Math Workout for the GMAT

Part III Content and Strategy Review

Chapter 9 Assorted Topics 2

PROBABILITY

The probability that a particular event will occur is expressed as a fraction (or a decimal or percent) between 0 and 1. For example, the probability that a flipped coin will come up heads is , and the probability that it will rain on a certain day might be 30%. However, it wouldn’t make sense to say that the probability of something was more than 1 (absolutely certain to occur) or less than 0 (no chance of occurring). For most GMAT problems involving probability, you should use the following formula:

To calculate the probability of an event, take the number of all potential outcomes and put it in the denominator of a fraction. Then take the number of favorable outcomes, those that match the event, and put it in the numerator of the fraction.

For example, suppose you have a box containing 20 chocolates, of which 10 are filled with fudge, 6 with creme, and 4 with caramel. If you choose a chocolate at random and bite into it, there are 20 potential outcomes, one for each chocolate. The probability of choosing one with fudge filling is or . Suppose you like fudge and caramel but hate creme filling. There are 10 + 4 = 14 chocolates that have fudge or caramel, so the probability of getting a chocolate you like is or .

Note that this fraction method applies only when you have an equal chance of getting each of the possible outcomes. In other situations, such as with a lopsided coin, the question will provide you with the necessary probability.

Independent Events

Many problems involve the probability of several events occurring. To find the probability of a series of events, find the probability for each separate event and then multiply them together. Suppose a question asks for the probability of getting heads on three coin flips in a row. The chance of getting heads on each separate coin flip is , so the probability of getting three in a row is × × = .

In the previous example, the events were independent, meaning that the result on one coin flip did not affect the probabilities for other coin flips. When events are independent, you can calculate the probability for each separate event without regard for what happens with other events. Try this next example.

1.Two dice, each with six sides numbered 1 through 6, are tossed. What is the probability that both dice will come up with either a 1 or a 2 ?

The probability that the first die will result in a 1 or a 2 is 2 out of 6, or = . The probability is the same for the second die. To get the probability that both events will occur, just multiply the two probabilities: × = . So the answer is (C).

Dependent Events

In some cases, a series of events are dependent. The result of one event affects the probabilities for later events. This often happens if things are used up during some process. For dependent events, always assume that earlier events had the desired result when you determine the probability for each later event.

Suppose there are 4 marbles in a jar, 3 blue and 1 red. What is the probability of drawing 2 blue marbles from the jar? Well, the probability that the first marble is blue is . However, there are now only 3 marbles left, 2 blue and 1 red (assume that you successfully picked a blue marble the first time). The probability that the second marble is blue is , so the probability of picking 2 blue marbles is × = . Now try the following example.

2.Frank has a box containing 6 doughnuts: 3 jellies and 3 cinnamon twists. If Frank eats 3 doughnuts, chosen randomly from the box, what is the probability that he eats 3 jelly doughnuts?

The probability that the first doughnut is a jelly is . Assuming he gets a jelly, the probability that the second doughnut is a jelly is , because there are 2 jellies left out of 5 doughnuts. Assuming he got 2 jellies already, the probability that the third is a jelly is , because there is 1 jelly left out of 4 doughnuts. To find the probability that all three events happened, multiply the probabilities: × × = . The answer is (D).

“Not” Probability

Sometimes it’s easier to calculate the probability that something will not happen than to calculate the probability that it will happen. If the probability that something will happen is x, then the probability that it won’t happen is 1 – x. For example, suppose there is a 40% chance that it will rain tomorrow. That means there is a 1 − 0.4 = 0.6 or 60% chance that it will not rain tomorrow. Suppose you need to find the probability that you will get at least 1 head when you flip a coin 5 times. The hard way to do it would be to find the probabilities of getting exactly 1 head, exactly 2 heads, exactly 3 heads, exactly 4 heads, and exactly 5 heads, and then add them all up. The easy way is to find the probability that you do not get at least 1 head. That means you get 5 tails. The probability of that is simply × × × × = . So the probability that you will get at least 1 head is 1 − = . Try the following example.

3.A six-sided die, with sides numbered 1 through 6, is rolled three times. What is the probability that the sum of the three rolls is at least 4 ?

This problem is easier if you start with the probability that the sum of the rolls is less than 4. The only way that can happen is to get a 1 on each of the three rolls. The probability of getting a 1 is , so the probability of getting a 1 three times is × × = . The probability that you will get at least 4 is 1 − = . The correct answer is (E).

FUNCTIONS

Function problems usually contain some bizarre symbols or strange terms. A question might test a function such as x # y = x² + y² + 2xy or use an odd term such as hyper-prime. However, a function problem is really just an exercise in following directions. The question must define the function for you, either as a formula or as a description of how to manipulate the numbers. Just plug the numbers you’re given into that definition. Here’s an example.

1.If a @ b = 3a² + 2b − 1, then 2 @ 3 =

Just plug in a = 2 and b = 3 into the formula the question provides. You get 2 @ 3 = 3(2)² + 2(3) − 1 = 12 + 6 − 1 = 17. The answer must be (B).

Sometimes the GMAT indicates a function by using strange terms rather than a symbol. You may even see something that has a real-life use, such as the net present value of an annuity or some other financial function. Don’t worry. The question will provide the directions, either as a formula or a step-by-step description. All you have to do is plug in the numbers and do the calculation.

RATES

A number of GMAT questions deal with how fast someone can complete a task (for example, painting a wall or building widgets) or how fast someone travels. These problems revolve around that person’s rate. The formula for rate can be expressed a couple of ways.

Just fill in the numbers the question provides and solve for the other one. Look at this example.

1.If Albert can travel 200 miles in 4 hours, how many hours will it take Albert, traveling at the same constant rate, to travel 350 miles?

The key number is usually the rate. Find it if the question doesn’t state it. Albert’s rate (or speed) is = 50 per hour. He needs to travel 350 miles. Plug those numbers into the formula to get 350 = 50 × Time, which is solved as Time = = 7 hours. The answer is (C).

If several people are working together, just add their individual rates to find their combined rate. For example, if Joe can paint 10 square feet per hour and Jack can paint 8 square feet per hour, then they paint 18 square feet per hour when they work together. This principle also applies if two people are traveling toward each other. Add their rates to find how quickly they close the gap. Suppose Alice and Brenda are 300 miles apart and begin driving toward each other. If Alice drives at a speed of 40 miles per hour and Brenda drives at a speed of 60 miles per hour, they get closer at a rate of 40 + 60 = 100 miles per hour. So they will meet in = 3 hours. Try the following example:

2.Mark can process 30 insurance claims per hour. Bruce can process 15 claims per hour. Mark starts working on a batch of 555 insurance claims. Two hours later, Bruce begins working with Mark until the batch is finished. How many hours did Mark spend working on the batch?

If Mark works alone for 2 hours, he processes 2 × 30 = 60 claims, leaving 555 − 60 = 495 claims to go. When Bruce joins Mark, they work at the combined rate of 30 + 15 = 45 claims per hour, so they need = 11 hours to finish the batch. So Mark spends a total of 2 (alone) + 11 (with Bruce) = 13 hours working on the batch. The correct answer is (C).

If the task or the distance is undefined, you may want to use the Plugging In technique. Make up a number for the amount of work or the distance and solve the problem from there. Try the following example:

3.Working alone, Bud can complete a particular task in 6 hours. Lou, working alone, can complete the same task in 8 hours. If Bud and Lou work together, how many hours will it take them to complete the task?

The task is undefined, so make up a number for it. Suppose the task is 48 units (because it works well with 6 and 8). Bud can complete the task in 6 hours, so his individual rate is = 8 units per hour. Lou’s rate is = 6 units per hour. Combined, they work at a rate of 8 + 6 = 14 units per hour. To complete 48 units, they will need = = 3 hours. The correct answer is (D). Note that the answer is not = 7 hours or × 7 = 3 hours. Don’t fall for those obvious answer traps. Use the formula and work it out.

DIGIT PROBLEMS

In digit problems, one digit of a number is replaced by a symbol or letter. For example, the problem may contain a number such as 11,8#3, where # represents the missing digit. Usually, you need to find the value or the range of possible values for the missing digit. Remember that there are only 10 possible values for a single digit: 0 through 9.

You can usually solve a digit problem by applying the Plugging In The Answer technique. Try the numbers from the answer choices until you find one that fits. Look at the following example:

1.In the multiplication above, # represents a single digit. What digit does # represent?

Just try out the different answers until you find one that fits. If you plug in # = 2, the product is 1,020, which doesn’t fit. With # = 3, the product is 1,530. That doesn’t fit either. Plug in # = 4 and the product is 2,040. That doesn’t work. With # = 5, the product is 2,550. That works, so the answer is (D).

COMBINATIONS AND PERMUTATIONS

Problems involving combinations and permutations deal with the number of different ways to arrange a group of things or the number of different selections you can make from a group of things. One critical distinction to make is between problems in which the order of the things doesmatter and problems in which the order of the things does not matter.

Suppose you’re scheduling speakers for a seminar. You have three people—A, B, and C—available but you need only two different people, one for the morning and one for the afternoon. Your choices are A then B, A then C, B then A, B then C, C then A, and C then B. That’s a total of six different pairs. Notice that order does matter. You want to count “A then B” and “B then A” as two different pairs.

Now suppose that you’re choosing fruits to make a health shake. You need two different fruits and you have three from which to choose—A, B, and C. You could choose A and B, A and C, or B and C. That’s only three different pairs (not pears). You don’t count “A and B” and “B and A” as two different pairs because they’re just going into the blender together anyway.

When order matters, the different arrangements are called permutations. When order doesn’t matter, the different arrangements are called combinations. The names aren’t that important, but you do need to understand the difference and be able to identify which one the question wants.

There are three basic patterns you will see on the GMAT. You may also see questions that mix these three patterns together, but the key is to understand how to handle each of the three basic models: combination from different sources, permutation from the same source, and combination from the same source.

Combination (Different Sources)

When you have a single decision to make, the number of possible outcomes is easy to determine: It’s simply the number of options for that one decision. For example, if you need to choose a shirt to wear and you have 12 different shirts, then there are 12 possible outcomes.

When you have more than one decision to make, you find the number of possible outcomes by multiplying the numbers of options for the decisions together. Suppose you need to choose a shirt and a pair of pants, and you have 12 different shirts and 5 different pairs of pants. Multiply the number of options for shirts and the number of options for pants to get 12 × 5 = 60 possible outcomes. Try this example.

1.Nick is enrolling in classes for next quarter. He must choose a math class, a language class, a science class, and a humanities class. If there are 3 math classes, 2 language classes, 1 science class, and 5 humanities classes available, how many different groups of classes could Nick choose?

Nick has to make four decisions—one for each type of class—so multiply those four numbers together: 3 × 2 × 1 × 5 = 30. The correct answer is (E).

The previous example involves choosing items from different sources. The math class Nick chooses must come from the math list, not the language list or one of the others. With these different source questions, just multiply the number of options for each decision. The result is the number of possible combinations. Nick didn’t have to decide the order of the classes, just which ones would be included in the group.

Permutation (Same Source)

In some problems, you will need to choose several items from the same list. For example, you may need to choose 3 books to read from a list of 6 books and determine in what order to read them. For each book that you must select, you are making a decision. However, the number of options changes with each decision because you have already used up some of the possible selections. When selecting the first book, you have 6 options. However, when you select the second book, you’ve already crossed one off the list, so there are only 5 options left. Then there are 4 options for the third book. So, the number of possible outcomes is 6 × 5 × 4 = 120 possible reading lists.

Note that this process results in the number of permutations, not combinations. If the six books are A, B, C, D, E, and F, you are counting ABC as separate from CAB. That’s what you want if you must determine the order in which you will read the books.

Try this next example.

2.Mandy is judging a talent contest involving 5 dogs. She must select the top 3 dogs and award a blue ribbon to the most talented dog, a red ribbon to the second-most talented, and a white ribbon to the third-most talented. How many different arrangements of ribbon-winners are possible?

This is a permutation problem because the order of the dogs matters. Awarding A the blue ribbon, B the red, and C the white is different from awarding B the blue, A the red, and C the white. Mandy makes three selections: the first dog, the second dog, and the third dog. In selecting the first dog, Mandy can choose any of the 5 dogs. In selecting the second, she has only 4 dogs left, and 3 dogs for the third choice. Thus, she has 5 × 4 × 3 = 60 possible permutations of ribbon-winners. The correct answer is choice (E).

So far, the method for the calculation of combinations from different sources and the method for the calculation of permutations from the same source are very similar—just multiply the number of options for the choices together. When you learn to find combinations from the same source, it gets a bit trickier. It’s important to correctly identify which type of problem you’re dealing with, so you can use the proper method for calculating the answer.

Combination (Same Source)

Suppose Mandy didn’t need to determine the ranking among the top 3 dogs, but rather merely had to choose the best 3 to receive doggie treats. She is still selecting 3 from a list of 5, but now the order of the 3 doesn’t matter. This is a combination chosen from the same source.

The number of combinations will always be smaller than the number of permutations in selecting a certain number of items from some list. Think of the process of selecting a permutation as first choosing one of many possible groups and then putting that particular group in order. With a combination, you would stop after selecting the group. The number of permutations is the number of combinations multiplied by the number of ways to arrange a particular group.

Suppose you must choose a group of songs from a list and then choose the order in which to play those songs. Say there are 10 possible groups of songs that you could choose. If there are 6 ways to arrange a given group of songs, you will end up with 10 × 6 = 60 possible arrangements, or permutations.

To find the number of combinations, you will work backward through the above process. First, pretend you are finding the number of permutations. Then, divide that number by the number of ways to arrange a particular group. This process of “factoring out the order” will leave you with the number of possible groups, or combinations. See how this method works on the following problem.

3.Eric is ordering a pizza. He wants three different toppings on his pizza, and the available toppings are pepperoni, sausage, black olives, green peppers, and mushrooms. How many different combinations of toppings are available?

This is a combination problem, because Eric doesn’t care about the order in which the toppings are selected, just which ones are in the group selected. To find a combination, start by finding the permutation. If he were selecting them in order, he would have 5 choices for the first topping, 4 for the second, and 3 for the third. That gives you 5 × 4 × 3 = 60 possible permutations. Of course, that’s not the number you want, because it treats pepperoni-sausage-mushrooms as different from mushrooms-sausage-pepperoni.

Next, determine how many ways a group of three toppings can be arranged. This is equivalent to choosing 3 from a list of 3. There are 3 options for the first choice, 2 for the second, and only 1 for the third, so 3 toppings can be arranged in 3 × 2 × 1 = 6 different ways.

Finding the combination is the same as taking the permutation number, 60, and factoring out the number of ways to order a given group, 6. So the number of combinations is = = 10. The correct answer is choice (B).

MEDIAN, MODE, AND RANGE

In Chapter 4, you learned how to find the average of a group of numbers. The GMAT also tests some close cousins of the average, namely the median, mode, and range of a set of numbers.

To calculate the median of a group of numbers, first arrange the numbers in order, either ascending or descending. The median is simply the middle number. If there is an even number of things in the group, there won’t be a single middle number. In that case, the median is the average of the two middlemost numbers.

The mode of a group of numbers is the number that appears the most times. If every number is different, there is no mode.

The range of a set of numbers is the difference between the greatest number in the set and the least number in the set. Just subtract.

Suppose you have this set of numbers: 1, 5, −5, 3, 1, 2, and 8. To find the median, arrange them in order: −5, 1, 1, 2, 3, 5, and 8. The number in the middle is 2, so the median is 2. The mode in this set is 1 because it shows up twice while each of the other numbers shows up once. The range of the set is 8 – (−5) = 13.

Comprehensive Assorted Topics 2 Drill

Answers can be found on this page.

Remember!

For Data Sufficiency problems in this book, we do not supply the answer choices. The five possible answer choices are the same every time.

1.Alex drives to and from work each day along the same route. If he drives at a speed of 80 miles per hour on the way to work and he drives at a speed of 100 miles per hour on the way from work, which of the following most closely approximates his average speed in miles per hour for the round trip?

2.When a certain coin is flipped, the probability that it will land on heads is and the probability that it will land on tails is . If the coin is flipped three times, what is the probability that all three results are the same?

3.Joe is choosing books at the bookstore. He has a list of 7 books that he would like to buy, but he can afford to buy only 3 books. How many different groups of books could Joe buy?

4.In the addition above, #, @, and & each represent a distinct, positive digit. If & is even, what is the value of # ?

(1) # and @ are even and & = 6

(2) # < @

5.If Max can complete a job in 4 hours and Nick can complete the same job in 6 hours, how many fewer hours do Max and Nick working together need to complete the job than Max alone needs to complete the job?

6.If a set of numbers consists of 10, 15, 0, 3, and x, and the range of the set is 30, what are the possible values for the median of the set?

7.If the area of a number is defined as the difference between that number’s greatest and least prime factors, what is the area of 100 ?

8.Carol is having friends over to watch home movies. She has 6 reels of home movies, but she and her friends have time to watch only 3 of them. Carol must decide which reels to watch and in what order. How many different orderings does Carol have from which to choose?

9.A jar contains 5 marbles: 3 red and 2 blue. If two marbles are drawn randomly from the jar, what is the probability that they will be different colors?

10.Oscar is running in a straight line away from Nancy at the rate of 20 feet per second. Nancy is chasing Oscar at the rate of 25 feet per second. If Oscar has a 100-foot head start, how long, in seconds, will it take Nancy to catch Oscar?

11.A cupboard holds 10 cans, of which 3 contain pumpkin puree. If a group of 4 cans is randomly selected from the cupboard, what is the probability that the group includes the 3 cans containing pumpkin puree?

12.Mo is planning a dinner party and must create a menu consisting of 1 soup, 1 entree, 3 different side dishes, and 2 different desserts. If Mo has 3 possible choices for the soup, 2 for the entree, 5 for the side dishes, and 3 for the desserts, how many different menus can Mo create?

13.From a group of 10 students, 7 girls and 3 boys, a teacher must choose 2 girls and 2 boys to present book reports. How many different arrangements of students, in order, are possible?

14.Each candle in a particular box is either round or square and either scented or unscented. If 60% of the candles are round, what is the probability that a candle selected randomly from the box will be unscented?

(1) If a candle is scented, it has an 80% chance of being round.

(2) If a candle is square, it has a 25% chance of being scented.

15.The greatest number in a set of 4 numbers is 70. What is the average (arithmetic mean) of the set?

(1) The median of the set is 25.

(2) The range of the set is 70.

16.Anthony, Brad, and Cris are inviting 2 friends each to a party. If Anthony has 4 friends from which to choose, Brad has 2, and Cris has 5, how many different groups of 6 invitees are possible?

17.Ric begins walking up a mountain trail, ascending at a constant rate of 200 feet per hour. Sixty minutes later, Josie begins walking down the same trail, starting at a point 1,700 feet higher than Ric’s starting point. If Josie descends at a constant rate of 300 feet per hour, how many feet will Ric have ascended when the two meet?

18.If x & y = x² − 2y, what is the value of a & 2 ?

(1) 3 & a = 13

(2) The value of a is either 2 or −2.

19.Jack is making a list of his 5 favorite cities. He will choose 3 cities in the United States from a list of 5 candidates. He will choose 2 cities in Europe from a list of 3 candidates. How many different lists of cities, ranked from first to fifth, can Jack make?

Challenge!

Take a crack at this high-level GMAT question.

20.If f(n) = xaⁿ, what is the value of f(5) ?

(1) f(0) = 3

(2) f(4) = 48

ANSWERS AND EXPLANATIONS

Comprehensive Assorted Topics 2 Drill

1.BThe distance is undefined, so make one up. Suppose his trip is 400 miles each way. On the way to work, he travels at 80 miles per hour and completes the trip in = 5 hours. On the way from work, he travels 100 miles per hour and completes the trip in = 4 hours. For the round trip, he travels 400 + 400 = 800 miles in 5 + 4 hours. That’s an average speed of = 88.89 miles per hour. (B) is the correct answer.

2.BIt doesn’t really matter which way the first flip ends up, so don’t count that probability. The probability that the second flip is the same as the first is . The probability that the third flip matches is also . So the probability that the last two flips match the first is × = . Choose (B).

3.DThis is a single-source combination problem. For the first of his 3 chosen books, Joe can pick from 7 titles; he has 6 choices for the second book, and 5 choices for the third book. So multiply: 7 × 6 × 5 = 210. But wait, you’re not done yet! Don’t forget to divide 210 by the number of ways you can arrange the group of 3 books he selected: 3 ways for his first choice, 2 ways for his second choice, and 1 way for his third choice, or 3 × 2 × 1 = 6. The correct answer is 35, so choose (D).

4.CLook at Statement (1). If both # and @ are even, positive digits that add up to 6, then one of them is 2 and one of them is 4. However, you can’t tell which one is which. You can’t answer the question, so narrow the choices to (B), (C), and (E). Look at Statement (2). This isn’t much help by itself. There are several possible values for #, @, and &. You can’t answer the question, so eliminate (B). Try Statements (1) and (2) together. From Statement (1), you know that # and @ are 2 and 4. From Statement (2), you know that # is the smaller of the two. So # = 2 and @ = 4. You can answer the question, so choose (C).

5.AThe task is undefined, so make up a value, say 48 units. Max can complete 48 units in 4 hours, so his rate is = 12 units per hour. Nick can complete 48 units in 6 hours, so his rate is = 8 units per hour. Combined, Max and Nick work at the rate of 12 + 8 = 20 units per hour and can complete the job in = 2.4 hours. That’s 4 − 2.4 = 1.6 fewer hours than Max alone needs. Choose (A).

6.EFor the range to be 30, either x is the highest number and x = 30 or x is the lowest number and x = −15. If x = 30, the set is 0, 3, 10, 15, and 30, so the median would be 10. If x = −15, the set is −15, 0, 3, 10, 15, so the median would be 3. So the possible values for the median are 3 and 10. Choose (E).

7.CThe prime factorization of 100 is 2 × 2 × 5 × 5. So the greatest prime factor is 5 and the least is 2. That’s a difference of 5 − 2 = 3. Choose (C).

8.AOrder is important, so set up a permutation (same source). Carol must choose 3 reels. She has 6 options for the first reel, then 5 options for the second, and 4 options for the third. That gives you 6 × 5 × 4 = 120 possible arrangements (or permutations).

9.EThe probability that the first marble will be red is . Assuming the first one is red, the probability that the second marble will be blue is because there are 4 marbles left, 2 of which are blue. So the probability of getting red then blue is × = . However, getting blue then red would also be acceptable. The probability that the first marble is blue is . Assuming that the first marble is blue, the probability that the second marble is red is , because 3 of the remaining 4 marbles are red. So the probability of getting blue then red is × = . Because either of these patterns is acceptable, add the probabilities to find the probability that the two marbles are different colors: + = = . Choose (E).

10.DIf Oscar is traveling 20 feet per second and Nancy is traveling 25 feet per second, then Nancy is gaining on Oscar at the rate of 25 − 20 = 5 feet per second. So it will take her = 20 seconds to close the 100-foot gap. Choose (D).

11.BThere are 4 ways to select 3 pumpkins and 1 other: PPPO, PPOP, POPP, and OPPP. For PPPO, the probability is × × × = = . For PPOP, the probability is × × × = . For POPP, the probability is × × × = . For OPPP, the probability is × × × = . Since any of the four ways is acceptable, the overall probability is + + + = = . Choose (B).

12.BBreak this complex problem into several sub-problems. Mo has 3 possibilities for the soup. She has 2 possibilities for the entree. She needs to choose a group of 3 side dishes from a list of 5. That’s a combination, so she has = 10 possible groups of side dishes. She will choose 2 of 3 possible desserts. That’s another combination, so she has = 3 possible groups of desserts. Soups, entrees, side dishes, and desserts are a different source problem, so multiply her options together: 3 × 2 × 10 × 3 = 180 possible menus. Choose (B).

13.DBreak this complex problem into several sub-problems. Choosing 2 boys from a list of 3 is a combination. The teacher has = 3 possible pairs of boys. Choosing 2 girls from a list of 7 is another combination. The teacher has = 21 possible pairs of girls. Choosing a pair of boys and a pair of girls is a different source problem. The teacher has 3 × 21 = 63 possible combinations. However, the question asks for an arrangement, so you must figure out how many ways each group could be arranged. Arranging 4 students from a list of 4 gives you 4 × 3 × 2 × 1 = 24 ways to arrange each possible group, or 24 × 63 = 1,512 possible arrangements. Choose (D).

14.CYou know that 60% of the candles are round, so 40% are square. With Statement (1), you know that 80% of the scented candles are round, so 20% are square. However, you can’t determine the split between scented and unscented, so you can’t answer the question. Narrow the choices to (B), (C), and (E). With Statement (2), you learn that 25% of square candles are scented, so 75% are unscented. That means 40% × 75% = 30% of all candles are square and unscented and 40% × 25% = 10% of all candles are square and scented. However, you don’t know how the round candles are split between scented and unscented. Eliminate (B). With both statements together, you know that 10% of all candles are square and scented. Those represent 20% of scented candles, so solve the equation 20% × s% = 10% to get s = 50% of all candles are scented, so 50% are unscented. That tells you the probability, so choose (C).

15.CStart by seeing which pieces of the puzzle you need. You know there are four numbers, so you need the total in order to find the average. With Statement (1), you can determine that the middle two numbers add up to 50, because the median of an even set is the average of the two middle numbers. You also know that the biggest number is 70; however, you’re missing information about the smallest number. Statement (1) is insufficient, so narrow the choices to (B), (C), and (E). With Statement (2), you can tell that the smallest number must be 0, since the range is equal to the largest number minus the smallest number. With Statement (2) alone, however, you don’t know anything about the middle numbers. Eliminate (B). With both statements, you know that the largest number is 70, the smallest is 0, and the two in the middle sum to 50. Although you don’t know the middle numbers individually, you know the total of all four numbers is 0 + 50 + 70 = 120, so the average must be 30. Choose (C).

16.BBreak this problem into pieces. Anthony must choose a group of 2 friends from a list of 4. That’s a combination, so he has = 6 possible groups. Brad has = 1 possible group of invitees. Cris has = 10 possible groups. Combining their lists is a different source problem, so there are 6 × 1 × 10 = 60 possible groups of invitees. Choose (B).

17.CBreak this problem into two phases: the first when Ric is the only walker and the second when both are walking. The first phase is 1 hour, so Ric travels 200 × 1 = 200 feet up. Thus, he is now 1,700 − 200 = 1,500 feet below Josie. When they are walking together, they travel at a combined rate of 200 + 300 = 500 feet per hour, because after each hour they are 500 feet closer together. At this rate, it takes them = 3 hours to meet. Ric has walked a total of 4 hours, so he has traveled 200 × 4 = 800 feet. Choose (C).

18.DThe missing piece of the puzzle is the value of a to plug into the formula for the function. With Statement (1), you can set up the equation 3² − 2a = 13, which you can solve for a = −2. That’s sufficient, because you could plug that into the formula to find a & 2. Narrow the choices to (A) and (D). With Statement (2), you have two possible values for a. However, since the formula tells you to square a, it doesn’t matter whether it is positive or negative. Thus both values of a lead to the same value for a & 2 = 2² − 2(2) = 0. You can answer the question, so choose (D).

19.DBreak this problem into sub-problems. Jack must choose a group of 3 U.S. cities and a group of 2 European cities, and then place them in order. Choosing 3 U.S. cities from a list of 5 is a combination; he has = 10 possible groups of U.S. cities. Choosing 2 European cities from a list of 3 is also a combination; he has = 3 possible groups of European cities. Combining the U.S. and European lists is a different source problem; he has 10 × 3 = 30 possible groups of 5 cities. For each group, there are 5 × 4 × 3 × 2 × 1 = 120 possible ways to rank them. So there are 30 × 120 = 3,600 possible arrangements. Choose (D).

20.EFirst evaluate Statement (1). If you plug 0 into the function, you get f(0) = xa⁰, or f(0) = x. Since you also know from the statement that f(0) = 3, you now know that x = 3. However, without knowledge of a, you cannot solve. So, narrow your answers to (B), (C), and (E). Statement (2) by itself does not allow you to solve for anything since plugging it into the function yields f(4) = xa⁴ = 48. You do not know the x or a values. Eliminate choice (B). Now, try the statements together. From Statement (1), x = 3, so Statement (2) can now be read as f(4) = 3a⁴ = 48. At first glance, this looks sufficient to solve for a, but you must consider that a could be either negative or positive. Therefore, the combination is not sufficient. Choose (E).