### Plugging in: GRE and GMAT Quant strategy

Quant Strategy: plugging in values

Avoid messy algebraic equations

Plugging in values when there are unknowns in the questions or the answer options is a fairly straightforward strategy that can help you navigate high-complexity questions on the GRE and GMAT Quant sections. Especially those that can send you into a spiral by making you solve complex algebraic equations.

The premise of plugging in values is that it’s generally easier to work with numbers (ideally whole numbers and not fractions) than to work with multiple unknown variables (Xs, Ys and Zs).

Let’s look at a very simple example to see how plugging in a value can help deal with tricky situations.

## Plugging in – Illustrative Example

### John is X years old. Bob is 5 years younger than John. In terms of X, how old will Bob be in Y years?

A. Y (X – 5)
B. X (Y – 5)
C. X + Y – 5
D. X + Y
E. (X / 5) + Y

This is not a tough question; we could most likely do this algebraically. But that doesn’t seem like an elegant way to solve, so why even bother when we can use arithmetic instead of algebra!

But how do we do that?

Instead of “X” and “Y” let us substitute integers and follow the rules of the question. Let’s assign a value for X; X = 10 and for Y; Y = 20 (why those numbers? Because those are both divisible by 5 and they don’t violate any of the ‘rules’ that the question has laid out). Now, once we’ve written down our numbers, we need to solve the problem using those numbers.

Here’s the question restated with numbers instead of variable:

John is 10 years old. Bob is 5 years younger. How old will Bob be in 20 years?

Let’s break this down:

– If John is 10 and Bob is 5 years younger, then Bob is currently 5 years old.

– In 20 years, Bob will be 25 years old.

Hence 25. Now let’s push those values of X and Y (that we plugged in to the question) to the answer options.

A. 20 (10 – 5) would equal 100; ELIMINATE
B. 10 (20 – 5) is equal to 150; ELIMINATE
C. 10 + 20 – 5 equals 25; SELECT! This is the answer!

Do we stop here?

In this question there’s no possibility of the variables standing for negative or fractional numbers. You should, therefore, stop with C and not move any further.

If, on the other hand, there is possibility for the variables to stand for negative or fractional numbers, you must check all answer options.

Can you apply what you’ve learned about plugging in to solve the following question without resorting to (messy) algebra?

### A mall’s parking garage has space only for certain number of cars. If 1 / 5 of places are left empty, and 2 / 5 of the places are used by compact cars, then non-compact cars take up what fraction of filled spaces in the garage?

A. 1/3
B. 2/5
C. 1/2
D. 3/5
E. 4/5

Let’s plug in values: assume that total number of spaces in the garage is 25: it’s easier to calculate with 25 as there are 5’s in denominator for 1/5 and 2/5 …

Hence, total space in garage = 25

Then Empty places in garage = 1/5 of 25 = 1/5 x 25 = 5
– Filled places in garage = 25 – 5 = 20
– Places used by compact cars = 2/5 of 25 = 2/5 x 25 = 10

Therefore, places used by non-compact cars = 20 – 10 = 10

Then non-compact cars as fraction of filled spaces = 10/ 20 = 1/2