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<< table of contents GMAT home page GRE home page 800score.com home Mathematics For easier use, print out this guide Math for 14 year olds Most of the math on the GMAT is set to what would be called junior high school level in the United States. It involves simple algebra and geometry. The irony is that most students could have done better on the GMAT Math section ten years earlier when the material was fresher in their minds. To help you, we have math chapters dedicated to Algebra, Percentages/Ratios, Geometry, and Arithmetic. Test writers must only use pre-high school level math because if they used higher level math, it would create an advantage for math majors. The result is that there is no calculus or trigonometry on the exam. For test writers this creates a challenge, "How do you create questions using math for 14 year olds that Harvard grads will get wrong?" The answer is that they primarily use: 1) trick questions 2) extremely difficult Word Problems 3) questions that require extensive deductive reasoning (Geometry and Data Sufficiency) Word Problems are the focus of the test, and for that reason, the Math section is more of a reading test than strictly a "math" test (for those British/Commonwealth students "maths" is "math" in the U.S.). Test writers devise extremely complex and contorted Word Problems that can fool 95% of GMAT students using only basic math principles. That is why we have a chapter dedicated exclusively to Word Problems strategy. Foreign students should pay particular attention to this section because Word Problems will give them the most trouble. We focus on Word Problems in chapter 7. Note: these chapters are particularly dense with material. Take them slowly and digest the information thoroughly before progressing to the next chapter. Scrap Paper We discussed scrap paper earlier in the guide. Doing calculations in your head may increase errors, particularly under the pressure of test day. This is no-win situation because copying questions also invites errors. Because you cannot write on the screen, every question has to be partially recopied onto scrap paper. Students should always practice with unlimited scrap paper (just like on test day) in order to get used to this awkward process of recopying questions. This chapter is broken into two parts: I. General Math Strategies II. Basics I. General Math Strategies A. Backsolving B. Plug-In C. Possible Range Strategy A. Backsolving (otherwise known as "reverse-solving") This is the most important strategy for the math section! Backsolving involves inserting answer choices to solve problems. This is often preferable to trying to translate a complicated question into an algebraic equation. Inserting answer choices into a Word Problem can also make a complicated question more understandable. Backsolving helps you eliminate or choose a sample answer. Remember, math questions on the GMAT use very simple principles, so the test writers have to make the questions as complicated and intimidating as possible. Some questions are actually written so that backsolving is the only effective way to solve the problem. How to Backsolve: 1. Decide if the problem is too complicated to solve algebraically (this should take only a few seconds). 2. Insert the middle answer--the one that would be in the middle of potential answers if it were on a number line. 3. If a smaller number would work, choose answer choices 1 or 2; if a larger number would work, choose 4 or 5. (The answer choices are usually arranged from lowest to highest value--answer choices 1 through 5). 4. Eliminate down to one answer and choose. Try backsolving on this question: When the positive integer Z is divided by 24, the remainder is 10. What is the remainder when Z is divided by 8? a) 1 b) 2 c) 3 d) 4 e) 5 Solution Notice that this question seems to defy a quick algebraic solution. The best way to address this problem is to backsolve, to take potential answers and feed them into the question until one answer works. Pick a number for Z such that Z/24 has a remainder of 10; that is, a number 10 greater than a multiple of 24. 24 is a multiple of 24, so let Z = 24 + 10 = 34. When 34 is divided by 8 you get 34/8 = 4 with a remainder of 2. Not satisfied? Try 58 (which is 2 × 24 with a remainder of 10). 58 divided by 8 gives a remainder of 2. It appears that (B) is the correct answer. Strategy: Plug-in is an even more effective strategy for double checking your answers. When you arrive at an answer, plug in the answer to test that it works or plug in more numbers. Remember that you have nearly two minutes to do each question on the GMAT CAT. This gives you plenty of time to double check yourself; however, pacing is still extremely important, as you read in Chapter One. B. Plug-In Sometimes the best approach to backsolving is not to put in sample answer choices, but to insert numbers that prove/disprove the question. The numbers you choose for backsolving should fit the question's parameters. For example, if the question asks for an integer, you should insert integers. Usually try plugging in a few different numbers (positive, negative, etc.). 1. Decide if the problem is too complicated to solve algebraically (this should take a only few seconds). 2. Insert sample numbers into the equation. Depending on the question, try negative numbers, positive numbers, odd/even and fractions. 3. Eliminate down to one answer and choose. Plug-in Example If n is an even integer, which of the following must be an odd integer? a) 3n - 2 b) 3(n + 1) c) n - 2 d) n/3 e) n/2 Solution (B) Every time you have variables in the answer choices, you should plug in. Make n equal to 2. If n is 2, then 3(n + 1) = 9. Since our target is an odd integer, this answer choice works. Try a few more numbers to double check. For example, 2 may work with choice (e) to make an odd number (1), but it will not work with any other even numbers. C. Ballpark Strategy (otherwise known as the possible range strategy) This strategy allows you to answer questions quicker than doing the calculations, and it is an effective tool to double-check your answers. Using the Ballpark Strategy, you find an answer by what could reasonably be in the range of the answer. This is particularly useful when the possible answers are scattered over a large range. Try the Ballpark Strategy here: If 0.303z = 2,727, then z = a)9,000 b)900 c)90 d)9 e)0.9 Solution (A) Because the answer choices are so far apart, you can ballpark this problem. Think about it: .303 is close to 1/3. 1/3 of z = 2,727, then what answer could possibly be correct? You don't even have to do the math. 2,727 is about 1/3 of 9,000; therefore, the answer must be 9,000, according to the Ballpark Strategy (note that there are no other answers even in the 9,000 range. Or, you could multiply both sides by 1000 to eliminate the decimal points, then divide 2,727,000 by 303 and get the same answer. Strategy: Ballparking is an effective strategy for double checking your answers. When you arrive at an answer, make sure it is in the ballpark of what the answer could be. II. The Basics This chapter contains a basic review of math concepts. Most students should skim through this section. It is vital that you know all the terminology in this section since the questions on the test will assume you know it. A. Integers B. Positive / Negative Rules C. Fractions D. Equivalent Fractions E. Multiplying and Dividing Fractions F. Adding and Subtracting Fractions G. Decimals H. Adding and Subtracting Decimals I. Multiplying and Dividing Decimals J. Averages and Medians A. Integers A positive number is a number greater than zero, such as + 5 (usually written as 5). A negative number is a number less than zero, such as -5. A whole number is a number that does not include a decimal part or a fractional part. It is also called an integer. The absolute value of a number, written as | -5 |, is the magnitude of the number; the absolute value of + 5 is equal to the absolute value of -5, written as |+5 | = |-5 | = 5. Zero is an integer B. Positive / Negative Rules What happens when positive and negative numbers are added/subtracted/multiplied? Multiplication/Division Positive × Positive = Positive, for example 2 × 2 = 4 Positive × Negative = Negative, for example 2 × -2 = -4 Negative × Negative = Positive, for example -2 × -2 = 4 Addition/Subtraction Adding a negative number is the same as subtraction. 4 + (-5) = 4 - 5 = -1 Subtracting a negative number is the same as addition. 4 - (-5) = 4 + 5 = 9 An even number is an integer that is divisible by 2 (4, 6, 8, 10). An odd number is an integer not divisible by 2 (3, 5, 7, 9). A prime number is a positive integer that has exactly two different positive divisors, 1 and itself. For example, 2, 3, 5, 7. The number 1 is not a prime number since it has only one positive divisor. Consecutive numbers are a set of numbers in which each member of the set is the successor of its predecessor. Examples: even consecutive numbers: 4, 6, 8, 10 odd consecutive numbers: 3, 5, 7, 9 prime consecutive numbers: 3, 5, 7, 11, 13. Adding, Subtracting and Multiplying odd/even: The following is true of even and odd whole numbers (use example numbers in your mind to illustrate (note that zero is an even number): Even + Even = Even exg. 4 + 4 = 8 (even) Odd + Even = Odd, exg. 3 + 4 = 7(odd) Odd + Odd = Even exg. 3 + 3 = 6 (even Even × Even = Even exg. 2 × 2 = 4 (even) Even × Odd = Even exg. 2 × 3 = 6 (even) Odd × Odd = Odd exg. 3 × 3 = 9 (odd) Even - Even = Even exg 16 - 8 = 8 (even) Even - Odd = Odd exg. 16 - 5 = 11 (odd) Odd - Odd = Even exg. 9 - 5 = 4 (even) Even/Odd Example If k is an odd integer, state whether each of the following is odd or even: a) k + k + k b) k × k × k c) k + 2k d) 2k × k Solution a) (k + k) is even. Thus (k + k) + k is an even plus an odd, which is odd. b) k × k is odd. Thus (k × k) × k is an odd times an odd, which is odd. c) k + 2k is an odd plus an even, which is odd. d) 2k is even. An even times an odd is even. Strategy 1: If you do not remember the rules, merely plug in by selecting an odd number (e.g., 3) or an even number and perform the required operation. Strategy 2: Get used to plugging in numbers. On most math questions, you will have to use the either plug in strategy or backsolve. A factor is an integer that divides another number resulting in a whole number. Consequently, a number can be expressed as a multiple of each of its factors. The number 24 has factors 1, 2, 3, 4, 6, 8, 12, 24. Note that 24 is a multiple of any one of its factors, i.e., 24 is a multiple of 8. The least common multiple (LCM) of several numbers is the smallest integer which is a common multiple of the several numbers. LCM example Write the LCM (Least Common Multiple) of 6 and 12. Solution The factor 2 × 3 is used only once because it occurs in both numbers. Thus, the LCM is (2 × 3 × 2) = 12. Or, what is the smallest number that both 6 and 12 go into? 12. 12 is the lowest number divisible by both 6 and 12. EXAMPLES Example 1 Compute each of the following: a) 2 - 3 b) 4 + 6 - 10 c) (-5 + 2)(-3) Solution a) 2 - 3 = -1 b) 4 + 6 - 10 = 10 - 10 = 0 c) (-5 + 2)(-3) = (-3)(-3) = 9 Example 2 State all the factors of 63. Solution The number 63 can be divided by 1, 3, 7, 9, 21, and 63. Hence, these are its factors. Example 3 Write 63 as a product of prime numbers. Solution The number 63 is obviously not divisible by 2, but is divisible by 3. Hence, 63 = 3 × 21 = 3 × 3 × 7 Both 3 and 7 are prime numbers. We do not include 1 as a prime number. Example 4 Find the LCM (Least Common Multiple) of the three numbers 20, 30 and 50. Solution Write each number as a product of prime numbers: 20 = 2 × 2 × 5 30 = 2 × 3 × 5 50 = 2 × 5 × 5 The factor 2 × 5 occurs in all three numbers; thus it is used only once. The LCM is LCM = (2 × 5) × 2 × 3 × 5 = 300 C. Fractions A fraction is one number divided by another number. It is division, such as 3/5. The numerator is the top number, and the bottom number is the denominator. The denominator represents the number of equal parts into which an entity has been divided; the numerator represents the number of parts that are selected. For example, if a garden is divided into 5 equal plots, 3 of the plots is 3/5 of the garden. A fraction that has 0 as its denominator (e.g., 5/0) is infinitely large and undefined (not a real number or an integer). If the numerator is zero, then the fraction equals zero (e.g. 0/5 = 0). If the fraction has a numerator equal to the denominator (e.g., 5/5), the fraction is equal to 1. A mixed number is a number that is an integer plus a fraction. The number 4 2/3 is the integer 4 plus the fraction 2/3. Any mixed number can be written as a fraction, and any fraction greater than 1 can be written as a mixed number. To express 4 2/3 as a fraction we multiply 4 × 3, add the numerator to this product, 12 + 2 = 14, and divide by the denominator: 4 2/3 = 14/3. To convert the fraction 17/5 into a mixed number, divide by the denominator (17 divided by 5 is 3 with 2 remaining), and add the remainder over the denominator: 17/5 = 3 2/5. Example 5 Convert 4 5/7 into a fraction. Solution We multiply 4 × 7 and obtain 28. Add 5 to this and obtain 33. Put this over 7, and we find 4 5/7 = 33/7 Example 6 Convert 79/9 into a mixed number. Solution Divide 79 by 9 and obtain 8 with 7 remaining. Now add to the 8 the fraction 7/9 (8 7/9). D. Equivalent Fractions A fraction that has a common factor in both numerator and denominator is equal to the fraction with the common factor canceled. The fraction 6/10 is equivalent to the fraction 3/5 since the common factor 2 occurs in both numerator and denominator of 6/10. In fact, the following fractions are all equivalent: 3/5 = 6/10 = 9/15 = 12/20. A fraction that has no common factors in the numerator and denominator is said to be expressed in lowest terms. A fraction with a negative numerator or denominator is equivalent to a negative fraction, that is, -3/5 = 3/-5 = - (3/5). If both numerator and denominator are negative, the fraction is positive, that is, -3/-5 = 3/5. Example 7 Express 26/16 as a mixed number in lowest terms. Solution The mixed number is found by dividing by 16 giving 1 with 10 remaining. Hence 26/16 = 1 10/16 = 1 5/8 E. Multiplying and Dividing Fractions To multiply fractions, cancel out any common factors that appear in both numerators and denominators. Then multiply all numerators to form one numerator and all denominators to form one denominator. This final fraction may then be written as a mixed number, if desired. To divide fractions, say (x/y)/(a/b), invert the divisor (the fraction a/b) and multiply the two fractions, i.e., (x/y)/(a/b) = (x/y) × (b/a). For instance, (5/6) × (7/2) = (7× 5)/(6 × 2) = 35/12 F. Adding and Subtracting Fractions To subtract one fraction from another, we simply add a negative fraction to a second fraction. Consequently, the rules for adding and subtracting are the same. The first step is to write the fractions such that each fraction has the same denominator. Then add or subtract the numerators. Then simplify. 1/3 + 1/4 = 4/12 + 3/12 = 7/12 To write all fractions with the same denominator, a quick choice is to multiply all denominators together. Hoever, this may give a rather large denominator. To avoid a large denominator, we could find the least common denominator (LCD); it is the least common multiple (LCM) of all the denominators. G. Decimals A decimal fraction is a fraction whose denominator is a power of 10 but not a factor of the numerator. For example, the fraction 3/100 is written in decimal form as 0.03; the fraction 223/10,000 is 0.0223. Note that the number to the right of the decimal point is the numerator of the decimal fraction, and the denominator is 10 raised to the power n where n is the number of places to move the decimal point. To write 0.0031 as a decimal fraction, 31 is the numerator, and because we moved the decimal point 4 places, the denominator is 10 = 10,000; consequently, the decimal fraction is 31/10,000. For decimals .1 = one tenths .01= one one-hundredths .001 = one one-thousandths Rule #1: If you multiply a decimal by 10, you would move the decimal 1 to the right; 100, move 2 to the right; 1000, move 3 to the right, and so on. For instance, 100,000 × 0.0054 = 540. You see that there are 5 zeros in 100,000; therefore, we moved the decimal 5 places to the right. Rule # 2: When you divide by 10, you move the decimal once to the left; 100, move 2 to the left; 1000, move 3 to the left. Example 8 Write 31.2/(100,000) as a decimal. Solution When we divide by 100,000, which is 10 , we move the decimal point 5 places to the left. Then 31.2 = 0.000312. A mixed decimal is the sum of an integer and a decimal fraction, much like the mixed fraction. The integer 5 added to the decimal fraction 35/100 is 5 + 35/100 = 5 + 0.35 = 5.35. The mixed decimal, or a decimal fraction, is usually simply called a decimal. (Note: the leading 0 in 0.35 is simply convention; it does not have mathematical significance: 0.35 = .35.) Example 9 Write the fraction 649/100 as a decimal. Solution The numerator 649 can be written as the decimal "649.0". When we divide by 100, we move the decimal point 2 places to the left so that 649/100 = 6.49 It is not necessary to write any zeros after the 9; that is, 6.490 is equivalent to 6.49. Example 10 Express 0.075 as a fraction in lowest terms. Solution The decimal is expressed as a fraction .075 = 75/1000 The denominator and numerator are factored resulting in 75/1000 = (25 × 3) / (25 × 40) = 3/40 H. Adding and Subtracting Decimals To add or subtract decimals, we write the decimals in a column with the decimal points aligned vertically. First, combine the decimals with a plus sign. Next, combine the decimals with a minus sign. Then subtract the sum of the negative decimals from the sum of the positive decimals. In performing these tasks, we add zeros to the right of the decimal point so that each number has an entry in each column. For example, if we subtract 3.021 from 5, we write 5 as 5.000 so that there is an entry in each of the 3 places to the right of the decimal in both numbers. Example 11 Add 5 + 2.783 + 3.04. Solution Write the decimals in a column, with zeros added if none exist: 5.000 2.783 + 3.040 10.823 Example 12 Compute 6.98 + 3.217 + 3 - 3.637 Solution Add the positive decimals and the negative decimals: + 6.980 + 3.217 + 3.000 - 3.637 9.56 Subtract the sum of the negative decimals from the sum of the positive decimals: 13.197 - 3.637 9.560 I. Multiplying and Dividing Decimals Multiply two decimals just like you would multiply two integers. The number of decimal places in the product is then equal to the total of the decimal places in the two decimals. To divide two decimals, move the decimal point in the divisor (the number doing the dividing) to the right so that the divisor is an integer. Move the decimal point in the dividend to the right the same number of places. Now perform the division, placing the decimal point in the answer directly above the decimal point in the dividend. Example 13 Compute 0.05 × 12. Solution Set up the multiplication as though the decimals were integers: 12 × .05 = .60 The answer must have a total of 2 decimal places: .60. J. Averages and Medians The average, or arithmetic mean, is the sum of a set of numbers divided by the total number of elements in the set. Arithmetic mean is used on the GMAT because it is a more precise term than average. If all the numbers in a set are arranged in ascending or descending order, the middle number is the median. The median is different from the arithmetic mean. Half of the people in a country earn more than the median income, and half earn less. The average income does not split the people into a top half and bottom half. For example, if 5 people have weekly incomes of $200, $300, $500, $13000 and $6000, the median is $500, but the average is $4000. If a large number of people earn very little and a few earn a huge amount, the average would be quite impressive, but the median would be surprisingly low. Example 14 Ten students on an exam scored 20, 30, 30, 25, 30, 35, 80, 60, 40, and 90. Calculate the average and the median. Solution The average is the sum of all the numbers divided by 10: average = (20 + 25 + 3 × 30 + 35 + 40 + 60 + 80 + 90) / 10 = 44. The median is, for the case of an even number of entries, the average of the two middle numbers when arranged in order: median = (30+35)/2 = 65/2 = 32.5 More questions are available in our practice tests. Email any questions or suggestions to 800score.com. << BACK to table of contents << table of contents GMAT home page GRE home page 800score.com home Algebraic Expressions For easier use, print out this guide Option #1: Print out Algebraic Expressions section to use offline (14 pages) Option #2 CLICK HERE to view Algebraic Expressions Chapter online. Browse through the sections of this chapter >> This chapter is divided into three parts: I. Simplifying Rules II. Complex Expressions with Exponents III. Manipulating Complex Expressions I. Simplifying Rules A. Exponent Rules B. Simplifying Expressions A. Exponent Rules This section is intended as a basic review of algebra; skim the material and review what is necessary. When attempting to solve algebra problems, it is important that you strictly follow the laws concerning exponents. Rule I: Add the exponents when multiplying two powers of the same base: 33 =3 Rule II: Multiply the exponents when obtaining the power of a power: (3 ) =3 Rule III: Subtract the exponents when dividing a power of a specified base by another power of the same base: 3 /3 = 3 a negative exponent is the equivalent of (1/the number) raised to the power. For example, 3 = 1 / 3 or 1 / 9. Rule IV: The power of a product of factors is written by raising each factor to the specified power. In general, (abc) = a b c Rule V: The power of a fraction is written by raising the numerator and the denominator to the specified power. This is expressed by (a/b) = a /b B. Simplifying Expressions Rule I: Perform multiplications and divisions before you perform additions and subtractions. The expression x + 2y/3 is not the same as (x+2y)/3. Rule II: Combine all like terms in an expression. The expression 2x + x - y + 4y is simplified by combining the like terms resulting in 3x + 3y . Rule III: Perform operations inside parentheses first. The expression (x + 2y)/3 is not the same as x + 2y/3. With the parentheses present, we add first, then divide; with no parentheses, we use Rule I and divide first, then add. Rule IV: Eliminate inner parentheses first and the outermost parentheses last. In the expression x(x + 2(3x + 4) -3), we remove the inner parentheses first obtaining x(x + 6x + 8 - 3); then we combine like terms giving x(7x + 5). We may then remove the last parentheses providing 7x + 5x. Often brackets and braces are used if two or three sets of parentheses are needed. II. Complex Expressions with Exponents A. The Products of Monomials and/or Polynomials B. Quadratic Equations C. Factoring D. Division of Algebraic Expressions A. The Products of Monomials and/or Polynomials To multiply a binomial (a polynomial with two terms like, x + 3) by a binomial use FOIL: First: Mulitply the first terms of each binomial. Outside: Multiply the outside terms (the outer extreme terms of the expression). Inside: Multiply the inside terms (the inner extreme terms of the expression). Last: Multiply the last terms of each binomial. Then add them making sure to combine like terms. Example 1 (x + 3)(x - 5) = Solution F: x × x = x O: -5 × x = -5x I : 3 × x = 3x L: +3 × -5 = -15 Therefore, the answer is the sum of F + O + I + L: x - 5x + 3x - 15 then combine (-5x and +3x) x - 2x - 15 Example 2 (2a + b)(a - 2b) = 2a -4ab + ab - 2b =2a - 3ab - 2b B. Quadratic Equations A quadratic equation may be written in the standard form: ax + bx + c = 0 where a, b, and c are constants. If either b or c is zero, the equation is relatively easy to solve. If neither b nor c is zero, we will consider only those equations in which the quadratic expression on the left side can be factored. Try this sample using the techniques from the section above: x - 2x - 15 Here we do "reverse foil." FIRST: The first number x is easy. It must be the product of two numbers, which in this case are most likely x and x. (x )(x ) LAST: The last number -15 is trickier: The second number in each binomial must multiply to be 15, so we are looking for two factors:(15, 1) and (3,5). One of the two numbers must be negative. The middle number is -2x. The middle number is the combination of the OUTSIDE, INSIDE steps. Note that among the factors, 3 and 5 can sum to -2x, here is how: (x + 3)(x - 5) OUTSIDE = -5x INSIDE = 3x Total= -2x Thus, (x + 3)(x - 5) must be the binomial solution because FOIL will produce x - 2x - 15. Strategy: Quadratic equations usually have two answers. For example X = 25 has two answers, -5 and +5. The test writers often generate trick questions by exploiting the fact that many students forget that these equations have two answers. Data sufficiency questions will often turn on this issue. Example 3 Solve x + 4x = 0 Solution The left side of this equation can be factored so that we can write it as x(x + 4) = 0. This equation is satisfied if either factor is zero. Consequently we write, x=0 x + 4 = 0. The solution is x = 0, x = -4. Example 4 Solve x - 5x + 4 = 0. Solution This quadratic equation has all terms present, so we expect that the left side can be factored. (x - 4)(x - 1) = 0 We set each factor equal to zero and solve for x: x-4=0 x-1=0 The solution is x = 4, 1. Example 5 Solve 2x - 3 = (4 / x) + x. Solution First, put the equation in standard form by multiplying both sides by x. This equation is factorable so we write 2x - 3x = 4 + x x - 3x - 4 = 0 (x - 4)(x + 1) = 0 Set each factor equal to zero and solve for x: x - 4 = 0; x = 4 x + 1 = 0; x = -1 The solution is x = 4, -1. To multiply a monomial by a monomial, multiply the numerical coefficients and then follow the laws of exponents with the same base. For example, (2rs )(-3r s )= 2(-3)(r x r )(s x s ) = -6r s To multiply a polynomial by a monomial, multiply each term of the polynomial by the monomial. This is illustrated by 4x(3x - 2xy + y ) = 12x - 8x y + 4xy To multiply two polynomials, multiply one of them by each term of the other and then combine like terms. The following illustrates the process: (2x - y)(2 + x - 3y) = 2x(2 + x - 3y) - y(2 + x - 3y) = 4x + 2x - 6xy - 2y -xy + 3y = 4x - 2y + 2x - 7xy + 3y C. Factoring Before we proceed with the rules of division, we need to introduce the notion of factoring. If an algebraic expression can be written as the product of two other algebraic expressions, the two other expressions are called factors. Sometimes there may be more than two factors. Factoring an expression will often allow us to simplify a problem, so it is important when solving many algebra problems. There are several algebraic expressions that occur frequently. You should be able to recognize their factors immediately. Five common algebraic factoring types: 1. ax + ay = a(x+y) 2. x - y = (x + y)(x - y) 3. x + 2xy + y = (x + y) 4. x - 2xy + y = (x-y) 5. x + (a + b)x + ab = (x + a)(x + b) Example 6 Factor 3a b - 15a b Solution Note that 3a b is common to both terms. Factor it out and obtain 3a b - 15a b = 3a b(b - 5a ). The two factors are 3a b and (b - 5a ). Example 7 Factor 4x - 9. Solution This is the difference of two perfect squares: (2x) and 3 . Using factoring expression No.2 (see above), we see that 4x - 9 = (2x - 3)(2x + 3) NOTE: No. 2 is the most common expression that you will need to factor, so be sure to recognize it quickly. Example 8 Factor y + 6y + 9. Solution The number 9 is 3 , so we try type No.3 and observe that: y + 6y + 9 = (y + 3) . Note that this can be factored using type No.5. What two numbers add together to give 6 and multiply together to give 9? The answer is 3 and 3 so that y + 6y + 9 = (y + 3)(y + 3). Example 9 Factor r - 8r + 16. Solution The number 16 is 4 so we try type No.4 and observe that r - 8r + 16 = (r - 4) . This also can be factored using type No.5. What two numbers add together to give -8 and multiply together to give 16? The answer is - 4 and - 4 so that r - 8r + 16 = (r - 4)(r - 4). D. Division of Algebraic Expressions There are two common divisions that you may be asked to perform. The first involves dividing an algebraic expression by a monomial, such as (3x + 6ax + 15xa )/3x We first recognize the numerator as a type No.1 expression and factor out 3x. The factor 3x then cancels out with the denominator and we have (3x + 6ax + 15xa )/3x = [3x(1 + 2ax + 5a )]/3x = 1 + 2ax + 5a The second type of division involves divisions such as (5x - 15x + 10) / (5x -10) We write the denominator as 5(x -2) and the numerator as 5(x - 3x + 2). The numerator is a type No.5 expression: What two numbers added together give - 3 and multiplied together give 2? They are -2 and -1. Hence, we can write [5(x - 3x + 2)] / [5(x - 2)] = [(x - 2) (x -1)]/[x - 2] = x - 1. III. Manipulating Complex Expressions A. Multiplying and Dividing Algebraic Fractions B. Addition of Algebraic Fractions C. Equations D. Equivalent Equations E. Linear Equations F. Inequalities G. Simultaneous Equations A. Multiplying and Dividing Algebraic Fractions When multiplying algebraic fractions, we search for common factors: factors that are common to both numerators and denominators. These factors are canceled, and the resulting fraction simplified, if possible. When dividing algebraic fractions, we invert the divisor (the fraction we're dividing by) and then multiply, searching for common factors. Example 1 Perform the division. 3a - 4 divided by 3ab - 4b 2 - 3a 4 - 9a Solution First, invert the divisor and express the division as the multiplication. 3a - 4 x 4 - 9a 2 - 3a 3ab - 4b Next, factor any factorable expressions. 4 - 9a = (2 - 3a)(2 + 3a) 3ab - 4b = b(3a - 4) The given division is now written and the common factors (in red) canceled. 3a - 4 x 2 - 3a (2 - 3a)(2 + 3a) = 2 + 3a b(3a - 4) b B. Addition of Algebraic Fractions When adding (or subtracting) algebraic fractions, we proceed as in arithmetic: o o o Find the lowest common denominator (LCD) of the fractions. Write each fraction using the LCD of the fractions. Add (or subtract) the numerators. The denominator will be the LCD. Simplify the resulting fraction. Example 2 Express as a single fraction: 2x + 1 - x - 3 2a 3a Solution The lowest common denominator is 6a. Each fraction is written with its denominator as 6a. This gives 2x + 1 - x - 3 2a 3a = 3(2x+1) - 2(x-3) 6a 6a = 6x + 3 - 2x - 6 6a 6a = 6x + 3 - 2x + 6 6a = 4x + 9 6a C. Equations An equation is a mathematical statement that two algebraic expressions are equal. The letters in the equation are the unknowns. An equation may be thought of as asking the question, "What numerical value of the unknown satisfies the equation?" (Satisfies means to make both sides equal.) The value of the unknown is called the solution of the equation, or root of the equation. We solve an equation by finding the numerical value of all of its roots. There are equations that do not have real roots. For example , the equation x = -4 does not have a real root; the square of any real number, whether positive or negative (e.g., + 2 or -2) is positive. Hence, there is no real root to x = -4. D. Equivalent Equations Two equations that have the same roots are said to be equivalent. To solve an equation, we often put the equation in a form that is more easily solved with one or both of the following operations: o o Add or subtract the same term to or from each side of an equation. Multiply or divide each side of an equation by the same number or algebraic expression. E. Linear Equations (linear equations do not contain squares or square roots) To solve a linear equation that contains fractions, first remove all fractions by multiplying both sides by the lowest common denominator of all the fractions. Example 3 x/3 + x/2 = 5 Solution Multiply both sides by the LCD which is 6: (3 x 2) 2x + 3x = 30 5x = 30 x=6 Then move all terms containing the unknown to one side and all terms that do not contain an unknown to the other side. Factor out the unknown from all terms that contain the unknown; finally, divide each side by the coefficient of the unknown. Example 4 Solve a(a + t) = b - bt for t Solution To solve for t, first remove the parentheses: a + at = b - bt Move "- bt" to the left side and "a " to the right side: bt + at = b - a We can factor t out of the two terms on the left side; we can also factor the right side to obtain t(b + a) = (b + a)(b - a). Divide both sides by (b + a) and find the solution to be t = b - a. F. Inequalities An inequality is simply a comparison of two quantities or expressions. The following symbols with their meanings are used: > is greater than < is less than >= is greater than or equal to <= is less than or equal to If we place all numbers on a number line with negative numbers to the left of zero and positive numbers to the right of zero, then if A is to the right of B we state that A > B; if A is to the left of B then A < B. Consequently, we conclude that -5 > -7 and -3 < 2. -7 -5 -3 0 2 We also form inequalities using algebraic symbols. The inequality 3x + 2 > x - 6 is solved just like an algebraic equation is solved. We subtract 2 from each side of the inequality so that 3x > x - 8. Then subtract x from each side so that 2x > - 8. Divide by 2 and obtain x > - 4. This is the solution. Any number greater than - 4 satisfies the inequality. There are several rules that we must follow when manipulating inequalities: o o o o The same number or algebraic expression can be added or subtracted from each side of an inequality. The same positive number (or positive algebraic expression) can multiply or divide each side of an inequality. Both sides of the same type of inequality can be added and the inequality remains. (If x < y and w < z, then x + w < y + z.) If a negative number (or negative algebraic expression) multiplies or divides each side of an inequality, the inequality sign must be reversed. (Be sure to remember this; it often leads to errors!) Example 5 Solve the inequality 2x - 2 > x - 5. Solution The inequality is treated in much the same manner as an algebraic equation. Add 2 to each side: 2x > x - 3. Subtract x from each side and the solution is x > -3 NOTE: Retain the same inequality symbol throughout the solution unless an operation reverses the symbol. Example 6 Solve the inequality 3r + 5 > 6r - 7. Solution First, subtract 5 from each side: 3r > 6r - 12 Next, subtract 6r from each side: -3r > -12. Now divide each side by (-3) and reverse the inequality. The solution is r < 4. Rather than working with the negative signs, we could have added 7 to each side of the original inequality to obtain 3r + 12 > 6r. Then subtract 3r from each side and write 12 > 3r so that the solution is 4 > r. This is equivalent to the above solution except that the symbol r is on the right side rather than the more conventional left side. G. Simultaneous Equations Two unknowns in two equations are solved by either of two methods. 1. Substitution 2. Addition or Subtraction. 1. The substitution method: a. Solve the first equation for one unknown in terms of the other unknown and substitute the result into the second equation. b. Solve the resulting equation for the unknown. c. Substitute the value of this known unknown into either original equation and solve for the second unknown. Example 7 Solve for x and y. x-y=2 2x + y = -5 using the substitution method. Solution Using the substitution method, the first equation gives x=2+y Substitute this into the second equation and solve for y: 2(2 + y) + y = -5 4 + 2y + y = -5 3y = -9 y = -3 Substitute this value back into the first equation and solve for x: x+3=2 x = -1 The solution is x = -1, y = -3. 2. The addition or subtraction method: a. Multiply one equation by a properly chosen number so that one of the unknowns has the same coefficient in both equations. b. Add or subtract the equations so that one of the unknowns is eliminated. c. Solve the resulting equation for the remaining unknown. d. Substitute the value of this known unknown into either original equation and solve for the second unknown. Example 8 Solve for x and y x-y=2 2x + y = -5 This time use the addition or subtraction method. Solution We simply add the two equations since this will eliminate y: x-y=2 + 2x + y = - 5 3x = -3 x = -1 Substitute this value back into the first equation and solve for y: (-1) - y = 2 -y = 3 y = -3 Have any more questions or suggestions, email 800score.com. <<go back to table of contents << table of contents GMAT home page GRE home page 800score.com home Word Problems For easier use, print out this guide Option #1: Print out Word Problems section to use offline (22 pages) Option #2 View the Word Problems Chapter online. Browse through the sections of this chapter >> Word Problems use simple math concepts and apply them in a contorted and complicated manner. The usual strategy to solve a Word Problem is to express the question as a mathematical equation letting x, or some other letter, represent the quantity that we wish to determine. The five step process for Word Problems: 1. Quickly read the question and the answer choices to get a feel for what the question is specifically asking. 2. Read the question again (on the GMAT you have nearly two minutes per math question, so there is time to spare as long as you budget it properly--read Chapter One for pacing information). 3. Translate the equation to paper and translate the question into an expression with variables. 4. If you get a mental block or see a shortcut, use Backsolving (take numbers and feed them into the question--either answer choices or numbers you choose). 5. If that does not work, start eliminating answers that are outside of the ballpark; guess and move on. I. Review of Word Problem Concepts A. Percentages B. Interest, Discount, and Markups C. Progressions D. Uniform Motion E. Work F. Ratio and Proportion G. Grouping and Counting H. Tables, Charts, and Graphs (Data Interpretation) I. Symbols A. Percentages The word percent is abbreviated by the symbol % and is a fraction whose denominator is 100. 26% is equivalent to the fraction 26/100. To change a decimal number to a percent, we simply multiply by 100; the number 0.321 is equivalent to 32.1%. If a percentage is given, move the decimal two places to the left to express its equivalent decimal form. Example 1 Convert 4% into a decimal and a fraction in lowest terms. Solution To convert 4% into a decimal, we move the decimal point two places to the left: 4% = 0.04 To express 4% as a fraction, we divide by 100: 4/100 = 1/25 Hence, 4% = 0.04 = 1/25 Example 2 If the price of a stock falls from $50 to $40, what is the percentage of decrease? Solution First, subtract the numbers resulting in the decrease: 50 - 40 = 10. Then divide by the original amount: (50 - 40) / 50 = 10 / 50 = .2 Convert to a percentage by moving the decimal point two places to the right: % decrease = 20% Example 3 An employee is to mark up a piece of jewelry 120%. If it cost $100, what should its selling price be? Solution The amount of the markup is 1.2 × 100= $120 The selling price is then $100 + $120 = $220 Example 4 A college bookstore purchases trade books on a 40% margin, i.e., it purchases a trade book for 40% less than its retail price. What is the percentage markup based on its wholesale price? Solution Since the retail price is not given, the percentage markup that we seek must be the same for all trade books. Therefore, let the retail price of a trade book be $100 (rather than the symbol x). Then the bookstore's purchase price is 100 - 100 × 0.4 = 100 - 40 = $60 If a book sells for $100 and costs $60, its percentage markup is %markup = (100 - 60) / 60 × 100 = 40 / 60 x 100 = 66% Example 5 Kathy buys a bike for $240 after a 40% markdown. What was the original price? Solution Let P be the original price. Then P - P × 0.4 = 240 0.6P = 240 divide both sides by .6 therefore, P = $400 Example 6 Find the number of residents of a city if 20% of them, or 6,200 people, ride bicycles. Solution Let R be the number of residents. The equation that represents the verbal statement is 0.2R = 6,200. R = 6200/.2 = 62000/2 = 31,000 people. Example 7 Kent pays 20% taxes on income between $10,000 and $20,000 and 30% on income over $20,000. The first $10,000 is tax free. If he pays $14,000 in taxes, what was his income? Solution Let Kent's income be L. Then the total tax is 0.2(20,000 - 10,000) + 0.3(L - 20,000) =14,000 2,000 + 0.3L - 6,000 = 14,000 0.3L = 14,000 + 4,000 = 18,000 L = 18,000/.3 = $60,000 Example 8 How many gallons of pure water must be added to 100 gallons of a 4% saline solution to provide a 1% saline solution? Solution Let x be the gallons of pure water to be added. There are 100 × 0.04 = 4 gallons of salt and 96 gallons of pure water in a 4% saline solution. The total number of gallons will be x + 100. The amount of salt will remain constant. Hence, 0.01(x + 100) = 4 0.01x + 1 = 4 0.01x = 3 x = 3/.01 = 300 gallons B. Interest, Discount, and Markups The interest, I ,earned on the amount, P, of money invested depends on the interest rate, i, and the time, T, the money is invested. This is represented by the equation I = PiT The interest would be the dollars earned (or paid), the interest rate is always the annual interest rate (unless otherwise stated), and the time is measured in years. Simple interest means that the interest, I, is determined using the total time period, e.g. 10 years, rather than compounding the interest, that is, adding the interest, I, to the amount, P, after each year. Discount is the percent reduced on the price of an item. Markup is the amount of increase when the cost of an item is increased a certain percent. The following examples will illustrate this concept. For markups and discounts, calculate: New Price - Original Original If the value is negative, that is the amount of the discount. If the number is positive, that is the amount of the markup Example 9 A student invests $1,000 at 10% for the summer (3 months). How much interest does the student earn? Solution The interest is calculated to be I = PiT = 1000(0.10)(3/12)= $25 We have expressed the 10% interest rate as 0.10 and the 3 months as 3/12 of a year since the interest rate is assumed to be an annual rate. Example 10 A professor retires with a retirement fund of $400,000. If she is paid monthly interest of $3,600, what is the interest rate? Solution The interest rate is assumed to be an annual rate. The annual interest income is $3,600 (12) 50 so that I = PiT 3600(12) = 400,000(i) I = 3600(12) / 400,000= (3.6(3)(4))/4(100) = 10.8/100 = 0.108 or 10.8% Example 11 A pair of aerobic shoes is marked $120 and is discounted to $90. What is the percent discount? Solution The percent discount is based on the initial cost. It is % discount = ((120-90)/120) × 100 30/120 × 100 = 25% Example 12 A pair of running shoes is purchased at wholesale for $90 and is sold to a store for $120. What is the percent markup? Solution The percent markup is based on the original cost. It is % markup ((120 - 90) / 90) × 100 = 30 / 90 × 100 = 33 1/3% Example 13 An investment of $1,000 is placed into a particular account at the beginning of each year at a simple interest of 8%. How much money is in the account after 5 years (no compounded interest)? Solution First, note that 1,000 is placed in EACH year, that is 5,000 is invested. The first $1,000 will earn interest for 5 years for a total of $80 × 5 = $400. Its value will be $1400. The second $1000 will earn interest for 4 years for a total of $320. Its value will be $1,320. The third $1,000 will be worth $1,240. The progression is $1,400, $1,320, $1,240, $1,160, $1,080. The money in account is the total of those five numbers: $1,400 + $1,320 + $1,240 + $1,160 + $1,080 = $6,200 C. Progressions J. Progressions Sequences and Progressions An ordered list of numbers is called a sequence. For example, a sequence of positive even numbers would be: 2, 4, 6, 8, 10, … Within a sequence, each individual member is called a term. The terms are defined by their position in the sequence. For example, in the above sequence, 2 is the first term, 4 is the second term, and 6 is the third term. The ellipsis symbol (…) indicates that the sequence continues beyond the terms that are written. For the above sequence, the next term would be 12, then 14, and so on. Even though these two terms (and those beyond) are not written out, we know that they are terms within the sequence because the ellipsis indicates that the sequence continues forever (to infinity). A common kind of sequence problem that you may encounter on the GMAT is one in which you are given a few terms of a sequence and asked to define the next term. In the above example, given that the sequence was defined as consisting of positive even numbers, it was easy to deduce that the next two terms in the sequence are 12 and 14. Often however, sequences will be more complicated. Given more complex sequences, how can you determine the next term in a sequence of numbers? The key to solving these problems is to determine the relationship between the terms in the sequence that you are given. This relationship can be described in terms of a progression, a function or manipulation that can be applied to each individual term of a sequence that will generate the next term in that sequence. The types of progressions typically found on the GMAT can often be described as being either arithmetic or multiplicative. Let's now look at examples of each of these. Arithmetic progressions Simply stated, in an arithmetic progression, a fixed amount is added to each term in order to generate the next term. An important consequence of this is that the difference between any two consecutive terms will remain constant for the entire sequence. Here is an example of an arithmetic sequence: 0, 3, 6, 9, 12, 15,… What is the constant value, or fixed amount, being added to each term to generate the next? We can figure this out by subtracting any term in the sequence from the next term. If we subtract 6 from 9, the difference is 3. We see that this difference is the same if we subtract 9 from 12, 12 from 15, 0 from 3, or 3 from 6. Once you determine the difference between the terms, it is easy to generate the next terms in the sequence: 0, 3, 6, 9, 12, 15, 18, 21, … It may also be helpful for some problems to translate the sequence into a general form. In this case, we could represent this as n, n+3, (n+3) +3, … Keep in mind, however, that this is useful only in thinking about the general trend, that is, what is the nature of the relationship between the terms. It does not, however, substitute for the actual sequence, since it gives us no information as to the starting point. The general equation representation of n, n+3, (n+3) +3, … could represent 0, 3, 6, 9,… or it could represent 1, 4, 7, 10,… or it could be 0.5, 3.5, 6.5, 9.5, …, etc. Keeping that caveat in mind, when solving problems involving more complex sequences, it is often useful to jot down the general formula describing that sequence. Let's look at another example. What are the next two terms of the following sequence? 2, 4.5, 7, 9.5, 12, 14.5, … It's usually easiest to start with the first terms in the series, as those are typically the smaller numbers, and thus easier to manipulate quickly. If we subtract 2 from 4.5, we get 2.5. If we subtract 4.5 from 7, the difference is also 2.5. Going through all the given pairs of terms in the sequence, we can confirm that this relationship holds true, that each term can be generated through the addition of 2.5 to the preceding term. (Note: if this relationship were not to hold true for each pair of terms, we would decide that this is not an arithmetic progression, and we would then test whether it were instead a multiplicative progression, described below.) Once we determine the constant value of 2.5, it is straightforward to generate the next two terms in the sequence, 17 and 19.5. In this case, the general form of the sequence would be n, n + 2.5,… Let's try another sequence. What are the next two terms in the following sequence? 19, 12, 5, -2, -9, … Again, to determine the next two terms, we first must determine the constant value that is the difference between pairs of terms. If we subtract 19 from 12, we have a difference of -7. If we subtract 12 from 5, again we have -7, and likewise for the remaining terms of the sequence. Once we know this value, we can quickly ascertain the next two terms, -16 and -23. How could we write this in a general form? n, n + (-7),… This example should have also demonstrated to you another important feature of arithmetic progressions: the constant values may be either positive or negative. Keep this in mind when solving these problems. Multiplicative Progressions In a multiplicative progression, the ratio of consecutive terms is constant. Rather than adding a constant value to a term in order to generate the next term as we do for arithmetic progressions, for multiplicative progressions, we multiply each term by a constant value to generate the next term. For example, let's consider the following sequence: 1, -3, 9, -27, 81, … First, let's determine the constant value separating sets of consecutive terms. Depending on your personal preference, you can think about it in terms of division or multiplication. What number must we multiply 1 by in order to get -3 as a product? Alternatively, -3 divided by what number results in 1? The answer (in both cases) is -3. Next, confirm that the constant value can be multiplied to each of the terms in the sequence to generate the next (that is, -3 x -3 = 9; 9 x -3 = 27; and -27 x -3 = 81). In general form, this would be n, n x (-3),… Let's look at another example of a geometric progression: 64, 32, 16, 8, 4, … What is the next term in this sequence? First, determine the constant value. 64 multiplied by what number results in a product of 32? Or, what number must 32 be divided by to result in 64? The answer is 0.5, or ½. Again, confirm that each term can be multiplied by this constant value to generate the following term. Once you have established that you are correct in identifying this as a geometric progression, you can generate the next term by multiplying the constant value of 0.5 or ½ by 4, resulting in 2. In general form: n, n x ½,… More complex sequences Now, not all sequences will be either arithmetic or multiplicative progressions. Some sequences will be defined by more complex functions than either addition or multiplication alone. For example, imagine a sequence of numbers in which each term (except the first two) is generated by the addition of the two preceding terms: 1, 1, 2, 3, 5, 8, 13, … This is actually a well-studied sequence of numbers called the Fibonacci series. After the first two numbers, we can represent each term of the series as n = (n-1) + (n-2), or n is the sum of the two preceding numbers. (By using subscripts, we are indicating the previous terms, where n-1 is the term immediately preceding n and n-2 is the term immediately preceding n-1.) Alternatively, you may see a sequence in which each subsequent term is derived from both arithmetic and multiplicative processes. For example, the following sequence begins with 2, and all subsequent terms are generated by adding 1 to the preceding term, and then multiplying that sum by 2: 2, 6, 14, 30, 62, … The general representation of this, not including the first term, would be n = ((n-1) + 1) x 2, or n equals the sum of the preceding term and 1, multiplied by 2. For sequences of this type, it is much more difficult to determine the relationship between the terms at first glance. However, it is rare that you would actually be asked to do this on the GMAT . Instead, you may see a problem in which they define the relationship for you (as in giving you the rule to add 1 to the preceding term, and then multiply the sum by 2), and then ask you to generate the next term of the sequence. Sample problems Let's now look at some sample sequence problems that you might find on the actual exam. Example 33 Except for the first two numbers, every number in the sequence 1, -2, -2, 4, … is the product of the two immediately preceding numbers. What is the seventh term of this sequence? (A) -8 (B) 32 (C) -32 (D) 256 (E) -256 Answer explanation: (D) We are given the rule to use in order to generate the next terms of the sequence: multiply the two immediately preceding numbers to generate the next. We already have the first four terms, and need to generate three more. -2 x 4 = -8, this is the fifth term in the sequence. 4 x -8 = -32, this is the sixth term. -8 x -32 = 256, this is the seventh term, and the correct answer, choice D. Example 34 The fifth term in a sequence of numbers is 19 and each number after the first number in the sequence is 3 less than the number immediately preceding it. What is the second number in the sequence? (A) 31 (B) 30 (C) 28 (D) 13 (E) 10 Answer explanation: (C) This is an example of an arithmetic progression in which we are given only one term and asked to determine another. We are told that each term is 3 less than the previous term. A good technique to use in solving this is to draw out five blanks to represent the terms of the sequence, and fill in the fifth one, the one term we know, with 19: We are told that each term is 3 less than the term immediately preceding it. Does that mean that the fourth term, the one immediately preceding 19, will be 3 more or 3 less? (Be sure to read carefully!) The fourth term will be three more than 19, or 19 + 3, or 22. In this way, we can work backwards to generate each of the first four terms, resulting in a sequence that looks like this: Now we can see that the second term is 28, choice C. Example 35 What is the next term of the sequence -3, 6, -12, 24,…? (A) -48 (B) 48 (C) -64 (D) 64 (E) -144 Answer explanation: (A) This is a multiplicative progression generated by multiplying each term by the constant value of -2 in order to generate the next term. (-3 x -2 = 6; 6 x -2 = -12, etc.) To generate the next term, multiply 24 by -2 to get -48, choice A. Example 36 In a sequence of integers, A, B, C, D, E…, the value of each integer except the first is equal to two more than the product of the previous integer and 2. If E equals 14, what is the value of B? (A) -14 (B) -8 (C) 0 (D) 4 (E) 8 Answer explanation: (C) This is one of the more complex sequences in which each subsequent term is derived from both arithmetic and multiplicative processes. Like that previous problems, it may be useful to create a little diagram to hold the places for the terms as you figure them out. We are told that the fifth term, E, equals 14, so we can fill in that information as follows: We are told that the value of each integer is equal to two more than the product of the previous integer and 2. A general representation of this would be n = ((n-1) x 2) + 2. It is extremely important to be very precise in your interpretation of the description of the relationship between the terms of the sequence. In this case, a number is multiplied by 2, then 2 is added to that product in order to generate the next term. Now, if we use this general formula and substitute our known term, 14, for n, we can derive the preceding term as follows: n = ((n-1) x 2) + 2 14 = ((n-1) x 2) + 2 Subtracting 2 from both sides, we get 12 = ((n-1) x 2) Dividing both sides of the equation by 2, we get 6 = n-1 Therefore, position D is 6. Knowing the value of D, we can now apply the formula again to solve for C. Once we have C, we can solve for B. Finally, we end up with the sequence as follows: Now we can see that B = 0, choice C. Summary for sequence problems: If you are asked to solve for a term in the sequence, determine what the relationship is between the terms of the sequence. Is the sequence an arithmetic progression, multiplicative progression, or a more complex sequence defined within the problem? If you think it's an arithmetic progression, determine the common value and make sure that each term of the sequence can be generated through the addition of that common value to the preceding term. Likewise, if you think it's a multiplicative progression, determine the common value and confirm that each term can be multiplied by this constant value to generate the following term. If it helps, write out a more general representation of the sequence or the formula used to determine the next term of a sequence using n to represent any given term. When you are asked to solve for the value of a specific term, it may be useful to make a little diagram to ensure that you are solving for the correct position. D. Uniform Motion When an object is moving at a constant speed (or velocity), the object is traveling with uniform motion. The distance, D, that the object will travel in time, T, depends on the velocity, V. It is expressed mathematically as D = VT If we desire the distance in miles, we usually express the velocity in miles per hour (mph) and time in hours. If the distance is in kilometers, then the velocity would be in kilometers per hour (kph). Example 15 A biker travels 60 miles in 2.5 hours. Determine the biker's average speed. Solution The equation relating distance, velocity and time provides 60 = V(5/2) Divide both sides by 5/2 to solve for V. V = (60)2/5 = 24 miles/hour Example 16 A car travels between two cities 400 miles apart in 7 hours. The return trip takes 9 hours. Find the average speed of the car. Solution The total distance is 2(400) = 800 miles. The total time is 7 + 9 = 16 hours. The average speed is found from D = VT: 800 = V(16) V = 800/16 = 50 miles/hour Example 17 A police officer, traveling at 100 miles per hour, pursues Philip who has a 30 minute head start. The police officer overtakes Philip in two hours. Find Philip's speed. Solution Let x miles per hour be Philip's speed. The distance traveled by the officer equals the distance traveled by Philip: 2 × 100 = (2 + 30/60)x 200 = (2 + 0.5)x 200 = 2.5x therefore x = 80 mph E. Work The amount of work, W, accomplished in time, T, depends on the rate, R, at which the work is being accomplished. Work problems are quite similar to the problems of uniform motion. The equation we use is W = RT Try to solve by determining the rate per time period (usually per hour or per minute). The rate, R, is most often expressed as the job to do divided by the time, where W = 1 job. For example, a tractor plows 1/10 of a field each hour; the job is one field, so the rate is 1/10 of a field per hour. If it takes x tractors to do one job in 1 hour, then each tractor works at a rate of 1/x of the job per hour. If it takes x tractors 4 hours to do one job, then each tractor works at one quarter of the previous rate, or at the rate of 1/4x of the job per hour. In general, if it takes x tractors y hours to do one job, the rate that each tractor works is 1/(x × y) of the job per hour. Example 18 It takes 3 men 8 hours to paint a house. How long will it take 5 men to paint the same house? Solution The per hour rate at which each man works is R = 1/(3 x 8) = 1/24 houses per hour The rate for 5 men is (5R). The work is 1 house. Our equation gives us 1 = 5/24T Therefore, T = 24/5 = 4.8 hours or 4 hours and 48 minutes. NOTE: 0.8 hours is 0.8 × 60 = 48 minutes. Example 19 Michelle can input a day's invoices into the computer system in 40 minutes, and John can input the same invoices in 60 minutes. How long will it take both of them, working simultaneously, to input the invoices? Solution Michelle's rate for doing the job is 1/40 of the job per minute. John's rate is 1/60 of the job per minute. Let the time they work be T. Then the sum of the work that Michelle does and the work that John does must equal one job: 1=(1/40)T + (1/60)T This is most easily solved by multiplying by 40(60): 40(60) = [40(60)]/40 × T + [40(60)]/60 × T 2400 = 60T + 40T T = 24 minutes Example 20 Kelly and Shelley can type the manuscript in 8 hours. Kelly can type the manuscript alone in 20 hours. How long would it take Shelley to type the manuscript? Solution The rate that Kelly works is 1/20 of the job per hour. Let the rate that Shelley works be R. To do one job in 8 hours we have 1 = 1/20(8) + R(8) To solve for R, multiply by 20: 20 = 8 + 20R(8) 12 = 8(20)R Therefore, R = 12/[8(20)] = 3/40 of the job per hour. To type the entire manuscript alone, Shelley takes T = W/R = 1/(3/40) = 40/3 = 13 1/3 or 13 hours and 20 minutes. F. Ratio and Proportion A ratio is a fraction that compares two numbers. The ratio of x to y is written as x : y or x/y. Ratios are usually used to compare quantities of the same type, for example, the ratio of the length of a Toyota to the length of a Cadillac. We would not form the ratio of the length of a Toyota to the cost of a Cadillac. A proportion states that two ratios are equal. Two ratios involve four numbers: two numerators and two denominators. Most often, one of these four numbers is not known; it is found by equaling the two ratios, such as 2 = 15 6 x The unknown x is then found by cross multiplying: 2x = 15(6) therefore, x = 45 Two quantities are directly proportional if one is a constant times the other: x = cy (where c is a constant). They are inversely proportional (or indirectly proportional) if one is a constant divided by the other: x = c/y, or equivalently, xy = c. If a quantity is stated to be proportional to another, the word directly is implied, so if x is stated to be proportional to y, it means x = cy. To decide if two quantities are directly or inversely proportional, we ask the question, "Do the quantities both increase or decrease or does one increase while the other decreases?" If they both increase or decrease, they are directly proportional; if one increases while the other decreases, they are inversely proportional. To solve an equation that represents a direct proportion, such as x = cy, we set up the equation as x1 = y1 x2 y2 where the subscript 1 refers to the first situation and the subscript 2 to the second situation. If the equation results from an inverse proportion, such as xy = c, we have x1 = y2 x2 y1 To solve problems involving proportions, 3 of the 4 numbers will usually be known and the problem will be to calculate the fourth. Example 21 Calculate x if 4:15 = 16:x Solution The equation is written in a more obvious form as 4 = 16 15 x 4x = 16(15) therefore, x = 60 Example 22 The ratio of two numbers is 4:1, and their sum is 40. Find the two numbers. Solution This is expressed mathematically as x/y = 4 x + y = 40. The first equation can be written in the form x = 4y. This is substituted into the second equation to yield 4y + y = 40. 5y = 40, therefore, y = 8. Since x = 4y, we find that x = 4(8) = 32. The two numbers are 8 and 32. Example 23 If an airplane travels 1,200 miles in 2.5 hours, how far will it travel in 10 hours? Solution This represents a direct proportion: both the distance traveled and time increase. Consequently, if we let x = distance the airplane will travel, we have 1200 = 2.5 x 10 12,000 = 2.5x x = 4,800 miles Example 24 What is the ratio of 2/3 to 5/4? Solution The ratio is (2/3)/(5/4) which is equal to 2/3 × 4/5 =8/15. G. Grouping and Counting Overlapping Groups When a question relates to overlapping groups, try diagramming the problem with overlapping circles. This will make it easy to account for the overlap. Example 25 If, in a certain school, 20 students play soccer, 10 play basketball, and 7 play both, how many students play basketball, soccer or both? (A) 20 (B) 22 (C) 23 (D) 25 (E) 29 Solution Using the diagram above we have deduced some new facts: 3 play only basketball 13 only play soccer 7 play both total of 23 players Possible Range Questions When questions ask for a possible range, be sure to examine the lowest and highest possibilities. Example 26 A cabinet contains 3 to 5 bottles, each of which contains 30 to 40 mushrooms. If 10 percent of the mushrooms are flawed, what is the greatest possible number of flawed mushrooms in the cabinet? (A) 51 (B) 40 (C) 30 (D) 20 (E) 12 Solution There are, at most, 5 bottles, each of which contains at most 40 mushrooms; so, there are, at most, 5 × 40 = 200 mushrooms in the drum. Since 10 percent of the mushrooms are flawed, there are at most 20 (20 = 10% × 200) flawed mushrooms. The answer is (D). Count Inclusively When doing counting problems, always be sure to count the first and last of the range of items. Example 27 Fence posts are being placed at 20 foot intervals along a road 1000 feet long. If the first fence post is placed at one end of the road, how many fence posts are needed? (A) 49 (B) 50 (C) 51 (D) 52 (E) 53 Solution The average test taker would simply take 1000 and divide it by 20 to get 50. However, to get the right answer, you must include the first and last choice. Since the road is 1000 feet long and the fence posts are 20 feet apart, there are 1000/20 = 50, or 50 full sections in the road. If we ignore the first fence post and associate the fence post at the end of each section with that section, then there are 50 fence posts (one for each of the fifty full sections). Counting the first fence post gives a total of 51 fence posts. The answer is (C). H. Tables, Charts, and Graphs (Data Interpretation) Graphs and charts show the relationship of numbers and quantities in visual form. By looking at a graph, you can see at a glance the relationship between two or more sets of information. If such information were presented in written form, it would be hard to read and understand. Here are some things to remember when doing problems based on data interpretation: 1. Take your time and read carefully. Understand what you are being asked to do before you begin figuring. 2. Check the dates and types of information required. Be sure that you are looking in the proper columns, and on the proper lines, for the information you need. 3. Check the units required. Be sure that your answer is in thousands, millions, or whatever the question calls for. 4. In computing averages, be sure that you add the figures you need and no others, and that you divide by the correct number of years or other units. 5. Be careful in computing problems asking for percentages. 6. (a) Remember that to convert a decimal into a percent you must multiply it by 100. For example, 0.04 is 4%. (b) Be sure that you can distinguish between such quantities as 1% (1 percent) and .01% (one one-hundredth of 1 percent), whether in numerals or in words. (c) Remember that if quantity X is greater than quantity Y, and the question asks what percent quantity X is of quantity Y, the answer must be greater than 100 percent Example Set #28: Table Chart Examples 1-5 are based on this Table Chart. The following chart is a record of the performance of a baseball team for the first seven weeks of the season. Games Won Games Lost Total No.of Games Played First Week 5 3 8 Second Week 4 4 16 Third Week 5 2 23 Fourth Week 6 3 32 Fifth Week 4 2 38 Sixth Week 3 3 44 Seventh Week 2 4 50 1. How many games did the team win during the first seven weeks? (A) 32 (B) 29 (C) 25 (D) 21 (E) 50 Choice B is correct. To find the total number of games won, add the number of games won for all the weeks, 5 + 4 + 5 + 6 + 4 + 3 + 2 = 29. 2. What percent of the games did the team win? (A) 75% (B) 60% (C) 58% (D) 29% (E) 80% Choice C is correct. The team won 29 out of 50 games or 58%. 3. According to the chart, which week was the worst for the team? (A) second week (B) fourth week (C) fifth week (D) sixth week (E) seventh week Choice E is correct. The seventh week was the only week that the team lost more games than it won. 4. Which week was the best week for the team? (A) first week (B) third week (C) fourth week (D) fifth week (E) sixth week Choice B is correct. During the third week, the team won 5 games and lost 2, or it won about 70% of the games that week. Compared with the winning percentages for other weeks, the third week's was the highest. 5. If there are fifty more games to play in the season, how many more games must the team win to end up winning 70% of the games? (A) 39 (B) 35 (C) 41 (D) 34 (E) 32 Choice C is correct. To win 70% of all the games, the team must win 70 out of 100. Since it won 29 games out of the first 50 games, it must win (70 - 29) or 41 games out of the next 50 games. Example Set #29: Interpreting Graphs Answer the following questions based on the graph above. 1. During what two-year period did the company's earnings increase the most? (A) 95-97 (B) 96-97 (C) 96-98 (D) 97-99 (E) 98-00 Reading from the graph, the company's earnings increased from $5 million in 1996 to $10 million in 1997, and then to $12 million in 1998. The two-year increase from '96 to '98 was $7 million-clearly the largest on the graph. The answer is (C). 2. During the years 1996 through 1998, what were the average earnings per year? (A) 6 million (B) 7.5 million (C) 9 million (D) 10 million (E) 27 million The graph yields the following information: Year Earnings 1996 $5 million 1997 $10 million 1998 $12 million To figure out the average, add (5 + 10 + 12)/3 = 9. The answer is (C). 3. In which year did earnings increase by the greatest percentage over the previous year? (A) 96 (B) 97 (C) 98 (D) 99 (E) 2000 To find the percentage increase (or decrease), divide the numerical change by the original amount. Year Earnings % increase from year before 1995 8 n/a 1996 5 decrease 1997 10 100% 1998 12 20% 1999 11 decrease 2000 8 decrease The largest number in the right-hand column, 100%, corresponds to the year 1997. The answer is (B). 4. If the company's earnings are less than 10 percent of sales during a year, then the Chief Operating Officer will get a 50% pay cut. How many times between 1995 and 2000 did the Chief Operating Officer take a pay cut? (A) None (B) One (C) Two (D) Three (E) Four Calculating 10 percent of the sales for each year yields Year, 10% of Sales (millions), Earnings (millions). Year 10% of sales Earnings is 10% of sales greater than earnings? 1995 .10 x 80 = 8 8 no cut 1996 .10 x 70 = 7 5 cut 1997 .10 x 50 = 5 10 no cut 1998 .10 x 80 = 8 12 no cut 1999 .10 x 90 = 9 11 no cut 2000 .10 x 100 = 10 8 cut Comparing the right columns shows that earnings were less than 10 percent of sales in 1996 and 2000. The answer is (C). I. Symbols On some questions the test will create new functions. You can identify these questions by the symbols that are used--triangles, squares, ampersand, etc.). These questions are generally easy as long as you don't get confused by the symbols. Simply take the function and plug in the numbers. Look at this: If a # b = a + b, then what is 2 # 3? Explanation 2 # 3 would equal 2 + 3, or 5. This is a function. In english "a # b = a + b" means (pretend you are a computer) take the first number (a) and add it to the second number (b) Example 31 If a # b = a + b, then what is (2 # 3) # 2? Solution Solve inside of the parentheses first. 2 # 3 would equal 2 + 3, or 5. Then (5) # 2 = 5 + 2 or 7. Example 32 (harder) If for numbers x, y, z the function # is defined as x # y = xy - x then x # (y # z) = Solution The first step to solving x # (y # z) is to solve inside the parenthesis (y # z), then after we have solved what is in the parenthesis the second step is to do x # (what is in the parenthesis), then the third step is to solve the equation using the symbol. Step 1 (solve the parenthesis-- y # z) 1a) if x # y = xy - x (as stated in the question stem) 1b) then y # z = yz - y ( you get this by substituting y for x and z and y) Step 2 (insert the parenthesis value) the original question asks x # (y # z), we have already solved y # z, which according to 1b above y # z = yz - y So, in the original equation x # (y # z), substitute yz - y for y # z. Now, x # (y # z) = x # (yz - y) (NOTE: WHEN SOLVING QUESTIONS WITH SEVERAL NUMBERS AND OPERATIONS, ALWAYS MULTIPLY AND DIVIDE BEFORE YOUR ADD AND SUBTRACT, FOR EXAMPLE 7 + 3 x 2 = 13, NOT 20. DO NOT SIMPLY GO FROM LEFT TO RIGHT, MULTIPLY AND DIVIDE BEFORE YOU ADD AND SUBTRACT.) So, we are now dealing with x # (yz - y) Step 3 (apply the # to the final equation) The # symbol means x # y = xy -x. (Essentially, take the first number--here x--, multiply it by the second number--here y--and then subtract the first number. Let's apply that to the equation at the end of step 2 x # (yz - y) = x(yz-y) - x then factor it out the x's. = xyz - xy - x = x(yz - y - 1) Help, I still don't get it! To play with this question, try inserting numbers such as 1 and 3 for x and y to see how x # y = xy - x or 1 # 3 = 1(3)- 1 1#3=2 So now look at x # (y # z) = 1. Set x =1 , y =3, z = 2 and plug them in 1 # (3 # 2) 2. Do the parenthesis first (as is the rule always). To make a problem less intimidating, break it into smaller component parts: (3 # 2) = (3)(2) - (3) 3#2=6-3 3#2=3 3. Go back to the original question and plug in the value you solved for the parenthesis. 1 # (3 # 2), plug in the value for (3 # 2) to get 1 # (3) = (1)(3) - (1) So, 1 # (3) = 2 4. Now that you have applied the function, you have the result x # (y # z) = 5. Set x =1 , y =3, z = 2 and plug them in and the result is 2. Now, for fun, look at the answer we came up with before x # (y # z) = x(yz - y - 1) See, now plug in x = 1, y=3, z=2 for x(yz - y - 1), what do you get 1 [(3)(2) - 3 - 1] = 2, so we can assume we got the right answer. When doing complex functions or algebra, plug in numbers to better understand the equation and to check if you got the right answer. This takes time, though, so use this judiciously. Have any more questions or suggestions, email 800score.com << go back to table of contents << table of contents GMAT home page GRE home page 800score.com home Probability (optional) Option #1: Print out Probability section to use offline (9 pages) Option #2 View the Probability Chapter online. Browse through the sections of this chapter >> Probability questions are becoming increasingly common. Probability questions tend to be bundled among the difficult questions, so high scorers will commonly encounter them. If you are a low scorer and are pressed for time, consider skipping most of the material past "Simple Probability." This is a computer-adaptive test, and low scorers aren't likely to encounter the most difficult probability question types. A. Simple Probability B. Probability of Multiple Events C. Independent and Dependent Events D. Mutually Exclusive Events E. Conditional Probabilities F. Combinations A. Simple Probability In general, the probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. Probability= (# of favorable outcomes) / (# of possible outcomes) Example 1 What is the probability that a card drawn at random from a deck of cards will be an ace? Solution In this case there are four favorable outcomes: (1) the ace of spades (2) the ace of hearts (3) the ace of diamonds (4) the ace of clubs. Since each of the 52 cards in the deck represents a possible outcome, there are 52 possible outcomes. Therefore, the probability is 4/52 or 1/13. The same principle can be applied to the problem of determining the probability of obtaining different totals from a pair of dice. Example 2 What is the probability that when a pair of six-sided dice are thrown, the sum of the numbers equals 5? Solution There are 36 possible outcomes when a pair of dice is thrown. Consider that if one of the dice rolled is a 1, there are six possibilities for the other die. If one of the dice rolled a 2, the same is still true. And the same is true if one of the dice is a 3,4,5, or 6. If this is still confusing, look at the following (abbreviated) list of outcomes: [(1,1),(1,2),(1,3),(1,4),(1,5),(1,6); (2,1),(2,2),(2,3)… (3,1),(3,2),3,3)… (4,1)…(5,1)…(6,1)…. The total number of outcomes is 6 × 6 = 36. Since four of the outcomes have a total of 5 [(1,4),(4,1),(2,3),(3,2)], the probability of the two dice adding up to 5 is 4/36 = 1/9. Example 3 What is the probability that when a pair of six-sided dice are thrown, the sum of the number equals 12? Solution We already know the total number of possible outcomes is 36, and since there is only one outcome that sums to 12, (6,6--you need to roll double sixes), the probability is simply 1/36. NOTE: The material from here on through the end of the section is dense and intended only for medium to high scorers. Because this is a CAT (computer-adaptive test), it is relatively unlikely that lower scorers will encounter these questions, and, if they are short of time, they are better off putting their time into other sections. B. Probability of Multiple Events For questions involving single events, the formula for simple probability is sufficient. For questions involving multiple events, the answer combines the probabilities for each event in ways that may seem counter-intuitive. The following strategy is excellent for acquiring a better feel for probability questions involving multiple events or for making a quick guess if time is short. We will focus on questions involving two events. If two events have to occur together, generally an "and" is used. Take a look at Statement 1: "I will only be happy today if I get email and win the lottery." The "and" means that both events are expected to happen together. If both events do not necessarily have to occur together, an "or" may be used as in Statement 2, "I will be happy today if I win the lottery or have email." Consider Statement 1. Your chances of getting email may be relatively high compared to your chances of winning the lottery, but if you expect both to happen, your chances of being happy are slim. Like placing all your bets at a race on one horse, you've decreased your options, and therefore you've decreased your chances. The odds are better if you have more options, say if you choose horse 1 or horse 2 or horse 3 to win. In Statement 2, we have more options; in order to be happy we can either win the lottery or get email. The issue here is that if a question states that event A and event B must occur, you should expect that the probability is smaller than the individual probabilities of either A or B. If the question states that event A or event B must occur, you should expect that the probability is greater than the individual probabilities of either A or B. This is an excellent strategy for eliminating certain answer choices. These two types of probability are formulated as follows: Probability of A and B P(A and B) = P(A) × P(B). In other words, the probability of A and B both occurring is the product of the probability of A and the probability of B. Probability of A or B P(A or B) = P(A) + P(B). In other words, the probability of A or B occurring is the sum of the probability of A and the probability of B (this assumes A + B cannot both occur). If there is a probabiilty of A and/or B occuring, then you must subtract the overlap. Look at the following examples. Example 4 If a coin is tossed twice, what is the probability that on the first toss the coin lands heads and on the second toss the coin lands tails? a) 1/6 b) 1/3 c) ¼ d) ½ e) 1 Solution First note the "and" in between event A (heads) and event B (tails). That means we expect both events to occur together, and that means fewer options, a less likely occurrence, and a lower probability. Expect the answer to be less than the individual probabilities of either event A or event B, so less than ½. Therefore, eliminate d and e. Next we follow the rule P(A and B) = P(A) × P(B). If event A and event B have to happen together, we multiply individual probabilities. ½ × ½ = ¼. Answer c is correct. NOTE: Multiplying probabilities that are less than 1 (or fractions) always gives an answer that is smaller than the probabilities themselves. Example 5 If a coin is tossed twice what is the probability that it will land either heads both times or tails both times? a)1/8 b)1/6 c)1/4 d)1/2 e)1 Solution Note the "or" in between event A (heads both times) and event B (tails both times). That means more options, more choices, and a higher probability than either event A or event B individually. To figure out the probability for event A or B, consider all the possible outcomes of tossing a coin twice: heads, heads; tails, tails; heads, tails; tails, heads. Since only one coin is being tossed, the order of heads and tails matters. Heads, tails and tails, heads are sequentially different and therefore distinguishable and countable events. We can see that the probability for event A is ¼ and that the probability for event B is ¼. We expect a greater probability given more options, and therefore we can eliminate choices a, b and c, since these are all less than or equal to ¼. Now we use the rule to get the exact answer. P(A or B) = P(A) + P(B). If either event 1 or event 2 can occur, the individual probabilities are added: ¼ + ¼ = 2/4 = ½. Answer d is correct. NOTE: We could have used simple probability to answer this question. The total number of outcomes is 4: heads, heads; tails, tails; heads, tails; tails; heads, while the desired outcomes are 2. The probability is therefore 2/4 = ½. The following chart summarizes the "and's" and "or's" of probability: Probability Formula Expectation P(A and B) P(A) × P(B) Lower than P(A) or P(B) P(A or B) P(A) + P(B) Higher than P(A) or P(B) C. Independent and Dependent Events The types of events that we have discussed so far are all independent events. By independent we mean that the first event does not affect the probability of the second event. Coin tosses are independent. They cannot affect each other's probabilities; the probability of each toss is independent of a previous toss and will always be 1/2. Separate drawings from a deck of cards are independent events if you put the cards back. An example of a dependent event, one in which the probability of the second event is affected by the first, is drawing a card from a deck but not returning it. By not returning the card, you've decreased the number of cards in the deck by 1, and you've decreased the number of whatever kind of card you drew. If you draw an ace of spades, there are 1 fewer aces and 1 fewer spades. This affects our simple probability: (number of favorable outcomes)/ (total number of outcomes. This type of probability is formulated as follows: If A and B are not independent, then the probability of A and B is P(A and B) = P(A) × P(B|A) where P(B|A) is the conditional probability of B given A. Example 6 If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces? Solution Event A is that the first card is an ace. Since 4 of the 52 cards are aces, P(A) = 4/52 = 1/13. Given that the first card is an ace, what is the probability that the second card will be an ace as well? Of the 51 remaining cards, 3 are aces. Therefore, p(B|A) = 3/51 = 1/17, and the probability of A and B is 1/13 × 1/17 = 1/221. The same reasoning is applied to marbles in a jar. Example 7 If there are 30 red and blue marbles in a jar, and the ratio of red to blue marbles is 2:3, what is the probability that, drawing twice, you will select two red marbles if you return the marbles after each draw? Solution First, let's determine the number of red and blue marbles respectively. The ratio 2:3 tells us that the total of 30 marbles must be broken into 5 groups of 6 marbles, each with 2 groups of red marbles and 3 groups of blue marbles. Setting up the equation 2x + 3x = 5x =30 employs the same reasoning. Solving, we find that there are 12 red marbles and 18 blue marbles. We are asked to draw twice and return the marble after each draw. Therefore, the first draw does not affect the probability of the second draw. We return the marble after the draw, and therefore, we return the situation to the initial conditions before the second draw. Nothing is altered in between draws, and therefore, the events are independent. Now let's examine the probabilities. Drawing a red marble would be 12/30 = 2/5. The same is true for the second draw. Since we want two red marbles in a row, the question is really saying that we want a red marble on the first draw and a red marble on the second draw. The "and" means we should expect a lower probability than 2/5. Understanding that the "and" is implicit can help you eliminate choices d and e which are both too big. Therefore, our total probability is P(A and B) = P(A) ×. P(B) = 2/5 × 2/5 = 4/25. Now consider the same question with the condition that you do not return the marbles after each draw. The probability of drawing a red marble on the first draw remains the same, 12/30 = 2/5. The second draw, however, is different. The initial conditions have been altered by the first draw. We now have only 29 marbles in the jar and only 11 red. Don't panic! We simply use those numbers to figure our new probability of drawing a red marble the second time, 11/29. The events are dependent and the total probability is P(A and B) = P(A) ×. P(B) = 2/5 × 11/29 = 132/870 = 22/145. If you return every marble you select, the probability of drawing another marble is unaffected; the events are independent. If you do not return the marbles, the number of marbles is affected and therefore dependent. D. Mutually Exclusive Events Another type of probability deals with mutually exclusive events. What do we mean by mutually exclusive events? And what does it mean for two events not to be mutually exclusive? Consider the following example of drawing cards: Example 8 What is the probability that a card selected from a deck will be either an ace or a spade? a)2/52 b)2/13 c)7/26 d)4/13 e)17/52 Solution First, identify events A and B and notice the "or" in between them. That means a greater probability than either A or B individually. Therefore, we expect the answer to be greater than 4/52(ace)=1/13 or 13/52(spade)= 1/4. Eliminate a and b. The tricky part of this question lies in the fact that when we figure probability, we are really just counting, and sometimes, we count twice. In this case we have counted the ace of spades twice. If you don't see this, consider what the 4 in 4/52 stands for: ace of hearts, ace of diamonds, ace of clubs, ace of spades. The 13 in 13/52 stands for all the spades: 1,2,3…King, Ace(of spades). Therefore if we just combined the probabilities by the rule for P(A or B) = P(A) + P(B) we would be over counting. We have to subtract 1/52, the ace of spades that was counted twice. Our answer becomes 4/52 + 13/52 1/52 = 16/52 = 4/13. Another way to think about the question is to just count aces and spades; that is, use simple probability. There are 13 spades in a deck and 3 aces other than the ace of spades already included in the 13 spades. Therefore, there are 16 desired outcomes out of a total of 52 possible outcomes, or 16/52 = 4/13. In the above example, events A and B are not mutually exclusive. Figuring the probability for event A includes part of the probability of event B, and we must therefore subtract out this "overcounted" probability to get the correct answer. The following example illustrates mutually exclusive events: Example 9 What is the probability that a card selected from a deck will be either an ace or a king? a)1/169 b)1/26 c)2/13 d)4/13 e)8/13 Solution The question asks for either an ace or a king. Since there are four kings and four aces in a deck, the probabilities for event A and event B are the same, 4/52 = 1/13. Our answer must be more than this, so eliminate a and b. Do kings and aces have anything to do with each other? Is there such a thing as an ace of kings or a king of aces? No, so we don't have to worry about having over-counted; the events are mutually exclusive. The probability is straightforward: P(A or B) = P(A) + P(B) = 1/13 + 1/13 = 2/13. C is correct. Again we could have used simple probability. Count the total number of kings and aces (4+4) and divide by the total number of cards in a deck: 8/52 = 2/13. E. Conditional Probabilities A conditional probability is the probability of an event given that another event has occurred. Example 10 What is the probability that the total of two dice will be greater than 8 given that the first die is a 6? Solution This can be computed by considering only outcomes for which the first die is a 6. Then, determine the proportion of these outcomes that total more than 8. All the possible outcomes for two dice are shown in the section on simple probability. There are 6 outcomes for which the first die is a 6: (6,1),(6,2),(6,3),(6,4),(6,5),(6,6), and of these, there are four that total more than 8. The probability of a total greater than 8 given that the first die is 6 is therefore 4/6 = 2/3. 1. Probability of A and B If A and B are independent, then the probability that events A and B both occur is p(A and B) = p(A) × p(B). In other words, the probability of A and B both occurring is the product of the probability of A and the probability of B. What is the probability that a coin will come up with heads twice in a row? Two events must occur: a heads on the first toss and a heads on the second toss. Since the probability of each event is 1/2, the probability of both events is: 1/2 × 1/2 = 1/4. Now consider a similar problem: someone draws a card at random out of a deck, replaces it, and then draws another card at random. What is the probability that the first card is the ace of clubs and the second card is a club (any club)? Since there is only one ace of clubs in the deck, the probability of the first event is 1/52. Since 13/52 = 1/4 of the deck is composed of clubs, the probability of the second event is 1/4. Therefore, the probability of both events is 1/52 × 1/4 = 1/208. What's the probability of A and B (2 of 2) if A and B are not independent? If A and B are not independent, then the probability of A and B is p(A and B) = p(A) × p(B|A) where p(B|A) is the conditional probability of B given A. If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces? Event A is that the first card is an ace. Since 4 of the 52 cards are aces, p(A) = 4/52 = 1/13. Given that the first card is an ace, what is the probability that the second card will be an ace as well? Of the 51 remaining cards, 3 are aces. Therefore, p(B|A) = 3/51 = 1/17, and the probability of A and B is 1/13 × 1/17 = 1/221. 2. Probability of A or B If events A and B are mutually exclusive, then the probability of A or B is simply: p(A or B) = p(A) + p(B). What is the probability of rolling a die and getting either a 1 or a 6? Since it is impossible to get both a 1 and a 6, these two events are mutually exclusive. Therefore, p(1 or 6) = p(1) + p(6) = 1/6 + 1/6 = 1/3 If the events A and B are not mutually exclusive, then p(A or B) = p(A) + p(B) - p(A and B). The logic behind this formula is that when p(A) and p(B) are added, the occasions on which A and B both occur are counted twice. To adjust for this, p(A and B) is subtracted. Example 11 What is the probability that a card selected from a deck will be either an ace or a spade? Solution The relevant probabilities are p(ace) = 4/52 p(spade) = 13/52 The only way an ace and a spade can both be drawn is to draw the ace of spades. There is only one ace of spades, so p(ace and spade) = 1/52. The probability of an ace or a spade can be computed as p(ace)+p(spade)-p(ace and spade) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13. F. Combinations Suppose that a job has two different parts. There are m different ways of doing the first part, and there are n different ways of doing the second part. The problem is to find the number of ways of doing the entire job. For each way of doing the first part of the job, there are n ways of doing the second part. Since there are m ways of doing the first part, the total number of ways of doing the entire job is m x n. The formula that can be used is Number of ways = m x n For any problem that involves two actions or two objects, each with a number of choices, and asks for the number of combinations, the above formula can be used. Example 13 William wants a sandwich and a drink for lunch. If a restaurant has 4 choices of sandwiches and 3 choices of drinks, how many different ways can he order his lunch? Solution Since there are 4 choices of sandwiches and 3 choices of drinks, using the formula Number of ways = 4(3) = 12 Therefore, the man can order his lunch 12 different ways. Now, what if the combinations available decrease after each combination is taken? Example 14 There are five meal options in the cafeteria of a certain school. Assuming that a different meal must be eaten each day, and each different type of meal must be eaten once before any type of meal can be eaten a second time, how many different orders of meals can a student eat in the first five days? Solution The answer is 120. There are five types of meals, so the total number of possibilities is 5!. (the "!" stands for factorial), or 5 x 4 x 3 x 2 x 1 = 120. Since a different meal is assigned to every day, you must reduce the amount that you multiply by on a daily basis from 5 to 4 to 3 to 2 to 1. If you like, assign the letters A, B, C, D and E to the five meals and see how many different orders you can create. What if two or more samples are chosen at a time? If we have objects a, b, c, d and want to arrange them two at a time--that is, ab, bc, cd, etc. (We have four combinations taken two at a time). If you have four different combinations taken two at a time, you can write this as 4 C 2, which can be written as 4C2=4x3 2x1 Examples 8C3=8x7x6 3x2x1 10 C 4 = 10 x 9 x 8 x 7 4x3x2x1 Have any more questions or suggestions? Email 800score.com << go back to table of contents Note#2: You cannot calculate (1/6) (5/6) + (5/6) (1/6) which undercounts the possiblities. Simply add combinations where it cannot be true. << table of contents GMAT home page GRE home page 800score.com home Data Sufficiency Option #1: Print out Data Sufficiency section to use offline (9 pages) Option #2 View the Data Sufficiency online. Browse through the sections of this chapter >> This section is broken into 4 parts: I. Introduction II. Strategies for Solving Data Sufficiency Questions III. Data Sufficiency Trick Questions IV. More practice questions I. Introduction In this chapter, we will review strategies for the Data Sufficiency questions and go over trick questions test designers write to fool you on these questions. The Data Sufficiency questions (typically 1/3 of all the math questions) do not require the test taker to find a solution. Instead, the Data Sufficiency questions require the test takers only to find if each of the statements provides enough information for solving the question. Data Sufficiency question instructions look like this: Directions: In each of the problems, a question is followed by two statements containing certain data. You are to determine whether the data provided by the statements are sufficient to answer the question. Choose the correct answer based up on the statement's data, your knowledge of mathematics, and your familiarity with everyday facts (such as number of minutes in an hour or cents in a dollar). (international students: 100 cents to the dollar). Choose choice A) if statement (1) by itself is sufficient to answer the question, but statement (2) by itself is not; B) if statement (2) by itself is sufficient to answer the question, but statement (1) by itself is not; C) if statements (1) and (2) taken together are sufficient to answer the question, even though neither statement by itself is sufficient; D) If either statement by itself is sufficient to answer the question; E) If statements (1) and (2) taken together are not sufficient to answer the question, requiring more data pertaining to the problem. Note: Diagrams accompanying problems agree with information given in the questions, but may not agree with additional information given in statements (1) and (2). All numbers used are real numbers. The Data Sufficiency questions are broken into the stem (the top question and two statements). You answer the question by determining if the information in the two statements is sufficient to answer the question. Lets look at an example to clarify this. (stem) What is the sum of a + b? (statement) (1) A = 5 (statement) (2) B = 10 Explanation: Statement (1) tells you that A is 5. This is not enough information to answer the question. Statement (2) alone is also not enough to answer the question. However, if you combine the two statements, knowing that A=5 and B=10, then you can determine the solution to the question II. Strategies for Solving Data Sufficiency Questions 1. Memorize the Data Sufficiency answer choices. The directions and answer choices for Data Sufficiency questions never change. Memorize them so that you have no problems on test day. There is no excuse for walking into test day without these five answer choices perfectly memorized! A) Statement (1) by itself is sufficient to answer the question, but statement (2) by itself is not. B) Statement (2) by itself is sufficient to answer the question, but statement (1) by itself is not. C) Statements (1) and (2) taken together are sufficient to answer the question, even though neither statement by itself is sufficient. D) Either statement by itself is sufficient to answer the question. E) Statements (1) and (2) taken together are not sufficient to answer the question, requiring more data pertaining to the problem. Some students confuse C and D. Read each answer choice carefully. 1. Note that A requires that B not be sufficient, and vice versa with B. 2. C stipulates that A and B cannot be able to answer the question alone. This means that although A and B together may be able to answer, the answer is not C if either one can answer the question alone. 3. The answer is D if both can answer the question independently, even if both can answer the question together. What does it mean that a statement is "sufficient"? Sufficient does not mean that a statement is right or true, just that you can use the statements to derive an answer. Many beginning students err and think a statement is not sufficient if it proves a statement false. 2. Methodically progress through the two statements It takes mental discipline to progress through the Data Sufficiency questions. The test writers deliberately build tricks to each question. There are two basic questions that you must ask yourself on every Data Sufficiency question: Step 1: Can you answer the question using the information from statement (1) only? Step 2: Can you answer the question using the information from statement (2) only? Step 3: If the answer to both of these questions is "no," then you ask yourself a third question: can you answer the question if you combine the information from both statements? Example Does 3 + x = 1? 1) x is positive 2) x is an odd number Solution (1) alone is sufficient, because it proves that 3 + x cannot equal 1. 3 plus a positive number cannot equal 1. Thus, statement (1) is sufficient because it establishes that the statement is false. (2) Statement (2) is also sufficient, because it proves 3 + x cannot equal 1. 3 plus an odd number cannot equal 1. Therefore, it is sufficient. Since both statements are sufficient, the answer must be D. 3. Data Sufficiency process of elimination strategies In Step 2, as you progress through each statement, you may eliminate questions. Just solve for one of the statements and you are halfway done. Statement 1 is insufficient: automatically eliminate choices A and D, which require (1) to be sufficient. Statement 1 is sufficient: automatically eliminate choices B, C and E, which require (1) to be insufficient. Statement 2 is insufficient: automatically eliminate choices B and D, which require (1) to be sufficient. Statement 2 is sufficient: automatically eliminate choices A, C and E, which require (1) to be insufficient. 4. Analyze questions in terms of sufficiency. Do not think in terms of "what is the exact value," "is this true or false?" Instead, review questions in terms of one question "is there enough information to answer the question?" Look at each statement and ask yourself if it provides enough information to arrive at a conclusion. III. Data Sufficiency Trick Questions The tricks used in Data Sufficiency questions come from a narrow pool of tricks. Learn to identify each of the tricks, and you will be in a strong position to answer each question. In approaching each question, apply the three step method we discussed earlier: Step (1) Look at the question stem. Step (2) Look at each statement individually. Step (3) Then look at both statements in combination. It is important that you have the discipline to stick to this approach. The Data Sufficiency questions tend to be trick questions, particularly the difficult ones, and straying from this basic strategy will increase the chances of you being fooled. Remember that standardized tests are based on the premise that you can separate students into groups of ability. In order to do this, the less capable students must get questions wrong. To make sure less capable students get low scores, the tests are deliberately designed with trick questions specifically made to fool you. Selected Trick Questions AMNESIA TRICK How many adults ride bicycles in city A if all adults in City A either ride bicycles or drive cars? (1) 85% of the 10,000 adults in city A drive cars. (2) 8500 adults in city A drive cars. (A) Statement (1) is sufficient since if 8,500 drive cars then 1,500 ride bicycles. Statement (2) is not sufficient since we do not know the total population; it cannot be assumed from (1). The trick here is that 1 alone can answer the question. Although 1 and 2 together may answer the question, the answer is still A. The unskilled reader will carry over the information from statement 1 when reading statement 2 and not catch the flaw with statement 2 (it does not tell you the population). Trick #2: note that the question doesn't tell you the total population of City A, but the total population is not relevant since the question only asks for "Adults". This question shows how you must have discipline and stick to the 3 step process. (1) Read the stem (2) Read each statement individually (3) If both statements cannot answer the question alone, then look at both statements together. Before you try to combine statement 1 and 2, make sure each answer can/cannot answer the question. When you first read statement 2, temporarily forget what you read in statement 1 so that you may evaluate if (2) alone is sufficient. Hence the name of the trick question: "Amnesia." Get temporary amnesia after reading statement 1 and don't use statement 1's information when you first evaluate statement 2 (because you need to see if statement 2 is sufficient alone. DELAY TRICK How much was a certain Babe Ruth baseball card worth in January 1991? (1) In January 1997 the card was worth $100,000. (2) Over the ten years 1987-1997, the card steadily increased in value by 10% each 12 months. (1) alone is obviously insufficient. To use (2) you need to know what the card was worth at some time between 1987 and 1997. So (2) alone is insufficient, but by using (1) and (2) together you can figure out the worth of the baseball card in January 1991. The trick here is not to do the calculations. If you tried to actually calculate the value in January 1991, you have fallen into the trap. All that matters is that sufficient information is available. The test designers make these questions to make you waste time so that you do not finish the test on time. This is called the DELAY trick because it causes you to be delayed and lose valuable time if you do unnecessary calculations. BACKSOLVE Is the two-digit integer, with digits r (first) and m(second), a multiple of 7? (1) r + m = 13 (2) r is divisible by 3 With statements 1 and 2 we may determine that the two digit number is not a multiple of 7. Using statement (1) Try all the two digit numbers that sum 13: 94, 85, 76, 67, 58, 49. Of those, only 49 is divisible by 7. So, using statement 1, rm may or may not be a multiple of 7; it is insufficient. (2) Is not sufficient because there are many numbers with r that are divisible by 3 and that are multiples of 7 (35, 63, 98). Combined, there are NO possible numbers rm that are divisible by 7 and satisfy statements 1 and 2. The answer is NO, rm is not a multiple of 7. Using statements 1 and 2 we may deduce this. Using statement 2, however, 49 is not a multiple of 3. So, combining the two statements proves that rm is not a multiple of 7. In other words, we've used the two statements to deduce that rm is not a multiple of 7. This looks like a very intimidating question. As a rule, when you encounter a highly intimidating question such as this one, you should plug in possible answers. This question defies an algebraic solution, so it must be solved through backsolving. RED HERRING TRICK Billy sells twice as many $20 tickets as Tim, and Tim sells three times as many $10 tickets as Billy. How many tickets did Billy sell? Tickets are either $10 or $20. (1) Tim sold a total of 35 tickets (2) Together Billy and Tim sold 70 tickets for $1000 1 is not sufficient. Let x = the number of $20 tickets sold by Tim and y = the number of $10 tickets sold by Billy. Then Billy sold 2x ($20 tickets) + y ($10 tickets) Tim sold x ($20 tickets) + 3y ($10 tickets) (2) implies 70 = x + 2x + y + 3y and 1000 = 20(x + 2x) + 10 (y + 3y)- divide this equation by 20 to simplify. Subtract these two equations 70 = 3x + 4y -50 = -3x - 2y 20 = 2y may be solved for x and y and subsequently y = 10 and x = 10, Billy sold 2(10) + 10 = 30 tickets. The trick here is that 1 is completely unnecessary and a distraction. The information in 1 may help answer the question, but it is unnecessary; 2 can do it alone. International students: A "red herring" is an American/English phrase for something that is a distraction to the issue. In this case, the first statement is a distraction. SUPER STATEMENT TRICK What is the average (arithmetic mean) of 3x and 12z? (1) x + 4z = 20 (2) x + z = 8 Yes, combining A and B will solve the question, but A can do it alone. The trap is C. Students will know that the two statements together can solve the question. SUPER STATEMENT questions involve questions where together both statements can solve a question, but carefully examined, one statement may solve it alone. The given information asks for the average of 3x and 12z, which is (3x + 12z) / 2, or 3(x + 4z)/2. Statement 1 tells us the value of x + 4z, (x + 4z)/3. So you can solve the average formula directly without using the second statement. x + 4z = 20, so 3x + 12z = 60, meaning that the average = 30. You may use statement 2 to solve the problem, but statement 1 can do it itself (thus disqualifying choice C, which requires both 1 and 2 to be insufficient). HINT: on difficult Data Sufficiency questions, the statements usually have more value than it appears at first glance. IV. More practice questions Example 1 Is x > 4? 1) x squared = 9 2) x is an integer Solution (1) implies that x = +/- 3 (+/- means positive or negative). Both +3 and -3 are less than 4, so the answer is "NO" and (1) is sufficient, that is NO, x is not greater than 4. A "NO" answer is equally acceptable as a "YES" answer. It is only necessary that there is sufficient information to answer the question. (2) does not provide any necessary information. The correct response is A. Example 2 What is x - y? 1) x + y = 8 2) x - 2y = 2 Solution (1) is not sufficient since (x - y) is the quantity desired. Likewise, (2) is not sufficient. But (1) and (2) together provide us with 2 equations and two unknowns from which x - y can be determined. The correct response is C. (We may solve the problem by subtracting (2) from (1): 3y = 6, therefore y = 2 and x = 6, so that x - y = 6 - 2 = 4. This calculation is, however, unnecessary.) Example 3 How old is Gloria? 1) Gloria's age is four times Alex's age plus Becky's age. 2) Becky was Alex's age fifteen years ago. Solution (1) is obviously not sufficient as is (2). Can the question be answered with (1) and (2)? Let x be Gloria's age, y be Alex's age, and z be Becky's age. (1) states that x = 4y + z. (2) states that z 15 = y. These two equations contain three unknowns; consequently, we cannot determine x. More information is needed and the correct response is E. Example 4 A student group sold only donuts and GMAT books to raise funds. How many GMAT books were sold? 1) 30% of the 90 items sold were GMAT books. 2) 63 donuts were sold. Solution (1) is sufficient since 30% of 90 is 0.3 x 90 = 27. (2) is not sufficient since we do not know the total number of items sold. So the correct response is A. A note of caution: Never let information in (1) influence your decision regarding the information in (2). In this example we cannot assume that 90 items were sold when deciding if (2) provides sufficient information. This is the Amnesia trick that undisciplined test takers will always fall into. Remember to look at each statement individually before comparing the two. << BACK to table of contents