GMAT Blog | Variables in the Answer Choices – Part II

In the last article, we learned the algebraic approach and the input-output approach for solving questions with variables in the answer choices (VIACs). In that article, I ended by saying that, while the input-output approach is often easier and faster, it is not without its issues. To help understand these potential issues, please use the input-output approach to solve the following problem:

Townville has X residents, and Y of them are females. If half of Townville’s male residents are smokers, and 1/Z of the female residents are smokers, which of the following represents the total number of Townville’s residents who are NOT smokers?

A) X – Y + Y/Z

B) X - Y/Z - Y

C) 2X – Y - ZX

D) X/2 + Y/2 – Y/Z

E) X + 2Y – 4Z

For the input-output approach, we’ll select some “nice” values for the variables, and then use those chosen values to answer the question. We’ll then plug (input) those same values into the answer choices, and see which one yields (outputs) the same answer to the question.

Before we select some numbers to work with, let’s talk about what makes a value “nice.” Nice values are typically small, because selecting large values can result in complicated calculations. For example, imagine the work involved if we say there are 188 residents in Townville (X = 188), there are 51 females (Y = 51), and that 1/3 of the women smoke (Z = 3). By selecting these large values, we’ve made checking the answer choices a very tedious task that is prone to errors.

On the other we must be careful not to choose values that are too convenient. For example, let’s say there are 3 residents in Townville (X = 3). Let’s also say that there is 1 female (Y = 1), and that 1/1 of “them” smoke (Z = 1). While these values make it very easy to check the answer choices, it’s possible that we’ll run into some problems. To see what I mean, let’s use these selected values to answer the question.

First, if 1 of the 3 residents is a female, then the other 2 residents are males. Since half of the males smoke, there must be 1 male smoker.

Next, since 1/1 of the 1 female is a smoker, we can conclude that there must be 1 female smoker.

So, in total, there are 2 smokers in Townville, which means there is 1 resident who does not smoke.

In other words, when we INPUT X = 3, Y = 1 andZ = 1, the correct OUTPUT is 1 non-smoker.

At this point, we’ll plug the three chosen values into the five answer choices to see which one yields an output of 1.

A) X – Y + Y/Z: 3 – 1 + 1/1 =3

B) X - Y/Z - Y: 3 – 1/1 – 1 = 1

IMPORTANT: Before we conclude that B is the correct answer, we must evaluate the three remaining answer choices.

C) 2X – Y – ZX: 2(3) – 1 – (1)(3) = 2

D) X/2 + Y/2 – Y/Z: 3/2 + 1/2 – 1/1 = 1

E) X + 2Y – 4Z: 3 + 2(1) – 4(1) = 1

So, answer choices B, D and E all have the correct output of 1, which means we must select another set of values for X, Y and Z, then solve the question, and then evaluate answer choices B, D and E, to see which one yields the correct output. As you can imagine, doing all of this will take time.

What went wrong?

The problem is that the test-makers anticipated that we might choose super easy values to work with, and they penalized us the best way they could: they made us spend extra time solving the question. For this question, it’s likely that we were penalized for choosing values that yielded a scenario in which 1/1 of the 1 female is a smoker. It’s bad enough that 1/1 is a silly fraction, but also notice that our chosen values (X = 3, Y = 1 and Z = 1) yielded an output of 1. That’s too many 1’s. If you repeat values, it’s difficult to identify the affect each variable has on the resulting output.

Some Guidelines

While there’s no guaranteed strategy for selecting numbers that will yield only one matching output among the answer choices, there are some guidelines that you can follow that will help minimize the number of times you’ll be forced to test a second set of values when solving VIAC questions.

Small numbers: Small numbers are easier to work with, but it’s a good idea to avoid selecting 0 and 1 as values. The test-makers know that these are popular choices, so they’ll often (but not always!) create two or more answer choices that yield the same output when either 0 or 1 are chosen as input values.

Prime numbers: There are fewer relationships between prime numbers, and this reduces the likelihood that a set of selected values will yield more than matching output. Having said that, it can be difficult finding prime numbers that fit the conditions specified in the question. So, don’t spend too much time trying to prime numbers work.

Different numbers:If you must select more than one input value, choose different values. Duplicate input values can often yield more than matching output.

Okay, with these guidelines in mind, let’s answer the question again.

A) X – Y + Y/Z

B) X - Y/Z - Y

C) 2X – Y - ZX

D) X/2 + Y/2 – Y/Z

E) X + 2Y – 4Z

So, let’s say there are 5 residents in Townville (X = 5), and 2 of them are males. This means there are 3 females (Y = 3), and . . . . . . STOP!

If there are 3 females, and 1/Z of them are smokers, then it’s going to be difficult to choose a nice Z-value. If we make Z = 1, then we have the weirdo fraction 1/1, and the test-makers may penalize us for that. On the other hand, we could make Z = 3, but we now have 3 as two of our selected values, and we may be penalized for this duplication. Also notice that if we input X = 5, Y = 3 and Z = 3 the output is 3 (yet another 3!). At this point, we might still go ahead with these values (at the risk of getting more than one matching output), or we might select another set of values. It's your choice.

Me? I’m going to select another set of values.

So, let’s say there are 11 residents in Townville (X = 11). Let’s also say that there are 9 females (Y = 9), and that 1/3 of them smoke (Z = 3).

When we use these values to solve the question, we find that there are 7 residents who do not smoke. In other words, when we INPUT X = 11, Y = 9 andZ = 3, the OUTPUT is 7 non-smokers.

At this point, we’ll plug the three chosen values into the five answer choices to see which one yields and output of 7.

A) X – Y + Y/Z: 11 – 9 + 9/3 =5

B) X - Y/Z - Y: 11 – 9/3 – 9 = -1

C) 2X – Y – ZX: 2(11) – 9 – (3)(11) = -20

D) X/2 + Y/2 – Y/Z: 11/2 + 9/2 – 9/3 = 7

E) X + 2Y – 4Z: 11 + 2(9) – 4(3) = 17

Since D is the only answer choice that correctly yields the output 7, it must be the correct answer.

Extra practice

Here are some questions to practice the input-output technique:

- http://www.beatthegmat.com/what-mistake-am-i-making-in-this-substitution-last-sunday-a-t276293.html

- http://www.beatthegmat.com/car-dealer-sales-t274136.html

- http://www.beatthegmat.com/ps-rate-times-t276107.html