# GMAT Math Cheat Sheet: Formulas and Tips for Success

An individual who is creating a study plan for the GMAT knows that math must be a part of the equation. Though many people love all sorts of math, there are some who become worried about the Quantitative portion of the exam.

If you’re concerned about the math questions on the GMAT, it can be useful to become more familiar with the specific content in this section. Find out about the types of problems in the Quantitative section and consider some GMAT geometry formulas. Also, check out a gathering of tips on how to prep in an effective way:

What is in the Quantitative Section?
Data Sufficiency and Problem-Solving are the two types of questions in the Quantitative section. The Problem-Solving questions are multiple-choice and test your skills in algebra, basic arithmetic, and geometry. The basic arithmetic questions involve decimals, positive and negative integers, fractions, percentages, and averages. The problems you find in this section are on par with the level of material taught in high school math classes. Though many of the questions on the exam involve basic arithmetic, it’s helpful to have a GMAT formula sheet to refer to when preparing for algebra and geometry problems.

GMAT Formulas for the Math Section
Your GMAT math formulas cheat sheet should include the Pythagorean Theorem. This formula helps you to find the measurement of the third side of a right triangle when given the measurements of the other two sides. Another item on your GMAT math cheat sheet should be A = 1/2 bh, which is the formula for finding the area of a triangle. Distance = rate*time is a very helpful formula to know, too. Find the area of a rectangle in fast fashion by using the formula A = lw. The formula A = s2 will help you discover the area of a square.

Moving Beyond Memorization

A GMAT math formulas cheat sheet is an effective study tool, but it’s equally important to know which formula to apply to a problem, so you should spend time practicing problems that employ each of those formulas. This way, on test day, you’ll be familiar with the formulas and feel comfortable using them. The easiest way to do this, of course, is to let us help you.

The expert instructors at Veritas Prep partner with students to help them learn and to practice these formulas for the Quantitative section. We hire tutors who have excellent teaching skills as well as GMAT scores in the 99th percentile. When you study with us, you know you’re learning from the best! Our instructors work through practice math problems with you to ensure that you understand how to solve them in the most efficient way.

Get the Timing Right
Test-takers are given 75 minutes to tackle the 37 questions in the Quantitative section. This sounds like a long time, but if you get hung up on one question for several minutes, you could end up running out of time for this section. In order to avoid this, you should take timed practice tests. Taking timed tests allows you to establish a rhythm for solving problems and answering questions. Once you establish a rhythm, you don’t have to be so concerned about running out of time before you finish all of the problems.

More Tips for Mastering the Quantitative Section
Studying with a GMAT math cheat sheet is one way to prepare for the test. Another way to save test time and make questions more manageable is to eliminate answer options that are clearly wrong – this allows your mind to focus only on the legitimate choices. Estimating the answer to a problem as you read through it is another way to save test time and arrive at answers more quickly.

Our GMAT curriculum teaches you how to approach questions on the separate math topics within the Quantitative section. Our strategies give you the tools you need to problem-solve like a business professional! We are proud to provide both online and in-person courses that prepare you for the GMAT. Veritas Prep instructors offer solid instruction as well as encouragement to individuals with the goal of acing the GMAT and getting into a preferred business school. Let us partner with you on the road to GMAT success! Contact us to talk with one of our course advisers today.

# 2 Tips to Make GMAT Remainder Questions Easy

Several months ago, I wrote an article aboutBecause the concepts of division, quotients, and remainders show up so often on the GMAT, I thought it would be useful to revisit this frequently-tested topic. (After all, there always remains at least something unsaid about remainders, right?) When you encounter quotient/remainder problems on the GMAT, at times, it will be helpful to know the kind of division terminology we’re taught in grade school – in particular the quotient + remainder formula you’ll see detailed below – while at other times, you will simply want to select simple numbers that satisfy the parameters of a Data Sufficiency statement.

To ensure that you’re prepared for all types of quotient/remainder problems on the GMAT, let’s explore each of these division-related scenarios in a little more detail. A simple example can illustrate the important division terminology: if we divide 7 by 4, we’ll have 7/4 = 1 + 3/4.

7, the term we’re dividing by something else, is called the dividend. 4, which is doing the dividing, is called the divisor. 1, the whole number component of the mixed fraction, is the quotient. And 3 is the remainder. This probably feels familiar even if the terminology takes a little reminding to come back to you.

In the abstract, the classic remainder formula is: Dividend/Divisor = Quotient + Remainder/Divisor. If we multiply through by the Divisor, we get another helpful variant of the remainder formula: Dividend = Quotient*Divisor + Remainder.

Simply knowing this terminology and the remainder equation will be sufficient to answer the following official GMAT question:

When N is divided by T, the quotient is S and the remainder is V. Which of the following expressions is equal to N?

A) ST
B) S + V
C) ST + V
D) T(S+V)
E) T(S – V)

In this division problem, N – which is getting divided by something else – is our dividend, T is the divisor, S is the quotient, and V is the remainder. Plugging the variables into our remainder equation of Dividend = Quotient*Divisor + Remainder, we get N = ST + V… and we’re done! The answer is C.

(Note that if you forgot the remainder formula, you could also pick simple numbers to solve this problem. Say N = 7 and T = 3. 7/3 = 2 + 1/3.  The Quotient is 2, and the remainder is 1, so V = 1. Now, if we plug in 3 for T, 2 for S, and 1 for V, we’ll want an N of 7. Answer choice C will give us an N of 7, 2*3 + 1 = 7, so this is correct.)

When we need to generate a list of potential values to test in a data sufficiency question, often a statement will give us information about the dividend in terms of the divisor and the remainder.

Consider, for example, the following question: when x is divided by 5, the remainder is 4. Here, the dividend is x, the divisor is 5, and the remainder is 4. We don’t know the quotient, so we’ll just call it q. In equation form, it will look like this: x = 5q + 4. Now we can generate values for x by picking values for q, bearing in mind that the quotient must be a non-negative integer.

If q = 0, x = 4. If q = 1, x = 9. If q=2, x = 14. Notice the pattern that emerges with our x values: x = 4 or 9 or 14, or 19… In essence, the first allowable value of x is the remainder. Afterwards, we’re simply adding the divisor, 5, over and over. Without much math at all, you could continue this cycle indefinitely: 4, 9, 14, 19, 24, 29, etc. This is a handy shortcut to use in complicated data sufficiency problems, such as the following:

If x and y are integers, what is the remainder when x^2 + y^2 is divided by 5?

1) When x – y is divided by 5, the remainder is 1
2) When x + y is divided by 5, the remainder is 2

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient
(C) Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
(D) EACH statement ALONE is sufficient
(E) Statements (1) and (2) TOGETHER are NOT sufficient

In this problem, Statement 1 gives us potential values for x – y. Remember the pattern discussed above: x – y must be 1 greater than a multiple of 5. If we begin with the remainder (1) and continually add the divisor (5), we know that x – y = 1 or 6 or 11, etc. If x – y = 1, we can say that x = 1 and y = 0. In this case, x^2 + y^2 = 1 + 0 = 1, and the remainder when 1 is divided by 5 is 1. If x – y = 6, then we can say that x = 7 and y = 1. Now x^2 + y^2 = 49 + 1 = 50, and the remainder when 50 is divided by 5 is 0. Because the remainder changes from one scenario to another, Statement 1 is not sufficient by itself.

Statement 2 gives us potential values for x + y. Once again, let’s use that pattern: x + y must be 2 greater than a multiple of 5. If we begin with the remainder (2) and continually add the divisor (5), we know that x + y = 2 or 7 or 12, etc. If x + y = 2, we can say that x = 1 and y = 1. In this case, x^2 + y^2 = 1 + 1 = 2, and the remainder when 2 is divided by 5 is 2. If x + y = 7, then we can say that x = 7 and y = 0. Now x^2 + y^2 = 49 + 0 = 49, and the remainder when 49 is divided by 5 is 4. Because the remainder changes from one scenario to another, Statement 2 is also not sufficient on its own.

Now that we find ourselves in the classic C or E scenario, let’s test them together – simply select one scenario from Statement 1 and one scenario from Statement 2 and see what happens. Say x – y = 1 and x + y = 7. Adding these equations, we get 2x = 8, or x = 4. If x = 4, y = 3. Now x^2 + y^2 = 16 + 9 = 25, and the remainder when 25 is divided by 5 is 0.

Now remember this: for us to pick a non-E answer on Data Sufficiency, we must know that the value will stay the same in any scenario allowed by the question and statements. To be safe, let’s try another scenario. Say x – y = 6 and x + y = 12. Adding the equations, we get 2x = 18, or x = 9. If x = 9, y = 3, and x^2 + y^2 = 81 + 9 = 90. The remainder when 90 is divided by 5 is, again, 0. No matter which values we select, this will be the case – we can prove definitively that the remainder is 0. Together, the statements are sufficient, so the correct answer is C.

Now let’s summarize some important takeaways regarding GMAT quotient/remainder problems and the ever-important remainder formula. You’re virtually guaranteed to see remainder questions on the GMAT, so you want to make sure you have this concept mastered. First, make sure you feel comfortable with the remainder formula: Dividend = Divisor*Quotient + Remainder. Second, if you need to select values, you can simply start with the remainder and then add the divisor over and over again. If you internalize these two ideas, remainder questions will become considerably less daunting.

Remember, also, that division is something you were once quite good at as a primary school student, so do not let the terminology intimidate you as an adult! The GMAT thrives on abstraction in these problems, so if you find yourself distracted by terminology or abstraction, simply try using small numbers to remind yourself how the operation works. The remainder equation Dividend = Divisor*Quotient + Remainder is an important one, but if you blank on it you can reconstruct it. Try, as we did at the beginning, 7 divided by 4. The result of that is 1, remainder 3. And the quotient is 1 because 4 goes into 7 one time, leaving 3 left over. So you can reconstruct the equation: to get back to 7, multiply the divisor (4) by the 1 time it went in to 7, and then add back the remaining 3: 7 = 4(1) + 3. Once you’ve stripped away the abstraction in quotient/remainder problems, the remainder is a concept you’ve been quite adept with your whole life!

Interested in more practice with GMAT division problems and the remainder equation? Check out some of our other articles on this frequently-tested GMAT topic, or try your hand at some practice questions via the Veritas Prep GMAT Question Bank or practice tests.

# A 750+ Level GMAT Geometry Question

Today we will discuss a pretty advanced GMAT question, because we can still use our basic GMAT concepts to find the answer. It may seem like we will need trigonometry to handle this question, but that is not so. In fact, the question will look familiar at first, but will present unforeseen problems later on.

While going through this exercise, we will learn a few tips and tricks which will be useful in our mainstream GMAT questions, hence, it will add value to our GMAT repertoire (especially in elimination techniques). Let’s go on to the question now:

In triangle ABC, if angle ABC is 30 degrees, AC = 2*sqrt(2) and AB = BC = X, what is the value of X?

(A) Sqrt(3) – 1

(B) Sqrt(3) + 2

(C) (Sqrt(3) – 1)/2

(D) (Sqrt(3) + 1)/2

(E) 2*(Sqrt(3) + 1)

What we see here is an isosceles triangle with one angle as 30 degrees and other two angles as (180 – 30)/2 = 75 degrees each.

The side opposite the 30 degrees angle is 2*sqrt(2). One simple observation is that X must be greater than 2*sqrt(2) because these sides are opposite the greater angles (75 degrees).

2*sqrt(2) is a bit less than 2*1.5 because Sqrt(2) = 1.414. So 2*sqrt(2) is a bit less than 3. Note that options (A), (C), and (D) are much smaller than 3, so these cannot be the value of X. We have already improved our chances of getting the correct answer by eliminating three options! Now we have to choose out of (B) and (E).

Here is what is given: Angle ABC = 30 degrees, and AC = 2*sqrt(2). We need to find the value of X. Now, our 30 degree angle reminds us of a 30-60-90 triangle in which we know the ratio of the sides – given one side, we can find the other two.

The problem is this: if we drop an altitude from angle B to AC, the angle 30 degrees will be split in half and we will actually get a 15-75-90 triangle, instead. We won’t have a 30-60-90 triangle anymore, so what do we do now? Let’s try to maintain the 30 degree angle as it is to get the 30-60-90 triangle, and drop an altitude from angle C to AB instead, calling it CE. Now we have a 30-60-90 triangle! Since BCE is a 30-60-90 triangle, its sides are in the ratio 1:sqrt(3):2. Side X corresponds to 2 on the ratio, so CE = x/2.

Area of triangle ABC = (1/2)*BD*AC = (1/2)*CE*AB

(1/2)*BD*2*sqrt(2) = (1/2)*(X/2)*X

BD = X^2/4*Sqrt(2)

Now DC = (1/2)AC = 2*sqrt(2)/2 = sqrt(2)

Let’s use the pythagorean theorem on triangle BDC:

BD^2 + DC^2 = BC^2

(X^2/4*Sqrt(2))^2 + (Sqrt(2))^2 = X^2

X^4/32 + 2 = X^2

X^4 – 32*X^2 + 64 = 0

X^4 – 16X^2 + 8^2 – 16X^2 = 0

(X^2 – 8)^2 – (4X)^2 = 0

(X^2 -8 + 4X) * (X^2 – 8 – 4X) = 0

Normally, this would require us to use the quadratic roots formula, but let’s not get that complicated. We can just plug in the the two shortlisted options and see if either of the factors is 0. If one of the factors becomes 0, the equation will be satisfied and we will have the root of the equation.

Since both options have both terms positive, it means the co-efficient corresponding to B in Ax^2 + Bx + C = 0 must be negative.

x = [-B +- Sqrt(B^2 – 4AC)]/2A

-B will give us a positive term if B is negative, so we will get the answer by plugging into (X^2 – 4X – 8):

Put X = Sqrt(3) + 2 in X^2 – 4X – 8 and you do not get 0.

Put X = 2*(Sqrt(3) + 1) in X^2 – 4X – 8 and you do get 0.

This means that X is 2*(Sqrt(3) + 1), so our answer must be (E).

To recap:

Tip 1: A greater side of a triangle is opposite a greater angle.

Tip 2: We can get the relation between sides and altitudes of a triangle by using the area of the triangle formula.

Tip 3: The quadratic formula can help identify the sign of the irrational roots.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

# Quarter Wit, Quarter Wisdom: Should You Use the Permutation or Combination Formula?

A recurring question from many students who are preparing for GMAT is this: When should one use the permutation formula and when should one use the combination formula?

People have tried to answer this question in various ways, but some students still remain unsure. So we will give you a rule of thumb to follow in all permutation/combination questions:

You never NEED to use the permutation formula! You can always use the combination formula quite conveniently. First let’s look at what these formulas do:

Permutation: nPr = n!/(n-r)!
Out of n items, select r and arrange them in r! ways.

Combination: nCr = n!/[(n-r)!*r!]
Out of n items, select r.

So the only difference between the two formulas is that nCr has an additional r! in the denominator (that is the number of ways in which you can arrange r elements in a row). So you can very well use the combinations formula in place of the permutation formula like this:

nPr = nCr * r!

The nCr formula is far more versatile than nPr, so if the two formulas confuse you, just forget about nPr.

Whenever you need to “select,” “pick,” or “choose” r things/people/letters… out of n, it’s straightaway nCr. What you do next depends on what the question asks of you. Do you need to arrange the r people in a row? Multiply by r!. Do you need to arrange them in a circle? Multiply by (r-1)!. Do you need to distribute them among m groups? Do that! You don’t need to think about whether it is a permutation problem or a combination problem at all. Let’s look at this concept more in depth with the use of a few examples.

There are 8 teachers in a school of which 3 need to give a presentation each. In how many ways can the presenters be chosen?

In this question, you simply have to choose 3 of the 8 teachers, and you know that you can do that in 8C3 ways. That is all that is required.

8C3 = 8*7*6/3*2*1 = 56 ways

Not too bad, right? Let’s look at another question:

There are 8 teachers in a school of which 3 need to give a presentation each. In how many ways can all three presentations be done?

This question is a little different. You need to find the ways in which the presentations can be done. Here the presentations will be different if the same three teachers give presentations in different order. Say Teacher 1 presents, then Teacher 2 and finally Teacher 3 — this will be different from Teacher 2 presenting first, then Teacher 3 and finally Teacher 1. So, not only do we need to select the three teachers, but we also need to arrange them in an order. Select 3 teachers out of 8 in 8C3 ways and then arrange them in 3! ways:

We get 8C3 * 3! = 56 * 6 = 336 ways

Let’s try another one:

Alex took a trip with his three best friends and there he clicked 7 photographs. He wants to put 3 of the 7 photographs on Facebook. How many groups of photographs are possible?

For this problem, out of 7 photographs, we just have to select 3 to make a group. This can be done in 7C3 ways:

7C3 = 7*6*5/3*2*1 = 35 ways

Here’s another variation:

Alex took a trip with his three best friends and there he clicked 7 photographs. He wants to put 3 of the 7 photographs on Facebook, 1 each on the walls of his three best friends. In how many ways can he do that?

Here, out of 7 photographs, we have to first select 3 photographs. This can be done in 7C3 ways. Thereafter, we need to put the photographs on the walls of his three chosen friends. In how many ways can he do that? Now there are three distinct spots in which he will put up the photographs, so basically, he needs to arrange the 3 photographs in 3 distinct spots, which that can be done in 3! ways:

Total number of ways = 7C3 * 3! = (7*6*5/3*2*1) * 6= 35 * 6 = 210 ways

Finally, our last problem:

12 athletes will run in a race. In how many ways can the gold, silver and bronze medals be awarded at the end of the race?

We will start with selecting 3 of the 12 athletes who will win some position in the race. This can be done in 12C3 ways. But just selecting 3 athletes is not enough — they will be awarded 3 distinct medals of gold, silver, and bronze. Athlete 1 getting gold, Athlete 2 getting silver, and Athlete 3 getting bronze is not the same as Athlete 1 getting silver, Athlete 2 getting gold and Athlete 3 getting bronze. So, the three athletes need to be arranged in 3 distinct spots (first, second and third) in 3! ways:

Total number of ways = 12C3 * 3! ways

Note that some of the questions above were permutation questions and some were combination questions, but remember, we don’t need to worry about which is which. All we need to think about is how to solve the question, which is usually by starting with nCr and then doing any other required steps. Break the question down — select people and then arrange if required. This will help you get rid of the “permutation or combination” puzzle once and for all.