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Breaking Down a GMATPrep Percent Problem

by Stacey Koprince on October 3rd, 2010
7 comments
Stacey is a GMAT Instructor living in Montreal. Click here to read more articles from Manhattan GMAT and to learn more about Manhattan GMAT's classes.
Posted in
  • Arithmetic
  • GMAT Math
  • Percents

Math-ChalkboardThis week, we’re going to tackle a challenging GMATPrep® problem solving question from the topic of Percents.  (The GMATPrep® software can be downloaded for free at MBA.com)

Let’s start with the problem.

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

The correct answer is C.

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.

If you liked this article, let Stacey Koprince know by clicking Like.

RELATED ARTICLES

7 comments

  • Joanna on October 3rd, 2010 at 5:47 am

    The thing I found most difficult about this problem was figuring out what "100 units of country R’s currency" meant! I first represented it with a variable (100c) in my formula. Glacing at the answer choices I knew that there was not going to be another variable in there so I was hoping that it would cancel out during the calculations somehow. After a while when it didn't seem to be going away I just made an assumption (luckily a correct one) that it just meant the number 100.

    Reply to this comment
    • Ishaan Singhal on October 3rd, 2010 at 2:11 pm

      joanna by saying '100 units' is actually like saying '100 bucks' you know when u are not sure about the unit of measurement of a particular thing we use units in that case.
      For eg:- A rectangle has dimensions of 4x6 and a triangle has the dimensions of 2x3x4. find the difference of the area of the two figures.
      Don't bother solving this, the main idea is the question doesn't give any detail if it is cm/mm/km/inch etc. but still the numerical can be solved.

      I hope you understand.

      Thanks
      Ishaan

    • Joanna on October 3rd, 2010 at 11:29 pm

      Yep, it makes sense when you just keep it simple, but the fact that I'd never seen such wording before made my brain perk up and think that it was something more complex. I guess I just have to remember to start out with keeping things simple.

  • renu on October 3rd, 2010 at 6:18 am

    I was able to crack this question. However, it took 3.5 mins to get to the answer. I always try to do algebra and though I know its time consuming. I will have to tune myself into doing 'Plugging in the numbers' method to save more time.

    Thank you for the question Stacey.
    Your posts always help.

    Regards,
    Renu

    Reply to this comment
  • piku datta on October 4th, 2010 at 11:10 am

    once understood the language its easy.
    2 % of annual income gives (2/100)I = I/50----------------(1)

    2nd part is average of 100 and 1% of I which gives
    (100+I/100)/2 which simplify to give 50+ I/200-------------(2)

    adding 1 and 2

    I/50 + 50 + I/200
    = 50 + I/40............. ans. (c)

    Reply to this comment
  • A Hemalatha on October 6th, 2010 at 7:44 pm

    If x is 25% more than y, then by what percentage is y less than x? Please explain me.

    Reply to this comment
    • Stacey Koprince on October 7th, 2010 at 12:05 pm

      Good question. When we're given words and have to translate into math, it can often be helpful to try real numbers in order to understand what's going on.

      Let's start with the first part: "if x is 25% more than y"
      Hmm. Okay, let's pick a number for y. Try 100.
      "If x is 25% more than 100..."
      What does this mean now?

      25% more than 100 = 100 plus 25% of 100 = 100 + (100*0.25) = 100 + 25 = 125. So x = 125.

      Okay, so y = 100 and x = 125. The question is:
      "by what percentage is y less than x?"

      substitute the real numbers:
      "by what percentage is 100 less than 125?"

      And here's where we hit a little snag. In your example, the language is a little loose - the test wouldn't word it quite this way. What's our starting point? X? That question would be something like "y represents a percentage decrease of what amount?" (with x given as the starting point).

      If that's the question (x is our starting point, y represents a percentage decrease of what?), then we'd do this:
      difference / starting point = 25/125 = 1/5

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