorse. Algebra is another approach that works well here: X = wholesale cost of the first car Y = wholesale cost of the second car Total profit (or loss) = ? (always write down your goal!) Equation 1: The dealer sold the first car for $20,000 and made a 25% profit on it (25% more than 100%, so 125% of the cost): 20,000 = 1.25*x Equation 2: The dealer sold the second car for $20,000 and lost 20% on it (leaving 80% of the cost): 20,000 = .8*y Convert those decimals into fractions and solve: Eq. 1: 20,000 = (5/4)*x (4/5)*20,000 = x x = 16,000 Eq. 2: 20,000 = (4/5)*y (5/4)*20,000 = y y = 25,000 So the dealer bought the first car for $16,000 and the second car for $25,000, for a total cost of $41,000. The dealer then sold the two cars for $20,000 each, or $40,000 total. That’s a loss of $1,000 overall. The answer is C. Key Takeaways -Use your natural number sense when comparing two simple quantities—ask yourself which number is bigger and which is smaller, or faster vs. slower, or did more work vs. did less work. Then look at the answers and see if you can eliminate some choices. -Be wary of relying on your natural number sense when the relationships between numbers are more complex, such as percent change or combined rates problems.