Plug and play method

One of the most effective strategies for answering GRE questions is to plug and play. Rather than solving questions mathematically, it’s often possible to take a shortcut by plugging in numbers to see if the answer choices match the expected results.

Plug and play for problems involving percentages

Here’s an example of a GRE question that is well-suited for this technique:

A particularly busy 4-way stop experiences $c$ car crashes every $d$ days. If the surveillance camera captured only $p$ percent of the car crashes, then which of the following represents the total number of crashes recorded in $2d$ days?

A. $2cp/100$
B. $2dcp$
C. $2dp/100c$
D. $dcp/100$

You could write out the situation described in the question as an equation that equals the number of crashes recorded in $2d$ days, but it may be easier to simply plug in your own values for $c$, $d$, and $p$ and solve for the number of recorded crashes.

Pick some easy numbers to work with. Good numbers are simple but unique. You want to make the math easy while ensuring you don’t mix up the values. Let’s pick $2$ crashes each day, with $50\%$ of the crashes recorded. Our variables look like this:

$$\begin{array}{rl}c& =2\\ d& =1\\ p& =50\end{array}$$

Let’s solve the question with these values. If there are $2$ crashes every day, and $50\%$ of them are recorded, that means there is $1$ recorded crash every day. So in $2$ days, we would expect $2$ recorded crashes. Having solved the question, we can see which answers match our solution of $2$.

A. $2cp/100=2(2)(50)/100=2$ TRUE
B. $2dcp=2(1)(1)(50)=100$ FALSE
C. $2dp/100c=2(1)(5)/100(2)=0.05$ FALSE
D. $dcp/100=(1)(2)(50)/100=1$ FALSE

By choosing extremely simple values, it becomes much easier to determine the solution. When we plug in these values, only choice A gives the correct result of $2$, meaning it is the correct answer to this question.

Plug and play for quantitative comparison problems

When you see a quantitative comparison question with a variable (and possibly a constraint for that variable) and quantities that depend on the variable, the plug and play method is likely to be a good strategy.

For example, consider this question:

Given: $x>3$

Quantity A: $x^2$
Quantity B: $2x$

When deciding on values to plug in, pick values for $x$ within the constraint of $x>3$. The number $4$ is a good choice since it is the smallest integer that satisfies the constraint. When $x$ is $4$, Quantity A is $4^2=16$ and Quantity B is $2\times 4=8$. We have thus proven that A is greater than B in this one scenario.

But that’s just in one scenario! You can never be sure if you have the correct answer if you’ve picked only one number. You must use the plug and play method for two or three numbers to try to prove that another scenario is possible. If you come across conflicting results (e.g. A is greater in one scenario but B is greater in another) your final answer becomes D: it is impossible to know the relationship between Quantity A and Quantity B. If you suspect that D is the answer upon initially looking at the problem, try to prove the two conflicting scenarios indicating that D is the answer.

In our question above, if we plug in $5$ or $6$ for $x$ we will notice that Quantity A will grow much faster than Quantity B. Thus, we can be sure that Quantity A is always greater than Quantity B, and choice A is the answer to this question.

Let’s use this method on a more complicated quantitative comparison problem.

$x>0$

Quantity A: $x^4/3^4$
Quantity B: $4^3/3^x$

Try it yourself, and then read on for the answer and walkthrough!

The question states that $x>0$, so let’s start by plugging in a small, easy number greater than zero: $x=1$.

$$\begin{array}{rlrlr}A:x^4/3^4& =& 1^4/3^4& =& 1/81\\ B:4^3/3^x& =& 4^3/3^1& =& 64/3\end{array}$$

So $B>A$ when $x=1$.

Let’s see if we can get a different result by picking a larger number: $x=5$.

$$\begin{array}{rlrlr}A:x^4/3^4& =& 5^4/3^4& =& 625/81\\ B:4^3/3^x& =& 4^3/3^5& =& 64/729\end{array}$$

So $A>B$ when $x=5$.

Because we have conflicting results, the correct answer to this quantitative comparison question is D. The relationship cannot be determined.

Even if you chose to plug in $x=2$, you would still find that Quantity A is greater than Quantity B. Using $x=1$ gives a unique result because it stays the same when multiplied by itself many times. This helps show why it’s always important to plug in numbers that you believe have a good chance of giving unique results. They will vary depending on the question, but in general, these values might be good ones to try:

$$-10,-5,-1,0,1,5,10$$

Bringing it all together: question walkthrough video

Here’s a video going through one of our practice questions to demonstrate these ideas in action: