Word problems
Statistics
Rates
Probability
Study this chapter to learn about:
Word problems
Statistics
Rates
Probability
Students of math at almost every level experience some intimidation when confronted with lengthy word problems. The words in the situation often seem to conceal the math necessary to get to the answer. This whole chapter is devoted to ways to get past this difficulty. Word problems fall into several predictable categories, and this chapter exploits this predictability by providing you with a framework to implement each time the GRE throws a certain type of word problem at you. This point will be reiterated throughout the chapter, but it is worth mentioning now: your ultimate goal is to convert the words into algebra. This is the most difficult and most important component of these questions. Be sure to give yourself sufficient time to convert the words into algebra. Once you have done so, it is simply a matter of implementing the algebraic concepts that have been covered thus far in the book.
Word problems tend to be intimidating for many test-takers. One of the most common sentiments students express is concern over where to start. You are given a sentence or several sentences and expected to somehow develop mathematical relationships from these sentences. In the following pages, you will see step-by-step approaches for dealing with these situations. Keep in mind that your ultimate goal should always be the following:
Once you’ve created relationships, you can then use your algebra skills to solve the problem. Let’s look at a typical word problem and a step-by-step methodology for creating these algebraic relationships.
The sum of the lengths of two pipes is 70 feet. The length of the longer piece is 20 feet more than the length of the shorter piece. What is the length of the shorter piece?
Step 1: Identify your unknowns. An unknown is any quantity with an unspecified value. An unknown can be something like Bob’s age, Jack’s height, the number of people in a room, and so on. In the preceding question, there are two unknowns: the length of the shorter pipe and the length of the longer pipe.
Step 2: Assign variables to the unknowns. Since your ultimate goal is to derive algebraic relationships, you should express your unknowns as variables: Let l = the length of the longer pipe and let s = the length of the shorter pipe. You can use other letters as well, but it is helpful to use letters that help you remember which unknown the variable refers to (in this case, you can use l for “longer” and s for “shorter”).
Step 3: Identify relationships among the unknowns. This is the final step in going from words to algebra. Once you identify a relationship, you can create an equation or inequality and start solving for your variables. Words such as is, equals, is greater, and is less are helpful indicators of relationships. In the previous example, there are two relationships among the variables:
Relationship 1: The sum of the lengths of two pipes is 70 feet. Since sum means addition, you should interpret this information to mean the length of the shorter pipe + length of the longer pipe = 70
Relationship 2: The length of the longer piece is 20 feet greater than the length of the shorter piece. The word is indicates a relationship between l and s. On your paper, write down:
l = s
Now translate the wording “20 feet greater.” Since the longer piece is 20 feet more than the shorter piece, it must be true that s alone is not enough to equal l. Thus to make the two sides of the equation equal, you must add 20 to s:
l = s + 20
Step 4: Solve for the unknown. At this point, you have two algebraic relationships:
l + s = 70
l = s + 20
Since the question asked for the length of the shorter piece, the final step is to use substitution to solve for s. Substitute (s + 20) for l in the first equation: (s + 20) + s = 70. Solve for s:
2s + 20 = 70
2s = 50
s = 25
Some word problems will require you to recognize the following relationship:
(price/unit) × (number of units) = total price
Though it might be intimidating, this is a translation that most people use every day. To illustrate this, look at the following example:
If Bob purchased 15 $30 shirts, how much money did he pay for all the shirts?
SOLUTION: Plug the values into the given formula. The price per shirt is $30, and the number of shirts is 15. Thus the total price is 30 × 15 = $450.
Now look at a more GRE-like example:
Bob spends a total of $140 at a certain shop, where he purchases a total of 20 shirts and ties. If the price of each shirt is $10, and the price of each tie is $5, how many shirts does he buy?
Step 1: Assign variables. Since you are not given a value for the number of shirts or the number of ties, let s = the number of shirts and t = the number of ties.
Step 2: Identify relationships. There are two relationships in the question:
Relationship 1: Bob purchases a total of 20 shirts and ties. Thus t + s = 20.
Relationship 2: The total amount Bob pays for the shirts and ties is $140. This amount will be the sum of the amount he paid for the shirts and the amount he paid for the ties. If each shirt costs $10, then he paid 10s for all the shirts. If each tie costs $5, then he paid 5t for all the ties. The algebraic relationship will be 10s + 5t = 120.
Step 3: Combine the equations to solve.
Since you are asked to solve for s, you should use the first equation to write t in terms of s: t = 20 – s. Next, substitute (20 – s) for t in the second equation:
10s + 5(20 – s) = 120
Solve for s:
A common type of word problem that gives many students difficulty concerns age. The approach toward these questions is very similar to what you have looked at so far, though you will need to keep a couple key facts in mind. Let’s look at an example:
Bob is 13 years older than Jack. In 3 years Bob will be twice as old as Jack. How old is Jack?
Step 1: Assign variables. Let j = Jack’s current age and let b = Bob’s current age.
Step 2: Identify relationships.
Relationship 1: Bob is 13 years older than Jack
b = j + 13
Relationship 2: In 3 years Bob will be twice as old as Jack. Bob’s age in 3 years will be b + 3. Jack’s age in 3 years will be j + 3. Thus you can construct the following equation: b + 3 = 2(j + 3)
Step 3: Combine the equations to solve.
Since the question asks to solve for j, you should substitute (j + 13) for b in the second equation.
Common Translations for Word Problems
1. The sum of Kathy’s salary and Janet’s salary is $200,000. If Kathy’s salary is $50,000 less than Janet’s salary, what is Janet’s salary?
$50,000
$75,000
$125,000
$140,000
$175,000
2. Bob breaks up a 220-mile trip into three parts. If the first part of the trip is 20 miles longer than the second part of the trip, and the third part of the trip is 50 miles longer than the second part of the trip, how many miles was the second part of the trip?
50
60
70
80
100
3. A certain store sells only shirts and ties. The price of a shirt is $40 and the price of a tie is $20. If Bob pays $340 for 10 items from this store, how many shirts did he purchase?
3
4
5
6
7
4. A manufacturer purchased three equally priced computers and five equally priced monitors for a total of $4,000. If the price of each computer was 5 times the price of each monitor, how much did each computer cost?
200
400
600
800
1,000
5. An $800 bill is to be split equally among 10 friends. If x of the friends do not pay their share and the remaining friends split the bill evenly, how much will each of the remaining friends pay, in terms of x?
6. Bob is 10 years older than Jack. In 2 years, Bob will be double Jack’s age. How old will Bob be in 5 years?
8
18
22
23
27
7. Telecharge charges $0.75 for the first minute of a call and $0.25 for each minute thereafter. If total cost for a Telecharge call is $5.75, how many minutes long was the call?
18
19
20
21
22
8. The price of Stock A is $5 greater than the price of Stock B. If the price of Stock A increases by $10, the new price will be double the price of Stock B. What is the price of Stock B?
10
15
20
25
30
9. Bob can spend a maximum of $56 on hamburgers and shakes. Hamburgers cost $5 apiece, and shakes cost $4 apiece. If Bob purchases both hamburgers and shakes, what is the maximum number of hamburgers he can purchase?
7
8
9
10
11
10. The height of a tree increases by x feet each year. At the start of a certain year, the height of the tree is 10 feet. If the height of the tree is 60 feet after 8 years, what is the value of x?
6.25
7.75
8
8.25
9
11. In an auditorium with 300 students, the number of boys is more than double the number of girls. What is the maximum number of girls who can be in the auditorium?
97
98
99
100
101
For this question, write your answer in the box.
12. Were Jack 6 years older, he would be double Bob’s age. Jack is currently 1 year older than Bob. How old is Bob?
13. A group of marbles is arranged into x rows and y columns. If the number of columns is 7 times the number of rows, and there are 343 total marbles, how many rows are there?
7
14
49
62
74
Each of the following questions consists of two quantities, Quantity A and Quantity B. You are to compare the two quantities. You may use additional information centered above the two quantities if additional information is given. Choose
if Quantity A is greater
if Quantity B is greater
if the two quantities are equal
if the relationship between the two quantities cannot be determined
1. B Let k = Kathy’s salary and j = Janet’s salary. The first relationship is that k + j = 200,000. The next relationship is that Kathy’s salary is $20,000 less than Janet’s salary. Expressed algebraically, this relationship is: k = j – 20,000. Next, combine the equations to solve for j. To do so, substitute (j – 20,000) for k in the first equation: (j – 20,000) + j = 200,000. Solve for j:
2. A Let f = the number of miles of the first part of the trip. Let s = the number of miles of the second part of the trip. Let t = the number of miles of the third part of the trip. The sum of the three parts is 220 miles, so f + s + t = 220. The first part of the trip is 20 miles longer than the second part, so f = s + 20. The second part of the trip is 50 miles longer than the first part, so t = s + 50. To solve for s, substitute (s + 20) for f and (s + 50) for t in the original equation: (s + 20) + s + (s + 50) = 220. Solve for s:
3. E Let s = the number of shirts Bob purchases. Let t = the number of ties Bob purchases. If Bob purchases 10 items, then s + t = 10. The next relationship is the more difficult one. The $340 that Bob pays is the sum of the amount he pays for shirts and the amount he pays for ties.
The amount he pays for the shirts is ($/shirt) × (number of shirts) = 40s.
The amount he pays for the ties is ($/tie) × (number of ties) = 20t. So 40s + 20t = 340. You now have two equations:
To solve for s, use substitution. First, manipulate the first equation to express t in terms of s: t = 10 – s. Next, substitute that expression for t in the second equation: 40s + 20(10 – s) = 340. Solve for s:
4. A Let c = the price of each computer and m = the price of each monitor.
The manufacturer thus paid a total of $3c for the computers and $5m for the monitors. The sum of these values equals $4,000: 3c + 5m = 4,000. The other algebraic relationship you can create from the information is c = 5m. There are now two equations:
To solve for m, substitute 5m in for c in the first equation: 3(5m) + 5m = 4,000. Distribute and combine like terms: 20m = 4,000. Solve for m: m = 200.
5. D The question asks you to determine the amount paid per friend. This can be expressed as the following ratio: total bill/number of people paying. The numerator of this ratio will be 800. The denominator will be the total number of friends splitting the bill after x do not pay. If there were 10 friends originally, and x do not pay, the number of people paying is 10 – x. Thus the ratio will be 
6. D Let b = Bob’s current age and j = Jack’s current age. You are asked to solve for b + 5. Use the assigned variables to construct algebraic relationships. The first sentence provides a relationship between Bob’s current age and Jack’s current age: b = j + 10. The second sentence provides a relationship between their ages in 2 years: b + 2 = 2(j + 2). Use substitution to solve for b:
Step 1: Manipulate the first equation to write j in terms of b: b – 10 = j
Step 2: Substitute b – 10 for j in the second equation:
In 5 years, Bob will be 23.
7. D Since you are asked to solve for the length of the call, you should assign a variable: l = length of the call. The total charge of $5.75 consists of two parts: the charge for the first minute and the charge for all subsequent minutes. The charge for the first minute is $0.75. The charge for all subsequent minutes will be cost/minute × number of minutes. You are told that cost/minute is $0.25. If l is the length of the entire phone call, then the number of minutes charged at $0.25 will be l – 1. Thus you can construct the following equation: 0.75 + 0.25(l – 1) = 5.75. Solve for x:
8. B Assign variables: Let a = the price of stock A, and let b = the price of stock B. The information in the problem provides two relationships:
“The price of stock A is $5 greater than the price of stock B”: expressed algebraically: a = b + 5
“If the price of stock A increases by $10, the new price will be double the price of stock B.”: a + 10 = 2b
Now use substitution to solve for b:
Substitute b + 5 for a in the second equation: (b + 5) + 10 = 2b. Solve for b:
9. D Let h = the number of hamburgers Bob purchases, and let s = the number of shakes Bob purchases. The amount Bob spends on hamburgers is 5h, and the amount he spends on shakes is 4s. Since the maximum that he can spend is $56, the sum of these two quantities must be less than or equal to 56: 5h + 4s ≤ 56. Since the question asks for the maximum number of hamburgers that Bob can purchase, you should minimize the number of shakes that he purchases by letting s = 1:
Since h must be an integer, the maximum value for h will be the greatest integer less than 
Thus the greatest value for h will be 10.
10. A The final height of the tree will be the sum of the original height and the increase after 8 years. If the height of the tree increases by x feet each year, then the tree will have increased by 8x feet after 8 years. Thus you can create the following equation: 10 + 8x = 60. Solve for x:
11. C Let b = the number of boys and g = the number of girls.
Relationship 1: b + g = 300
Relationship 2: b > 2g
Substitution with inequalities can be difficult. Think of what would be the case if b = 2g. Substitute 2g for b in the first equation and arrive at 2g + g = 300 → 3g = 300 → g = 100. Since 2g < b instead of 2g = b, the number of girls must be less than 100. The greatest integer less than 100 is 99 (you only consider integers since the number of girls must be a whole number).
12. 7 Let j = Jack’s current age and b = Bob’s current age.
Relationship 1: j + 6 = 2b
Relationship 2: j = b + 1
To solve for b, substitute b + 1 for j in the first equation:
13. A Rows × columns = total marbles. Thus xy = 343. You also know that y = 7x. To solve for x, substitute 7x for y in the first equation:
1. C Let k = Kate’s current age, and let s = Sara’s current age. Based on the prompt, you can create the following relationship: k = s + 3. Since the quantities compare Kate’s age and Sara’s age, make them comparable by substituting (s + 3) for k in Quantity A. The value for Quantity A will be (s + 3) – 4 = s – 1. The value for Quantity B will be s – 1. Thus the two quantities are equal.
2. B Since the columns ask you to compare the number of chairs in the auditorium to x2, you should attempt to express the number of chairs in terms of x. To determine the total number of chairs, do 
Thus the number of chairs = x(x–1) = x2 – x. Quantity A is thus x2 – x and Quantity B is x2. Since x must be positive (there can’t be a negative number of chairs or rows), x2 must be greater than x2 – x.
3. A Let b = Bob’s weight and j = Jack’s weight. Use these variables to create the equation b + j = 50. If Bob weighs more than Jack, then Bob’s weight must account for more than half of their sum. Algebraically,
Thus Quantity A is greater.
4. B Let j = the number of Jupiter bars sold and b = the number of Big Slugger bars sold. The revenue for the Jupiter bars is 2j and the revenue for the Big Slugger bars is 1b. Since the total revenue on the two bars is $30, it follows that 2j + b = 30. To maximize b, minimize j. Since the retailer sold both candy bars, the minimum value for j is 1. Thus 2(1) + b = 30 → b = 28. The maximum for b is 28. Quantity B is greater.
5. D The given information provides no relationship between Jack’s age and Tanya’s age.
6. D Let s = the price of a sedan and t = the price of a truck. Thus s + 3,000 = 2t. You are given one relationship between two variables. Without additional information, you cannot determine the relationship between the variables.
7. C Let s = the price of each stereo and m = the price of each MP3 player. The total price of the three stereos is thus 3s, and the total price of each MP3 player is thus 2m. Since the total prices are equal, 3s = 2m. Manipulate to solve for s/m:
Statistics refers to the properties of a set of data. For the purposes of the GRE, you can think of data as numerical pieces of information. For example, the number of students in a class is a data point, as is the average grade for a class, or the range of scores in a class. Though statistics is a broad field within mathematics, on the GRE, you will be expected to understand the following statistical concepts:
Mean
Median
Mode
Range
Standard deviation
Let’s look at a set of data to understand these concepts:
The test scores for seven students in a class are 72, 90, 72, 83, 81, 63, and 94. The median refers to the data point that has an equal number of data points greater than it and less than it. In other words, the median is the middle value in a set when the data points are listed in increasing order.
To determine the median, list the data points from least to greatest.
Listed in increasing order, the data points in the preceding list will read: 63, 72, 72, 81, 83, 90, 94. Since there are seven data points, the median is the fourth data point.
In this case, the middle value is 81.
When the number of terms in a set is even, the median will be the average of the two middle terms.
What is the median of 2, 8, 10, 11, 12, and 14?
SOLUTION: Since there are six data points, the median will be the average of the middle two values. In this case, the middle two values are 10 and 11.
To determine the median, find the average of 10 and 11. 
The range refers to the positive difference between the largest and smallest value in a set. In the earlier example, the range is 94 – 63 = 31.
The mode is the data point that appears most frequently in a set. In the earlier example, the mode is 72.
The average of a set is the sum of the data points/the number of data points. Put more simply:
where A = average of the set, S = sum of the set, and N = number of data points in the set.
The average of the set in the earlier example is 
Of all statistical topics, averages are tested most frequently. Unfortunately, most average questions will not be as simple as the preceding one. Usually, the GRE will give you the average for a set and expect you to solve for the values of one or more data points in a set or the number of items in a set. Regardless of how the question is framed, you will always use the average formula: 
If a company’s average yearly revenue over a 10-year period was $550,000, what was the company’s total revenue during that period?
SOLUTION: Since you want to solve for the sum, manipulate the average formula to isolate the sum:
Substitute the given values for A and N: 550,000 × 10 = $5,500,000.
Sometimes you will have to use the average formula multiple times in a question:
The average height of eight students is 64 inches. If the average height of seven of the students is 62, how tall is the eighth student?
SOLUTION: Use the average formula to determine the sum of the heights of all eight students:
Next, use the average formula to determine the sum of the heights of seven of the students:
Let the height of the eighth student = h. You know that h + (the sum of the heights of the seven other students) = (the sum of the heights of the eight students). Thus h + 432 = 512 → h = 80.
An even more difficult example will use three averages in the question:
A company’s average daily revenue over a 10-day period was $40,000. If the average daily revenue for the first 4 days was $25,000, what was the average daily revenue for the last 6 days?
SOLUTION: revenue for the first 4 days + revenue for the last 6 days = total revenue
Total revenue = A × N = $40,000 × 10 = $400,000
Revenue for the first four days = A × N = $25,000 × 4 = $100,000
Revenue for the last 6 days = A × 6 = 6A, where the average daily revenue for the last 6 days.
Thus:
When determining the average of two data points, the average will always fall in the middle of the two data points. For example, the average of 30 and 40 = 
But what if one of the data points appears more often than the other data point? For example: What is the average of 30, 30, and 40? In this case, the average = 
Now that you have added an additional data point of 30 to the set, the average is weighted closer to 30 than it is to 40.
The preceding example represents a weighted average. You will have a weighted average any time the frequency of a data point pulls the average closer to that data point than to the other data point. In the preceding example, the fact that there were more 30s than 40s meant that the average was skewed more toward 30 than toward 40.
Quantitative Comparison Strategy
Weighted averages are tested most frequently in Quantitative Comparison questions. For these questions, it is important to keep in mind the mandate that you must minimize calculations! Oftentimes, one of the quantities will be the average of the set if the number of data points were equal. Think about which data point the average is skewed toward, and you will cut back on time spent calculating.
A student took 10 exams for his biology course. His average on 6 of the exams was 80. His average on the other 4 exams was 90.
Solution: Look at the columns! Notice that 85 is the average of 80 and 90. If 80 and 90 appeared with equal frequency, then the student’s average for the course would be 85. However, six of the data points correspond with the average of 80 and four of the data points correspond with the average of 90. The average for the course will thus be closer to 80 than to 90 and therefore less than 85. The correct answer is B.
Though it is tested less commonly, you should also know how to calculate a weighted average. Let’s use the preceding example:
A student took 10 exams for his biology course. His average on 6 of the exams was 80. His average on the other 4 exams was 90. What was his average for the 10 exams?
Approach 1: Use the average formula. For the 4 courses in which he averaged 90, the sum is 90(4). For the 6 courses in which he averaged 80, the sum is 80(6). The average for all 10 courses is thus 
Approach 2: Use the weights.
Note that 
The fractions represent the weight of each data point as a piece of the total weights of the set. From this, you can extrapolate the following formula:
Note that this approach works when the number of data points is represented as a ratio or percent as well!
For the first 10 days of a 30-day period, a stock’s average return was $6.00. For the last 20 days of the 30-day period, the stock’s average return was $12.00. What was the stock’s average return for the 30-day period?
SOLUTION: The two data points are $6.00 and $12.00. The frequency of the $6.00 data point is 
The frequency of the $12.00 data point is 
The weighted average is thus 
Earlier, the section defined the median as the middle data point when the data points are in increasing order. Though calculating the median for small sets simply requires that you put the data points in increasing order and find the middle value, it becomes trickier when the set has a large number of data points. For example, if you want the median of 51 data points, it would be too time-consuming to list all of them. Instead, you should recognize that the median will be the 26th data point. Why? Because there will be 25 data points below this value and 25 data points above this value.
A certain test was administered to 29 students. 6 students scored 60, 8 students scored 73, and 15 students scored 79. What was the median score for the 29 students?
SOLUTION: The median is the 15th-largest (or smallest) score. The smallest 6 data points have a value of 60. The next 8 data points have a value of 73. These two groups account for the first 14 data points. The next data point will thus equal the median. That data point occurs in the last set. Since all the data points in the last set are 79, the median is 79.
Standard deviation refers to how far from the mean the numbers in a set typically fall. Two sets can have the same mean and the same number of items, but completely different spreads from the mean. The concept of standard deviation is employed to represent this spread. The greater the average spread of the data, the greater the standard deviation. Look at the following three sets, all of which have the same mean:
In the first set, all the data points are equal, so there is no spread from the mean. The standard deviation is thus zero. In the second set, the distance between successive numbers is one, so the standard deviation is greater than in the first set. In the third set, the distance between successive numbers is two, so the standard deviation is even greater than in the second set.
You might be wondering what the standard deviations of the first and second sets are. Fortunately, the GRE almost never tests the exact formula for standard deviation. It does, however, expect you to understand the concept and to be able to compare the standard deviation of different sets of numbers.
Which of the following sets has the greatest standard deviation?
{4, 4, 4, 4, 4}
{3, 4, 4, 4, 5}
{3, 4, 4, 4, 8}
{0, 2, 4, 4, 12}
{1, 4, 4, 4, 7}
SOLUTION: Look at how the choices compare to A, which has a spread of zero. The further the values in a set move away from 4, the greater the standard deviation. The values in Choice D spread the most from 4. Thus the correct answer is D.
1. Bob’s average for three tests was 83. If he scored 72 and 94 on the first two tests, what was his score on the third test?
81
82
83
84
85
2. Janet’s average for four tests was 75. If she scored 63 and 83 on the first two tests, what was her average for the last two tests?
73
75
77
79
81
3. The average of a and b is 70. The average of a and c is 100. What is b – c?
–60
–30
15
30
60
4. The set of numbers 2, 3, 5, 9, 11, x is in increasing order. If the average of the set equals the median, what is x?
12
14
15
30
42
5. In 2012, a store’s average monthly revenue was $20,000. For the first 10 months of the year, the store’s average monthly revenue was $18,000. If the revenue in December was 50% greater than the revenue in November, what was the store’s revenue in November?
24,000
30,000
36,000
48,000
56,000
For this question, write your answer in the box.
6. Three crates have an average weight of 70 pounds and a median weight of 90 pounds. What is the maximum possible weight, in pounds, of the lightest box?
For this question, indicate all of the answer choices that apply.
7. A real estate agent is selling 5 homes priced at $200,000, $240,000, $300,000, $320,000, and $350,000. If the price of the $200,000 home increases by $20,000, which of the following must be true?
The median price will increase.
The median price will decrease.
The average price will increase.
The standard deviation will decrease.
The range of the prices will decrease.
8. The preceding table shows a business’s spending over a 25-day period, in thousands of dollars. For each number amount earned in the first row, the second row gives the number of days that the business spent that amount. What was the median amount of money that the business spent during the 25-day period?
$100,000
$120,000
$130,000
$170,000
$200,000
9. If a is the standard deviation of b, c, and d, what is the standard deviation of b + 3, c + 3, and d + 3, in terms of a?
a
a + 3
a + 6
a + 9
3a
10. A certain company has only directors and managers. The company has 300 directors and 200 managers. If the average age of the directors is 42 and the average age of the managers is 38, what is the average age of all the company’s employees?
39.8
40
40.2
40.4
41
11. A student’s average score for 10 tests was x. The grader then discovered that there was an error on two of the tests. On one of the tests, the student’s score should have been 10 points lower. On another test, the student’s score should have been 20 points higher. After the grader corrects these mistakes, what is the student’s new average, in terms of x?
x – 1
x
x + 1
x + 10
x + 20
12. What is the average of n, n + 1, n + 7, n + 9, and n + 13, in terms of n?
n + 2
n + 3
n + 6
n + 8
n + 15
Each of the following questions consists of two quantities, Quantity A and Quantity B. You are to compare the two quantities. You may use additional information centered above the two quantities if additional information is given. Choose
if Quantity A is greater
if Quantity B is greater
if the two quantities are equal
if the relationship between the two quantities cannot be determined
At a certain company, the average annual salary of the college graduates is $73,000, and the average annual salary of the people who did not graduate from college is $52,000.
1. C The sum of Bob’s three test scores = A × N = 83 × 3 = 249. His score for the two given tests = 166. His score for the third test is thus 249 – 166 = 83.
2. C The sum of Janet’s four test scores = A × N = 75 × 4 = 300. The sum of her first two test scores = 63 + 83 = 146. The sum of her last two test scores thus equals 300 – 146 = 154. Her average for these last two tests = 
3. A Use the average formula for both sets.
Now set up a system of equations to solve for b – c:
4. A Since the set has an even number of items, the median is the average of the two middle terms: 
The average of the set is 
You are told that the average equals the median, so:
5. A the revenue for the entire year = average monthly revenue × number of months = $20,000 × 12 = $240,000. The revenue for the first 10 months = average monthly revenue × number of months = $18,000 × 10 = $180,000. The revenue for the last two months is thus $240,000 – $180,000 = $60,000. Let n = the revenue in November. Since the revenue in December was 50% greater than the revenue in November, the revenue in December = 1.5n = 
n. Solve for n:
6. 30 The sum of the weights of the crates = A × N = 70 × 3 = 210. Let the weight of the lightest box = l and the weight of the heaviest box = h. Thus
To maximize the weight of the lightest box, you must minimize the weight of the heaviest box. Since the heaviest box is to the right of the median, its weight can be no less than the median weight. Thus the least weight for the heaviest box = the median = 90. Substitute 90 for h:
7. C, D, and E The median value of the original set is $300,000. The increase in the price of the smallest home is too small to affect the median. Eliminate Choices A and B. The $20,000 increase means that the sum of the set will increase. If the sum of a set increases and the number of items stay the same, then the average must increase. Choice C is true. The increase in the smallest price decreases the spread of all the data points. Thus the standard deviation will decrease. Choice D is true. The increase in the smallest price will decrease the distance between the smallest and greatest values in the set. The range will thus decrease. Choice E is true.
8. D The median refers to the middle data point when all the data points are in increasing order. Since there are 25 data points, the median will be the 13th largest data point (since there are 12 data points below this value and 12 above it). The bottom 12 data points are in the $100,000; $120,000; and $130,000 columns. The 13th data point is thus in the next column: $170,000.
9. A Standard deviation is a measure of the typical spread of the data points from the mean. If all of the data points in a set are increased by the same value, the spread between the data points will not change. Thus the standard deviation will not change. Since the standard deviation of the original set was a, the standard deviation of the new set is also a.
10. D Since the frequency of the data points differs, this is a weighted average. Let’s use this formula: weighted average = (frequency of data point 1/total number of data points) × data point 1 + (frequency of data point 2/total number of data points) × data point 2. In the question, the data points are the average ages for the two groups. Let the average age for the directors be data point 1 and the average age for the managers be data point 2. Thus the weighted average = 
11. C Since the choices have variables, you can plug in numbers.
Step 1: Let x = 100.
Step 2: Answer the question when x = 100. Originally, the sum of the scores is thus A × N = 100 × 10 = 1,000. After the corrections, one of the scores will decrease by 10 points, and one of the scores will increase by 20 points. The net change is thus (20 – 10) = 10. The new sum is 1,010. The new average is thus 
The goal is 101.
Step 3: Substitute 100 for x in the choices, and identify which choice matches the target of 101.
12. C The sum of the terms is n + (n + 1) + (n + 7) + (n + 9) + (n + 13) = 5n + 30. Since there are 5 terms, the average of the set is 
1. A The average for a set must fall between the average of each subset. Thus the average annual salary must be greater than $52,000.
2. A Don’t calculate! Look at how the average of 6, 6, 8, and 10 compares to 7. The average of those numbers is 
Quantity A has four numbers with an average of 7 (7, 7, 7, and 7) and four numbers with an average greater than 7. Its average must therefore be greater than 7.
3. D Without knowing where the average age for the entire school falls in relation to 16.3 and 16.8, it is not possible to determine the ratio of boys to girls.
4. A Express Quantities A and B algebraically:
Since you are told that a > b, Quantity A is greater.
5. B To maximize the range, minimize the smallest value and maximize the greatest value. Since the terms are positive integers, the minimum value for an item in the set is 1. To maximize the greatest value, minimize the rest of the terms. In this case, that would mean making them all equal to 1. Therefore:
The maximum possible range is 96 – 1 = 95. Quantity B is greater.
6. A The difference between the two sets is the 0 in the second set. Since 0 is less than the average of 2,19, 34, and 60, the average of the second set must be smaller than the average of the first set. Quantity A is greater.
7. B Express Quantity A algebraically:
The numerators of the two quantities are the same. Since the denominator of Quantity B is smaller, that fraction is larger.
8. B In Quantity B, each term in Quantity A is doubled. The spread of the set will thus be doubled. Since the spread in Quantity B is greater, the standard deviation of Quantity B is greater.
Rate questions always come in one of two forms: distance or work. In both cases, there will be a constant relationship between the rate, time, and work or distance:
The primary difference in the two equations is the following:
The rate × time = distance formula is almost always the best way to solve distance equations. In simpler questions, you will be given two components of the formula and asked to solve for the third.
If a car travels at a constant rate of 50 miles per hour, in how many hours will the car have traveled 325 miles?
Step 1: Set up the rate × time = distance chart:
Step 2: Solve for t:
When solving rate questions, be sure that you use the same unit throughout the question. If the rate is expressed per minute, then you must express time in minutes, not seconds or hours.
Running at a constant rate, Bob travels 5 miles in 1 hour. If Bob travels for 20 minutes, how many miles does he travel?
Step 1: Set up the r × t = d chart:
rate (mi/hr) × time (hr) = distance (miles)
Bob’s rate is distance/time = 5 miles/1 hour, and his time is 20 minutes. It might be tempting to simply substitute these values into the original formula, but this would be incorrect. Why? Because the units do not match up. The rate is expressed per hour, whereas the time is expressed in minutes. Before plugging the values into the formula, you should convert the time from 20 minutes to 
hour.
Step 2: Once you have done so, you can plug the values into the formula:
For distance questions, the r × t = d table is particularly useful for problems that involve multiple rates or multiple travelers. In these situations, you will generally want to create two rows on the r × t = d table, fill them in with the appropriate values or variables, and use the resulting expressions to identify a relationship between the distances.
Bob and Jack start at opposite ends of a 200-mile track. Bob travels at a constant rate of 50 miles per hour and Jack travels at a constant rate of 25 miles per hour. If they start traveling toward each other at the same time, in how many hours will they meet?
Step 1: Put the given information into the r × t = d table:
Note that there are two rows—one for Jack and one for Bob. You are asked to solve for the amount of time they travel, so let t represent time. Since Bob and Jack start and end at the same time, they will each have traveled for t hours. In terms of t, Bob travels 50t miles and Jack travels 75t miles.
Now you must identify the relationship between these distances. Since they are traveling toward each other, Bob will travel some of the 200 miles and Jack will travel the remaining distance. Thus the distances the two travel must add up to 200. Algebraically, you can represent this relationship as:
Step 2: Solve for t:
Train A and train B start at the same point and travel in opposite directions. Train A travels at a constant rate of 80 miles per hour, and train B travels at a constant rate of 60 miles per hour. If train A starts traveling 2 hours after train B, how many miles will train A have traveled when the two trains are 720 miles apart?
Step 1: Put the given information into the r × t = d table. The rate for trains A and B are given as 80 miles per hour and 60 miles per hour, respectively. To solve for the number of miles Train A will have traveled, you need to determine how many hours Train A traveled. Let t = the number of hours Train A travels. Since Train A started 2 hours before Train B, and the two trains stop traveling at the same time, Train B must have traveled t – 2 hours. Thus in terms of t, Train A’s distance is 80t and Train B’s distance is 60(t – 2).
As in the previous example, it is essential to identify the relationship between the two trains’ distances. For the two trains to end up 720 miles apart, Train A must cover some portion of the 720 miles and Train B must cover the rest. Thus the sum of their distances is 720. Expressed algebraically, the equation is:
Step 2: Now solve for t:
You are asked to solve for the number of miles Train A traveled, so plug 6 in for Train A’s time:
Look at the following question:
Sarah goes on a 600-mile trip. She travels at a constant rate of 50 miles per hour for the first 300 miles of the trip, and at a constant rate of 100 miles per hour for the last 300 miles of the trip. What is Sarah’s average speed in miles per hour for the entire trip?
70
75
80
85
Solution: Many test-takers are tempted to average the rate of 50 and 100 to arrive at an average speed of 75. Though perhaps intuitive, this would be incorrect. Why? Because Sarah spent more time traveling at a rate of 50 miles per hour than at a rate of 100 miles per hour. Her overall rate will thus be closer to 50 than to 100. Based on this logic alone, you can immediately eliminate C, D, and E.
To actually calculate her average speed, you should use the rate formula: 
The total distance is 600 miles, so you need to calculate the total time. To determine the total time, you should determine the time for each part of the trip, and then add these up. Use the r × t = d formula to determine the time for each part of the trip:
Solve for x:
Solve for y:
The total time is x + y = 6 + 3 = 9. The average speed for the trip is thus 
The correct answer is Choice A.
The second type of rate problem involves work. In contrast to distance problems, work problems are concerned with some output per unit of time. An output can be something produced (such as widgets, cups, cars, etc.) or a job done (such as mowing a lawn, cooking a meal, writing a paper, etc.). In both cases, you want to use the work formula, though as you will see, the way you express work will differ. Let’s look at a simple work question:
If a machine produces pencils at a constant rate of 1,500 pencils per hour, in how many hours will the machine have produced 6,750 pencils?
SOLUTION: Put the given values into the rate × time = work (RTW) table. The machine’s rate is 1,500 pencils/hour, and its output (work) is 6,750 pencils. Since you are trying to solve for time, assign a variable: t
Solve for t:
Let’s look at another example, this time with work represented as some job done instead of units produced:
Working at a constant rate, Bob can mow 3 same-sized lawns in 5 hours. How many hours will it take Bob to mow 2 same-sized lawns?
SOLUTION: Set up the RTW table. Remember that rate 
so Bob’s rate will be 
Solve for t:
In certain work questions, you will be given the individual rates of two or more elements and will be asked about what happens when they work together.
Working alone at a constant rate, Bob can mow 1 lawn in 3 hours. Working alone at a constant rate, Jack can mow 1 same-sized lawn in 8 hours. If Bob and Jack work together but independently at their respective constant rates, how many hours will it take them to mow half a same-sized lawn?
SOLUTION: Bob’s rate is 
Jack’s rate is 
To determine how long it takes them to mow the lawn when they work together, you need their combined rate. The combined rate is the sum of their individual rates. In this case, their combined rate is 
This means that, when they work together, Bob and Jack can mow 
of a lawn in 1 hour. Now input this rate into the r × t = w formula:
Note that you input 
for the total work done, since the question is asking for the time necessary for them to mow half the lawn.
Solve for t:
For this question, write your answer in the box.
1. Susan takes a 45-mile trip. She travels the first 15 miles in 40 minutes. She travels the remaining 30 miles at double the rate at which she traveled the first 15 miles. What is her average speed for the entire trip in miles per minute?
2. Boris ran the first 12 miles of a 24-mile race in 1.8 hours. If his rate for the entire race was 6 miles per hour, how many hours did it take him to run the last 12 miles?
2
4
6
3. Working at a constant rate, a certain hose can fill a pool in 10 minutes. Working at a different constant rate, another hose can fill the same pool in 15 minutes. If the two hoses work together but independently to fill the pool after it is half full, how many minutes will it take the hoses to fill the pool?
3
4.5
5
6
12.5
4. Four machines, working independently at the same constant rate, can fill a production lot in 3 hours. How many of these machines would be needed to fill the lot in 30 minutes?
12
24
36
40
48
5. Working at a constant rate, a cook can create a batch of pasta in 20 minutes. Another cook, working at a constant rate, can create the same batch of pasta in 10 minutes. If the two cooks work together but independently, how many hours will it take them to create 3 batches of pasta?
2
6. A hose fills an empty tank with water at a rate of 15 liters/second. At the same time, water drains from the tank at a rate of 10 liters/second. If the tank has a capacity of 200 liters, in how many seconds will the tank be half full?
4
8
20
25
40
7. Working together at their respective constant rates, Bob and Sam can mow a lawn in 12 hours. If Bob’s rate is twice Sam’s rate, how many hours would it take Bob, working alone, to mow the lawn?
15
18
24
32
36
8. Working alone at a constant rate, Machine A can produce 15 widgets in 2 minutes. Working alone at a constant rate, Machine B can produce 20 widgets in 3 minutes. If the two machines start working at the same time, after how many minutes will Machine A have produced 40 more widgets than Machine B?
12
24
36
48
60
9. The birthrate of a certain species is approximately 500 organisms per day. The death rate of the species is approximately 300 organisms per day. If the population has 50,000 organisms at the beginning of a certain year, approximately how many organisms will be in the population at the end of the year?
73,000
123,000
159,500
232,500
293,000
For Questions 10 and 11, write your answer in the box.
10. Bob and Jack start traveling at the same time at constant rates and in the same direction. If Bob’s rate is 2 miles per hour greater than Jack’s, after how many hours will Bob and Jack be 
of a mile apart?
11. Two scientists, working at the same constant rate, can fill 8 equally sized beakers in 10 minutes. How many scientists would be needed to fill 24 of these beakers in 5 minutes?
Each of the following questions consists of two quantities, Quantity A and Quantity B. You are to compare the two quantities. You may use additional information centered above the two quantities if additional information is given. Choose
if Quantity A is greater
if Quantity B is greater
if the two quantities are equal
if the relationship between the two quantities cannot be determined
Sally travels the first 30 miles of a 60-mile trip at 50 miles per hour and the last 30 miles of the trip at 40 miles per hour.
Dennis can cook x cakes in 3 hours, and he can cook y pies in 0.5 hours, where x > y.
Traveling at a constant rate, Jake ran a lap in x minutes.
Car A and Car B start at opposite ends of a 200-mile track and begin driving toward each other at the same time. When the cars meet, Car A is 80 miles from its starting point.
Car A starts a race 10 miles behind Car B and ends the race 15 miles ahead of Car B.
A tank that is half full is filled with gas at a constant rate of 3 liters per minute.
Working at the same constant rate, 8 machines can produce 7 computers in 6 hours.
1. 
To determine average speed, use the rate formula: 
You are told that the distance for the trip is 45 miles, so the formula now reads 
You must determine the time spent on the trip. To do so, determine the time for the first part of the trip and for the second part, and add up these quantities. You are told that the time for the first part of the trip is 40 minutes. For the second part of the trip, use the formula: 
The distance for the second part of the trip is 30 miles. The rate for the second part of the trip is twice the rate for the first part. The rate for the first part is 
so the rate for the second part of the trip is 
Plug these values into the formula for time and arrive at 
The time for the second part of the trip is also 40 minutes, so the time for the entire trip is 80 minutes. Now you can plug this value into the rate formula: 
Reduce and arrive at 
2. C Boris’s rate for the entire 24-mile race is 6 miles per hour. Plugging these values into the r × t = d formula, you arrive at
Thus the whole race took him 4 hours. If it took him 1.8 hours to run the first part of the race, then the second part of the race took him 4 – 1.8 = 2 hours.
3. A To get the combined rate of the two hoses, you should add up their rates. 
so the rate of the first hose is 
and the rate of the second hose is 
Add up these rates: 
Since the pool is already half full, the work that the two hoses need to do is only Now plug these values into the r × t = w formula.
4. B Let r be the rate of each machine. If 4 work together, their combined rate is 4r. You are told that these 4 machines will fill the production lot in 3 hours. Plug these values into the r × t = w formula:
The rate of each machine, in production lots/hour, is 
. Now plug this value into the r × t = w formula to determine how many machines, y, would be necessary to fill the lot in 30 minutes. Since your rate is in lots/hour, express 30 minutes as 0.5 hours.
Solve for y:
5. B 
so the rate of the first cook, in batches per minute, is 
, and the rate of the second cook, in batches per minute, is 
To get their combined rate, add up these rates: 
To determine how long it will take the cooks to produce 3 batches, use the r × t = w formula:
Solve for t: t = 20 minutes. The question asks for how many hours it will take, so convert minutes to hours. 20 minutes is of an hour.
6. C The rate at which the tank fills will be the difference between the rate of the hose filling the tank and the rate at which the tank empties: 15 liters/second – 10 liters/second = 5 liters/second. Since the question asks for the time at which the tank will be half full, the work is 100 liters. Now use the r × t = w formula to solve the problem:
Solve for t: t = 20 seconds.
7. B Let s = Sam’s rate. Bob’s rate is thus 2s, and their combined rate is s + 2s = 3s. Now use the r × t = w formula to find the value for s.
Bob’s rate is twice Sam’s rate, so Bob’s rate is 
Now plug this rate into the r × t = w formula.
8. D Use the r × t = w table to determine the number of minutes it will take Machine A to produce 40 more widgets than machine B. Machine A’s rate is widgets/minute = 
and Machine B’s rate is widgets/minute = 
Since the machines start at the same time, use t to represent each of their times. Plug these values into the table.
Represented algebraically, Machine A’s work is thus 
and Machine B’s work is 
You know that Machine A produces 40 more widgets, so 
Now solve for t:
9. B To answer this question, you need to determine the net increase in the population during the year. If 500 organisms were born each day and 300 organisms died each day, then the population increased at a constant rate of 200 organisms/day. For the entire year, the population increase will be 200 organisms/day × 365 days = 73,000. The approximate total population will thus be 50,000 + 73,000 = 123,000.
10. 
Use the r × t = d table. Let j = Jack’s rate and j + 2 = Bob’s rate:
If they end up of a mile apart, then Bob’s distance must be mile more than Jack’s. Expressed algebraically, you arrive at (j + 2)(t) = jt. Solve for t:
11. 12 Use the r × t = w formula, where x is the rate of each scientist:
Thus
Now that you know the rate of each scientist, you need to determine how many scientists are needed to fill 24 beakers in 5 minutes. Let y = the number of scientists. Substitute the given information into the r × t = w table:
1. B Since Sally spends half of the distance of the trip traveling 50 miles per hour, and the other half of the distance of the trip traveling 40 miles per hour, she must spend more time traveling 40 miles per hour than 50 miles per hour. Her average speed must therefore be closer to 40 than to 50. Since 45 is the midpoint of 40 and 50, her average speed must be less than 45.
2. D Dennis cooks cakes at a rate of 
He cooks pies at a rate of 
Use the r × t = w table to arrive at a value for each of the quantities.
If x = 6 and y = 2, then the two quantities are equal. If x = 6 and y = 1, then Quantity A is greater. A relationship cannot be determined.
3. C Jake’s rate is 
Double that rate would be 
. To solve for how long it would take him to run a lap at this rate, use the r × t = d table:
The two quantities are equal.
4. B If Car A is 80 miles from its starting point when the cars meet, then Car B is 120 miles from its starting point. Since the cars traveled for the same time and Car B traveled a greater distance, Car B’s rate must have been greater.
5. D From the given information, you can infer that Car A traveled 25 more miles than Car B. However, Car A could have traveled a total of 26 miles or a total of 31 miles. Thus the relationship between the quantities cannot be determined.
6. D Without knowing the capacity of the tank, you cannot determine the amount of time it will take to fill the tank.
7. C The combined rate of the eight machines is computers/hours 
. Note that a rate of 
reduces to 
Thus the number of machines necessary to produce 21 computers in 18 hours equals the number of computers necessary to produce 7 computers in 6 hours.
Probability refers to the likelihood that a given event will occur. To calculate the probability of a single event, use the following formula:
The formula is illustrated in the following example.
A certain jar has 3 red marbles, 5 black marbles, and 7 white marbles. If a marble is to be selected at random from the jar, what is the probability that a black marble will be selected?
SOLUTION: In the example, an “outcome” is considered the event of selecting a marble. Since there are 15 marbles to select, there are 15 total outcomes. The “desired outcomes” refers to the number of events that satisfy the outcome you want. Since there are 5 black marbles, there are five “desired outcomes.” Using the formula, the probability of selecting a black marble is thus: desired outcomes/total outcomes 
Imagine hearing that there is a 0.6 chance of rain. What would be the chance that it will not rain? 0.4. Why? Because the two events are mutually exclusive. Either it will rain or it won’t, so the probability of the events must add up to 1. This example highlights the following rule: when events are mutually exclusive, the sum of their probabilities is 1.
For this question, write your answer in the box.
A certain event has two possible outcomes, a and b. If the probability of a is 
, and the probability of b is 
what is the value of p?
SOLUTION: Since the event has only two outcomes, the probability of a and the probability of b must add to 1. Thus 
To solve for p, multiply across the equation by 15:
In the previous example, you looked at the probability of a single event occurring (selecting a black marble). Tougher probability questions will concern the probability that multiple events will occur. These questions will take two forms: probability questions with “and” and probability questions with “or.”
If a probability question concerns the likelihood that one event or another event will happen, add up the probabilities of each event. For example:
In a certain bookcase, 10 books are about science, 8 are about literature, 5 are about history, and 2 are about psychology. If a book is selected at random, what is the probability that the book is about science or history?
0.08
b 0.2
c 0.3
d 0.4
e 0.6
SOLUTION: Two different events, a science book or history book, would satisfy the desired outcome. Thus to determine the number of desired outcomes, you should add the number of science books and the number of history books: 10 + 5 = 15. The total number of outcomes is the sum of the books: 10 + 8 + 5 + 2 = 25. The probability of selecting a science book or history book is thus 
The correct answer is E.
If a question concerns the probability that an event occur multiple times, multiply the probabilities of each event.
If a fair-sided coin is flipped three times, what is the probability the coin will land on heads all three times?
0.125
0.25
0.5
0.75
1.5
SOLUTION: On any given flip, the probability that the coin will land on heads is 0.5. Since the question concerns the probability that the coin will land on heads all three times it is flipped, you should multiply these probabilities. 0.5 × 0.5 × 0.5 = 0.125. The correct answer is A.
To calculate the number of ways to combine groups, calculate the product of the number of elements in each group.
Bob’s wardrobe contains 5 shirts, 7 pairs of pants, and 4 ties. If he wants to select an outfit that consists of 1 shirt, 1 pair of pants, and 1 tie, how many different outfits can he select?
16
20
28
35
140
SOLUTION: Since the question concerns the combination of different groups, find the product of the number of elements in each group. 5 × 7 × 4 = 140. The correct answer is E.
One application of the fundamental counting principle concerns questions that ask you for the number of ways the items in a set can be ordered. These are called permutation questions and usually use the words orderings or arrangements. To answer such questions, you should use the slot method. Let’s use the following example to illustrate how the slot method works:
For this question, write your answer in the box.
How many five-digit locker codes can be created from the digits 0–9, inclusive, if no digit can repeat?
SOLUTION: The order of the digits is relevant to the question, so consider it a permutation and use the slot method.
Step 1: Set up a number of slots corresponding to the number of items that are being selected. Since the code is five digits, you will set up five slots:
___ ___ ___ ___ ___
Step 2: Starting with the first slot, put in the number of possible choices for each slot. Since you are selecting from 10 digits, there are 10 possibilities for the first slot. Since the digits cannot repeat, whichever number occupies the first slot cannot occupy the second slot. Thus there are 9 possibilities for the second slot. By the same reasoning, there are 8 possibilities for the third slot, 7 possibilities for the fourth slot, and 6 possibilities for the fifth slot.
The slots should thus look like the following:
Step 3: Multiply across.
1. A restaurant offers a special that consists of an entrée, a main course, and a dessert. If customers can choose from 4 entrées, 5 main courses, and 3 desserts, how many different meals can be chosen?
3
12
20
40
60
For this question, write your answer in the box.
2. S is the set of all positive integers from 1–100, inclusive. If a number is to be randomly drawn from S, what is the probability that the number chosen will be a multiple of 4?
3. Next week, the probability of rain on each day will be 0.6. What is the probability that it will rain on Monday, but not on Tuesday?
0.1
0.16
0.24
0.36
0.42
4. A fair, six-sided die is to be rolled 3 times. What is the probability that the die will land on a prime number each time?
0.125
0.25
0.5
0.75
0.9
5. 20 of the 80 employees at a certain organization are doctors. If two employees are to be selected at random, which of the following is closest to the probability that neither employee selected is a doctor?
0.2
0.4
0.6
0.7
0.8
6. A certain jar contains 21 marbles, 7 of which are red. If two marbles are to be selected, with replacement, what is the probability that both marbles will be red?
7. How many three-letter arrangements can be selected from the letters ABCDE?
15
20
60
120
240
8. 10 people enter a competition in which the winner earns a gold medal, the second-place finisher earns a silver medal, the third-place finisher earns a bronze medal, and no other awards are given. In how many different ways can the medals be awarded?
512
720
729
800
1,000
9. Set A consists of the integers 2, 3, 5, 7, and 9. Set B consists of the integers 1, 4, 6, 8, and 10. If a number in Set A is to be multiplied by a number in Set B, how many different products are possible?
5
10
15
20
25
Each of the following questions consists of two quantities, Quantity A and Quantity B. You are to compare the two quantities. You may use additional information centered above the two quantities if additional information is given. Choose
if Quantity A is greater
if Quantity B is greater
if the two quantities are equal
if the relationship between the two quantities cannot be determined
The probability that a certain event will occur is x.
The probability that the event will not occur is y.
Y is the set of all integers from 1–10,000 inclusive. A number is to be randomly selected from Y.
A fair-sided coin is to be flipped 5 times.
If a coin is flipped three times, the probability that it will land on heads all three times is 0.064.
1. E Use the fundamental counting principle: 4 × 5 × 3 = 60.
2. First, calculate how many multiples of 4 there are from 1–100, inclusive. Recall that the formula to calculate the number of items in an evenly spaced set is:
Thus the number of multiples of 4 from 1–100, inclusive, is 
There are thus 25 desired outcomes. Since there are 100 total outcomes, the probability of selecting a multiple of 4 is 
3. C Since the question concerns the probability that multiple events will occur, multiply the probabilities of each desired event. The probability of rain on Monday is 0.6. The probability that it won’t rain on Tuesday is 1 – 0.6 = 0.4. Thus the probability that it will rain on Monday, but not on Tuesday, is 0.4 × 0.6 = 0.24.
4. A Since the question concerns the probability that multiple events will occur, multiply the probabilities of each event. Since there are three prime numbers from 1–6 (2, 3, and 5), the probability that the die will land on a prime number on a single roll is 
= 0.5. The probability that the die will land on a prime number on all three rolls is thus 0.5 × 0.5 × 0.5 = 0.125.
5. C Since the question concerns the probability that multiple events will occur, multiply the probabilities of each event. The probability that the first employee selected is not a doctor is 
After the first employee is chosen, there are 79 total employees and 59 total nondoctors left. Thus the probability that the second employee selected will also be a doctor is 
. Since the question says “closest to,” you can estimate as roughly 
The approximate probability that neither employee is a doctor is thus 
The closest answer is C.
6. A Since multiple events must be satisfied to yield a desired outcome, you should multiply the individual probabilities of each desired event. The probability that the first marble chosen is red is 
The probability that the second marble chosen is red is also 
7. C The word arrangement indicates that this is a permutation question and that you should thus use the slot method. You should set up three slots. There are five possibilities for the first slot, four possibilities for the second slot, and three possibilities for the third slot:
Finally, multiply across: 5 × 4 × 3 = 60.
8. B The order in which the awards are given yields different outcomes, so this is a permutation question. Since three awards will be given, set up three slots. There are 10 possibilities for the first slot, 9 possibilities for the second slot, and 8 possibilities for the third slot:
Finally, multiply across: 10 × 9 × 8 = 720.
9. E Use the fundamental counting principle. Multiply the number of items in Set A by the number of items in Set B. 5 × 5 = 25.
1. B Since x and y are mutually exclusive events, their sum = 1. Since x and y are fractions, their product is less than 1. Thus Quantity B is greater.
2. A Remember to compare instead of calculating. Since there are more multiples of 3 than there are multiples of 5, the probability of selecting a multiple of 3 is greater than the probability of selecting a multiple of 5. Quantity A is greater.
3. C Although you may be tempted to figure the math out here, you can solve it much more quickly by recognizing a key point. For at least one coin flip to land on heads, it must be the case that not all of the coin flips land on tails. The values in the two quantities are thus equal.
4. A If x is the probability that the coin will land on heads on one flip, then x3 = 0.064 → x = 0.4. If there is a 0.4 chance the coin will land on heads on a given flip, then there is a 0.6 chance that the coin will land on tails on a given flip. Quantity A is greater.
