GMAT Quantum Math Notes

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 integers extends 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 odd and even integers. Even ± Even = Even Even ± Odd = Odd

Odd ± Odd = Even

Even × Even = Even

Even × Odd = Even

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GMATQuantum Math Notes

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Odd × Odd = Odd

OddEven = 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 two successive 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 the numbers 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 odd integers, 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. k • The sum of k consecutive integers, where k is even, is always divisible by . For example, consider the set of 10 2 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 10x + 5, which n can also be written as 5(2x + 1), and is always a divisible by 5. In general, if x is the th term in the sequence of an 2 n even number of consecutive integers, then the sum of the n consecutive integers is given by (2x + 1). 2

• 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 15 which 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 multiple of k. Example: Consider 10, 11, 12, 13 as the four consecutive integers, their product 10 × 11 × 12 × 13 is divisible by 4 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 consecutive integers, 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. A positive integer that is greater than 1 and is not prime is called composite. 2 13 31 53 73

3 17 37 59 79

5 19 41 61 83

7 23 43 67 89

11 29 47 71 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 not divisible 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 will always 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 three consecutive odd integers, at least one of them is divisible by 3, and therefore all sets of three consecutive odd integers will 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 · · · · · n then 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 two numbers 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 can be written uniquely as the product of two or more prime numbers. The factorization in to the prime numbers is unique except 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 only one way and in the most general case, N can be written as: N = pr11 pr22 pr33 · · · pri i The divisors or factors of N are numbers of the form n = ps11 ps22 ps33 · · · psi i

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: r

(1 + p1 + p21 + · · · + pr11 )(1 + p2 + p22 + · · · + pr22 ) · · · (1 + pi + p2i + · · · + pi j ) is a divisor of N , and that no other number is a divisor. The total number of terms in this expression is thus equal to the 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 find the 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: n n n + 2 + 3 + ··· Highest power of p = p p p 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: 100 100 100 100 Highest power of 3 = + + + = 33 + 11 + 3 + 1 = 48 3 32 33 34

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

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listed below.

Units Digit 0 1 2 3 4 5 6 7 8 9

Repeating pattern (0) (1) (2, 4, 8, 6) (3, 9, 7, 1) (4, 6) (5) (6) (7, 9, 3, 1) (8, 4, 2, 6) (9, 1)

For example, the table below shows the repeating pattern of the units digit of consecutive integral powers of 3 and 7: 31 = 3 32 = 9 33 = 27 34 = 81 35 = 243 36 = 729 37 = 2187 38 = 6561

71 = 7 72 = 49 73 = 343 74 = 2401 75 = 16807 76 = 117649 77 = 823543 78 = 5764801

• The last digit of any number that ends in 0, 1, 5, or 6, will always remain unchanged on repeated multiplication with itself. Example: Last digit of 5555 is 5, and that of 6666 is 6. • When a number is divided by 10, the remainder is the same as the last digit of that number. Example: What is the remainder when 777 is divided by 7? The last digit of 7 repeats in a cycle of 4, and there are 19 full cycles, with one remainder, therefore, the last digit of 777 is 7, which is also the remainder when 777 is divided by 10. • The units digit of the fifth power of a number is the same as the units digit of the original number. For example, the units digit of (387)5 is 7 and is same as that of the original number, 387 in this case. • The units digit of the square of an integer has to be one of the following numbers: {0, 1, 4, 5, 6, 9}. The table below shows the pattern: Units Digit of n 1 2 3 4 5 6 7 8 9

Units Digit of n2 1 4 9 6 5 6 9 4 1

Units Digit n3 1 8 7 4 5 6 3 2 9

Units Digit of n4 1 6 1 6 5 6 1 6 1

Units Digit of n5 1 2 3 4 5 6 7 8 9

• The number n5 − n has a units digit of zero, in other words when n5 − n is divided by 10, the remainder is always zero. For example, 25 − 2 = 32 − 2 = 30, which leaves a remainder of zero when divided by 10. The table above shows that the the units digit of n5 is identical to the units digit of n, and there difference leaves a units digit of zero, which is a number that is always divisible by 10. • If a and b are consecutive integers and a > b, then the units digit of a5 − b5 is always 1. Here a5 has the same units digit as a, and b5 has the same units digit as b, and therefore a5 − b5 has the units of one, because a and b are consecutive integers.

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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 number ab, 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 will be nine times the difference between the two digits. For example, if n = 64, then 64 − 46 = 18 = 9(6 − 4). In other words, if n = tu, where t is the tens digit and u is the units digit, then n = 10t + u, and the number formed by reversing the 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 always divisible 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 units digit. If we reverse the digits, the new number formed is m = cba = 100c + 10b + a. The difference between the two numbers 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 m divides n. The following table lists statements that are equivalent to m divides n. m divides n m is a factor or divisor of n n is divisible by m n 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 The number of positive integers that are less than or equal to n and divisible by an integer k j nMultiples: k is equal to , where the floor function or the greatest integer function, ⌊x⌋ is defined as the largest integer less than k or to x. For example, the number of positive integers less than or equal to 20 that are divisible by 3 is equal to equal 20 = ⌊6.67⌋ = 6. 3

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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 divisible by all of them. • To obtain GCF and LCM, first obtain the prime factorization of all the numbers, that is break them down into their prime constituents. Example: Given 48 and 120, find the GCF and LCM. The prime factorization of 48 is 24 × 3 and that of 120 is 23 × 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, in this 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), and 28 = (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 quotient and r emainder, respectively, such that: n = mq + r and 0≤r
and

0≤r
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 remainders are: 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 divided by m is equal to the remainder when r1 + r2 is divided by m. For example, when 26 and 46 are divided by 7, the remainders 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 divided by m is equal to the remainder when (r1 )(r2 ) is divided by m. For example, when 26 and 46 are divided by 7, the remainders are 5 and 4, respectively. If we multiply the two remainders, 5 × 4 = 20, and divide by 7, the remainder is 6, 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 divided by 5. For example, the remainder when 27 is divided by 5, is the same as the remainder when 7 is divided by 5, which is 2. The reason for this rule is that, a number like, 64578 can be expressed as: 64578 = 60000 + 4000 + 500 + 70 + 8 The tens and higher powers of 10, are all divisible by 5, irrespective of the place value, and the remainder only depends on 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 31 %

2 5

0.4

40%

1 2

0.5

50%

3 5

0.6

60%

2 3

0.666. . .

66 32 %

3 4

0.75

75%

4 5

0.8

80%

1

1

100%

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Common Squares, Cubes, and Square Roots Squares of Integers

Cubes of Integers

1 × 1 = 12 = 1

1 × 1 × 1 = 13 = 1

3 × 3 = 32 = 9

3 × 3 × 3 = 33 = 27

5 × 5 = 52 = 25

5 × 5 × 5 = 53 = 125

2 × 2 = 22 = 2

2 × 2 × 2 = 23 = 8

4 × 4 = 42 = 16

4 × 4 × 4 = 43 = 64

6 × 6 = 62 = 36

6 × 6 × 6 = 63 = 216

7 × 7 = 72 = 49 8 × 8 = 82 = 64 9 × 9 = 92 = 81

10 × 10 = 102 = 100 11 × 11 = 112 = 121

10 × 10 × 10 = 103 = 1000

12 × 12 = 122 = 144 13 × 13 = 132 = 169 14 × 14 = 142 = 196 15 × 15 = 152 = 225 16 × 16 = 162 = 256 20 × 20 = 202 = 400 25 × 25 = 252 = 625

Square Root √ 1=1 √ 4=2 √ 9=3 √ 16 = 4 √ 25 = 5 √ 36 = 6 √ 49 = 7 √ 64 = 8 √ 81 = 9 √ 100 = 10 √ 121 = 11 √ 144 = 12 √ 169 = 13 √ 196 = 14 √ 225 = 15 √ 256 = 16 √ 400 = 20 √ 625 = 25

Fractions • Reduced Fraction: The fraction m/n is called a reduced fraction if and only if the greatest common divisor of m and 15 n is 1. For example, is not a reduced fraction, because there greatest common divisor is 3. In general the answer 21 choices on the GMAT are always written in the reduced fraction form, therefore it is important to always simplify your answer and to rewrite in a reduced fraction form. In the example above: 15 5×3 5×3 = = = 21 7×3 7×3

5 7 |{z}

Reduced Fraction

• Adding and Subtracting Fractions: The fastest way to add any pair of fraction is to first multiply the denominators to form the new denominator. Subsequently, cross multiply the numerators and denominators and add to form the new numerator. Consider the following example 2 5 (2 × 7) + (3 × 5) 14 + 15 29 + = = = 3 7 3×7 21 21 In general to add fractions: a c (a × d) ± (b × c) ad ± bc ± = = b d b×d bd

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• Multiplying Fractions To multiply two fractions, multiply the numerators of each of the fraction to form the new numerator of the resulting fraction, and multiply the denominators of each of the fraction to form the new denominator of the resulting fraction. For example, 3 5 3×5 15 × = = 4 7 4×7 28 When multiplying a fraction by a whole number, follow the procedure illustrated below: 3×

2 3 2 3×2 6 = × = = 5 1 5 1×5 5

• Dividing Fractions To divide two fractions, invert the bottom fraction, and then multiply it with the fraction on the top. For example, 3 4 = 3 × 7 = 3 × 7 = 21 5 4 5 4×5 20 7 When dividing a fraction by a whole number, follow the procedure illustrated below: 2 2 5 = 5 = 2×1 = 2 3 3 5 3 15 1 5 8 • Comparing Fractions Is > ? To answer this question, multiply 5 by 11, which is 55, and place it on the same 7 11 side as the fraction that has 5 as the numerator. Subsequently, multiply 8 by 7, and place it on the right side, which is 5 8 56. The side that has the larger product is the one that is larger, in this example < . 7 11 √ √ 1 1 1 1 • If 0 < x < 1 then x > x > x2 > x3 . Example: x = , then x = , x2 = , x3 = . 4 2 16 64 √ √ • If x > 1 then x3 > x2 > x > x. Example: x = 1.21, then x = 1.1, x2 = 1.44.

Decimals • Decimals: Multiplying and Dividing by powers of 10 Multiplying a decimal by 10 moves the decimal place one place to the right. For example, 0.03 × 10 = 0.3. Similarly, 0.03 dividing a decimal by 10 moves the decimal place one place to the left. For example, = 0.003. 10 • Multiplying two decimals: To multiply two decimals, carry out the multiplication of the numbers disregarding the decimal points. For example, to multiply 0.04 × 0.005, multiply 4 and 5 to obtain 20. Count the total number of digits to the right of the decimal point in each of the numbers being multiplied, and add these up. In the example given, there are 2 digits to the right of the decimal point in 0.04, and 3 digits to the right of the decimal point in 0.005, their sum total is 5. Obtain the product of the two numbers which is 20, and then move the decimal point five digits to the left of 0 in 20, which yields 0.00020 = 0.0002. ✿×10000 ✯ 4 30000 × 4 30000 × 100 30000 × 100 30000 4 30000 = = = = = 10000 × 4 = 40000 • 3 1 0.75 0.75 × 100 3 ❃ 75 3✕ •

0.001 0.001 × 1000 1 1 = = = 8 = 10−8 100000 100000 × 1000 100000000 10

• When simplifying expressions with decimals in the numerator and denominator, move the decimals to the left the same number of times in the numerator and denominator. For example: 0.0001 0.0001 × 100000 10 = = = 10 (0.01)(0.001) (0.01 × 100)(0.001 × 1000) (1)(1)

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GMATQuantum Math Notes • Caution: 1, 000, 000 = 106 and not 105 , 106 means 10 multiplied 6 times.

• The decimal equivalent of a fraction m/n, written in the reduced fraction form with no common factors except 1, can 5 be expressed as a terminating decimal if, and only if, n has no prime factors other than 2 or 5. For example, is a 8 3 5 5 5×5 625 7 = = 0.625, whereas 7/24 = 3 is not a terminating decimal because terminating decimal = 3 = 3 8 2 2 × 53 1000 2 ×3 3 is one of the prime factors of the denominator.

Radicals √ √ • Definition of : The symbol a, or radical a, by definition means the non-negative square root of a. Therefore, √ √ √ √ 9 = 3, and 9 6= −3. In other words, 7 + 9 = 10 is considered valid, whereas 7 + 9 6= 4. √ √ • Irrational Numbers:√Real numbers like 2, 3, that cannot be written as a ratio of integers, are called √ irrational √ with itself gives 2, in other words ( 2)( 2) = 2. numbers. The symbol 2 refers to the number, which when multiplied √ √ We know that (1)(1) = 1, and (2)(2) = 4, therefore 1 < 2 < 2, and it turns out that 2 = 1.414 . . .. On the GMAT you may be required to use an approximate value for square root of common numbers. The table below lists the important approximations that are frequently tested. Number

Approximation

√ 2

1.4

√ 3

1.7

√ 5

2.2

√ 6

2.4

• Adding and Subtracting Radicals √ √ √ √ √ √ √ √ 3 5+2 5= 5+ 5+ 5+ 5+ 5= 5 5 √ √ √ √ √ √ 5+2 5= 5+ 5+ 5=3 5 √ √ √ √ 5 + 2 5 = 5(1 + 2) = 3 5 √ √ √ √ 7 − 3 7 = 7(1 − 3) = −2 7 • Multiplication of Radicals: If m > 0 and n > 0, then p √ √ ( m)( n) = (m)(n) Examples:

p √ √ √ ( 9)( 4) = (9)(4) = 36 = 6 √ √ √ √ (3 5)(2 5) = (3 × 2)( 5 × 5) = 6 × 5 = 30 √ √ √ √ √ (2 3)(3 2) = (2 × 3)( 2 × 3) = 6 6

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• Division of Radicals: If m > 0 and n > 0, then

r √ 9 9 3 = Examples: √ = 4 2 r 4 √ p √ √ √ 48 48 √ √ = = 8 = (4)(2) = ( 4)( 2) = 2 2 6 6

r √ m m √ = n n

• Simplifying Radical Expressions: On the GMAT, the answer choices that contain a radical typically need to be rewritten such that there are no radical terms in the denominator. For example, if after working through a problem 6 you obtain an answer of √ , then you need to rewrite it in a form without any square root terms in the denominator. 3 This simplification can be obtained in the following two ways: 2 √ √ √ √ √ 6 3 6 6× 3 6 3 6✕ 3 √ = √ ×√ =√ √ = = =2 3 3 3 3 3 3 3× 3 √ √ The other method is to use the definition of ( 3)( 3) = 3:

√ √ √ √ 1 √ 6 (2)(3) 2( 3)( 3) 2( 3)( ❃ 3) √ = √ √ = = =2 3 1 √❃ 3 3 3 3 • Simplifying using Conjugates: The denominator of any expression of the form a √ b+ c √ can be simplified by multiplying the numerator and denominator by b − c, the conjugate of the denominator. 2 √ Example: Simplify 2+ 3 √ √ √ 2 2(2 − 3) 4−2 3 √ = √ √ = =4−2 3 (4 − 3) 2+ 3 (2 + 3)(2 − 3) • Common Mistakes in Radical Operations √ √ √ √ √ √ √ • (2 2)(4 2) 6= 6 2, instead (2 2)(4 2) = (2)(4)( 2)( 2) = (8)(2) = 16 √ √ √ √ √ √ √ • (3 3)(3 3) 6= 9 3, instead (3 3)(3 3) = (3)(3)( 3)( 3) = (9)(3) = 27 √ √ √ √ • a2 − b2 6= a − b. For example: 132 − 122 = 169 − 144 = 25 = 5 6= 13 − 12 = 1.

Exponents • Definition of Exponents: The exponent symbol, an , is defined as: an = a | × a × a{z× · · · × a} multiplied n times

The a in the exponent term an is called the base and n is called the power of a. In addition, a1 = a.

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• Negative Exponents: The negative exponent, a−n , is defined as: a−n =

1 1 = n a a × a× a× ··· × a | {z } multiplied n times

Example: 2

−3

1 1 1 = 3 = = 2 2×2×2 8

−2 1 1 1 22 = = × = 22 = 4 1 2 1 1 22

• Multiplication of Exponents: When multiplying two exponents with the same base the following rule applies: am × an = a × a × a × · · · × a × a × a × a × · · · × a = a × a × a × · · · × a = am+n | {z } | {z } | {z } multiplied m times

multiplied n times

multiplied m + n times

• Division of Exponents: When dividing two exponents with the same base the following rule applies: multiplied m times m

a an

}| { z a × a× a× a × ···× a× a m−n = =a | × a × a{z× · · · × a} = a a × a × a × · · · × a | {z } multiplied m − n times

multiplied n times

• Number Raised to Zero Power: Any number, other than zero, when raised to the power zero is equal to one. We can use the following argument to arrive at this result: 36 = 1 = 36−6 = 30 36 Other examples are: 80 = 1, (−3)0 = 1. • Power of a Product: The power of a product of two numbers can be simplified according to the rule: (a × b)n = an × bn The basis of the rule can be explained by the following steps: (a × b)n = (a × b) × (a × b) × · · · × (a × b) = a × a × a{z× · · · × a} × b| × b × b{z× · · · × }b = an × bn | {z } | multiplied n times

multiplied n times

multiplied n times

• Power of a Quotient: The power of a quotient of two numbers can be simplified according to the rule: a n an = n b b The basis of the rule can be explained by the following steps:

multiplied n times

a n b

• Power of a Power:

=

z }| { a× a× a × ···× a an ··· = = n b} b| × b × b{z× · · · × }b b {zb

a a a

|b

b

a

multiplied n times

multiplied n times

(am )n = am×n = an×m = (an )m multiplied m times multiplied m times

z }| { (a ) = a × a × · · · × a = a × a · · · × a | {z } | m n

m

m

multiplied n times

m

multiplied m times

z }| { z }| { a × a · · · × a · · · a × a · · · × a = am×n {z } multiplied n times

14


• Sum of Powers: There is no general exponent rule when adding powers of numbers that have the same base, however, there are cases where simplification is possible using other rules of arithmetic. In general, if you see a question on the GMAT that asks you to add terms with the same bases, the best approach is to factor the largest common term, and in most cases the resulting terms will collapse to something simple. 24 + 24 + 24 + 24 = 24 (1 + 1 + 1 + 1) = 24 (4) = 24 (22 ) = 24+2 = 26 222 + 222 = 222 (1 + 1) = 222 (2) = (222 )(21 ) = 222+1 = 223 333 + 333 + 333 = 333 (1 + 1 + 1) = 333 (3) = (333 )(31 ) = 333+1 = 334 1011 + 1012 + 1013 1011 (1 + 10 + 102 ) 1011 (1 + 10 + 102 ) 1011 = = = = 1011−6 = 105 2 6 7 8 6 2 6 10 + 10 + 10 10 (1 + 10 + 10 ) 106 10 (1 + 10 + 10 )

Common mistakes in applying Exponent Rules • (24 )(24 ) 6= 216 , instead (24 )(24 ) = 24+4 = 28 . • 108 − 107 6= 101 , instead 108 − 107 = 10(107) − 107 = 107 (10 − 1) = 9(107 ). 2

• (3x )2 = (32x ). Specifically, (3x )2 6= (3x ) • [3x−2 ]3 = [33(x−2) ] = 3(3x−6) . Make sure, [3x−2 ]3 6= 3(3x−2) •

222 + 411 222 + 411 = 411 6= 22 2 222 222 + 411 222 + (22 )11 222 + 22×11 222 + 222 222 (1 + 1) 222 (2) Instead, = = = = = =2 222 222 222 222 222 222

• 20600 = 20(20599 ) 6= 400599 Instead, 20600 = (202 )300 = 400300 • (22 )(2x ) 6= 22x Instead, (22 )(2x ) = 22+x

Fractional Exponents 1 am 1 1 Examples: 2−3 = 3 = = 0.125 2 28 2 1 1 1 3x = × = 3x 2 = −x 1 1 1 3 3x2

• a−m =

1 1 1 1 1 • An exponent of is the same as taking the square root of a number. For example, 16 2 × 16 2 = 16 2 + 2 = 161 , therefore 2 √ 1 16 2 = 16 = 4. 1 1 1 1 1 1 1 is the same as taking the cube root of a number. For example, 27 3 × 27 3 × 27 3 = 27 3 + 3 + 3 = 3 √ 1+1+1 1 1 1 1 1 27 3 = 271 , therefore 27 3 = 3 27 = 3. Other common examples: 8 3 = 2, 64 3 = 4, 125 3 = 5, (0.001) 3 = 0.1, √ 1 1 1 (0.064) 3 = 0.4, 1000 3 = 10, (−27) 3 = 3 −27 = −3.

• An exponent of

√ √ 1 1 1 • An exponent of is the same as taking the fourth root of a number. Examples: 16 4 = 4 16 = 2, 81 4 = 4 81 = 3, 4 √ √ 1 1 1 1 4 4 = 4 4 = (22 ) 4 = 22× 4 = 2 2 = 2.

15


Approximations: I have listed some common approximations that the GMAT test writers expect students to be able to perform. • (0.998)3 ≈ 1 0.97 1 ≈ = 0.5 1.98 2 √ • 15.89 ≈ 4

•

1.0012 − 1 1.0012 − 12 (1.001 + 1)(1.001 − 1) (2.001)(0.001) = = = ≈2 0.001 0.001 0.001 0.001 √ √ √ √ √ 1 • 0.000035 ≈ 0.000036 = 36 × 10−6 = ( 36)( 10−6 ) = 6 × (10−6 ) 2 = 6 × 10−3 •

Percentages • Percentage: The term percent means per hundred. Percents are ratios that are used to represent parts of a whole, 1 5 = . where the whole consists of 100 parts. 5 percent means 5 parts out of 100, or 5% = 100 20 Part 100 = Percent Whole • Percent Change: To compute the percent change when a positive quantity changes from an initial amount to another positive amount, we compute the amount of change and then divide this by the initial amount and multiply by 100. Final − Initial Percent Change = 100 Initial If the final quantity is less than the initial quantity, we end up with a percent decrease. In both cases, the denominator is 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 we are given xy = y, we cannot divide both sides by y, and conclude that x = 1. What if y = 0? In that case, x could take on any value. In general, bring all the terms to one side, equate to zero, and then factor. xy = y xy − y = 0 y(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 identities in problem statements and be able to turn one form into another.

16


Difference of Squares (a + b)(a − b) = a2 − b2 Examples: • 512 − 492 = (51 + 49)(51 − 49) = (100)(2) = 200 •

9992 − 1 9992 − 12 (999 + 1)(999 − 1) (1000)(998) = = = = 1000 998 998 998 998

•

x4 − 1 (x2 )2 − 12 (x2 + 1)(x2 − 1) (x2 + 1)(x + 1)(x − 1) = = = = (x2 + 1)(x + 1) x−1 x−1 x−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 + 9y 2 + 12xy • (x2 + y 2 )2 = (x2 )2 + (y 2 )2 + 2(x2 )(y 2 ) = x4 + y 4 + 2x2 y 2 2 1 1 • x+ = x2 + 2 + 2 x x 2 1 1 • x2 + 2 = x4 + 4 + 2 x x

Square of Difference (a − b)2 = a2 + b2 − 2ab Examples: • (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 + 4y 2 − 20xy • (x2 − y 2 )2 = (x2 )2 + (y 2 )2 − 2(x2 )(y 2 ) = x4 + y 4 − 2x2 y 2 2 1 1 • x− = x2 + 2 − 2 x x 2 1 1 2 • x − 2 = x4 + 4 − 2 x x

17


Quadratic Equations A quadratic equation has the form ax2 + bx + c = 0, where a, b, and c are real numbers, and a 6= 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 obtained by factoring: x2 − x − 6 = (x − 3)(x + 2) = 0 When 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 example of 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 quadratic equation 2x2 − 5x − 33 = 0.

11 and −3 as the two roots of the 2

Quadratic Formula The solutions to the quadratic equation ax2 + bx + c = 0 can also be obtained by using the formula: √ −b ± b2 − 4ac x= 2a Example; 2x2 − 5x − 33 = 0, we have a = 2, b = −5, and c = −33. The quadratic formula yields p −(−5) ± (−5)2 − 4(2)(−33) x= 2(2) √ 5 ± 289 5 ± 17 x= = 4 4 5 + 17 11 5 − 17 −12 The two solutions are x = = and x = = = −3. 4 2 4 4

18


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 quadratic formula 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 that the 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 < 0. Graphically this means that the corresponding parabola y = ax2 + bx + c does not intersect the x−axis. It either lies completely above or below the x−axis.

Inequalities Addition in Inequalities • If a > 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

b a > c 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 b > d c

19


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 satisfy this inequality is x ≥ 1. A proper way to obtain the solution is: 1 ≤1 x 1 ≥0 x x−1 ≥0 x

1−

Therefore, either x ≥ 1 or x < 0. In general, collect all the terms on one side by subtracting or adding, and then determine 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, if x ≥ 0 |x| = −x, if x < 0 Absolute Value: Geometric Interpretation The absolute value of the real number x can also be interpreted as the distance from the origin to the point x on the number line. For example, | − 3| = 3, means that −3 is 3 units away from the origin (x = 0) on the number line. Examples: | − 4| = 4, |0| = 0, −| − 3| = −(3) = −3, | − (−5)| = |5| = 5. The distance between a number, x, and number y on the number line is given by |x − y|. |x − y| = 5

y b

b

b

b

b

b

−3

−2

−1

0

1

2

Square Root and Absolute Value The algebraic characterization of absolute value is |x| = x −5 0 5 Therefore, for all real values of x,

√ x2 = |x|.

x b

b

b

b

3

4

5

6

√ x2 . For example: √

x2

|x|

p √ (−5)2 = 25 = 5

| − 5| = 5

p √ (5)2 = 25 = 5

|5| = 5

p √ (0)2 = 0 = 0

|0| = 0

20


Absolute Value Equalities • For any number a, • For any numbers a and b, • For any number a,

| − a| = |a| |a · b| = |a| · |b| |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, or in other words, a > 0, b < 0 or a < 0, b > 0. • |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, or in other words a > 0, b < 0 or a < 0, b > 0. • Note: The phrase “if and only if” implies that both statements are either true or both are false, meaning if we are given |a| + |b| = |a + b|, then ab ≥ 0, and if we are given ab ≥ 0, then |a| + |b| = |a + b|. These types of inequalities are commonly seen in data sufficiency problems.

Common Algebraic Mistakes •

a a a a+b a b 6= + , however, = + . b+c b c c c c

•

ab + c 6= a + c. We cannot strike out the b term that is present in the numerator and denominator. Instead, we can b divide each term in the numerator to yield: ab c ab c c ab + c = + = + =a+ b b b b b b

•

√ √ √ √ a2 + b2 6= a + b. For example, 32 + 42 = 9 + 16 = 25 = 5 6= 3 + 4.

• (a + b)n 6= an + bn • (−x)2 6= −(x2 ), instead (−x)2 = x2 .

21


Word Problems Translating Statements into Algebraic Relationships

If x =

Word Statement

Algebraic Translation

n is less than 15

n < 15

x less than 3

3−x

3 less than a number x

x−3

−8 is 5 less than x

−8 = x − 5 or x = −3

y is 5 less than twice the value of x

y = 2x − 5

Is x less than y?

Is x < y?

x is how much less than y?

What is the value of y − x?

z and y = 3z, then y is how many times x? 2

y = 3z = 3(2x) = 6x, y is six times x.

−8 is 5 more than x

−8 = x + 5 or x = −13

x is 10 more than y

x = y + 10

5 times the quantity (x2 + 3x)

5(x2 + 2x)

7 divided by x 7 divided into a number x

7 x x 7

Is x at least 5?

Is x ≥ 5?

x is positive and is at most 5

0
n is the square of an integer

n = k 2 , where k is some integer.

The sum of a and b is 7

a+b=7

The sum of a and b is at least 7

a+b≥7

n is an integer greater than 5 but less than 9

n can be 6, 7, or 8 200 x= y = 2y 100 200 x=y+ y = 3y 100

x is 200% of y x is 200% greater than y

22


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

Total Distance d1 + d2 = v¯ = d1 d2 Total Time + v1 v2

In the special case, where the person travels to a location and returns by the same route, d1 = d2 = d, and the average speed is independent of the length of the trip, and is given by: Average Speed = v¯ ==

2d 2d 2v1 v2 d+d = = = d d 1 1 v1 + v2 v1 + v2 + + d d v1 v2 v1 v2 v1 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 P0 is invested at a simple annual interest rate of r percent, then the value of the investment at the end of t years is given by the formula rt P = P0 1 + 100 The total simple interest I earned in that same time period is given by: I = P − P0 =

P0 rt 100

• Compound Interest: In the case of compound interest the interest is computed at the end of each compounding period, such as, annually, quarterly, or monthly. The principal P0 that earns an annual interest rate of r percent compounded annually for t years grows to P , where P is given by: r t P = P0 1 + 100 The total compound interest I earned in that same time period is given by: r t r t I = P − P0 = P0 1 + − P0 = P0 1 + −1 100 100

• 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 than in the case of annual compounding. The new principal P at the end of t years that earns r percent annual interest and is compounded n times annually is given by: r nt P = P0 1 + 100n The table below shows the appropriate compound interest formula based on the frequency of compounding. Compounding Frequency

n

Monthly

12

Quarterly

4

Semi-Annual

2

Compounded Principal 12t r P0 1 + 12 × 100 4t r P0 1 + 4 × 100 2t r P0 1 + 2 × 100

23


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 it takes machines A and B, working simultaneously, at their respective constant rates is given by 1

=

1 1 + tA tB

tA+B =

tA tB tA + tB

tA+B

In general, tA+B < tA , and tA+B < tB , which says that it takes less time for two machines to complete a job working together 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 tA < tA+B < 2 2

For example, if tA = 15, and tB = 10, then 5 < tA+B < 7.5, in this case we know tA+B = satisfies the given inequality.

150 15 × 10 = = 6, which 15 + 10 25

• Rate and Time Relationship: In work problems, the rate and time are inversely related. Therefore, if a problem states that the rate at which machine A does a certain job is greater than the rate at which machine B does a job, then we can conclude that tA < tB . Also, because of the inverse relationship between time and rate, if we increase the rate at which a certain machine does a job by say 25%, then the time to complete the job is reduced by 20%. • Work Equation: Multiple Machines/Person: The work equation can also be extended to multiple machines/person, for example, in case of three machines A, B, and C, the time required to complete the job if each machine were to work at their individual rates is given by 1 1 1 1 = + + tA+B+C tA tB tC where tA , tB , and tC is the time it takes machines A, B, and C to complete the same job individually.

Profit Loss Relationships • Gross Profit = Selling Price − Purchase Price • Selling Price = Cost + Mark up • Gross Profit = Revenue − Expenses

24


Geometry Parallel Lines and Intersections

lkm

y◦ x◦

Line l

x◦ y◦ y◦ Line m

x◦ x◦ y◦

Triangles Sum of the Internal Angles of a Triangle The three angles of a triangle add upto a straight line or 180◦. C

E

α

◦

γ◦

β◦

D DE k AB α◦ + β ◦ + γ ◦ = 180◦

B

β◦

α◦

A

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 straight line. C

a+b>c a

b a+c>b

b+c>a B A c 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 triangle must 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 < |a + b| The largest angle of a triangle is opposite the longest side. In the figure above, c > b, and ∠C > ∠B.

25


Isosceles Triangle A triangle that has at least two equal sides is called an isosceles triangle. Properties of an Isosceles Triangle

C

CB = CA

β◦ β◦

∠A = ∠B = α◦ CD bisects AB (BD = DA) B

α◦

α◦ D

A

∠ACD = ∠BCD = β ◦

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

Pythagorean Theorem a2 + b 2 = c2 c

a

B C b • Converse of Pythagorean Theorem: If the sum of the squares of the two shorter sides of a triangle is equal to the square 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 the longest side is less than 90◦ . A

Pythagorean Inequality a2 + b 2 > c2 c=9

a=6

62 + 82 > 92 or 100 > 81 ∠C < 90◦

B C b=8 • If a + b < c , where c is the longest side, and a and b are the shorter sides of a triangle, then the angle opposite the longest side is greater than 90◦ . 2

2

2

A

Pythagorean Inequality a2 + b 2 < c2 c = 11

a=6

62 + 82 < 112 or 100 < 121 ∠C > 90◦

C

b=8

B

26


45◦ − 45◦ − 90◦ Triangle An isosceles right triangle has two angles measuring each 45◦ , and one right angle measuring 90◦ , as shown in the figure √ below. The application of Pythagorean theorem to this triangle shows that the sides of the triangle are in the ratio of 1:1: 2. A

Pythagorean Theorem 45◦

AC 2 = x2 + x2 √ x 2

x

AC 2 = 2x2 √ AC = x 2 45◦ C

x

B 30◦ − 60◦ − 90◦ Triangle

◦ The figure below shows an equilateral triangle divided into two halves, each halve being a 30◦ − 60◦ − 90√ triangle. The application of Pythagorean theorem to this triangle shows that the sides of the triangle are in the ratio of 1: 3:2.

C Pythagorean Theorem

Triangle CAD and CBD are congruent AD = DB = x and AB = BC = 2x

30◦ 30◦

CD2 + x2 = (2x)2

√ x 3

2x

√ CD = x 3

60◦ A

CD2 = 4x2 − x2 = 3x2

60◦ D

x

B

Equilateral Triangles • An equilateral triangle is a triangle with three equal sides and three equal angles (60◦ ). √ s 3 • The height of an equilateral triangle of side s is h = 2 √ 2 s 3 • Area of an equilateral triangle with side s is 4 h2 • The area of an equilateral triangle in terms of its height, h, is √ 3 • The angle bisectors and the medians are identical for an equilateral triangle.

s • The radius of the circle circumscribing an equilateral triangle of side s, is R = √ 3

27

GMATQuantum Math Notes s • The radius of the circle inscribed in an equilateral triangle of side s, is r = √ 2 3

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

s

30◦ − 60◦ − 90◦ Triangle √ s :R=1: 3:2 2 s s r = √ and R = √ 2 3 3 r:

60◦

r

R

60◦

30◦ A

s 2

C

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 able to recognize it in different situations. 32 + 42 = 52

C

62 + 82 = 102

C

9 + 16 = 25 5

3

A

36 + 64 = 100

B

4

C

1.52 + 22 = 2.52

A C

A

2.5

2

3a

B

B

8

(3a)2 + (4a)2 = (5a)2 9a2 + 16a2 = 25a2

2.25 + 4 = 6.25 1.5

10

6

A

5a

4a

B

28


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 able to recognize it in different situations.

C

5

A

C 13

26

10

B

12

A

24

B

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 the figure below. The formula for the area of the triangle is: bh 2

Height, h

Area of a Triangle =

Base, b For the purpose of calculating the area of a triangle, any side of the triangle may be considered a base, and the height is then the length of the perpendicular drawn to the base from the vertex that is opposite to the base. Area of a Triangle = E

A

B

D

bh 2 F

C

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

29

GMATQuantum Math Notes ∠A = ∠D,

∠B = ∠E,

∠C = ∠F F

a b c = = d e f C

d

e a

b

E B D A 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:

F

Area of △ABC a2 b2 c2 = 2 = 2 = 2 Area of △DEF d e f C a

b A

d

e

c

B

D

f

E

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.

b

diameter, d = s

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

30


s √ 2

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

di a

m et er

,d

=

b

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 =

√

l 2 + w2 .

Regular Polygons • A regular polygon has equal sides and equal internal angles. A square is a four-sided regular polygon and an equilateral triangle is a regular three-sided polygon. The figure below shows a regular pentagon with each of the internal angles being equal to 108◦ . B

C

A

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 five sides is 180(5 − 2) = 180(3) = 540 degress. • Each interior angle of an equiangular or regular n-gon is equal to 180(n − 2) n In case of a hexagon, each interior angle is equal to • The central angle of a polygon is equal to

180(6 − 2) = 120 degrees. 6

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

31


√ s2 (3 3) • The area of a hexagon is equal to , where s is the length of one of the sides of the hexagon. A regular hexagon 2 √ s2 3 is made up of six equilateral triangles, and the area of one of the equilateral triangles is , and the area of the 4 hexagon is six times this value

120◦

60◦ 60◦ • The center of a regular polygon is the same as the center of its inscribed and circumscribed circle. The radius of the circumscribed 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.

30◦ − 60◦ − 90◦ Triangle √ s :r:R=1: 3:2 2 √ s 3 r= and R = s 2

R

30◦ r

60◦ 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 62 equal to 2 = 4, or 4 to 1. 3

6 3

32


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 figure below shows that irrespective of the location of point C on the circle, ∠C is always equal to 90◦ . C

b

A

B

O

n◦ • Length of an Arc: In a circle of radius r, the length l of an arc with a central angle of n◦ equals fraction of the 360◦ circumference of the circle. n◦ Length of an Arc = (2πr) 360◦ • Area of a Sector: The area of a sector of a circle that subtends an angle n◦ , expressed in degrees, is given by: n◦ (πr2 ) Area of a Sector = 360◦

Cubes A

C B

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.


33

• 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. These relationships 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 = A1 A2 A3 , 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 a rectangular 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 and bottom surfaces, and 2πrh is the lateral surface area. • The volume of a cylinder is πr2 h. • The volume of the largest rectangular solid that can be fitted inside a right circular cylinder of radius r and height h is 2r2 h. • The volume of the smallest rectangular solid that can accomodate a right circular cylinder of radius r and height h is 4r2 h. √ • 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 by the formula: p Distance = (x1 − x2 )2 + (y1 − y2 )2

The 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 the figure below.

34


(x2 , y2 )

y b

p (x 1

−

)2 x2

(y 1

+

−

b

x1 + x2 y1 + y2 , 2 2

y2 − y1

y-axis

)2 y2

(x1 , y1 ) b

x2 − x1

x

b

(0,0) Origin

x-axis

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

x2 + x1 y2 + y1 , 2 2

• The coordinates of the point that divides the line segment AB in the ratio r : s is given by: rx2 + sx1 ry2 + sy1 , r+s r+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 passing

Two-Point Form

y2 − y1 (x − x1 ) y − y1 = x2 − x1

Line passing through

Intercept Form

x y + =1 a b

a and b are the

through (x1 , y1 )

(x1 , y1 ) and (x2 , y2 )

x and y intercept

35


(x2 , y2 )

y b

Slope =

y2 − y1

y-axis y2 − y1 x2 − x1 (x1 , y1 )

x-intercept

b

(0, b)

x2 − x1 b

y-intercept x

b b

(a, 0)

(0,0) Origin

x-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: (−

b , 0) m

• 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 passing through (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, x y 2 and − = 7, all have a slope of and are all parallel to each other. 3 2 3 • Two lines perpendicular to each other with slopes m1 and m2 satisfy the following relationship: m1 m2 = −1

The lines y = 3x and y = −

x are perpendicular to each other. 3

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 the distance 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

GMATQuantum Math Notes y

r

(h, k)

y−k

(x, y)

b

x−h x

b

(0,0) Origin

• If the circle is centered at the origin (h, k) = (0, 0), then the equation of the circle simplifies to x2 + y 2 = r2 .

Statistics Mean, 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 x¯ and is defined as Arithmetic Mean : x ¯=

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 an odd number of numbers, then the median is the middle number, and if there are an even number of numbers, then the median 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 x ¯A and x ¯B , respectively, and let x ¯A > x¯B . The averages could be for any specific property of the group like age, height, or years of experience. Also, let NA and NB , 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 NA x¯A + NB x¯B NA NB x¯A+B = = x ¯A + x ¯B NA + NB NA + NB NA + NB The equation can also be rewritten in terms of the fraction of people in group A: x¯A+B = fA x¯A + (1 − fA )¯ xB

NA where fA = . NA + NB The average of the new combined group, x ¯A+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 x ¯ is the arithmetic mean of the set is given by s (x1 − x ¯)2 + (x2 − x ¯)2 + · · · + (xn − x ¯)2 s= n

37


• Impact on Mean and Standard Deviation: The table below lists how the mean, x ¯, 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

x¯ + a

Unchanged

Subtract a constant, a

x¯ − a

Unchanged

Multiply by a constant, a

a¯ x

as

Increase by p%

p 1+ x¯ 100

Divide by a constant, a

x ¯ a

1+

p s 100

s a

• Standard Deviation: Computational Formula: The computation formula for the standard deviation is a rearrangement of the general formula for the standard deviation. s x21 + x22 + x23 · · · + x2n s= − x¯2 n where {x1 , x2 , x3 , . . . , xn } are the n data points, and x¯ 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 independent of 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 and 8 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 word permutation, it is left to the student to figure out if the order in which the objects are arranged is important in finding the total number of arrangements. The number of permutations, denoted by n Pr , or the number of arrangements of n objects taken r at a time is given by n! n Pr = = n(n − 1)(n − 2) · · · · · · (n − r + 1) (n − r)!

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

Cr =

n! n(n − 1)(n − 2) · · · · · · (n − r + 1) = r!(n − r)! (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 permutations of n distinct objects in a circle is given by (n − 1)! = (n − 1)(n − 2) · · · 1.

38


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, the sequence {3, 7, 11, 15, . . . }, where each term is 4 larger than the previous term is an arithmetic sequence. If a is the first term 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)d In other words, the nth term of an arithmetic sequence can be found by adding the common difference n − 1 times to the first term.

Sum of an Arithmetic Sequence The sum Sn of the first n terms of an arithmetic sequence is given by First + Last n(a + an ) Sn = (Number of Terms) × = 2 2 where an = a + (n − 1)d is the nth term. The arithmetic mean(average) and the median of an arithmetic sequence are equal and 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 Last Term − First Term +1= +1 d Spacing

39


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 ratio n − 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 the remainder r. Then the nth term of the sequence is the same as the rth term. To find the 100th digit after the decimal point in the repeating decimal 0.2857142857142 . . ., divide 100 by 6, which gives a remainder of 4. Then the 100th digit is the 4th digit 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 the decimal expansion of 2/7, then the 60th term is the last term in the repeating group, 4 in this example.

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