Set your timer for 2 minutes… and… GO!

*Before being simplified, the instructions for computing income tax in country R were to add 2 percent of one’s annual income to the average (arithmetic mean) of 100 units of country R’s currency and 1 percent of one’s annual income. Which of the following represents the simplified formula for computing the income tax, in country R’s currency, for a person in that country whose annual income is I?

(A) 50 + I/200
(B) 50 + 3I/100
(C) 50 + I/40
(D) 100 + I/50
(E) 100 + 3I/100

This is an awfully wordy question. These can tie us up in knots sometimes. So what’s going on here? There’s some formula for computing income tax in country R, and we need to find that formula.

Quick Quiz:

Name something critical that you MUST do on a problem solving question after reading the question but before starting to solve the problem.

Brainstorm a little bit. There are multiple good answers to the quiz, but I’m looking for one in particular, so come up with a few ideas to see if you can hit the one that I want.

Here’s the answer: it’s critical to look at the answer choices before starting to solve. Why? Because the answer choices can actually help you decide the best way to solve the problem. In this problem, I immediately noticed that the answers contain variable expressions. Great! I can turn this algebra problem into an arithmetic problem if I want. Now I have to decide whether to try algebra or arithmetic, depending upon my strengths and weaknesses here.

We’ll try both ways in this article. When you’re studying, actually figure out where the line is for you “ when is it better for you to do algebra and when is it better for you to do arithmetic? The very general rule is that you do algebra on the easier (for you) problems and you do arithmetic on the harder (for you) problems. But where is the specific line for you? And is it different on different types of problems? (Perhaps you prefer to switch to arithmetic on lower level “ relatively speaking “ geometry problems but can handle straight algebra on higher level fraction or percent problems.) In order to figure that out, try both solution methods when studying and note which method worked better for you on that particular problem and why.

Okay, so let’s try the algebra first. We’re given the variable I, which represents a person’s annual income. I don’t know about you, but I can’t write a reasonable italic I on my scratch paper, and my regular I is going to look too much like a 1 to me. I’m going to substitute a different variable, one that I won’t mix up with some number when doing the problem (and I won’t mix up this substitute variable with some other variable in the problem because I is the only variable in the problem). It’s Country R, so I’m going to use R.

So R is a person’s annual income. Now I need to translate this language:

add 2 percent of one’s annual income to the average (arithmetic mean) of 100 units of country R’s currency and 1 percent of one’s annual income

Confusing. Let’s break that down into separate steps:

add 2 percent of one’s annual income to the next thing

average (arithmetic mean) of 100 units of country R’s currency and 1 percent of one’s annual income

Okay, I know how to take an average: you sum the different things you have and divide by the number of things you have. I have two things: 100 units and 1 percent of one’s annual income. I know that one’s annual income is R, so that part is really saying 1 percent of R

The average of 100 and 1 percent of R is (100 + 0.01R)/2. (The standard average formula is: average = sum / # of things.)

Then, what was that other part? Oh, yes, add 2 percent of one’s annual income. So that’s 0.02R and I should add it to my average above. That gives me:

0.02 R + (100+0.01 R)/2

Now I have to simplify this thing. I can split the right-hand term into two parts:

0.02 R + (100/2) + (0.01 R /2)

0.02 R + 50 + 0.01 R /2

Interesting. So D and E definitely aren’t right. If I have to guess, it’s between A, B, and C. (Why?)

(Because the first and third terms will always contain the variable; only the middle term is a plain number. And only A, B, and C have 50 as the plain number term.)

At this point, I’m going to use another thing I noticed in my answer choices: the variable term is represented as a fraction, not as a decimal. That gives me the idea to convert my decimals, which I’ve been using so far, into fractions.

How is 0.02 (or 2%, our original number) represented as a fraction? It’s 2/100. What about 0.01 (or 1%)? That’s 1/100.

NOTE: when I’m reviewing my work, I’d tell myself that I should have been using fractions, not decimals, right from the start, when I noticed that the answers were in fraction form.

Okay, back to our math. Now we’ve got:

2R/100 + 50 + R /(100*2) = 2R /100 + 50 + R /200 = 4R /200 + 50 + R /200=

5R /200 + 50 = R /40 + 50

What if we want to use arithmetic instead? First, I pick a number (or numbers) for the variables at play. In this case, we have only one: I. Because this is a percent problem, I want to pick something that will work nicely with percents. Ordinarily, I’d pick 100, but I’m wary here “ 100 was given as part of the problem already and I don’t want to hit the one drawback of this method: picking something that works with more than one answer. (As a rule, avoid picking 0, 1, or something that already shows up in the problem.)

I’m still going to need to take 1% and 2% of the number, though, so I need to pick something that will work nicely with those. I’m going to pick 200.

1% of 200 is 2. 2% of 200 is double that “ 4. Plug those numbers in:

add 2 percent of one’s annual income to (this one’s my 4)

average (arithmetic mean) of 100 units of country R’s currency and 1 percent of one’s annual income (so I need the average of 100 and 2)

The average of 100 and 2 is (100+2)/2 = 51. 51+4 is 55. That’s my target answer: 55.

Now, I test the answers, substituting 200 in wherever the answer says I.

(A) 50 + I/200: 50 + 200/200 = 50 + 1 = 51. Nope.

(B) 50 + 3I/100: 50 + 3(200)/100 = 50+6 = 56. So close! But this isn’t right either.

(C) 50 + I/40: 50 + 200/40 = 50 + 5 = 55. Bingo!

(D) 100 + I/50: Let’s check just in case 100 + something? No. Too big.

(E) 100 + 3I/100: Ditto. 100 + something is too big.

The correct answer (still!) is C.

Notice how A and B worked very nicely because I happened to pick 200? You know what I’m going to say next, right? That wasn’t random. Use those answer choices to help you decide how to solve!

## Key Takeaways for Problem Solving Percent Problems:

1. Determine that you have a percent problem: this occurs when some type of percentage is discussed; possibly, the problem will also discuss fractions or decimals.
2. Before you start solving, glance at the answer choices for all problem solving problems. Does the format give you any clues about possible solution methods or other decisions you need to make along the way? If the answers contain variable expressions, you can turn the algebra into arithmetic instead. If the answer contains fractions, but the problem talks about decimals or percents, look for an opportunity to switch things over! You can even (sometimes) use the answers to be smart about what numbers you pick to turn algebra into arithmetic.
3. Know when you tend to prefer the algebra and when you tend to prefer the arithmetic; you should know yourself well enough to be able to make this choice instantly (after you’ve read the problem and as soon as you notice that the answers allow you to pick a number)
4. As always, write out all of your work and, when studying, analyze your work. Even when you answer correctly, there may still be an easier or more efficient way to do the problem!

* GMATPrep questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.

#### Stacey Koprince

Stacey Koprince is an Instructor and Trainer as well as the Director of Online Community for Manhattan Prep. She's also a management consultant who specializes in corporate strategy. She has been teaching various standardized tests for more than fifteen years and her entire teaching philosophy can be summed up in five words: teaching students how to think.

### 9 responses to Breaking Down a GMATPrep Percent Problem

1. I wonder why this one is called a tricky question. I agree that this question is bit ‘wordy’ as compared to other questions. but, easy to go with – totally my opinion.

Thanks anyways for bringing such questions to the light.. Really helps.

2. Saket

For me it is natural to recognize that problems which seem basic in my eyes may appear extremely complex and difficult in the eyes of others!

But that’s just how you and me are different. You’re better at math and I’m more insightful.

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