With BBQ season nearing, it’s time for hotdogs and burgers. Well, sort of. Try the problem below.

A concession stand sells either hotdogs for $1.75 each or hamburgers for $4 each. If Charlie buys a total of 9 items from the concession stand for a total of 27 dollars, then how many hotdogs did he buy?

(A) 3

(B) 4

(C) 6

(D) 7

(E) 8

On the GRE, you will most likely see a long, tedious word problem. Granted, this problem wasn’t too bad. The question is, how did you go about trying to solve it?

Many begin trying to turn the information from the problem into an equation. Some are very adept at this and get the answer quickly. But, most students I’ve seen struggle as soon as they start thinking about the above problem algebraically. So, I want you to try something that may seem counterintuitive. I want you to turn off the algebra part of your brain. Instead, I want you to work with the answer choices by plugging them back into the problem.

Let’s start with answer choice (C). The reason we start with this answer is because it is in the middle. If it turns out that the total price is less than 27 dollars, then we will need to plug in a lower number. The reason we would choose a lower number is because hotdogs are cheaper; therefore, the fewer hotdogs we have the more burgers we have, and this brings up the price.

Plugging in 6 we get $1.75 x 6 = $10.50. That leaves us 3 hamburgers at $4 each = $12. Adding the two together gives us $22.50. That amount is too low, so we can eliminate answer choices (D) and (E), as well, because they will increase the number of hotdogs and lower the overall price.

Plugging in (B) 4, we get $7 on hotdogs. That leaves us 5 hamburgers or $4 x 5 = $20. Add up $7 + $20 and we get $27, which is the correct answer.

In the next segment on plugging in, I will show you a very different type of problem in which plugging in can be equally effective.

Check out Clemmonsdogpark’s other GRE math practice problems to try plugging in on different problem types!