GMAT Quantum Math Notes

8/16/2019 GMAT Quantum Math Notes

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GMATQuantum Math Notes

http://gmatquantum.com/

July 2013, Version 1.0

Number Theory

Integers• The set of integers consists of the whole numbers (1, 2, 3, . . . ) and their negatives, including zero. The set of integersextends infinitely in both positive and negative directions.

• Positive integer refers to all integers greater than zero. Example: 1, 2, 3, . . ..• Negative integer refers to all integers less than zero. Example: −1, −2, −3, . . ..• The set of non-negative integers is: 0, 1, 2, 3, . . ..• Zero is neither a positive nor a negative integer, but is an even integer.

Basic Operations

• Associative Law of Addition: (a + b) + c = a + (b + c).• Commutative Law of Addition: a + b = b + a.• Associative Law of Multiplication: a · (b · c) = (a · b) · c = a · b · c.• Commutative Law of Multiplication: a · b = b · a.• Distributive Law: a · (b + c) = a · b + a · c• Notice that the operation of subtraction does not follow the associative or commutative law.

Odd and Even Integers

• Numbers that are divisible by 2 are called even, and all other numbers not divisible by 2 are called odd. The general

form of even numbers is 2k and that of odd numbers is 2k + 1, where k is an integer.

• The following list summarizes the outcome of operations of sum, difference, product, and powers when applied to oddand even integers.

Even ± Even = EvenEven ± Odd = OddOdd ± Odd = EvenEven × Even = EvenEven × Odd = Even

1

http://gmatquantum.com/http://gmatquantum.com/


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GMATQuantum Math Notes 2

Odd × Odd = OddOddEven = Odd (Example: 34 = 81)

OddOdd = Odd (Example: 33 = 27)

EvenEven

= Even (Example: 24

= 16)EvenOdd = Even (Example: 25 = 32)

Consecutive Integers

• Consecutive integers are those integers that follow each other in a sequence, where the difference between any twosuccessive integers is 1. They can be algebraically represented by n, n + 1, n + 2, n + 3, .., where n is an integer.Example: −3, −2, −1, 0, 1, 2, 3.

• Consecutive even integers can be represented by 2n, 2n + 2, 2n + 4, ...• Consecutive odd integers can be represented by 2n + 1, 2n + 3, 2n + 5, ...

• If there are an odd number of consecutive integers, it is better to assign the middle number as n. For example, to

represent a set of seven consecutive integers, let n be the middle number, then the set can be represented as {n − 3,n − 2, n − 1, n, n + 1, n + 2, n + 3}. In this representation, it is easy to add up the terms algebraically, because thenumbers cancel out, to yield 7n.

• The sum of n consecutive integers, where n is odd, is always divisible by n. In general, the sum of n consecutive integers,where n is odd, is given by nx, where x is the middle integer in the set. For example, the sum of three consecutive oddintegers, x − 1, x, and x + 1, is equal to 3x, which is always divisible by 3, or in other words is a multiple of 3.

• The sum of k consecutive integers, where k is even, is always divisible by k2

. For example, consider the set of 10

consecutive integers where the 5th integer is represented by x. The ten consecutive integers can be represented as{x − 4, x − 3, x − 2, x − 1, x, x + 1, x + 2, x + 3, x + 4, x + 5}. The sum of the ten terms is equal to 10 x + 5, whichcan also be written as 5(2x + 1), and is always a divisible by 5. In general, if x is the

n

2th term in the sequence of an

even number of consecutive integers, then the sum of the n consecutive integers is given by n2 (2x + 1).• The number of integers that lie between a and b, inclusive of a and b, is given by b −a +1. Example: How many integers

are there between −4 and 3? Answer: 3 − (−4) + 1 = 8.• The product of two consecutive integers can be represented as: n(n + 1) = n2 + n or n(n − 1) = n2 − n.• The product of any two consecutive integers is always even because one of them has to be even, therefore the following

expressions n2 − n = n(n − 1) and n2 + n = n(n + 1) are always even for all values of n.• The product of three consecutive integers can be represented as (n − 1)n(n + 1) = n3 − n.• n3 − n which is the product of three consecutive integers is always divisible by 6, because at least one of the three

integers is even, and one of them is a multiple of 3.

• Among a set of k consecutive integers, exactly one integer is a multiple of k . Example: 12, 13, 14, 15, 16 contains 15which is a multiple of 5, similarly, a set of three consecutive integers will always have one integer that is a multiple of 3.

• The product of k consecutive integers is always divisible by k . The reason is that there is one integer that is a multipleof k . Example: Consider 10, 11, 12, 13 as the four consecutive integers, their product 10 × 11 × 12 × 13 is divisible by4 because 12 is a multiple of 4.

• In general, the product of k consecutive integers is always divisible by k!. For example, the product of four consecutiveintegers, such as 7 × 8 × 9 × 10 = 5040 = 24 × 210 is always divisible by 4! = 4 × 3 × 2 × 1 = 24.


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Divisibility Rules

• 2 The last digit is even, the number is divisible by 2.

• 3 If the sum of the digits is divisible by 3, the number is also.

• 4 If the last two digits form a number divisible by 4, the number is also.• 5 If the last digit is a 5 or a 0, the number is divisible by 5.• 6 If the number is divisible by both 3 and 2, it is also divisible by 6.• 8 If the last three digits form a number divisible by 8, then so is the whole number.• 9 If the sum of the digits is divisible by 9, the number is also.• 10 If the last digit of a number is 0, then the number is divisible by 10.

Prime Numbers

• Prime numbers: A prime number is any positive integer greater than 1 that has exactly two whole number factors,itself and the number 1. The number 1 itself is not a prime. The table below lists the prime numbers less than 100. Apositive integer that is greater than 1 and is not prime is called composite.

2 3 5 7 1113 17 19 23 2931 37 41 43 4753 59 61 67 7173 79 83 89 97

• 2 is the only even prime number.

• To test if a given number less than 100 is prime, divide the number by 2, 3, 5, and 7, if the number is not divisible

by any of these four prime numbers, then the number is prime. For example, the number 89 is prime because it is notdivisible by 2, 3, 5, or 7.

• 2 and 3 are the only pair of consecutive integers that are both prime, because any other pair of consecutive integers willalways have one number that is even, which will be divisible by 2, and therefore cannot both be prime.

• 3, 5, and 7 are the only three consecutive odd integers that are all prime numbers. Again, when we select a set of threeconsecutive odd integers, at least one of them is divisible by 3, and therefore all sets of three consecutive odd integerswill always have a multiple of 3, with the exception of 3, 5, and 7.

• The possible units digit of all prime number greater than 5 are 1, 3, 7, and 9.• If the number n! is defined as the product of all positive numbers 1 through n

n! = 1 · 2 · 3 · · · · · nthen the n − 1 numbers n! + 2, n! + 3, n! + 4, . . . n! + n are all composite, or in other words not prime.

• For example, the number 11! + 7 is not prime because we can factor a 7 and rewrite the number as product of twonumbers as shown below:

11! + 7 = (11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1) + 7 = 7[(11)(10)(9)(8)(6)(5)(4)(3)(2)(1) + 1]


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Prime Factorization

• Fundamental Theorem of Arithmetic: Every integer greater than or equal to 2 is either a prime number or canbe written uniquely as the product of two or more prime numbers. The factorization in to the prime numbers is uniqueexcept for the order in which they are written. For example, 120 can be written as 120 = (2)(2)(2)(3)(5) = (23)(31)(51).

• Number of Divisors of a Composite Number: Any composite number can be resolved into prime factors in onlyone way and in the most general case, N can be written as:

N = pr11 pr22 p

r33 · · · prii

The divisors or factors of N are numbers of the form

n = ps11 ps22 p

s33 · · · psii where 0 ≤ sj ≤ rj for all j = 1, 2, 3, . . . i

Since there are (rj + 1) choices for each rj , the number of divisors of N is

(r1 + 1)(r2 + 1)(r3 + 1) · · · (ri + 1)

Another way to think about this problem is that each term of the following product:

(1 + p1 + p21 + · · · + pr11 )(1 + p2 + p22 + · · · + pr22 ) · · · (1 + pi + p2i + · · · + prji )

is a divisor of N , and that no other number is a divisor. The total number of terms in this expression is thus equal tothe total number of divisors of N , including 1 and the number itself.

• Prime numbers have two factors, 1 and the prime number itself.• Numbers that have only three factors are square of a prime number. Let n = p2,where p is a prime number, the factors

of n are: 1, p, and p2.

• The number of factors of pm, where p is prime, and m is a positive integer is equal to m + 1. The factors are {1, p, p2, p3, . . . , pm−1, pm

}.

Largest Factor of n!

Highest power of prime factor that divides n!: What is the largest value of k for which 3k is a factor of 100!? To findthe highest power of a prime number p contained in n!, divide the number n repeatedly by p, p2, p3, ...to obtain the set of quotients that are greater than or equal to one. The highest power of p contained in the prime factorization of n! is given by:

Highest power of p =

n

p

+

n

p2

+

n

p3

+ · · ·

where the floor function or the greatest integer function, ⌊x⌋ is defined as the largest integer less than or equal to x.We can use the above expression to answer the original question, which gives us:

Highest power of 3 = 1003 + 10032 + 10033 + 10034 = 33 + 11 + 3 + 1 = 48Last Digit: Repeated Multiplication

• The problems on GMAT that deal with the last digit of a number resulting from repeated multiplication can be solvedeasily by observing the repeating patterns of the units digits of consecutive integral powers of numbers from 0 to 9, as


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Digits and Place Value

• Any number can be expressed in terms of their digits by using the base 10 expression. For example, a two digit numberab, where a is the tens digit and b is the units digit, can be written as:

Number ab = 10a + b• If the digits are reversed the number can be expressed as: Number ba = 10b + a• The same concept can be expanded to larger numbers, for example a three digit number

Number abc = 100a + 10b + c

• If we select any two digit number and reverse the digits, then the difference between the two numbers thus formed willbe nine times the difference between the two digits. For example, if n = 64, then 64 − 46 = 18 = 9(6 − 4). In otherwords, if n = tu, where t is the tens digit and u is the units digit, then n = 10t + u, and the number formed by reversingthe digits, m = ut = 10u + t, therefore, n − m = 10t + u − (10u + t) = 9t − 9u = 9(t − u).

• If we select any number and subtract the number formed by reversing its digits, then the resulting difference is alwaysdivisible by 9 and 11. Let n = abc = 100a + 10b + c, where a is the hundreds digit, b is the tens digit, and c is the unitsdigit. If we reverse the digits, the new number formed is m = cba = 100c + 10b + a. The difference between the twonumbers is given by n − m = (100a + 10b + c) − (100c + 10b + a) = 99a − 99c = 99(a − c)

Divisors and Multiples

Let m be a non-zero integer, and n be an arbitrary integer. If there is an integer, k , such that n = km, then we say that mdivides n. The following table lists statements that are equivalent to m divides n.

m divides nm is a factor or divisor of n

n is divisible by mn is a multiple of m

For example, if n = 18 and m = 6, then 6 divides 18, 6 is a factor of 18, 18 is divisible by 6, and 18 is a multiple of 6. These

statements are commonly used on the GMAT test.The following list of statements are based on the above definition of divisibility:

• 0 is divisible by any non-zero integer. For example, 0 is divisible by 5 and the outcome is 0.• Any non-zero number n is always divisible by itself.• Any number n including zero is divisible by 1 and −1.• If r is divisible by s and s is divisible by t, then r is divisible by t. If integer t is called the common divisor or common

factor of r and s. For example, 6 is a common divisor or common factor of 18 and 30.

• If r is divisible by t and s is divisible by t, then the expression r + s is divisible by t.

• If r is divisible by t and s is divisible by t, then the expression ra + sb is divisible by t for all integer values of a and b.

• If r is divisible by t, then rs is divisible by t for all integers s.• Number of Multiples: The number of positive integers that are less than or equal to n and divisible by an integer k

is equal ton

k

, where the floor function or the greatest integer function, ⌊x⌋ is defined as the largest integer less than

or equal to x. For example, the number of positive integers less than or equal to 20 that are divisible by 3 is equal to20

3

= ⌊6.67⌋ = 6.


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Greatest Common Factor and Least Common Multiple

• Greatest Common Factor: GCF of a set of two or more integers is the largest integer that divides the given set of integers.

• Least Common Multiple: The LCM of two (or more) nonzero integers is the least positive integer that is divisibleby all of them.

• To obtain GCF and LCM, first obtain the prime factorization of all the numbers, that is break them down into theirprime constituents.

Example: Given 48 and 120, find the GCF and LCM. The prime factorization of 48 is 24 × 3 and that of 120 is23×3 ×5. The GCF is obtained by multiplying the common factors by picking the lowest power of the common factors,in this case 23 × 3 = 24. The LCM is obtained by multiplying all the prime factors, and picking the highest power, inthis case 24 × 3 × 5 = 240.

• Also, the GCF and LCM of two numbers satisfies the following property:GCF × LCM = Product of the Two Numbers

• The integers m and n are called relatively prime if their greatest common factor is 1. For example. 15 = (3)(5), and28 = (2)(2)(7), are relatively prime.

• A pair of consecutive positive integers is always relatively prime, and their greatest common divisor or factor is 1.• The greatest common divisor or factor of consecutive even integers is 2.• The greatest common divisor of factor of consecutive odd integers is 1.• The greatest common factor of any integer n and 1 is 1.• The greatest common factor of any non-zero positive integer n and 0 is n.• If n is divisible by m, then the greatest common factor of n and m is equal to m.

Remainder Algorithm

• If n and m are positive integer, then there exist unique integers q and r, called the q uotient and r emainder, respectively,such that:

n = mq + r and 0 ≤ r < mThe above relationship can also be written as:

n

m = q +

r

m and 0 ≤ r < m

If r = 0, then n = mq , and n is a multiple of m.

• The remainder is defined to be a positive number. For example, if a number is divided by 5, the only possible remaindersare: 0, 1, 2, 3, 4.

• If the remainder when x and y are divided by m are r1 and r2, respectively, then the remainder when x + y is dividedby m is equal to the remainder when r1 + r2 is divided by m. For example, when 26 and 46 are divided by 7, theremainders are 5 and 4, respectively. If we add the two remainders, 5 + 4 = 9, and divide by 7, the remainder is 2,which is equal to the remainder when 26 + 46 = 72 is divided by 7.

• If the remainder when x and y are divided by m are r1 and r2, respectively, then the remainder when xy is dividedby m is equal to the remainder when (r1)(r2) is divided by m. For example, when 26 and 46 are divided by 7, theremainders are 5 and 4, respectively. If we multiply the two remainders, 5 × 4 = 20, and divide by 7, the remainder is6, which is equal to the remainder when 26 × 46 = 1196 is divided by 7.


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• When a number is divided by 5, the remainder is equal to the remainder when the last digit of the number is dividedby 5. For example, the remainder when 27 is divided by 5, is the same as the remainder when 7 is divided by 5, whichis 2. The reason for this rule is that, a number like, 64578 can be expressed as:

64578 = 60000 + 4000 + 500 + 70 + 8The tens and higher powers of 10, are all divisible by 5, irrespective of the place value, and the remainder only dependson the last digit of the number.

• When a number is divided by 10, the remainder is the same as the units digit of the number.

Arithmetic

Common Fractions and Percent Equivalent

Fraction Decimal Percentage

1

10 0.1 10%

1

8 0.125 12.5%

1

4 0.25 25%

1

3 0.333. . . 33 13%

2

5 0.4 40%

1

2 0.5 50%

3

5 0.6 60%

2

3 0.666. . . 66 23%

3

4 0.75 75%

4

5 0.8 80%

1 1 100%


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Approximations:

I have listed some common approximations that the GMAT test writers expect students to be able to perform.

• (0.998)3

≈ 1

• 0.971.98

≈ 12

= 0.5

• √ 15.89 ≈ 4

• 1.0012 − 1

0.001 =

1.0012 − 120.001

= (1.001 + 1)(1.001 − 1)

0.001 =

(2.001)(0.001)

0.001 ≈ 2

• √ 0.000035 ≈ √ 0.000036 = √ 36 × 10−6 = (√ 36)(√

10−6) = 6 × (10−6) 12 = 6 × 10−3

Percentages

• Percentage: The term percent means per hundred. Percents are ratios that are used to represent parts of a whole,

where the whole consists of 100 parts. 5 percent means 5 parts out of 100, or 5% =

5

100 =

1

20 . Part

Whole

100 = Percent

• Percent Change: To compute the percent change when a positive quantity changes from an initial amount to anotherpositive amount, we compute the amount of change and then divide this by the initial amount and multiply by 100.

Percent Change =

Final − Initial

Initial

100

If the final quantity is less than the initial quantity, we end up with a percent decrease. In both cases, the denominatoris always the initial value.

Algebra

Simplifying Equations

• In equations that involve variables, in general, do not divide each side of an equation by a variable. For example, if weare given xy = y, we cannot divide both sides by y , and conclude that x = 1. What if y = 0? In that case, x could takeon any value. In general, bring all the terms to one side, equate to zero, and then factor.

xy = y

xy − y = 0y(x

−1) = 0

which means either y = 0 or x = 1.

Common Algebraic Identities

GMAT places a heavy emphasis on the following three algebraic identities. You will be expected to recognize these identitiesn problem statements and be able to turn one form into another.


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Difference of Squares

(a + b)(a − b) = a2 − b2

Examples:

• 512 − 492 = (51 + 49)(51 − 49) = (100)(2) = 200

• 9992 − 1

998 =

9992 − 12998

= (999 + 1)(999 − 1)

998 =

(1000)(998)

998 = 1000

• x4 − 1

x − 1 = (x2)2 − 12

x − 1 = (x2 + 1)(x2 − 1)

x − 1 = (x2 + 1)(x + 1)(x − 1)

x − 1 = (x2 + 1)(x + 1)

• (x + 2y)2 − (x − 2y)2 = [(x + 2y) + (x − 2y)][(x + 2y) − (x − 2y)] = [x + 2y + x − 2y][x + 2y − x + 2y] = [2x][4y] = 8xy• (a4 − b4) = (a2 + b2)(a2 − b2) = (a2 + b2)(a + b)(a − b)

Square of Sums

(a + b)2 = a2 + b2 + 2ab

Examples of application:

• (3 + 5)2 = 82 = 64 = 32 + 52 + 2(3)(5) = 9 + 25 + 30 = 64• (2x + 3y)2 = (2x)2 + (3y)2 + 2(2x)(3y) = 4x2 + 9y2 + 12xy• (x2 + y2)2 = (x2)2 + (y2)2 + 2(x2)(y2) = x4 + y4 + 2x2y2

•

x + 1

x

2= x2 +

1

x2 + 2

•

x2 + 1

x2

2= x4 +

1

x4 + 2

Square of Difference

(a − b)2 = a2 + b2 − 2abExamples:

• (7 − 4)2 = 32 = 9 = 72 + 42 − 2(7)(4) = 49 + 16 − 56 = 9• (5x − 2y)2 = (5x)2 + (2y)2 − 2(5x)(2y) = 25x2 + 4y2 − 20xy

• (x2 − y2)2 = (x2)2 + (y2)2 − 2(x2)(y2) = x4 + y4 − 2x2y2

•

x − 1x

2= x2 +

1

x2 − 2

•

x2 − 1x2

2= x4 +

1

x4 − 2


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Quadratic Equations

A quadratic equation has the form ax2 + bx + c = 0, where a, b, and c are real numbers, and a = 0, for example x2−x−6 = 0.The values of x that satisfy a given quadratic equation are called roots. The roots of any quadratic equation can be obtainedby factoring:

x2 − x − 6 = (x − 3)(x + 2) = 0When x is either 3 or −2, the above quadratic equation is satisfied.

Factoring Quadratic Equations

• The quadratic equations that are tested on the GMAT can be factored by following these steps. I will use the exampleof 2x2 − 5x − 33 = 0.

• Multiply the coefficient of x2(2 here) and the constant(−33), which gives a value of −66.• Find two numbers that multiply to give −66 but add up to the coefficient of x, which is −5.• The two numbers are −11 and 6.• Rewrite the middle term as the sum of these two numbers −5x = −11x + 6x.

2x2 − 5x − 33 = 0

2x2 − 11x + 6x − 33 = 0

Factor the largest common term from the first two terms, and also from the last two terms.

x(2x − 11) + 3(2x − 11) = 0

Factor (2x − 11)

(2x − 11)(x + 3) = 0

• The roots are then obtained by solving 2x − 11 = 0 and x + 3 = 0, which gives 112

and −3 as the two roots of thequadratic equation 2x2 − 5x − 33 = 0.

Quadratic Formula

The solutions to the quadratic equation ax2 + bx + c = 0 can also be obtained by using the formula:

x = −b ± √ b2 − 4ac

2a

Example; 2x2 − 5x − 33 = 0, we have a = 2, b = −5, and c = −33. The quadratic formula yields

x = −(−5) ±

(−5)2 − 4(2)(−33)

2(2)

x = 5 ± √ 289

4 =

5 ± 174

The two solutions are x = 5 + 17

4 =

11

2 and x =

5 − 174

= −12

4 = −3.


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Use of Quadratic Formula on the GMAT

Here I list the cases where the quadratic formula comes in handy on the GMAT:

• The quadratic equations tested on the GMAT are meant to be solved by factoring, but one can use the quadraticformula as a back up plan in case one is not successful by following the factoring approach.

• Some quadratic equations have only one real solution, and this results when b2 − 4ac = 0. Graphically this means thatthe corresponding parabola y = ax2 + bx + c is tangent to the x−axis.

• Some quadratic equations have no real solutions, and this results when b2 − 4ac b, and c is any real number, then a + c > b + c.• Two different inequalities can always be added. This is the most common operation with inequalities on the GMAT.

If a > b and c > d, then a + c > b + d

In words, the sum of two larger quantities exceeds the sum of the two smaller quantities. For example, 7 > 4 and 6 > 2,and 7 + 6 > 4 + 2, or 13 > 6.

Subtraction in Inequalities

• If a > b, and c is any real number, then a − c > b − c.• If a > b and c > d, then a − d > b − c. Note that this is equivalent to adding these two inequalities, which gives

a + c > b + d and then rearranging to yield a − d > b − c.

• Never subtract two inequalities, in general if a > b and c > d, then a − c ≯ b − d. For example, 7 > 4 and 6 > 2,however, 7 − 6 ≯ 4 − 2, or 1 ≯ 2.

Multiplication in Inequalities

• For any real numbers a, b, and any positive number c.If a > b, then a · c > b · c

The converse of the above statement is also true.

• For any positive numbers a, b, c and d.If a > b and c > d, then a · c > b · d

Division in Inequalities

• For any real numbers a, b, and any positive number c.

If a > b, then a

c >

b

c

The converse of the above statement is also true.

• For any positive numbers a, b, c and d.

If a > b and c > d, then a

d >

b

c


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Rewriting Inequalities

• Do not multiply or divide by a variable if you don’t know whether the quantity is positive or negative. For example, if we are given 1/x ≤ 1, and if we multiply both sides by x, we incorrectly conclude that the only values of x that satisfythis inequality is x

≥ 1. A proper way to obtain the solution is:

1

x ≤ 1

1 − 1x ≥ 0

x − 1x

≥ 0

Therefore, either x ≥ 1 or x < 0. In general, collect all the terms on one side by subtracting or adding, and thendetermine the values of the variable which satisfy the given inequality.

Absolute Value

Absolute Value Definition

The absolute value of the real number x, denoted by |x| is defined to be x if x is positive or zero, and to be −x if x is negative.In other words,

|x| =

x, if x ≥ 0−x, if x


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Absolute Value Equalities

• For any number a,| − a| = |a|

• For any numbers a and b,|a · b| = |a| · |b|

• For any number a,|a2| = |a|2

Absolute Value Inequalities

• The solution set of |x| ≤ p, where p > 0 is − p ≤ x ≤ p.• The solution set of |x| ≥ p, where p > 0 is x ≤ − p and x ≥ p.• |a| + |b| = |a + b|, if and only if ab ≥ 0, or in other words, a ≥ 0, b ≥ 0 or a ≤ 0, b ≤ 0.

• |a| + |b| > |a + b|, if and only if ab 0, b ||a| − |b||, if and only if ab 0, b


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Average Speed

• Average Speed: When a person travels a given route of length, d1, at an average speed of v1, and travels a differentroute of length, d2, at an average speed of v2. The average speed for the entire trip is given by:

Average Speed = Total DistanceTotal Time

= v̄ = d1 + d2d1v1

+ d2v2

In the special case, where the person travels to a location and returns by the same route, d1 = d2 = d, and the averagespeed is independent of the length of the trip, and is given by:

Average Speed = v̄ == d + dd

v1+

d

v2

= 2d

d

1

v1+

1

v2

= 2dd

v1 + v2

v1v2

= 2v1v2v1 + v2

Simple and Compound Interest

• Simple Interest is computed only on the amount of the initial deposit, called the principal, for the entire time period.

If the amount P 0 is invested at a simple annual interest rate of r percent, then the value of the investment at the endof t years is given by the formula

P = P 0

1 +

rt

100

The total simple interest I earned in that same time period is given by:

I = P − P 0 = P 0rt100

• Compound Interest: In the case of compound interest the interest is computed at the end of each compoundingperiod, such as, annually, quarterly, or monthly. The principal P 0 that earns an annual interest rate of r percentcompounded annually for t years grows to P , where P is given by:

P = P 0 1 + r100t

The total compound interest I earned in that same time period is given by:

I = P − P 0 = P 0

1 + r

100

t− P 0 = P 0

1 +

r

100

t− 1

• Compound Interest and Effect of Compounding Frequency: If the compound interest is paid at monthly,quarterly, or in intervals that are shorter than one year, then the interest is added to the principal more frequently thanin the case of annual compounding. The new principal P at the end of t years that earns r percent annual interest andis compounded n times annually is given by:

P = P 0

1 +

r

100nnt

The table below shows the appropriate compound interest formula based on the frequency of compounding.

Compounding Frequency n Compounded Principal

Monthly 12 P 0

1 +

r

12 × 10012t

Quarterly 4 P 0

1 +

r

4 × 1004t

Semi-Annual 2 P 0

1 +

r

2 × 1002t


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Work Equations

• Work Equation: If it takes tA and tB hours for machines A and B , respectively, to do a job, then the time tA+B ittakes machines A and B , working simultaneously, at their respective constant rates is given by

1tA+B

= 1tA

+ 1tB

tA+B = tAtBtA + tB

In general, tA+B < tA, and tA+B < tB, which says that it takes less time for two machines to complete a job workingtogether than it takes either one to complete the job on its own.

The time tA+B also satisfies the following inequality:

Assume tA > tB then tB

2 < tA+B <

tA2

For example, if tA = 15, and tB = 10, then 5 < tA+B


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Geometry

Parallel Lines and Intersections

Line l

Line mx◦

x◦

x◦

x◦

y◦

y◦

y◦

y◦

l m

Triangles

Sum of the Internal Angles of a Triangle

The three angles of a triangle add upto a straight line or 180◦.

DE

DE

AB

α◦ + β ◦ + γ ◦ = 180◦

B

C

Aα◦ β ◦

γ ◦ β ◦α◦

Triangle Inequality Theorem

The sum of any two sides of a triangle is greater than the third side. If a + b = c, then points A, B , and C lie on a straightline.

B

C

A

a b

c

a + b > c

a + c > b

b + c > a

In questions that deal with the range of length of the third side, it is convenient to use the fact that a given side of a trianglemust be greater than the difference between the two other sides, and less than the sum of the other two sides.

Alternative Statement of Triangle Inequality |a − b| < c b, and ∠C > ∠B.


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Isosceles Triangle

A triangle that has at least two equal sides is called an isosceles triangle.

B

C

Aα◦ α◦

β ◦ β ◦

D

Properties of an Isosceles Triangle

CB = C A

∠A = ∠B = α◦

CD bisects AB (BD = DA)

∠ACD = ∠BCD = β ◦

Right Triangles

• Pythagorean Theorem: In any right triangle the square of the longest length, called hypotenuse, is equal to the sumof the squares of the two shorter sides.

C B

A

a c

b

Pythagorean Theorem

a2 + b2 = c2

• Converse of Pythagorean Theorem: If the sum of the squares of the two shorter sides of a triangle is equal to thesquare of the longest length or hypotenuse, then the triangle is a right triangle.

• If a2 + b2 > c2, where c is the longest side, and a and b are the shorter sides of a triangle, then the angle opposite thelongest side is less than 90◦.

C B

A

a = 6 c = 9

b = 8

Pythagorean Inequality

a2 + b2 > c2

62 + 82 > 92 or 100 > 81

∠C


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45◦ − 45◦ − 90◦ TriangleAn isosceles right triangle has two angles measuring each 45◦, and one right angle measuring 90◦, as shown in the figurebelow. The application of Pythagorean theorem to this triangle shows that the sides of the triangle are in the ratio of 1:1:

√ 2.

A

B C

45◦

45◦

x

xx√

2

Pythagorean Theorem

AC 2 = x2 + x2

AC 2 = 2x2

AC = x√

2

30◦ − 60◦ − 90◦ TriangleThe figure below shows an equilateral triangle divided into two halves, each halve being a 30◦ − 60◦ − 90◦ triangle. Theapplication of Pythagorean theorem to this triangle shows that the sides of the triangle are in the ratio of 1:

√ 3:2.

A B

C

60◦ 60◦

30◦30◦

D x

2xx√

3

Pythagorean Theorem

CD2 + x2 = (2x)2

CD2 = 4x2 − x2 = 3x2

CD = x√

3

Triangle CAD and CBD are congruent

AD = DB = x and AB = BC = 2x

Equilateral Triangles

• An equilateral triangle is a triangle with three equal sides and three equal angles (60◦).

• The height of an equilateral triangle of side s is h = s√ 32

• Area of an equilateral triangle with side s is s2√

3

4

• The area of an equilateral triangle in terms of its height, h, is h2

√ 3

• The angle bisectors and the medians are identical for an equilateral triangle.• The radius of the circle circumscribing an equilateral triangle of side s, is R = s√

3


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• The radius of the circle inscribed in an equilateral triangle of side s, is r = s2√

3

• The figure below shows the how the radius of the inscribed and circumscribed circles are related to the length of theside of an equilateral triangle.

A

B

C

R r

s

2

s

30◦ 60◦

60◦

30◦ − 60◦ − 90◦ Triangle

r : s

2 : R = 1 :

√ 3 : 2

r = s

2√ 3and R =

s

√ 3

3:4:5 Triangle

The right triangle with sides of 3, 4, and 5, and its multiples appears frequently on the GMAT, and it is important to be ableto recognize it in different situations.

A B

C

4

53

A B

C

8

106

32 + 42 = 52

9 + 16 = 25

62 + 82 = 102

36 + 64 = 100

A B

C

2

2.51.5

A B

C

4a

5a3a

1.52 + 22 = 2.52

2.25 + 4 = 6.25

(3a)2 + (4a)2 = (5a)2

9a2 + 16a2 = 25a2


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5:12:13 Triangle

The right triangle with sides of 3, 4, and 5, and its multiples appears frequently on the GMAT, and it is important to be ableto recognize it in different situations.

A B

C

12

135

A B

C

24

2610

Area of a Triangle

The area of a triangle is given by half the length of the base( b) multiplied by the corresponding height(h) as shown in thefigure below. The formula for the area of the triangle is:

Area of a Triangle = bh

2

Base, b

H e i g h t , h

For the purpose of calculating the area of a triangle, any side of the triangle may be considered a base, and the height isthen the length of the perpendicular drawn to the base from the vertex that is opposite to the base.

Area of a Triangle = bh

2

A

B

C D

E F

Similar Triangles

• Two triangles that have the same shape but not necessarily the same size are called similar triangles. Two triangles aresimilar if their vertices can be matched up so that the corresponding angles are congruent or, equivalently, the lengthsof corresponding sides have the same ratio.


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A B

C

c

ab

D E

F

f

de

∠A = ∠D, ∠B = ∠E, ∠C = ∠F

a

d =

b

e =

c

f

• The ratio of the area of two similar triangles is equal to the ratio of the squares of the corresponding sides. For example,consider the two similar triangles shown below:

A B

C

c

ab

D E

F

f

de

Area of △ABC Area of

△DEF

= a2

d2 =

b2

e2 =

c2

f 2

Square

• The perimeter of a square of side s is = 4s.• The area of a square of side s is = s2.

• The length of the diagonal of a square is = s√ 2.• A square circumscribing a circle has a side of length s = d, where d is the diameter of the circle.

d i a m e t e r , d

=

s


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• A square inscribed in a circle has a side of length s = d√ 2

, where d is the diameter of the circle.

d i a m

e t e r

, d =

s√ 2

Rectangles

• The perimeter of a rectangle of length l and width w is = 2(l + w).• The area of a rectangle is = lw.• The length of the diagonal of a rectangle is d = √ l2 + w2.

Regular Polygons

• A regular polygon has equal sides and equal internal angles. A square is a four-sided regular polygon and an equilateraltriangle is a regular three-sided polygon. The figure below shows a regular pentagon with each of the internal angles

being equal to 108◦

.

A

B

C

D E

• Sum of interior angles of an n-gon: S = 180(n − 2). For example, the sum of the internal angles of a pentagon with fivesides is 180(5 − 2) = 180(3) = 540 degress.

• Each interior angle of an equiangular or regular n-gon is equal to180(n − 2)

n

In case of a hexagon, each interior angle is equal to 180(6 − 2)

6 = 120 degrees.

• The central angle of a polygon is equal to 360◦

n , where n is the number of sides of the polygon.


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• The area of a hexagon is equal to s2(3

√ 3)

2 , where s is the length of one of the sides of the hexagon. A regular hexagon

is made up of six equilateral triangles, and the area of one of the equilateral triangles is s2

√ 3

4 , and the area of the

hexagon is six times this value

60◦

120◦

60◦

• The center of a regular polygon is the same as the center of its inscribed and circumscribed circle. The radius of thecircumscribed circle is the same as the length of the line segment joining the center of the polygon to any of its vertices.The figure below shows the relationship between the radius of the inscribed circle, r, radius of the circumscribed circle,R, and the length of the side of a regular hexagon, s.

r

R

s

30◦

60◦

30◦ − 60◦ − 90◦ Triangle

s

2 : r : R = 1 :

√ 3 : 2

r = s√ 3

2 and R = s

• The area of regular polygons (squares, equilateral triangles, regular hexagons) are in the same ratio as the square of the corresponding sides. For example, the ratio of the area of the large hexagon to the small hexagon, shown below, is

equal to 62

32 = 4, or 4 to 1.

3

6


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Circles

• The circumference of a circle of radius r is given by 2πr.

• The area of a circle of radius r is given by πr2.

• The area of a circle of diameter d is given by πd2

4 .

• A triangle inscribed in a circle with one of its sides being the diameter of the circle is a right angle triangle. The figurebelow shows that irrespective of the location of point C on the circle, ∠C is always equal to 90◦.

A B

C

O

• Length of an Arc: In a circle of radius r, the length l of an arc with a central angle of n◦ equals n◦

360◦fraction of the

circumference of the circle.

Length of an Arc =

n◦

360◦ (2πr)

• Area of a Sector: The area of a sector of a circle that subtends an angle n◦, expressed in degrees, is given by:Area of a Sector =

n◦

360◦(πr2)

Cubes

A

BC

D• A cube has six square faces, eight corners, and twelve edges.• The surface area of a cube of length a is = 6a2.• The volume of a cube is equal to a3.• The longest diagonal of a cube is equal to a√ 3.


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• The sum of the length of the edges of a cube = 12a.• If we know any one of the following quantities: length of a side of a cube, surface area, volume, length of the longest

diagonal of a cube, area of one face of a cube, then we can obtain the value of all of the other quantities. Theserelationships are useful in data sufficiency questions.

Rectangular Solids

• A rectangular solid has six rectangular faces, eight corners, and twelve edges.• The surface area of a rectangular solid of length l, width w, and height h = 2(lw + lh + wh).• The volume of a rectangular solid is: V = lwh.• The volume of a rectangular solid is: V = √ A1A2A3, where A1 = lw, A2 = wh, and A3 = lh, are the surface areas of

the three distinct faces of the rectangular solid.

• The longest diagonal of a rectangular solid = √ l2 + w2 + h2.

• If E is the sum of the length of the sides of a rectangular solid, E = l + w + h, and d is the longest diagonal of arectangular solid, then the surface area S is given by S = E 2 − d2.

Cylinders

• The surface area of a cylinder of radius, r, and height, h, is given by 2πr2 + 2πrh, where 2πr2 is the sum of the top andbottom surfaces, and 2πrh is the lateral surface area.

• The volume of a cylinder is πr2h.• The volume of the largest rectangular solid that can be fitted inside a right circular cylinder of radius r and height h is

2r2h.

• The volume of the smallest rectangular solid that can accomodate a right circular cylinder of radius r and height h is4r2h.

• The longest rod that can be accommodated in a right circular cylinder has a length of √ 4r2 + h2(Why?).

Coordinate Geometry

Points in Coordinate Plane

• Distance Formula: The distance between any two points A and B with coordinates (x1, y1) and (x2, y2) is given bythe formula:

Distance =

(x1 − x2)2 + (y1 − y2)2The above expression can be obtained by applying the Pythagorean theorem to the right triangle formed by drawing

lines parallel to the x-axis and y-axis where the distance between the two points is the hypotenuse, as shown in thefigure below.


34/39


x

y

(0,0)Origin

(x1, y1)

(x2, y2)

x2 − x1

y 2

−

y 1

x1 + x2

2 ,

y1 + y22

x-axis

y-axis

( x 1 −

x 2 ) 2

+ ( y 1

− y 2 ) 2

• Midpoint M of the line segment AB is given by: x2 + x1

2 ,

y2 + y12

• The coordinates of the point that divides the line segment AB in the ratio r : s is given by:rx2 + sx1

r + s ,

ry2 + sy1r + s

Coordinate Geometry (Lines)• Equation of a Line: The equation of a straight line describes the relationship between the x and the y coordinates of

all points that fall on the line. The table below summarizes the most common ways of describing the equation of a line:

Description Equation Terms

Slope-Intercept Form y = mx + b m = slope and b = y-intercept

Point-Slope Form y − y1 = m(x − x1) m = slope and passingthrough (x1, y1)

Two-Point Form y − y1 = y2 − y1x2 − x1 (x − x1) Line passing through

(x1, y1) and (x2, y2)

Intercept Form x

a +

y

b = 1 a and b are the

x and y intercept


35/39


x-intercept

(a, 0)

y-intercept

(0, b)

x

y

Slope = y2 − y1x2 − x1

(0,0)Origin

(x1, y1)

(x2, y2)

x2 − x1

y 2

−

y 1

x-axis

y-axis

• The y-intercept is the point where the line intersects the y -axis and is given by:

(0, b)

• The x-intercept is the x-coordinate of the point where the line intersects the x-axis and is given by:

(− bm

, 0)

• The equation of the straight line that is parallel to the y axis and at a distance of a units is given by x = a. For example,

the lines x = 5 and x = −3 are both vertical lines parallel to each other and the y -axis and passing through (5, 0) and(−3, 0), respectively.

• The equation of the straight line that is parallel to the x axis and at a distance of b units is given by y = b. For example,the lines y = 2 and y = −4 are both horizontal lines parallel to each other and the x-axis with a zero slope and passingthrough (0, 2) and (0, −4), respectively.

• Parallel lines have the same slope and do not intersect each other. The set of distinct lines 3y = 2x− 3 and 2x− 3y = 4,and

x

3 − y

2 = 7, all have a slope of

2

3 and are all parallel to each other.

• Two lines perpendicular to each other with slopes m1 and m2 satisfy the following relationship:

m1m2 = −1

The lines y = 3x and y = −x3

are perpendicular to each other.

Coordinate Geometry: Circles

• A circle in a coordinate plane consists of all set of points that are equidistant from a given point, called the center of the circle. The figure below shows a circle with the center located at (h, k). If (x, y) is any point on the circle, then thedistance between the point (x, y) and the center of the circle (h, k) is a constant and is equal to the radius of the circle.The application of Pythagorean theorem to the right triangle shown inside the circle yields:

(x − h)2 + (y − k)2 = r2


36/39


x

y

(0,0)

Origin

x − h

r

y

−

k

(x, y)

(h, k)

• If the circle is centered at the origin (h, k) = (0, 0), then the equation of the circle simplifies to x2 + y2 = r2.

StatisticsMean, Median and Mode

• Average(Arithmetic Mean) The average, sample mean, or arithmetic mean, of a set of n numbers {x1, x2, x3, . . . ,xn} is denoted by x̄ and is defined as

Arithmetic Mean : x̄ = x1 + x2 + x3 + · · · + xn

n

• Median: The median of a list of numbers is obtained by first ordering them from least to greatest. If there are anodd number of numbers, then the median is the middle number, and if there are an even number of numbers, then themedian is the average of the middle two numbers.

• Mode:The mode of a list of numbers is the most frequently occuring number in the list.

Weighted Average

Consider two different groups, A and B, that have averages of x̄A and x̄B, respectively, and let x̄A > x̄B . The averages couldbe for any specific property of the group like age, height, or years of experience. Also, let N A and N B, be the number of people in each group. If the groups are combined to form a new larger group, then the average of the new group is given by

x̄A+B = N Ax̄A + N Bx̄B

N A + N B=

N A

N A + N B

x̄A +

N B

N A + N B

x̄B

The equation can also be rewritten in terms of the fraction of people in group A:

x̄A+B = f Ax̄A + (1 − f A)x̄Bwhere f A = N A

N A + N B.

The average of the new combined group, x̄A+B, lies between the averages of the two groups.

Standard Deviation

• Standard deviation measures the dispersion of the data set from the mean. The sample standard deviation of a list of n numbers {x1, x2, x3, . . . , xn}, where x̄ is the arithmetic mean of the set is given by

s =

(x1 − x̄)2 + (x2 − x̄)2 + · · · + (xn − x̄)2

n


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• Impact on Mean and Standard Deviation: The table below lists how the mean, x̄, and standard deviation, s, of a set are changed by the list of operations on each element of a data set.

Operation on each element of a Data Set New Mean New Standard Deviation

Add a constant, a x̄ + a Unchanged

Subtract a constant, a x̄ − a Unchanged

Multiply by a constant, a ax̄ as

Increase by p%

1 + p

100

x̄

1 +

p

100

s

Divide by a constant, a

x̄

a

s

a

• Standard Deviation: Computational Formula: The computation formula for the standard deviation is a rear-rangement of the general formula for the standard deviation.

s =

x21 + x

22 + x

23 · · · + x2n

n

− x̄2

where {x1, x2, x3, . . . , xn} are the n data points, and x̄ is the arithmetic mean of the set.

Fundamental Counting Principle

The fundamental principle of counting states that if one event can occur in m ways, and a second event that is independentof the first event can occur in k ways, then the two events can occur in m × k ways. For example, if someone has 5 shirts and8 trousers, then they can create a total of 40 distinct outfits that include a selection of one shirt and one trouser.

Permutations

The ordered arrangement of a set of objects is called a permutation. The problem statements in GMAT do not use the wordpermutation, it is left to the student to figure out if the order in which the objects are arranged is important in finding thetotal number of arrangements. The number of permutations, denoted by nP r, or the number of arrangements of n objectstaken r at a time is given by

nP r = n!

(n − r)! = n(n − 1)(n − 2) · · · · · · (n − r + 1)

CombinationsThe selection of a set of objects without regard to the order is called a combination. The number of combinations, denotedby nC r, or the number of ways of selecting r objects from a collection of n objects without regard to order is given by

nC r = n!

r!(n − r)! = n(n − 1)(n − 2) · · · · · · (n − r + 1)

(1)(2)(3) · · · (r)

Circular Permutations

The ordered arrangement of a set of objects in a circle is called a circular permutation. The number of circular permutationsof n distinct objects in a circle is given by (n − 1)! = (n − 1)(n − 2) · · · 1.


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Probability

Probability is defined as the ratio of the number of favorable outcomes to the total number of equally likely outcomes.

Probability of an event E = Number of outcomes of event E or favorable outcomes

Total number of equally likely outcomes

Summary of facts on probability:

• The probability of an event that is certain to occur is 1.• The probability of an event that is certain not to occur is 0.• In general the probability of an event falls between 0 and 1.• If P (E ) is the probability that a certain event occurs, then 1 − P (E ) is the probability that an event will not occur.• P (A or B ) = P (A) + P (B) − P (A and B )

• P (A or B ) = P (A) + P (B) if A and B are mutually exclusive.

• P (A and B ) = P (A)P (B) if A and B are independent.

Sequences

Arithmetic Sequence

An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant. For example, thesequence {3, 7, 11, 15, ... }, where each term is 4 larger than the previous term is an arithmetic sequence. If a is the firstterm of an arithmetic sequence, and d is the common difference between the successive terms, then the resulting sequence is

Arithmetic Sequence: a, a + d, a + 2d, a + 3d, . . . , a + (n − 1)d

nth term of an arithmetic sequence : an = a + (n − 1)dIn other words, the nth term of an arithmetic sequence can be found by adding the common difference n − 1 times to thefirst term.

Sum of an Arithmetic Sequence

The sum S n of the first n terms of an arithmetic sequence is given by

S n = (Number of Terms) ×

First + Last

2

=

n(a + an)

2

where an = a + (n − 1)d is the nth term. The arithmetic mean(average) and the median of an arithmetic sequence are equaland are also equal to the average of the first and last term

Mean = Median = First + Last

2

Number of Terms in an Arithmetic Sequence

If we rearrange the expression for the nth term of an arithmetic sequence, an = a + (n − 1)d, we can express the number of terms in a sequence

Number of Terms in a Sequence, n = an − a

d + 1 =

Last Term − First TermSpacing

+ 1


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Arithmetic Sequences: Examples

• The first n positive integers 1, 2, 3, . . . , n.

Sum of first n positive integers = 1 + 2 + 3 + ... + n =

n(n + 1)

2

• Set of consecutive integers. Example: −3, −2, −1, 0, 1, 2, 3, 4, 5.• Set of consecutive even integers. Example: −4, −2, 0, 2, 4, 6, 8, 10.• Set of consecutive odd integers. Example: −3, −1, 1, 3, 5, 7, 9, 11.• Multiples of a given integer. Example: 7, 14, 21, 28, 35, . . .

Geometric Sequence

In a geometric sequence the ratio between any two consecutive terms is constant. For example, the sequence {4, 8, 16, 32,. .

}, where each term after the first is twice the previous term. If the first term is designated as b and the common ratio is

r, then the resulting sequence is:Geometric Sequence: b, br, br2, br3, . . . , brn−1

nth term of a geometric sequence : bn = brn−1

In other words, the nth term of a geometric sequence can be obtained by multiplying the first term by the common ration − 1 times.

Cyclical Sequences

A sequence that consists of terms that repeat in a cyclical pattern is called a cyclic or a repeating sequence. For example,the decimal equivalent of fraction 2/7 is 0.2857142857142 . . ., where the digits repeat after every six term.To find the nth term of a repeating sequence, divide n by the number of terms in the repeating group, j, and find theremainder r. Then the nth term of the sequence is the same as the rth term. To find the 100th digit after the decimal point

n the repeating decimal 0.2857142857142 . . ., divide 100 by 6, which gives a remainder of 4. Then the 100th digit is the 4thdigit in the repeating group {2, 8, 5, 7, 1, 4}, which is 7. If the remainder is zero, as in the case of the 60th term in thedecimal expansion of 2/7, then the 60th term is the last term in the repeating group, 4 in this example.

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