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Quantitative Methods Rule 1: We can multiply or divide all terms5 on both sides of an equation by the same letter or number without changing the relationship expressed by the equation or the value of the unknown variables. Let’s multiply both sides of Equation 1 by the number 5. a 6 (1) 5 a 5 6 5a 30 a 6 By multiplying each side of Equation 1 by five, we did not alter the basic relationship between the left-hand and right-hand side of the equation.6 In fact, we could have multiplied every term in the equation by any number or any letter without changing the relationship expressed by the equation. In every case, the unknown value, signified by the letter a in Equation 1, would still equal 6. We can also divide both sides of the equation by a number or letter without altering the relationship expressed by the equation. Recall Equation 1: a 6 (1) This time let’s divide both sides by the number one. a 6 a 6 1 1 Dividing by one does not change the relationship expressed in Equation 1 since the value of a remains unchanged. To make the example easy, we relied on a very basic rule. Any term (number or letter) multiplied or divided by one equals the original number or letter. So: 6 1 6 and 6 6 1 Stated differently: a 1 a and 5 6 a a 1 Term is the mathematical expression for an entry in an equation. Equation 1 only has two terms: the letter a on the left side of the equation and the number 6 on the right. A number in combination with (multiplied by) a letter is referred to as the coefficient of the letter (the variable the letter represents). In this case, 5 is the coefficient of a. Page 8 CFA Fundamentals 2e.indb 8 2012 Kaplan, Inc. 1/31/2012 5:13:51 PM

Quantitative Methods Even though we used the number one in the example, we could have divided every term in Equation 1 by any number or letter and the result would have been the same. In every case, the letter a would have the same value (i.e., 6 in this case). Rule 2: We can add (or subtract) the same letter or number from both sides of an equation without changing the relationship expressed by the equation or the value of the unknown. Let’s look again at Equation 1, and this time we will add and subtract a number from both sides. a 6 (1) Adding and subtracting one, a 1 6 1 or a–1 6–1 The best way to solve an algebraic equation with one unknown is to collect all terms with the unknown on the left of the equal sign (the left side of the equation) and to collect all known values on the right side of the equation. a 1 6 1 a 6 1–1 a 6 a–1 6–1 a 6–1 1 a 6 Every time you move a variable or number from one side of the equation to the other, its sign changes (i.e., a negative becomes positive or a positive becomes negative). Notice that in both previous cases, the “1”on the left-hand side of the equation changed signs when we moved the number to the right side of the equation. When we added one to the left side of the equation and then moved it to the right, it went from positive to negative. When we subtracted one from the left side and then moved it to the right, it went from negative to positive. In both cases the positive and negative ones on the right side of the equation canceled each other out (sum to 2012 Kaplan, Inc. CFA Fundamentals 2e.indb 9 Page 9 1/31/2012 5:13:51 PM

Quantitative Methods zero).7 In fact, if we add any number or letter to, or subtract any number or letter from both sides of the equation, we observe the same outcome. The additional numbers or letters will cancel each other out on the right side of the equation, and the value of the unknown will be left unchanged. Now that we have a few rules to make algebra easier for us, we’ll proceed to a few examples. Consider the following: If apples cost 0.25 each and you have 1.00, how many apples can you buy? Since each apple costs 0.25, if you multiply that price by the number of apples you buy, the total cannot be greater than 1.00. We will let the letter A represent the number of apples you buy. Assume you buy one apple. That means A 1 and you spend 1 0.25 0.25. If you buy two apples, A 2 and you spend 2 0.25 0.50. If A 3 you spend 0.75, and if A 4 you spend exactly 1.00. Thus the answer is 4. You can buy a maximum of 4 apples with your 1.00, as long as they cost 0.25 each. We can formalize the relationship between the amount of money available to purchase apples, the cost of apples, and the number of apples purchased in the following algebraic equation: A M 1.00 4 C 0.25 (2) where: A the number of apples you can buy (the unknown) M the maximum number of money you can spend ( 1.00) C the cost per apple ( 0.25) We could state the problem another way. How much would four apples cost if each apple costs 0.25. If apples cost 0.25 each, four apples will cost 1.00. This modification can be shown by rearranging Equation 2 into Equation 2a: A M C (2) M AC M 4 0.25 1.00 (2a) 7 The term sum simply means to add numbers together. In this case, we summed (added) one and negative one, which equals zero. Page 10 CFA Fundamentals 2e.indb 10 2012 Kaplan, Inc. 1/31/2012 5:13:51 PM

Quantitative Methods In transforming Equation 2 into Equation 2a, we have utilized Rule 1 by multiplying each side of Equation 2 by the letter C: CA CM CA M C We rewrite this equation as follows: M AC The letter M is the unknown, and by convention, we put the unknown variable on the left side of the equation. Of course, the result is exactly the same either way. The equation is equally valid in either direction, so it really doesn’t matter which way we write it. That also holds for the direction we multiply the variables. Notice that we rearranged the combination CA to AC. This is another simple trick, which makes absolutely no difference to the value of M. You may confirm this relationship by multiplying three times two and then multiplying two times three. You’ll get six in either case. This property holds, regardless of the numbers or variables used, as long as you’re multiplying. Now, let’s look at the same problem from another perspective. Let’s assume you already purchased eight apples for 2.40. What was the cost per apple? To solve this problem, let’s set up Equation 2 exactly as before: A M C (2) where: A the number of apples you bought (8) M the amount of money you spent ( 2.40) C the cost per apple (the unknown) This time, however, we know how much money you spent ( 2.40) and how many apples you bought (8). Again, we substitute values for letters wherever we can. The result is: 8 2.40 C We face a new challenge. How do you solve for C when it’s in the denominator (i.e., bottom) on the right side of the equation and you would like it in the 2012 Kaplan, Inc. CFA Fundamentals 2e.indb 11 Page 11 1/31/2012 5:13:52 PM

Quantitative Methods numerator (i.e., top) on the left side? To get the C out of the denominator on the right side of the equation, we’ll utilize Rule 1 and multiply each side of the equation by C: C 8 C 2.40 C The Cs on the right side of the equation cancel each other out. Remember, any C number or algebraic letter divided by itself equals one, so 1 . The equation C now reads: 8C 2.40 Now we again must utilize Rule 1 and divide each side of the equation by the number eight in order to isolate the unknown, C, on the left side of the equation: 8 C 2.40 2.40 C 0.30 8 8 8 You might have noticed we have designated “multiplied by” in two different ways in our example. First we used AC to mean A times C. Then we used C 8 to mean C times eight. Each of these ways is entirely appropriate. In fact, we could also have written A(C) or (A)(C) to indicate A times C. So there are at least four ways to write exactly the same thing. Here’s another algebra example related to apples. You have a total of 5.00 to spend. If the apples cost 0.79 per pound, how many pounds can you buy? Letting: P the number of pounds you can buy (unknown) C the cost per pound ( 0.79) M the total amount of money you can spend ( 5.00) The amount of money you can spend, M, divided by the price per pound, C, will yield the number of pounds you can purchase, P. Let’s set up the equation. P M 5.00 P 6.33 C 0.79 (3) Now let’s alter this problem to determine the maximum amount of apples you can buy for 5.00. We know you cannot spend more than 5.00, but you can certainly spend less if you prefer. We also know that the number of pounds you buy, P, Page 12 CFA Fundamentals 2e.indb 12 2012 Kaplan, Inc. 1/31/2012 5:13:52 PM

Quantitative Methods multiplied by the cost per pound, C, is the total amount you spend. Let’s set up that equation: P C M (3a) Equation 3a is interpreted as follows: “The number of pounds you buy times the cost per pound must be less than or equal to the total money you can spend.” In Equation 3a, notice we have replaced the equal sign with an inequality. In this case the sign does not say the left side of the equation must equal the right side. Instead it says the left side must be equal to or less than the right side.8 An inequality sign has some properties that are the same as an equal sign and other properties that are different. The following section demonstrates the similarities and dissimilarities. Let’s put some numbers in Equation 3a: P C M P 0.79 5.00 P 5.00 P 6.33 0.79 Notice we get exactly the same numerical answer as before. When we solved for P the first time we got exactly 6.33 pounds. Now, however, the answer to the equation tells us the amount purchased, P, can be a range of values. It can be any value equal to or less than 6.33 pounds. In other words, with just 5.00 to spend, you cannot buy more than 6.33 pounds, but you can certainly buy less than 6.33 pounds if you wish. A word of caution is appropriate here. Even though the sign didn’t really change the answer to our problem, it is interpreted quite differently from the sign. For instance, we used a 6 in the very beginning as an example of an equation. The equal sign is interpreted as saying a is equal to six and no other value. That is a strict equality. If we substitute for , the meaning of the equation changes dramatically because the two signs actually ask two different questions. The asks for a specific value. The sign asks for a range of values, including the maximum value (or upper limit). 8 It is helpful to think of the point of the inequality sign as pointing to the side of the equation that has the smaller value. 2012 Kaplan, Inc. CFA Fundamentals 2e.indb 13 Page 13 1/31/2012 5:13:52 PM

Quantitative Methods The equation a 6 is interpreted as saying the letter a can have any value as long as it is equal to or less than 6. The unknown, a, could have a value that would include all negative values. We could change the sign in our example to a sign because we wanted to know the absolute maximum we could buy, but we could buy a smaller amount if we wanted. Rule 3: Multiplying or dividing an inequality by a negative number changes the direction of the inequality. Remember that an inequality is denoted by the less than or equal to sign or greater than or equal to sign. Multiplying or dividing the inequality by a negative number changes the direction of the inequality. For example: 3 x 9 ( 3 x ) (9) x 3 3 3 Observe that in the preceding equation, the inequality sign changes direction after dividing both sides of the equation by negative three. Now let’s take the previous result and multiply by negative three: x –3 – 3(x) –3(–3) –3x 9 Notice that after multiplying by negative three, the equation is right back where it began. Parentheses. Parentheses are used to group together variables and numbers that should be considered together in the equation. Consider the following equation: 3(x 4) 15 (4) We interpret this equation as three times the quantity9 (x 4) equals 15. We can treat the terms in the parentheses as if they are a single term. That means we can divide both sides of the equation by three (using Rule 1) and solve for the unknown. 3( x 4 ) 15 ( x 4) 5 3 3 9 Quantity is simply the mathematical expression used to denote the parentheses and everything inside them. Page 14 CFA Fundamentals 01.indd 14 2012 Kaplan, Inc. 2/3/2012 8:34:49 AM

Quantitative Methods Once the three is no longer outside the parentheses on the left side of the equation, we can simply eliminate the parentheses and solve for the unknown. x 4 5 x 5–4 1 Let’s look at a more complex equation: 3 15 ( x 4) (4a) At first this may appear difficult, but it’s actually just another way of writing Equation 4. We can see this by multiplying both sides of Equation 4a by the quantity (x 4) as follows: ( x 4 ) 3 ( x 4 ) 15 3( x 4 ) 15 ( x 4) (4) To get the quantity (x 4) out of the denominator on the right side of the equation, we multiplied both sides of the equation by (x 4) as if it were a single x 4 term. That leaves 3(x 4) on the left side of the equation, and since on the x 4 right cancels to 1, we are left with only the 15 on the right side of the equation. The result is Equation 4, which we solved earlier. There is another way to solve Equation 4. As an alternative, we can multiply everything inside the parentheses by three (multiply through by three10) instead of dividing both sides of the equation by three. Note that we get exactly the same answer. This is because we are solving the equation in two different ways without changing the relationship of the equation in any way. 3(x 4) 15 (4) 3x 12 15 3x 15 – 12 (4b) 3x 3 x 1 10 This is also known as distributing. The three, in this case, gets distributed across every number or variable inside the parentheses. 2012 Kaplan, Inc. CFA Fundamentals 2e.indb 15 Page 15 1/31/2012 5:13:53 PM

Quantitative Methods Now let’s assume you begin with the form in Equation 4b. How would you go about solving it? Let’s factor out a number common to all the terms on the left side of the equation.11 To factor the left side of the equation we’ll find the largest number that divides evenly into both 3x and 12. The largest number is three, so we’ll factor three out of both terms. 3x 12 15 3(x 4) 15 Dividing both sides of the equation by three, 3 15 ( x 4) ( x 4) 5 3 3 x 4 5 x 5–4 x 1 It should be clear from the solution that factoring the left side of the equation did not alter the value of the unknown. Exponents. Algebraic equations become slightly more complicated when an exponent affects one or more of the variables as in Equation 5: x2 9 (5) Equation 5 is interpreted as “x squared12 equals nine.” The solution for the value of x is that number which when multiplied by itself (squared) equals nine. You may be able to determine that the answer is three, since 3 3 9. Just in case you’re confronted with more difficult equations involving exponents, let’s see how to solve Equation 5. In order to solve for a variable, the exponent associated with that variable must equal one.13 In this example, in order to change the value of the exponent to one, we can take the square root of both sides of the equation. Since we used easy numbers, the equation is easy to solve as follows: x 2 9 x1 31 x 3 11 Factors are simply numbers that when multiplied together yield the original number. The numbers 3 and 5 are factors of 15, since 3 5 15. Likewise, 2 and 4 are the factors of 8. A prime number is divisible only by itself and 1. The number 3 has no factors other than 1 and 3, so it is considered a prime number. 12 Squaring a number simply means multiplying it by itself. For instance 2 squared is 2 2 4. 13 Any number, variable, or quantity raised to the power of one is equal to itself. In other words, 51 5. Five raised to the power of one is equal to five. Page 16 CFA Fundamentals 2e.indb 16 2012 Kaplan, Inc. 1/31/2012 5:13:54 PM

Quantitative Methods Another way to designate square root is with the exponent ½. Let’s try solving the equation using that method. Instead of using the square root sign, or radical, we’ll use ½ as an exponent. (x2)1/2 – (9)1/2 We’re trying to find the square root of x2 and the square root of nine. When you see a term with more than one exponent, simply multiply the exponents together. In this example, when you multiply the exponents associated with the x variable, you get an exponent of 2 ½ 1 and x1 or simply x on the left of the equation. Since there is an implied exponent of one associated with any number or letter, we can say we actually have (91)1/2on the right side of the equation. Again we multiply the exponents and get 1 ½ ½, leaving us with 9½ on the right side of the equation: x 9 1/2 3 Since an exponent of ½ indicates square root, we’re back to our original statement; x equals the square root of nine. Parentheses and Exponents. The next logical step is to solve an equation in which we have a set of parentheses with an exponent. Remember that quantities within parentheses can be treated like a single term. Consider the following equation: (x 5)3 – 8 (6) Keep in mind that in order to solve for a variable, its exponent must equal one. In this example we can accomplish this by taking the cube root (third root) of both sides.14 Remember, we’re going to treat the quantity inside the parentheses just like a single term and multiply the exponents, as we did before. [(x 5)3]1/3 81/3 x 5 81/3 x 5 2 x –3 14 The cube, or third root of a number, is the number that, when multiplied by itself three times, equals the original number. For example, the cube root of 1,000 is 10, since 10 10 10 1,000. 2012 Kaplan, Inc. CFA Fundamentals 2e.indb 17 Page 17 1/31/2012 5:13:54 PM p

CFA candidates come from different educational backgrounds and many walks of life. What is the Structure of the CFA Program? CFA Institute, the governing body that creates and administers the CFA exams, is headquartered in Charlottesville, Virginia. In July of each year, CFA Institute :