1. For which of the following sets does the mean equal the median?

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1 1. For which of the following sets does the mean equal the median? I. {1, 2, 3, 4, 5} II. {3, 9, 6, 15, 12} III. {13, 7, 1, 11, 9, 19} A. I only B. I and II C. I and III D. I, II, and III E. None of the above Solution: D. The key here is to arrange the sets in ascending order; set I is done that way for you, but once you rearrange sets II and III you ll see that set II is also evenly spaced (consecutive multiples of 3), and that set III is evenly distributed about the mean/median (1 and 19 are 9 away from the median of 10; 7 and 13 are each 3 away; and 9 and 11 are each 1 away, meaning that they ll average out to 10, too). 2. For her customer feedback score, Jess wants to average at or above an 8.0 rating. For the month of June, she received the following scores: 7, 8, 8, 7, 9, 10, 6, 6, 8 and 7. By what percent did she fall short of her goal? (A) 4% (B) 5% (C) 8% (D) 10% (E) 12% Solution: B. Since Average = Sum of Terms / Number of Terms, a shortcut for average problems like this is to calculate the distance from the desired average, rather than adding up all the terms themselves. Meaning that, for example, 7 earns a -1 value because it s one less than the desired average. This way you can typically do the work in your head, or at least with smaller numbers on paper. Here, that would make the list: -1, 0, 0, -1, 1, 2, -2, -2, 0, and -1, which nets out to four points short of her goal. And since she wanted 8 points average across 10, that means she wanted a total of 80 points and missed by 4, so she was 5% short.

2 3. Rank the following sets in order of their standard deviations, from lowest to highest: Set A: {-10; -7; -4; -1; 2} Set B: {10,000; 10,000; 10,000; 10,001; 10,0002} Set C: {1; 2; 3; 4; 5} (A) Set A, Set B, Set C (B) Set C, Set B, Set A (C) Set B, Set A, Set C (D) Set B, Set C, Set A (E) Set C, Set A, Set B Solution: D. For most standard deviation questions, you ll rely almost exclusively on a conceptual understanding that standard deviation means how far do the data points usually fall from the mean. In Set B, three points are the mean itself (so no deviation) and the others only differ by 1, so they re extremely compact. In set C, 3 is the mean and the distances from that are 2, 1, 0, 1, and 2, so that s not quite as compact a set as B, but still relatively compact. And in Set A, you should see that the points are spread farther than just a couple points away, so without even doing any math you should see that it s the most-spread. So the answer has to be D. 4. Which of the following could be the mode of a set with values: {-2, 4, 7, 13, J}? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5

3 Solution: D. Since mode means most frequently occurring, the mode here has to match one of the other known values (otherwise there wouldn t be just one mode). Only 4 matches a value from the set, so the answer must be Which of the following could be the range of a set with values: : {-2, 4, 7, 13, J}? (A) -2 (B) 6 (C) 9 (D) 14 (E) 20 Solution: E. Since the range is defined as highest value minus lowest value, J won t change the range unless it s either the highest or the lowest if it s in between -2 and 13, then the range is just 13 (-2) = 15. The only value that s either 15 or greater is 20, so the answer must be If y represents the median of set M, what is y if set M consists of {-11, 15, -7, 2, 7, y}? (A) 1 (B) 2 (C) 7 (D) 10 (E) It cannot be determined Solution: B. 7. From the following list of five question stems, how many are permutations questions and how many are combinations questions? I. How many ways can ten students arrange themselves in a row of ten seats? II. How many groups of students can attend a conference if there are ten students and four passes to attend the conference? III. How many subcommittees of four are possible from a committee of 12 people?

4 IV. How many ways can Karl arrange his twelve books on a shelf? V. How many seating arrangements are possible for six couples sitting around a circular table (A) 1 permutation, 4 combination (B) 2 permutation, 3 combination (C) 3 permutation, 2 combination (D) 4 permutation, 1 combination (E) All five are permutation questions Solution: C. The first is an arrangement in which order matters, so it s a Permutation. The second is a situation in which order does not matter, so it s a Combination. The third is a situation in which order does not matter, so it s a Combination. The fourth is an arrangement in which order matters so it s a Permutation. And the fifth Is another arrangement in which order matters so it is a Permutation. The answer is thus C. 8. How many three-digit area codes are possible if codes cannot begin with the digits 0 or 1? (A) 80 (B) 100 (C) 336 (D) 800 (E) 1000 Solution: D. Because you cannot use 0 or 1 for the first digit, there are 8 possibilities. Then for the second digit all ten options are on the table, and again all ten are possible for the third. So the answer is (8)(10((10) = What is? (A) 2730 (B) 2735 (C) (D) (E) Solution: A. 15! Is 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1, and 12! replicates all the digits from 12 on down, so this fraction reduces to just 15*14*13. Here s where you can use your Arithmetic skills

5 to shortcut the rest of the problem since 15 is a multiple of 5 and 14 is a multiple of 2, then the answer must end in 0 (it s an even multiple of 5). Only answer choice A fits. 10. A group of six friends has come across three front-row tickets to a concert. How many seating arrangements are possible if the seats are all together in a row? (A) 6 (B) 30 (C) 120 (D) 720 (E) 1000 Solution: C. There are 3 positions, which you can call Left, Middle, and Right for sake of demonstration. For the left seat, there are 6 possibilities; for the middle, you can choose anyone but whomever is sitting in the left seat, so there are 5 possibilities. And for the right seat, you can select anyone but the first two who have already been seated, so there are 4 options. That makes the answer 6*5*4 = A linear strand of lights has seven individual sockets for bulbs. If Lynette has four yellow light bulbs and three blue light bulbs, how many unique displays can she create if she is committed to using all seven bulbs? A) 35 B) 42 C) 120 D) 350 E) 720 Solution: A. The formula for permutations with repeating elements is to account for the repeating elements of 4 and 3, so the fraction is:. Here N = 7 and you need, which reduces to

6 12. Stu has registered for a basketball tournament in which each team consists of four players. How many teams can he create if he s picking his three teammates from a group of 9 friends? (A) 28 (B) 84 (C) 144 (D) 504 (E) 1028 Solution: B. There are 9 candidates for three spots (since Stu is already known to be on the team, he s not in the pool of candidates and his spot is not open, so we don t count him). That makes the formula: = = = At a business school mixer, 30% of the attendees are first-year MBA students. If you pick one attendee at random, what is the probability that you pick someone other than a first-year MBA student? (A) 30% (B) 40% (C) 50% (D) 60% (E) 70% Solution: E. Given that this is a mutually exclusive, complementary situation you can be a first-year MBA or you can not be a first-year MBA student, and everyone is either one or the other. So the probabilities must add to 100%, meaning that the probability of not an first-year MBA is 70%. 14. In a dice game, each player rolls five six-sided dice simultaneously and tallies her score. When Marisa rolls, four of her dice land on the table but one bounces off underneath a chair. Of the four dice on the table, all four display the number 6. What is the probability that, after the die on the floor has been viewed, her score will be less than 30?

7 (A) 1/3 (B) 1/2 (C) 2/3 (D) 5/6 (E) 100% Solution: D. Marisa s current score is 24, meaning that only a six will give her a score of 30 or more. All five other options keep her score less than 30, so 5 out of 6 options means a 5/6 probabiilty. 15. In column A, please select the probability of getting heads, then tails on two flips of a coin. In column B, please select the probability of getting one heads and one tails on two flips of a coin. Column A Column B (A) 25% 25% (B) 25% 50% (C) 50% 25% (D) 50% 50% (E) 50% 75% Solution: B. When flipping two coins there are exactly four sequences that could happen: Heads, Heads; Heads, Tails; Tails, Tails; Tails, Heads. One of those four sequences gives you heads, then tails, but two give you the outcome of one of each ( you could also go tails, then heads). So ¼ of the options satisfy the first condition (25%) and 2/4 of the options satisfy the second (50%). 16. What is the probability of getting a matching pair on two flips of a coin? (A) Less than 25% (B) 25% (C) 50% (D) 75% (E) More than 75% Solution: C. When you re asked about the probability of getting a pair (but not a specific pair like a pair of heads ), the first option doesn t matter it s the probability of matching that one that counts.

8 Whatever you pick on the first flip, there s a 50% chance you ll match it on the second, so the answer is 50%. 17. Drawing from the first 20 positive integers, what is the probability of drawing either a prime number or a multiple of 5? (A) 2/5 (B) 9/20 (C) ½ (D) 11/20 (E) 3/5 Solution: D. From the first 20 integers, there are 4 multiples of 5 (5, 10, 15, and 20) and 8 prime numbers (2, 3, 5, 7, 11, 13, 17, and 19). But notice that you ve counted 5 twice, so you have to subtract one of those, making that 11 affirmative results out of 20, so an 11/20 probability you get what you want. 18. A bowl of chips contains 5 Doritos and 5 Fritos. If you reach into the bowl and pick two chips simultaneously and at random, what is the probability that you will select two Fritos? (A) 1/9 (B) 5/18 (C) 2/9 (D) 4/9 (E) 2/3 Solution: C. Remember there s really no such thing as simultaneous, so you want to calculate the probability of getting two Fritos on two draws. Since there s only one sequence that gets you there (Fritos, then Fritos), the probability is 5/10 * 4/9 = 2/ A jar contains 60 marbles, and each marble is either green, red, or blue. If there are twice as many blue marbles as red marbles and a total of 12 green marbles, what is the probability that Jon would randomly draw a blue marble when selecting one marble at random?

9 (A) 8/15 (B) 1/2 (C) 2/5 (D) 3/10 (E) 1/5 Solution: A. Your first task here is to perform the algebra to see how many of the 60 marbles are blue. Since B + R + 12 = 60, B + R = 48. And since B = 2R, then you can solve for R: 2R + R = 48, so R = 16 and B = 32. If 32 of the 60 marbles are blue, then you can reduce that fraction 32/60 to get 8/ Melissa flips three coins simultaneously. What is the probability that at least one of the coins lands on heads? (A) 1/8 (B) 3/8 (C) ½ (D) 5/8 (E) 7/8 Solution: E. With at least one probability, it s often easier to calculate the probability of none and subtract that from 100%. And there s only one way to not get at least one heads it s tails, tails, tails. The probability of that is ½ * ½ * ½ = 1/8, which means that the probability of everything but tails, tails, tails (which is at least one heads) is 7/8.