Plug and play method

One of the most effective strategies for answering GRE questions is to plug and play. Instead of solving a problem algebraically right away, you can often take a shortcut by plugging in your own numbers and checking which answer choice matches the result.

Plug and play for problems involving percentages

Here’s an example of a GRE question that works well with 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 translate the situation into an equation, but it’s often faster to plug in values for $c$, $d$, and $p$, compute the number of recorded crashes, and then see which answer choice matches.

Choose easy, distinct numbers

Pick numbers that are:

• Simple (so the arithmetic is quick)
• Different from each other (so you don’t accidentally swap values)

Let’s choose $2$ crashes every 1 day, and $50\%$ of crashes recorded. That means:

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

Solve using your chosen values

• If there are $2$ crashes per day and $50\%$ are recorded, then $1$ crash is recorded per day.
• Over $2$ days, that’s $2$ recorded crashes.

Now check which answer choice equals $2$ when you plug in $c=2$, $d=1$, $p=50$:

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

Only choice A matches the correct result, so A is the correct answer.

Plug and play for quantitative comparison problems

Plug and play is also useful for quantitative comparison questions, especially when:

• A variable is given (often with a constraint), and
• Both quantities depend on that variable

For example:

Given: $x>3$

Quantity A: $x^2$
Quantity B: $2x$

Plug in values that satisfy the constraint

Because $x>3$, you can start with $x=4$ (the smallest integer that works).

• Quantity A: $4^2=16$
• Quantity B: $2\times 4=8$

So in this case, $A>B$.

Don’t stop after one test

One value only shows what happens in one scenario. To be confident, test two or three values.

• If you get the same relationship each time, that relationship is likely always true.
• If you get conflicting results (for example, $A>B$ for one value but $B>A$ for another), then the correct answer is D: the relationship cannot be determined.

In this example, if you plug in $x=5$ or $x=6$, Quantity A continues to grow faster than Quantity B, so $A$ stays greater than $B$. The correct answer is A.

Let’s use the same 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 start with a small, easy value: $x=1$.

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

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

Now try a larger value to see whether the relationship can change. Let $x=5$.

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

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

Because the relationship changes depending on the value of $x$, the correct answer is D. The relationship cannot be determined.

Even if you plug in $x=2$, you’ll still find that Quantity A is greater than Quantity B. Using $x=1$ is especially helpful here because powers of $1$ stay at $1$, which can create a very different outcome than other positive values. This is why it’s important to choose test values that are likely to produce distinct behavior.

Depending on the problem, these values are often useful 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: