Word translations guide by Horacio Quiroga

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GMAT Word Translations Study Guide • • • • • • •

Combinatorics Probability Mixtures Rate (Motion) Rate (Work) Statistics Overlapping Sets

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December 29, 2012 Dear Students, Thank you for buying my Word Translations guide. In my opinion, it is second to none in the market. I am convinced of the need the market has for a good guide that covers all the types of Word Translations problems likely to appear in the GMAT. I believe that most guides in the market either do not cover the real types of problems that appear in the GMAT or explain those problems in an unclear way, or so I have been told by thousands (yes, thousands) of students I talked to during my years at ManhattanGMAT, a respectable company that in no way endorses this book or is affiliated with me. I am convinced that if you master all the problems in this guide you will have no problem with any Word Translations problem in the GMAT. I would love your feedback on the contents of this guide; specifically I would like to know whether my explanations and tips were of use to you. If you have any comments, you can contact me directly at: horacio.quiroga@gmail.com and at my YouTube channel: horsegmat Best of luck with the GMAT!! Sincerely, HoracioQuiroga

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About the author: Horacio has been teaching the GMAT since 2001; he has taught the test in four countries (the U.S, Argentina, Brazil, Kuwait, and Uruguay) in three languages: English, Spanish, and Brazilian Portuguese. Took the real GMAT nearly 20 times, literally, and scored several times into the 99th percentile (780 being his highest score). He taught the GMAT at ManhattanGMAT from 2004 to 2012 where he tutored one-on-one over 100 students and made around 10,000 (yes, ten thousand) half-an-hour long office hours calls from mid 2007 to early 2012, getting an average rating of “outstanding” (the top rating a tutor can get in ManhattanGMAT's very demanding scoring system) from his students; proof of this is the fact that he was called (in the introduction to MG’s Official Guide Companion) “our office hours ace” by Andrew Yang, a former ManhattanGMAT CEO. He is the author of “Horacio’s Hot List” of the same book. Horacio was also called "one of our most beloved 99th percentile scoring instructors" in the introduction to one article targeted to non-native speakers of English that he wrote for one of ManhattanGMAT's books: GMAT Roadmap. He wrote hundreds of GMAT quant problems for ManhattanGMAT, a lot of which appear in MG’s CAT exams and question banks. He plans to bring all his experience into the writing of a series of GMAT books. ManhattanGMAT does not endorse, nor is affiliated in any way with the owner or any content of this book. Horacio holds an MBA in Finance from Tulane University (turned down offers from two top ten schools because Tulane offered a full merit-based international scholarship and was very happy when he realized that Tulane’s business school is an excellent one in spite of its relatively low ranking). He also holds a Tulane MA in Political Science / Latin American Studies, which he took on a full, merit-based scholarship as well. He has been married to Alejandra for 17 years and they have a 16 year-old son, Rodrigo. Horacio is a great chess and backgammon player as well as a very mediocre golfer.

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Very important note to our readers:

Do read the explanations to our problems, even if you get the problems right, because it is in our explanations where you will be getting value for your money, they are full of tips and insights.

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TABLE OF CONTENTS Chapter Content 1

Page

Combinatorics o 1.1 Combinations o 1.2 Slots Problems o 1.3 Miscellaneous

6 7 13 17

Probability o 2.1 "Pure" Probability o 2.2 Probability Meets Combinatorics

20 21 27

3 4 5 6 7

Mixtures Rate (Work) Rate (Motion) Overlapping Sets Statistics Short Answers Explanations to Combinatorics 1.1 Explanations to Combinatorics 1.2 Explanations to Combinatorics 1.3 Explanations to "Pure" Probability 2.1 Explanations to Probability Meets Combinatorics 2.2 Explanations to Mixtures Explanations to Rates (work) Explanations to Rates (motion) Explanations to Overlapping Sets Explanations to Statistics

30 37 45 53 60 65 66 70 72 73 76 78 81 84 87 95

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Chapter 1 Combinatorics 1.1 - Combinations 1.2 - Slots 1.3 - Miscellaneous

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1.1 Combinatorics - Combinations We use combinations when we are picking smaller groups out of a larger group and order does not matter. Of course, there can be some constraints, but the principle remains always the same, we pick a smaller group out of a larger group and order does not matter. For example: How many groups of 4 people can be picked out of 7 people? See that here order does not matter at all; the only important thing is who goes into the group and who does not. The formula to answer the problem is extremely simple:

! !

= 35, where the 7

factorial on top (7 x 6 x ... x 2 x 1) represents all the people in the group, the 4! in de denominator de number of people we are picking and the 3! the number of people we are NOT picking. The math of combinations allows us to do some cancelation, see that what we got in the end is: × × ×( × × × ) ( × × × )( × × )

×( )×

= ( × )× = 35

TIP: always cancel the number in your numerator with the larger number in the denominator, that will save you time. Now practice some:

8! = 28, 2! 6!

5! 5 × 4 × (3 × 2 × 1) = = 10 3! 2! (3 × 2 × 1)(2 × 1) 9! = 84, 3! 6!

4! = 6, 2! 2!

6! = 15 2! 4!

Now to our problems, the first three are plain vanilla combinations problems and are good to illustrate the general concept of combinations, as we move down the line we will see harder and harder problems that cover all the types of combinations problems you are likely to find in the GMAT.

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A teacher will pick a three-student group from a group of five students, how many different three-student groups are possible?

How many different three-people groups are possible if each group is picked from a group of 7 people and Peter, who is in the seven-people group, must be in the threepeople group?

(A) 10 (B) 12 (C) 15 (D) 30 (E) 60 2 If Lisa's backpack can only hold 4 of her 7 blouses, how many different four-blouse sets can she pack? (A) 840 (B) 420 (C) 140 (D) 70 (E) 35 3 How many different handshakes are possible if there are 9 people in a room and each person shakes hands with everybody else exactly once?

(A) 18 (B) 17 (C) 16 (D) 15 (E) 14 5 A basket-ball coach will pick a five-player team from a group of 7 players. If Mary and Patricia are among the 7 players the coach can pick from, how many different teams that include Mary but not Patricia can the coach form? (A) 5 (B) 6 (C) 7 (D) 8 (E) 9

(A) 72 (B) 36 (C) 32 (D) 24 (E) 18

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Jane and Keisha are among the six executives Company XYZ can pick from to send a four-executive delegation to a business conference. How many different four-executive groups can Company XYZ send to the conference if both Jane and Keisha must be included?

An investment bank will form a six-banker committee. How many different committees can the bank form if each committee must be made up of 3 women and 3 men and there are 5 female and 6 male bankers available?

(A) 6 (B) 8 (C) 12 (D) 18 (E) 24

(A) 30 (B) 120 (C) 160 (D) 200 (E) 240 10

How many different four-student teams made up of two girls and two boys can a teacher form if she can choose from 4 girls and 5 boys?

Six Canadian and 6 Mexican delegates attend an international trade conference. A three-delegate committee will be selected from the 12 delegates. How many threedelegate committees can be formed if each possible committee must include at least one Mexican delegate?

(A) 80 (B) 60 (C) 40 (D) 20 (E) 12

A) 260 B) 240 C) 220 D) 200 E) 20

A pharmaceutical company will form a fivepeople research team made up of 2 biologists and 3 engineers. How many different teams can the company form if there are 6 biologists and 5 engineers available?

At Lincoln High School, there are 5 girls and 3 boys qualified to be part of the school's debate team. If a three-student team is selected to be sent to a debate tournament in New York City, in how many ways can the three-people team be formed if at least 1 boy must be included in the team?

(A) 30 (B) 120 (C) 150 (D) 240 (E) 300

(A) 10 (B) 18 (C) 46 (D) 56 (E) 63

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12 A factory will form a three-people technical committee. If there are 6 engineers and 5 production-line workers available, how many different committees can be formed if each committee must include at least one engineer? (A) 165 (B) 155 (C) 145 (D) 135 (E) 125

A chess tournament is attended by five three-player teams. How many different chess games are possible if each one is played by two players and every player will play against every other one except the two other players in his team? (A) 180 (B) 160 (C) 120 (D) 110 (E) 90 16

13 In a polygon, a diagonal is defined as the line that joins a vertex to any other vertex except the two adjacent vertices. How many diagonals does a seven-sided polygon have? (A) 49 (B) 42 (C) 28 (D) 21 (E) 14 14

To form a six-people case competition team, a business school professor is to choose 3 female and 3 male students from a group of male and female students. If there are 7 male students in the group and 140 different sixstudent teams are possible, how many female students are there in the group? A) 4 B) 7 C) 10 D) 14 E) 20

How many different handshakes are possible if six girls are standing on a circle and each girl shakes hands with every girl except the two girls standing next to her? (A) 12 (B) 11 (C) 10 (D) 9 (E) 8

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A laboratory has x different types of alcohol and 4 different types of chemical accelerant available. In order to make a chemical composite the laboratory mixes two types of alcohol with two types of chemical accelerant. If 60 different such chemical composites are possible, what is the value of x? (A) 3 (B) 4 (C) 5 (D) 6 (E) 7 18 A retailer is packaging different types of ties and belts to be sold at a Father's Day special sale event. The ties and belts are packed in boxes that include 3 ties and 2 belts and there are 24 different possible such tie-belt combinations. If the retailer can pick from 4 different types of ties, how many different types of belts are there?

If Peter is excluded from a group of B boys, 35 different three-boy groups can be formed. What is the value of B? (A) 5 (B) 6 (C) 7 (D) 8 (E) 9 20 The Springville Police Department has C police cars available. When one car breaks down, 28 different six-car combinations are available to patrol the streets of Springville. What is the value of C? (A) 11 (B) 10 (C) 9 (D) 8 (E) 7 21

(A) 3 (B) 4 (C) 5 (D) 6 (E) 7

At a twenty-member social club, what is the ratio of the number of different five-member groups that can be formed to the number of different six-member groups that can be formed? (A) 2:5 (B) 3:5 (C) 5:4 (D) 5:3 (E) 5:2

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22 A toy car manufacturer packs its cars into boxes that contain either 4 or 5 cars of different models. If the manufacturer makes 15 different types of models, what is the ratio of the number of possible different four-car boxes to the number of possible different five-car boxes? (A) 4:11 (B) 5:11 (C) 6:11 (D) 11:6 (E) 11:5

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1.2 Combinatorics – Slots First of all, in slots problems, order always matters, that is we can pick the same group of, say, people, and with that group form different “ways.” The reader will see that here we do not teach “permutations” as a category apart, for the simple reason that permutations are simply a special case of a slots problems. For example: A teacher will pick a three-student group out of a five student group to play a chess tournament. If one student will play the first board, another the second board, and another the third board, in how many ways can the teacher field her team? Notice that in this case order DOES matter because if the teacher picks, say, Mary, Peter, and Jamal, one possible team is Mary on the first board, Peter on the second, and Jamal on the third; another possible team could be Jamal on the first board, Mary on the second, and Peter on the third. So our teacher has 5 ways of picking the first board (because she has 5 students available), 4 for the second (now she has 4 available), and 3 for the third. To solve the problem, we use what mathematicians call the “general multiplication principle” and go: 5 x 4 x 3 = 60 ways of fielding the chess team. Another example: Lisa has 3 blouses, 4 skirts, and 3 pairs of shoes, how many different outfits can she wear? 3 x 4 x 3 = 36 possible outfits. Now we go to our problems:

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23 How many different three-letter codes can be formed using the letters A, B, and C? (A) 6 (B) 12 (C) 18 (D) 27 (E) 30 24 How many different three-letter codes can be formed using the letters A, B, and C, with no two letters being equal? (A) 6 (B) 12 (C) 18 (D) 27 (E) 30 25 How many different three-number passwords can be formed picking from the digits 1 through 5 if the first two numbers must be the same? (A) 25 (B) 45 (C) 60 (D) 360 (E) 625

How many different three-number passwords can be formed picking from the digits 1 through 5 if the first two numbers must be the same and the third one different from the first two? (A) 32 (B) 25 (C) 24 (D) 20 (E) 12 27 The Roman alphabet has 26 letters, 5 of which are vowels and 21 are consonants. How many different three letters if the first one must be a vowel, the second one a consonant, and the third one a vowel different from the first? (A) 2100 (B) 525 (C) 420 (D) 400 (E) 380 28 Five women will stand in line, in how many different ways can they stand? (A) 40 (B) 60 (C) 80 (D) 100 (E) 120

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Five women will stand in line, in how many different ways can they stand if Mary must stand on the left?

Two boys and three girls will sit on a row of five chairs, in how many ways can they sit if the two boys must sit at the ends of the row?

(A) 60 (B) 48 (C) 24 (D) 12 (E) 6

(A) 24 (B) 18 (C) 12 (D) 8 (E) 6

Five women will stand in line, in how many different ways can they stand if Mary must NOT stand on the left?

Four girls and two boys will stand in line, in how many ways can they stand if the two boys must stand next to each other?

(A) 12 (B) 24 (C) 48 (D) 96 (E) 120

(A) 620 (B) 480 (C) 240 (D) 120 (E) 60

Six boys will stand in line, in how many ways can they stand if Jamal or Anthony must stand on the left?

Four girls and two boys will stand in line, in how many ways can they stand if the two boys must NOT stand next to each other?

(A) 14,400 (B) 720 (C) 640 (D) 240 (E) 120

(A) (5!)(4) (B) (5!)(3) (C) (5!)(2) (D) (4!)(3) (E) (4!)(2)

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How many different five-digit codes can be picked from the digits 1 through 6 if the middle digit must be odd and no two digits might be the same?

How many odd positive integers n are there such that 40,000 <n< 100,000 and n is made up of the digits 0, 1, 2, 3, 4, and 5?

(A) 420 (B) 360 (C) 180 (D) 120 (E) 60

(A) 1296 (B) 1295 (C) 750 (D) 749 (E) 150 37 At a certain company, all telephone extension numbers are odd and each extension is made up of the digits1, 2, 3, 4, and 5, how many possible 3-digit extension numbers can the company have? (A) 125 (B) 75 (C) 70 (D) 60 (E) 50

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1.3 Combinatorics – Miscellaneous In the category of miscellaneous problems we include basically three types: 1) circular arrangements 2) repeated letter problems, and 3) problems that mix combinations with some other type of constraint. Repeated Letter Problems Repeated letter problems come under several disguises in the GMAT, but they are quite simple to solve (and still considered 700+ problems). For example: how many different eight letter codes can we get by using all the letters of the word PASSPORT? The formula is quite simple, we divide the factorial of the number of letters in the word by the product of the factorials of the repeated letters, in our case:

! ( !)( !)

= 10,080, the (2!)(2!) in the

denominator comes because we got 2 Ps and 2 Ss. Circular Arrangements When n people, or trees, or any other items are arranged in a circle, without a fixed position, the formula is (n - 1)! For example: In how many different arrangements can five students stand on a circle? Our n here is 5, so we got (5 – 1)! = 4! = 24 possible arrangements. Combinations Other Type of Constraints For example: A teacher will pick a group of 3 students out of 5, she will also assign a number from 1 trough 6 to be printed on the back of each t-shirt and all the numbers in any 3 people group will be the same, how many possible arrangement of people/numbers can we get? First of all we have a plain vanilla combination, we pick 3 people out of 5 and order does not matter, so

! ( !)( !)

= 10, then we multiply the number of possible groups of people by the

possible numbers they can wear on their backs (we multiply because per every way of of picking the group of people we will have so many ways of picking the number), so our answer is (10)(6) = 60.

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In how many different arrangements can six trees be planted on the circumference of a circular garden if two arrangements are considered different when the positions of the trees are different relative to those of the others?

A luxury car dealer has 7 cars for sale: 2 Ferraris, 3 Maserattis, and 2 Lamborghinis. She will line them up on her shop window from left to right. If cars of the same make are identical to each other, in how many ways can the dealer exhibit the cars on her window?

(A) 120 (B) 300 (C) 480 (D) 560 (E) 720

(A) 240 (B) 210 (C) 180 (D) 150 (E) 120

39 42 Mike has 5 marbles, each of a different color; in how many ways can he arrange his marbles on a circle he drew on the floor? (A) 120 (B) 96 (C) 48 (D) 24 (E) 12 40

How many different 6-digit codes can be made by re-arranging all the digits of the number 123,451 if the two 1s cannot be next to each other? (A) 720 (B) 360 (C) 240 (D) 180 (E) 120

How many different six letter codes can be formed by re-arranging all the letters of the word LITTLE? (A) 360 (B) 320 (C) 240 (D) 220 (E) 180

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A certain company will pick 2-letter and 3letter codes from the letters A, B, C, D, E, F, G, and H. The letters in each code may be repeated and order matters, meaning that ABC is not the same as BCA. How many such codes can the company pick?

A certain company will pick 2-letter and 3letter codes from the letters A, B, C, D, E, F, G, and H. The letters in each code may NOT be repeated and order matters, meaning that ABC is not the same as BCA. How many such codes can the company pick?

(A) 392 (B) 424 (C) 576 (D) 624 (E) 720

45 How many 7-digit codes can be made by rearranging all the digits of the number 1,234,567, if the thousands digit must be odd? (A) 1,080 (B) 1,440 (C) 2,160 (D) 2,880 (E) 3,120

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Chapter 2 Probability 2.1 - ”Pure” Probability 2.2 –Probability Meets Combinatorics

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2.1 - "Pure" Probability Probability is something that puts the fear of God into most GMAT test-takers, but it should not, it is something very mechanical once you get the hang of it. First of all, the basic "probability fraction" will always be

!"# $ % " " !"# $ %

, favorable

outcomes is, in plain English, the outcomes which are good for you; total outcomes of course are all the possible outcomes within a certain constraints. A very basic example is: what is the probability of getting heads if a coin is tossed twice? You got only one favorable outcome, which is heads, and two possible outcomes, head or tails, so the

probability is , or 0.5. Something very important to notice is that probability will always be a

number between 0, inclusive, or the probability that something will NOT happen, and 1, inclusive, or the probability that something will certainly happen. AND or OR Plain and simple, "and" in probability means multiplication, "or" means addition. Let's see some examples: Example 1: If two cards are picked at random, without replacement, from a deck of cards numbered from 1 to 10, what is the probability of picking a 2 and a 5. See that there are two possible favorable outcomes for us, either we pick a 2 and then a 5, or a 5 and then a 2; so our scenarios are: 2 and 5

= '& , again we multiply because we need a 5 and then a 2.

& '

, see that we multiply because we need a 2 and then a 5.

or 5 and 2

& '

Our job is not done yet, because we got '& or

= , which is the answer to our problem. '&

, we add the two scenarios and we get '& + '& =

Example 2: A die is rolled and then a card is picked from a deck of ten cards numbered from 1 to 10, what is the probability of rolling a 2 and then picking a 5?

Here we got only one possible scenario, first a 2 and then a 5, so we go & = & .

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Example 3: A die is rolled and then a card is picked from a deck of ten cards numbered from 1 to 10, what is the probability of rolling a 2 or picking a 5? Here we first roll our die and then pick a card, and because it is one thing or the other, we go

Binomial Distribution Binomial Distribution problems are among the hardest you can find in the GMAT but are quite easy to solve once you get the hang of them. Incidentally, it is one of the few topics tested in the GMAT that will be useful for your MBA. Example 4: Three balls are picked at random, without replacement, from a bag that contains 4 red and 5 blue balls, what is the probability of picking exactly one red ball? See that one possible scenario is picking a red ball first and then a blue and then a blue, or RBB for short, so we got

. But that is NOT the only possible scenario that satisfies us, we

could also get BRB or BBR, each of which will have a probability of

of happening. Then,

because it is one way, or the other, or the other, we add our three scenarios and we get

Now, in this case it was easy to list all three possible scenarios, it is quite intuitive to see that we either pick the red ball first, second, or third. What happens when there are several scenarios and it is not so easy to count them out with our fingers? Let's take a look at example 5. Example 5: A teacher will pick a group of 5 students out of a group of 5 girls and 5 boys, what is the probability of picking a group that includes exactly 3 girls? Clearly one example is GGGBB, and the probability of it happening is

& '

. Now

you see the problem, right? The scenarios could be GBGBG, or BBGGG, and who knows how many more, so we have to find a way to count them fast and efficiently, and that takes us to the repeated letter problem explained in 1.3 Combinatorics - Miscellaneous. To see how many possible scenarios we have we need to answer the question: how many different codes can we

form with the letters GGGBB? The answer will be ! ! = 10. So because we got ten possible

scenarios we go (10) ) * = ) * = .

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A die is rolled twice, what is the probability of rolling a 3 both times?

A die is rolled twice, what is the probability of rolling a 3 exactly once?

(A)

(B)

(C)

(D)

(E)

(A)

(B)

(E)

A coin is tossed three times, what is the probability of getting three heads?

A coin is tossed three times, what is the probability of getting exactly two heads?

(A)

(B)

(A)

(B)

(C)

(D)

(E)

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A die is rolled twice, what is the probability of rolling at least one 4?

A die is rolled three times, what is the probability of rolling the same numbers on the first and second rolls and a number different from the first two on the third roll?

(A)

(B)

(D)

(C)

(E)

(D)

(B)

(C)

(E)

51 53 A bag contains 4 red and 3 blue marbles, if two marbles are picked at random, what is the probability of picking at least one red marble? (A)

(B)

A die is rolled three times, what is the probability of rolling the same number on the three rolls? (A)

(B)

(C)

(D)

(E)

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A bag contains 6 red marbles and 4 blue marbles, if four marbles are picked at random, what is the probability of picking exactly two red marbles?

A teacher will pick at random 4 students from a group of 5 female and 4 male students, what is the probability of picking a 4 student group that includes exactly two female students?

(A)

(B)

(A)

(B)

(C)

(D)

(E)

(C)

(D)

(E)

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If the probability of picking at random a defective computer from a certain shipment is one-eighth, and 5 computers are picked at random, what is the probability of picking exactly 2 defective computers?

Two dice are rolled, what is the probability that the sum of the two rolls is 3 or 8?

(A) ) * ) * (60)

(A)

(B)

(C)

(D)

(D) ) * ) * (10)

(E)

(B)) * ) * (20) (C)) * ) * (10)

(E) ) * ) * 57

Two dice are rolled, what is the probability that the sum of the two rolls is 7? (A)

(B)

(A)

(B)

(D)

(E)

(C)

Two dice are rolled, if the sum of the two rolls is 9, what is the probability that a 3 turned up on one of the dice rolled?

(D)

(E)

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2.2 - Probability meets Combinatorics

A teacher will pick a 3-student group from a group of 4 boys and 4 girls, if one of the possible groups is picked at random, what is the probability of picking a group that includes exactly 1 boy and 2 girls?

At a certain company, a group of 3 people will be chosen from a group of 7 people that includes Peter and Michael. If one 3-people group is picked at random, what is the probability of picking a group that includes both Peter and Michael?

(A)

(B)

(A)

(B)

(C)

(D)

(E)

(C)

(D)

(E)

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At a certain company, a group of 4 people will be chosen from a group of 9 people that includes Jamal and Maria. If one 4-people group is picked at random, what is the probability of picking a group that includes Jamal but not Maria?

A bag contains 1 red, 1 blue, and 6 green balls. If four balls are picked at random, what is the probability of picking neither the blue nor the red ball?

(A)

(B)

(C)

(D)

(C)

(E)

(A)

(D)

(E)

64 A bag contains 4 red and 5 green marbles. If three marbles are picked at random, what is the probability of picking at least one red marble? (A)

(B)

(C)

(D)

(E)

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A teacher will pick a 3-student group from a group of 6 girls and 5 boys. If one 3-student group is picked at random, what is the probability of picking a group that includes at least one girl?

Congress will pick a 4-senator committee from a group of 6 Party A and 5 Party B senators. If one group is picked at random, what is the probability of picking a group that includes exactly 2 Party A and 2 Party B senators?

(A)

(B)

(C)

(D)

(B)

(E)

(A)

(D)

(C)

(E)

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Chapter 3 Mixtures

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Mixtures problems are considered difficult in the GMAT, but are quite easy to solve once you learn the method we will teach here. Example: At a certain class the average score for girls was 84 and the average score for boys was 80, if the class average was 81, what fraction of the students in the class are boys? We could go the algebra way (not recommended); call B the number of boys and G the number of girls, the equation we get is

+, & +,

= 81, in case you do not understand it, see that in the

numerator you got all the points and in the denominator the total number of people; so if you divide all the points by all the people you get the average for the whole class.Now we manipulate our equation and 84G + 80B = 81G + 81B (we multiplied both sides by G + B and distributed our 81 in); group variables on each side and 84G - 81G = 81B - 80B, so 3G = 1B. We need to find the ratio of boys to girls to answer our problem so we divide first both sides by G and get

3 = 1 + , then divide both sides by 1 and get

= + , so for every 3 boys we have 1 girl; say we

have 3 boys and 1 girl for a total of 4 students, the fraction of boys in the class is 3/4.

Now a much faster way of solving our problem, and this is the one we strongly recommend is the following: Positive distance from boys to average: 1 Positive distance from girls to average: 3 To get the part for the boys in the ratio we use the distance of the girls, that is 3. To get the parts for the girls we use the distance of the boys, that is 1. In other ways, we flip the distances to get

The ratio, so + = , so we say we have 3 boys and 1 girl, or a total of 4 students, so the fraction

of boys in the class is, of course, 3/4, much faster, huh?

Example: A 25% alcohol solution is mixed with a 40% alcohol solution to get a mixture which is 30% alcohol, what is the ratio of the 25% alcohol solution to the 40% alcohol solution in the mixture? Using our fast method: Positive distance from 25% to average: 5 Positive distance from 40% to average: 10

Call T the 25% solution and F the 40% solution, and we get that =

flip the distances. And the answer is 2:1.

= , remember that we

Example: Mixture X is 40% oil and mixture Y is 80% oil, if the two mixtures are put together to form 44 liters of a mix which is 50% oil, how many liters of mixture X are in the 50% mixture? 31 | P a g e GMATâ&#x201E;˘ is a registered trademark of the Graduate Management Admission Councilâ&#x201E;˘ (GMAC). GMAC does not endorse, nor is it affiliated in any way with the owner or any content of this book.

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Positive distance from X to average: 10 Positive distance from Y to average: 30 So the ratio of X to Y is 30 to 10, or 3 to 1 (again, we flip distances). Now we got a pretty straight ratios problem, where we go 3 parts of something plus 1 part of something equal 4 parts, since the total mix is 44 liters then each part is worth 11 (or 44 divided by 4 parts), since we want the number of liters of X and X has 3 parts worth 11 each, we got 3(11) = 33, and that is our answer. The ratio is everything (well, almost) in mixtures problems! Look at this example: In a certain MBA class the average male student is 26.8 years old, and the average female student is 25.4 years old, what is the average age of the whole class? (1) The ratio of male to female students is 3 to 2. (2) There are 180 male students in the class. The easy statement here is (2), so we start from there. This statement is not sufficient because if write the weighted average equation, calling F the number of ladies in the class, we got: ( . )( &),(

. )( ) &,

=?, and because we do not know the value of F we cannot put a number

where our question mark is. Now, statement (1) is sufficient because it gives us the ratio of guys to girls in the class, and that is sufficient, how so? Notice how as long as we respect the ratio we get the same weighted average. Female 3 6 9 12

Male 2 4 6 8

Weighted Average 26.24 26.24 26.24 26.24

Of course, I used my calculator to find the weighted average, I am just trying to prove a point here, but as long as we respect the ratio we find the the same average for the whole class, that is what you have to know for the GMAT. So the answer to the problem is A.

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Jamal bought some ham at $12 a pound and some cheese at $8 a pound. If the total weight of the ham plus the cheese was 2 pounds, and the average price of Jamal's purchases was $9, how many pounds of cheese did he buy? (A) 2 (B) 1.5 (C) 1 (D) 0.5 (E) 0.25 68 Pedro has some coins in his pocket, some are 5-cent coins and some are 25-cent coins. If the average value of each coin in Pedro's pocket is 10 cents and he has $8 in his pocket, what is the total value of his 5-cent coins? (A)$1 (B)$2 (C)$3 (D)$4 (E)$6

A certain company has only managerial employees and production-line workers; the average employee has 14 years of formal education. If the average number of years of formal education is 18 years for managerial employees and 12 years for production-line workers, and the company has a total of 90 employees, how many managerial employees does the company have? (A)20 (B)30 (C)40 (D) 50 (E) 60 70 At a charity fundraiser each female attendant donated $1,200 and each male attendant donated $800, if the average donation at the fundraiser was $900, what was the ratio of the number of female attendants to that of male attendants? (A) 1:3 (B) 1:2 (C) 1:1 (D) 2:1 (E) 3:1

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Jamal and Mary read a total of 100 books and no book was read by both of them. Jamal found 40% of the books he read boring and Mary found 50% of the books she read boring, and together they found 44% of the 100 books boring, what percent of the 100 books was read by Mary?

At School A 40 percent of the students wear white t-shirts and 60 percent of the students wear red t-shirts. At School B 50 percent of the students wear white t-shirts and 50 percent of the students wear green t-shirts. If the two student bodies are put together to form School C, 42 percent of the students wear white t-shirts. What percent of the students at School C came from School A?

(A)30% (B) 40% (C) 50% (D) 60% (E) 70% 72 At a certain party attended by 180 people, each woman ate 2 pieces of pizza and each man ate 3 pieces of pizza. If, on average, each person at the party ate 2.2 pieces of pizza, how many men were present at the party? (A) 12 (B) 18 (C) 36 (D) 72 (E) 144

(A) 20% (B) 40% (C) 50% (D) 60% (E) 80%

74 The average male Springville inhabitant has $12,000 in his savings account and the average female Springville inhabitant has $15,000 in her savings account. The average inhabitant, regardless of gender, in Springville has $13,000 in his or her savings account. If the average male Springville inhabitant has $800 in his checking account and the average female Springville inhabitant has $1,400 in her checking account, what is the average amount of money that each Springville inhabitant, regardless of gender, has in her checking account? (A) $1,400 (B) $1,200 (C) $1,100 (D) $1,000 (E) $900

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Keisha sells only oranges and apples at her fruit stand. If on Tuesday she sold the average apple for 80 cents and the average orange for 60 cents, what was the average price per piece of fruit at which Keisha sold her fruit on Tuesday?

In a certain swimming club, 82 percent of the female members and 86 percent of the male members have passed a swimming proficiency test, what fraction of the members of the club are female?

(1) She sold 125 oranges. (2) She sold three times as many oranges as apples.

(1) 83 percent of the members of the club have passed the proficiency test. (2) The club has 95 female members. 79

76 At a certain high-school, the average GPA of female students is 3.2 and the average GPA of male students is 3.0, what is the average GPA of the whole student body? (1) The ratio of the number of female students to that of male students is 1.5:1.3. (2) There are 530 male students at the school. 77 Michael mixed a liters of a 10-percent alcohol solution with b liters of a 20-percent alcohol solution to get c liters of a 14percent alcohol solution, what is the value of b? (1) a = 12

What percent of the combined assets of hedge funds Aggresinv and Conservinv are invested in biotechnology stocks? (1) Hedge fund Aggresinv has 70 percent of its total assets invested in biotechnology stocks and hedge fund Conservinv has 40 percent of its total assets invested in biotechnology stocks. (2) The Dollar value of the assets managed by Agressinv and Conservinv are in a ratio of 2:1. 80 Will at most 80 percent of the students at a certain high-school apply for college? (1) Of the students at the high-school, 76 percent of the male and 80 percent of the female will apply for college?

(2) c = 20 (2) There are 980 students at the highschool.

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What is the average income at Town A?

What fraction of the combined populations of Towns A and B live in Town A?

(1) The combined income of the people at Town A is $42,000,000. (2) 60 percent of the people at Town A have an average income of $ 36,000 and the rest have an average income of $55,000.

(1) The average income of the population of Town A is $3,000 higher than the average income of the populations of both towns combined. (2) The average income of the population of Town B is $2,000 lower than the average income of the populations of both towns combined.

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Chapter 4 Rate (Work)

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You are likely to get two work problems in the GMAT, and they go from quite simple to rather difficult ones. Let's start with the plain vanilla work problem: Example 1: John can do a job in 2 hours, Mike can do the same job in 3 hours, how long will it take them, working together at their respective constant rates, to finish the job? Forget about using the RT = W formula for this problem, it is just a plain waste of time. Simply use this formula

( )( )

of an hour, or an hour and twelve minutes, or 72 minutes, all of these

could be possible answers in the GMAT. But the important thing here is how we arrive at the answer, we basically multiply the two times in the numerator and add them in the denominator. Because, as we saw above, the GMAT might give you the answer in several units, you have to be very cool at moving from fractions of an hour to minutes and vice versa, as the table below shows: Fraction of Hours 1/12 1/10 1/6 1/4 1/3 1/2 2/3 3/4

Minutes 5 6 10 15 20 30 40 45

Example 2: Lisa can shovel the snow on her driveway in 20 minutes, John, whose driveway is the same size as Lisa's, can do the same job in 30 minutes, how long will it take them to shovel the snow in both driveways working together at their respective rates? ( &)( &)

It will take them ( &),( &) = 12 minutes to clean one driveway, so it will take them 24 minutes to clean both.

VERY IMPORTANT: the reciprocal of the time it takes both people or items to do the job is

the rate of work, something important for other problems, so the rate in Example 1 is and in

Example 2 is . Example 3: Pedro can paint his room in 4 hours and Jamal can paint the same room in 3 hours, if they worked together for one hour and Jamal left, how many minutes will it take Pedro working alone to finish painting the room?

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They can paint the whole room in

( )( )

= ( ),( )

so their joint rate is

formula they completed ) * (1) = ) * of the job, so Because Pedro's rate is

finish the job.

. Using the RT = W

of the job remains to be done.

(the reciprocal of 4), his equation for the rest of the job is ) * (/) =

) *, solve for T, and T =

= =1

, so it will take Pedro working alone 100 minutes to

VERY IMPORTANT: When two people or machines work together to complete one job, the ratio of the part of the job done by one machine to that done by the other is the reciprocal of the time it takes each person or machine individually to do the full job. Example 4: Machine A can complete a job in 2 hours and Machine B can complete the same job in 3 hours, if they finish the job working together at their respective rates, what is the ratio of the part of the job done by A to that done by B?

Answer is very simple, the reciprocal of the times, or . Example 5: If 4 workers can do a job in 3 hours, how long will it take 6 workers to do the same job? In this case, because we are working with reverse proportion (more workers means less time), we multiply corresponding things, so we go (4)(3) = (6)(x), and x = 2 hours.

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85 83 Pipe A can fill up a tank in 2 hours, pipe B can fill up the same tank in 4 hours, how many hours will it take both pipes, working simultaneously at their respective constant rates, to fill up the thank? (A)

If machine A can manufacture 500 widgets in x hours and machine B can manufacture 500 widgets in y hours, how many hours will it take both machines, working simultaneously at their respective constant rates, to manufacture 500 widgets? (1) x + y = 12

(2) 0,1 =

(B) 1 (C) 1

(D) 1

Jamal can do a certain job in x hours and Michael can do the same job in 3 hours. If working simultaneously at their respective

rates they can do the same job in 1 hours,

(E) 2

what is the value of x?

84 Peter can paint a wall in 3 hours and Jamal can paint the same wall in 6 hours, how long will it take both of them, working simultaneously at their respective rates, to paint the whole wall?

(A) 2 (B) 2.2 (C) 2.4 (D) 2.6 (E) 2.8

(A) 1 hour (B) 1.5 hours (C) 2 hours (D) 2.5 hours (E) 3 hours

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John can type a certain report in x hours and Michael can type the same report in y hours. If working simultaneously at their respective constant rates they can type the same report in 3 hours, what is the value of x?

If 5 combines can harvest a field in 8 hours, how long will take 4 of the same combines to harvest the same field?

(2) y = 8

(A) 13 (B) 12 (C) 11 (D) 10 (E) 9

(1) x + y = 12.8

If Jamal can do a job twice as fast as Pedro, and both of them, working simultaneously at their respective constant rates, can do the same job in 6 hours, how long will it take Pedro, working alone at his constant rate, to do the same job? (A) 18 (B) 14 (C) 12 (D) 10 (E) 9 89 If 8 workers can do a certain job in 3 hours, how many workers will be needed to complete the same job in 2 hours? (A) 9 (B) 10 (C) 11 (D) 12 (E) 13

If 3 pipes can fill a pool in 6 hours, how many more of the same pipes will be needed to fill the same pool in 2 hours? (A) 4 (B) 5 (C) 6 (D) 8 (E) 9 92 If 4 workers can do a job in 9 hours, how many more hours will it take 3 workers to do the same job, assuming that they work at the same rate? (A) 3 (B) 4 (C) 6 (D) 8 (E) 12

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Sandra can do a certain job in 6 hours while Keisha can do the same job in 8 hours. If Sandra worked alone at her constant rate for 2 hours and was then joined by Keisha. How long did it take both of them, working simultaneously at their respective rates, to finish the job?

Pipe A can fill a tank in 2 hours and tank B can fill the same tank in 3 hours. If both tanks work together for 1 hour and pipe A continues to work alone at its respective constant rate, how many minutes will it take pipe A to fill up the tank?

(A) 2 hours

(B) 2 hours

(D) 3 hours

(E) 3 hourS

(A) 40 (B) 30 (C) 20 (D) 15 (E) 10 95 Jose can do a job in 3 hours and Maria can do the same job in 5 hours. If they worked simultaneously at their constant rates for 1.5 hours and then Jose left, how long did it take Maria to complete the job working at her constant rate? (A) 1 hour (B) 1.2 hours (C) 1.4 hours (D) 1.6 hours (E) 1.8 hours

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96 Machine A can do a job in 5 hours and machine B can do the same job in 4 hours. Machines A and B, working simultaneously at their respective constant rates, finish the

job in 2 ' hours, what is the ratio of the fraction of the job done by Machine A to that done by Machine B? (A) 3:4 (B) 4:5 (C) 4:9 (D) 5:4 (E) 7:4

Lisa and Maria worked together at their respective rates to paint a wall. If Lisa and Maria, working alone at their respective constant rates, can paint the same wall in 4 and 7 hours, respectively, what fraction of the wall was painted by Lisa when they finished painting the wall working together? (A)

(B)

(C)

97 Leroy can shovel the snow on his driveway in 20 minutes and Anthony can shovel the snow of his driveway, which has exactly the same dimensions as Leroyâ&#x20AC;&#x2122;s and the same volume of accumulated snow, in 30 minutes. If they work together at their respective rates to shovel both driveways, what is the ratio of the fraction of the job done by Leroy to that done by Anthony? (A) 2:5 (B) 3:5 (C) 2:3 (D) 3:4 (E) 3:2

(D) (E)

99 It takes Machines A and B, working alone at their constant rates, 3 and 5 hours, respectively, to manufacture 400 steering wheels. If they work simultaneously at their constant rates to manufacture 800 steering wheels, what percentage of the job will have been done by Machine B? (A) 37.5% (B) 40% (C) 50% (D) 62.5% (E) 80%

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100

101

Paul can paint a certain wall in 3 hours, Jose can paint the same wall in 4 hours, and Mary can paint the same wall in 5 hours. If the three of them paint the wall working simultaneously at their respective constant rates, what fraction of the job was done by Mary?

(A)

(B) (C)

(D)

Pipe A can fill a tank in 3 hours, pipe B can fill the same tank in 5 hours, and pipe C can fill the same tank in 8 hours. If the three pipes fill the tank working simultaneously at their respective constant rates, what fraction of the job was done by pipe C? (A)

(B) '

(D) '

(E) '

(E)

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Chapter 5 Rate (Motion)

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horacio.quiroga@gmail.com As you most likely remember from high-school, the basic formula for motion problems is RT = D, where R is rate, T is time, and D is distance. There several cases that you must master in order to ace the GMAT, they might look difficult in the beginning, but they are pretty mechanical once you get the hang of them.

First Case, the Round or Half and Half Trip Example 1: Maria goes to school at an average speed of 4 miles per hour and bikes along the same road at an average speed of 6 miles per hour. What is her average speed for the full trip? Forget about the longer method suggested in other books and use the "harmonic average" formula, which is basically

( )( )( ) ( ),( )

= 4.8 miles per hour. Easy, huh? And this is considered a difficult problem in the

GMAT. By the way, see that our 2 is a constant in the formula, the 4 and the 6 are of course the out and back speeds, respectively. VERY IMPORTANT: this formula works when you average 2 rates and 2 rates only. Example 2: Jamal walks from his home to his girl-friend's at an average speed of 4 mph and comes back by the same route at an average speed of x mph. If he wants to average 6 mph for the whole trip, what is the value of x? We set up our equation

( )( )(0) ( ),(0)

= 6, so 8x = 24 + 6x, so 2x = 24, and x = 12.

TIP: Watch on YouTube (horsegmat is my channel) my explanations to Problem Solving 142 of the Quantitative Review 2 Guide, and Problem Solving Diagnostic 24 of the 12th Edition of the GMAT Official Guide. Both problems are hard problems made very easy by the use of the harmonic average formula.

Second Case and Third Case, Getting Away from Each Other or Going Towards Each Other: this two cases have one thing in common, in order to solve them, the first thing you do is ADD the rates of the two items. Example 3: Two trains start traveling towards each other at the same time from both ends of a 660 milelong track. If one train travels at 50 mph and the other one at 60 mph, how long will it take both trains to meet each other? Add the rates and you get a joint rate of 110 mph, with that we can write our RT = D equation, (110)(T) = 660. Solve for T and T = 6. Example 4: Two trains start traveling towards each other at the same time from both ends of a 660 milelong track. If one train travels at 50 mph and the other one at 60 mph, how far from its departure point will the train that travels at 60 mph will have travelled when the two trains meet? Add the rates and you get a joint rate of 110 mph, with that we can write our RT = D equation, (110)(T) = 660. Solve for T and T = 6. If our train travels at 60 mph it will have travelled (60)(6) = 360 miles.

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horacio.quiroga@gmail.com Example 5: Two trains start traveling towards each other from both ends of a 600 mile-long track. One train travels at 50 mph and the other one at 60 mph, if the train that travels at 60 mph started traveling one hour later, how long after the second train starts travelling will the two trains meet each other? In this case we have to consider that the train that left first travelled 50 miles (it travelled alone for one hour) while the other train stood idle, so the distance that they will cover together will be 600 - 50 = 550 miles. Now because they are going towards each other the joint rate is again 60 + 50 = 110. Using RT = D we get (110)(T) = 550, solve for T and T is 5. Example 6: Paul and Mary start getting away from each other from the same point along a straight line. Paul travels at 4 mph while Mary travels at 5 mph. If Mary started travelling one hour later, how long will it take them to be 40 miles apart? Their joint rate is 4 + 5 = 9. Paul had already walked 4 miles when Mary started walking, so they have to cover 36 miles together, using RT = D we get (9)(T) = 36, and T = 4

Fourth Case, Chasing Each Other: In this case the secret is to SUBTRACT both rates in order to get what I personally call the "getaway rate" or "catch-up" rate. Example 7: Train A leaves New Orleans and travels at a rate of 60 mph; an hour later, a second train leaves from the same station and along the same track travelling at a rate of 80 mph. How far from New Orleans will the second train catch-up with the first one? The catch-up rate in this case is 80 - 60 = 20 mph. Because the the first train travelled 60 miles while the second one stood idle, the second train will have to catch up 60 miles, using RT = D we get (20)(T) = 60, and T = 3 hours. Because the second train travels at a rate of 80 mph, it will catch up with the first train (80)(3) = 240 miles from New Orleans. Example 8: Train A leaves New Orleans and travels at a rate of 60 mph; an hour later, a second train leaves from the same station and along the same track travelling at a rate of 80 mph. How long will it take, from its departure point, for the second train to be 60 miles ahead of the first one? In this case, the second train has to travel 120 miles more than the first train (the 60 miles the second train travelled while the first one stood idle plus the 60 miles it has to get ahead). Because the catch-up/get away rate is 80 - 60 = 20 mph, using RT =D we get (20)(T) = 120, and T = 2.

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102

104

Two trains start traveling towards each other from opposite ends of a 220-mile long straight track. One train is traveling at 50 miles per hour and the other at 60 miles per hours. If the two trains started moving towards each other at the same time, how long did they travel until the met?

Car A starts traveling East from one end of a 330-mile long straight road at 3pm. Half an hour later car B starts traveling West on the same road. If car A travels at an average speed of 60 miles per hour and car B travels at an average speed of 40 miles per hour, at what time will both cars pass each other?

(A) 1 hour (B) 1.5 hours (C) 2 hours (D) 2.5 hours (E) 3 hours

(A) 6 pm (B) 6:30 pm (C) 7 pm (D) 7:30 pm (E) 8 pm

103

105

Two trains travel towards each other starting from each end of a 550-mile long straight track, if train X is traveling East at 70 miles per hour and train Y is traveling West at 50 miles per hour, and train Y started traveling one hour later than train X, how far had train X travelled when they met?

Two cars start traveling in opposite directions on a straight road. If one car travels at 80 miles per hour and the other one at 90 miles per hours, how long will it take for both cars to be 510 miles apart?

(A) 200 miles (B) 250 miles (C) 275 miles (D) 300 miles (E) 350 miles

(A) 2 hours (B) 2.5 hours (C) 3 hours (D) 4 hours (E) 4.5 hours

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106

108

Two cars started traveling in opposite directions on a straight road. Car A travelled West at an average speed of 50 miles per hour, car B travelled East at an average speed of 70 miles per hour. If car B started traveling one hour later than car A, how far had car A travelled when both cars were 410 miles apart?

Car A is traveling at an average speed of 60 miles per hours and is 100 miles behind car B, which is traveling at an average speed of 40 miles per hour. How long will it take for car A to catch up with car B?

(A) 240 (B) 230 (C) 210 (D) 200 (E) 150 107 Two trains started traveling in opposite directions on a straight track. Train A travelled West at an average speed of 60 miles per hour, train B travelled East at an average speed of 80 miles per hours. If train B started traveling half an hour later than train A, how many miles had train B travelled when both trains were 310 miles apart?

(A) 5 hours (B) 4 hours (C) 3 hours (D) 2 hours (E) 1 hours 109 Car A leaves New York and travels to Boston at an average speed of 80 miles per hour. Car B left New York half an hour earlier and also travelled to Boston but at an average speed of 60 miles per hour. How many hours after it starts traveling will car A catch up with car B? (A) 1 (B) 1.5 (C) 2 (D) 2.5 (E) 3

(A) 120 (B) 150 (C) 160 (D) 170 (E) 180

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110

112

Because of a storm alert, Horacio has to evacuate New Orleans and drives his car towards Houston at an average speed of 60 miles per hour. Two hours later Ricardo also leaves New Orleans and travels towards Houston on the same road but at an average speed of 90 miles per hour. How many miles from New Orleans will Ricardo catch up with Horacio?

Mike and Amanda start walking towards each other from opposite ends of a 12-mile long trail. If it will take Mike and Amanda 2 and 3 hours, respectively, to walk the full length of the trail, how far from his starting point was Mike when they passed each other?

(A) 300 (B) 320 (C) 340 (D) 360 (E) 380 111 Ricardo is evacuating New Orleans towards Houston and is already 30 miles away from New Orleans; he is 30 miles behind Horacio, who evacuated before Ricardo did. Ricardo is driving his car at a constant rate of 90 miles per hour and Horacio is driving his car at a constant rate of 60 miles per hour. How many miles from New Orleans will Ricardo be when he is 60 miles ahead of Horacio? (A) 360 (B) 330 (C) 300 (D) 270 (E) 240

(A) 8 miles (B) 7.2 miles (C) 6 miles (D) 4.8 miles (E) 4 miles 113 Bill started driving his car from X to Y at an average constant speed of 80 miles per hour. At the same time, Horacio started to drive his car also from X to Y at an average speed of 60 miles per hour. After driving for half an hour, Bill car broke down and he decided to wait for Horacio in order to get a ride to the nearest mechanic shop. How many minutes did Bill have to wait for Horacio? (A) 10 (B) 12 (C) 15 (D) 18 (E) 30

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114

116

At 7 am Cindy starts walking from home to school at an average speed of 6 miles per hour. At 7:30 am of the same day, Vanessa, who lives in the same house, starts walking to the same school, along the same road, at an average speed of 4 miles per hour. Thirty minutes after starting, Cindy gets tired and decides to sit down on the side of the road and wait for Vanessa. How many minutes will Cindy have to wait?

Paz drives half of the way of a road trip at an average constant speed of 60 miles per hour and the other half at an average constant speed of 90 miles per hour, what was her average speed for the full trip?

(A) 30 (B) 45 (C) 60 (D) 75 (E) 80 115 Megan walks from home to school at an average speed of 4 miles per hour and returns to her home along the same road at an average speed of 6 miles per hour. If Megan's school is 5 miles away from her home, what is her average speed for the full trip? (A) 4.4 mph (B) 4.6 mph (C) 4.8 mph (D) 5 mph (E) 5.2 mph

(A) 72 mph (B) 74 mph (C) 75 mph (D) 77 mph (E) it cannot be determined from the information given 117 Jamal drives from home to school at an average speed of 40 miles per hour and returns to his home along the same road at an average speed of x miles per hour. If Jamal's school is 12 miles away from his home and he wants to average 48 miles per hour for the full trip, what is the value of x? (A) 52 (B) 54 (C) 56 (D) 58 (E) 60

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118

120

Jose walks from home to school at an average speed of 2 miles per hour and returns to his home along the same road at an average speed of x miles per hour. If Jose's school is 4 miles away from her home, and he wants to average 3 miles per hour for the full trip, what is the value of x?

Elida walks from home to her job at an average speed of 6 miles per hour and returns to her home along the same road at an average speed of x miles per hour. If Elida's school is d miles away from her home, what is the average speed for the full round trip?

(A) 6 (B) 5 (C) 4 (D) 3 (E) 2

(1) d = 10

119 Maria walks from home to school at an average speed of a miles per hour and returns to her home along the same road at an average speed of b miles per hour. If Maria's school is 6 miles away from her home, what is, in terms of aandb her average speed for the full trip?

(A) 2,3 (B)

2,3

(2) x = 8 121 Peter and Keisha walk towards each other from opposite ends of a straight track. They meet exactly halfway from their starting points. If it took Peter 90 minutes to get to the meeting point and Keisha started walking half an hour after Peter did, what is the ratio of Peter's rate to that of Keisha? (A) 2:1 (B) 3:2 (C) 1:1 (D) 2:3 (E) 1:2

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Chapter 6 Overlapping Sets

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Overlapping sets problems are quite common in the GMAT and you are likely to see two of them in the test. They look confusing for the uninitiated, but using the double-entry chart method they become very easy to be solved. Example: At a certain school, 60% of the students are male and 40% of the students play the guitar, if 20% of the students are female students who play the guitar, what percent of the students are male students who do not play the guitar?

Notice how we enter the information into our chart: the numbers inside a square box are the data given in the problem, the numbers not in a box we obtain by subtracting from the total at the end of a row or a column the number given. For instance, if you take a look at the bottom row, we assume we got 100 students, and 60 is given as the total number of male students, because 60 + 40 = 100 we know that 40 is the total number (percentage in fact) of female students. Now, after discounting numbers we get that the number of male students who play the guitar are 20, and if we got a total of 60 male students, then 40 do not play the guitar and our answer is the number in a circle.

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122

124

At a certain class of 28 students, 15 students play soccer, 10 students lift weights, and 5 students both lift weights and play soccer, how many students in the class neither play soccer nor lift weights?

A certain company has 120 employees. If three-fourths of the employees at the company have a car, three-fourths have an MBA, and eight-ninths of those who have a car also have an MBA, how many employees neither have a car nor hold an MBA?

(A) 20 (B) 18 (C) 10 (D) 8 (E) 5 123 At a certain class of 100 students, 60 students are girls. If one-third of the girls and one-fourth of the boys take calculus, how many students in the class, do NOT take calculus? (A) 70 (B) 60 (C) 50 (D) 40 (E) 30

(A) 80 (B) 30 (C) 20 (D) 15 (E) 10 125 At a company event attended by 200 people, everybody has a watch, a ring, or both. If 190 people have a watch and 150 people have a ring, how many of the people at the event have both a watch and a ring? (A) 150 (B) 140 (C) 120 (D) 110 (E) 10

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126

128

Among a group of homeowners, 60 percent have a house an 50 percent have an apartment. If 10 percent have both a house and an apartment, what percent have an apartment only?

At a college party 40 percent of the people were women who were wearing a red t-shirt. If 20 percent of the women were not wearing a red t-shirt, what percent of the people at the party were women?

(A) 70% (B) 60% (C) 50% (D) 40% (E) 10%

(A) 20% (B) 30% (C) 40% (D) 50% (E) 60%

127

129

At a certain 80-student class, every student took a test or wrote a paper or both. If 25 percent of the students took a test and 90 percent wrote a paper, how many students took a test only?

At a certain company 20 percent of the applicants for a position were interviewed and got the job. If 40 percent of those who were interviewed did not get the job, what percent of the applicants were interviewed?

(A) 8 (B) 12 (C) 20 (D) 60 (E) 72

(A) 25% (B) 30%

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130

132

At an exclusive club, 35 percent of the members have an apartment in Manhattan and a house in The Hamptons. If 50 percent of those that have a house in The Hamptons do NOT have an apartment in Manhattan. What percent of the members have a house in The Hamptons?

At a certain 1200-student school 20 percent of the students take German and 60 percent take French. If 30 percent take neither German nor French, how many take French only?

(A) 96% (B) 90% (C) 80% (D) 70% (E) 60% 131 A certain company has 240 employees. Twenty percent hold a PhD; 40 percent have a Master’s Degree and ten percent have both. How many employees have neither a Master’s degree nor a PhD? (A) 50 (B) 60 (C) 80 (D) 120 (E) 140

(A) 600 (B) 480 (C) 120 (D) 50 (E) 10 133 A certain club has 360 members, of which 30 percent play backgammon, 40 percent play chess and 20 percent play chess only. How many members play backgammon only? (A) 10 (B) 20 (C) 32 (D) 36 (E) 72

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134

136

Among a group of 200 executives, 50 have leather jackets and 40 have black shoes. If the number of executives who have leather jackets only is twice the number of executives that have black shoes only, how many have neither?

At a 180-member gaming club, 60% of the members play backgammon and 50% play chess. If the number of members that play backgammon only is twice that of the members that play chess only, how many people play both?

(A) 160 (B) 150 (C) 140 (D) 130 (E) 120

(A) 40 (B) 50 (C) 60 (D) 72 (E) 76

135

137

At a 300-student school 80 students are female and 60 of those have blue t-shirts. If the number of male students who have blue t-shirts is one-half the total number of students who do NOT have blue t-shirts, how many students have blue t-shirts?

At a certain 65-member gaming club 28 of the members play backgammon, 23 play chess, and 24 play bridge. If 12 members do not play any game and the number of members that play the three games is twothirds that of the members that play nothing, how many members play exactly two games?

(A) 60 (B) 70 (C) 80 (D) 140 (E) 160

(A) 6 (B) 8 (C) 12 (D) 14 (E) 16

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138

141

Among a group of 73 people, 26 speak German, 27 speak Italian, and 34 speak Spanish. If 15 people speak none of the three languages mentioned above, and the number of people who speak the three languages is one-third that of the people who do not speak any of the three, what is the total number of people who speak one language only?

Jose has a box with 1200 rare coins. If 55% of the coins are Russian, how many of the coins in Jose’s box are Russian coins with a value not higher than 5 rubles?

(A) 15 (B) 19 (C) 24 (D) 34 (E) 36 139 When asked what brand of soap they prefer 70% of the people chose Eagle and 60% chose Nightingale. What percent of the people preferred Eagle but not Nightingale? (1) Fifty percent of the people asked said they prefer both brands.

(1) Thirty percent of the Russian coins in Jose’s box have a value higher than 5 rubles. (2) Twenty percent of the coins in Jose’s box are gold coins. 142 At a certain conference attended by 1300 Lawyers, 700 of the lawyers are partners at their respective firms and 800 have more than 8 years of experience. How many of the lawyers are partners with more than 8 years of experience? (1) One-hundred of the lawyers at the conference are not partners and have less than or 8 years of experience. (2) Four-hundred of the lawyers at the conference are partners and have less than or 8 years of experience. 143

(2) Twenty percent of the people asked said they prefer neither. 140 What percent of the people in a class are women who have a car? (1) Of the women in the group, 25 percent have a car.

At a college party, 70% of the women are wearing red t-shirts. If 60% of the people at the party are wearing red t-shirts, what is the ratio of women to men at the party? (1) Forty percent of the men at the party are wearing red t-shirts. (2) There are 120 people at the party.

(2) Of the men in the group, 20 percent have a car.

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Chapter 7 Statistics

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horacio.quiroga@gmail.com I will concentrate my intro to statistics on standard deviation, because it is the concept that in my opinion gives students their strongest headaches. Averages, median, and range will be dealt with in greater detail in our Arithmetic guide. Standard Deviation First of all, the GMAT will, to the best of my knowledge, not ask you to calculate the standard deviation of a set; what the test wants is mostly to see that you have a thorough understanding of the concept and a good gut feeling of what standard deviation is about. Let's start our discussion with a look at these three sets: (a) 3, 3, 3, 3, 3 (b) 1, 2, 3, 4, 5 (c) -10, -5, 3, 5, 10 Notice that our three sets have the same average, but clearly they are not equal; and here is where the concept of standard deviation comes in, it is a "measure of dispersion"; in other words, it gives you an idea of how spread apart a set is. Our three sets are ordered, in standard deviation's terms from least to most. Notice that set (a) has no deviation at all, set (b) has a little more of dispersion, and definitely set (c) has the greatest standard deviation. The formula for standard deviation: not that you will ever have to calculate it, but a good understanding of its components will help you understand the concepts. Standard Deviation = 7

∑(090̅ ); <

, where x is one element in the set, =̅ is the average of the set, and n is

the number of elements in the set.

VERY IMPORTANT: as you can see the the standard deviation is the square root of the variance; in other words, if you know the variance you know the standard deviation. This is a very important concept for data sufficiency problems. Let's calculate Standard Deviation for set (a): = 3 3 3 3 3

=̅ 3 3 3 3 3

?@AABCDEF DG (090̅ );

= − =̅ 0 0 0 0 0 Sum of (= − =̅ ) =

(= − =̅ )

0 0 0 0 0 0

= 0 = H4IJ4KLM

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horacio.quiroga@gmail.com Standard Deviation for (a): √0 = 0.

Now let's calculate Standard Deviation for set (b): = -10 -5 3 5 10

=̅ 3 3 3 3 3

?@AABCDEF DG (090̅ );

= − =̅ -13 -8 0 2 7 Sum of (= − =̅ ) =

(= − =̅ )

169 64 0 4 49 286

= 57.2 = H4IJ4KLM

Standard Deviation for (b): √57.2 = 7.6. Not surprisingly, set (b) has greater standard deviation, its dispersion is greater.

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horacio.quiroga@gmail.com 147 144 The heights of students at a certain college are symmetrically distributed to both sides of the mean x. If 34 percent of the distribution lies at more than one standard deviation s from the mean, what percent of the distribution is approximately more than x + s? (A) 94% (B) 68% (C) 46% (D) 32% (E) 16% 145 Which of the following pairs of integers, when added to a list of 50 integers with a mean of 7 and standard deviation S will yield a list of integers with a standard deviation more than S?

Set S is made up of all integers n such that n is a multiple of 5 and −20 ≪ K ≪ 60, which of the following must be true of Set S? I) If each term is multiplied by 3, its average, median, and standard deviation will change. II) The absolute value of the terms of set S will have the same standard deviation. III) If 6 is added to each of the terms in set S, its average (arithmetic mean) and median will change but not its standard deviation. (A) I only (B) I and II only (C) II and III only (D) III only (E) I and III only 148

(A) -8 and 8 (B) 6 and 7 (C) 7 and 7 (D) 8 and 8 (E) 8 and 9 146 Positive integers a, b, c, and d have a standard deviation S, what is the standard deviation of a 2, b - 2, c - 2, and d - 2? (A) 18S (B) 12S (C) 8S (D) S - 8 (E) S

If the starting line-up of basket-ball team A has players of heights 78, 80, 82, 84, and 86 inches, and the starting line-up of basket-ball team B has players of heights, v, w, x, y, and z. Which of the two teams has the largest standard deviation of the heights of its starting players? (1) Both teams have the same average (arithmetic mean) starting players' height. (2) The variance of the heights of starting players of team B is 3.4.

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horacio.quiroga@gmail.com 149

151

Each person at towns A, B, and C, makes $30,000, $35,000, and $40,000 a year, respectively, if the weighted average of the income of the three towns put together is $37,500, which of the following must be true?

Set S is a 7-positive integer set with an average of 25 and a median of 30. If n is in S, what is the least possible value of n?

(A) The population of town A is smaller than the population of town B. (B) The population of town B is smaller than the population of town A. (C) The three towns have equal populations. (D) The population of town C is greater than the population of town A. (E) The population of town A is greater than the population of town C.

150 A litter of puppies weighed 0.6, 0.8, 1.2, and 1.4 pounds at birth. Which of the following, if it happened one month after birth, would NOT change the standard deviation that the weights of the puppies had when they were born? (A) Each puppy gained 1 pound. (B) Each puppy tripled its weight. (C) Each puppy doubled its weight. (D) Each puppy squared its weight. (E) Each puppy increased its weight by 10 percent.

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5 152 A farmer counted the number of apples in each of 15 baskets. What was the standard deviation of the number of apples in the 15 baskets? (1) The number of apples in each basket was the same. (2) The median number of apples for the 15 baskets was 7, 153 Sets P and M consist of 7 positive integers and the average (arithmetic mean) of each set is 30. The integers 25, 30, and 35 are in each set, which set has the greater standard deviation? (1) Positive integer 32 is in set M. (2) Positive integer 20 is in set P.

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horacio.quiroga@gmail.com ANSWERS: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

A E B D A A B C D D C B E D E A C B D C

21 A 22 B 23 D 24 A 25 A 26 D 27 C 28 E 29 C 30 D 31 D 32 C 33 C 34 A 35 B 36 A 37 B 38 A 39 D 40 E

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

B C C A D A D C D C E B D A A C E C D B

61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

E 81 A 82 D 83 C 84 A 85 C 86 B 87 E 88 B 89 A 90 B 91 C 92 E 93 D 94 B 95 A 96 D 97 A 98 C 99 A 100

B C B C B A B A D D C A C C A B E D A D

101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120

A C E B C D C A B D C B A B C A E A A B

121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140

D D A C B D A D C D D A D C D D A D D E

141 142 143 144 145 146 147 148 149 150 151 152 153

A D A E A E E B E A A A E

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Explanations to Section 1.1 Combinatorics - Combinations

is a plain vanilla combinations problem, the teacher is picking groups of 3 people out of 5 and order does not matter. So we

anymore) for our 4 free spots, therefore the question became: how many groups of 4 people

use our combinations formula

Problems 4, 5, and 6 belong to the same type of problem.

1- (A)This

! ! !

= 10. Our

numerator is 5! because we have a total of 5 people, our denominator is (3!)(2!) because we pick 3 people and we do NOT pick 2. Problems 1, 2, and 3 belong to the same type of problem.

2 - (E)

Again, a plain vanilla combinations problem, Lisa will pick 4 out of 7 blouses and we don't care about the way she puts them into her backpack. So we go

! ! !

= 35. Problems 1,

can we pick out of 5? So we go

! ! !

=5 .

6 - (A) Jane and Keisha already took 2 spots so we got 2 spots left. Also, because we already picked Jane and Keisha, we got 4 people to pick from, therefore the question is: how many groups of 2 people can we pick out of 4, or !

! !

= 6 . Problems 4, 5, and 6 belong to the

same type of problem.

2, and 3 belong to the same type of problem.

7 - (B) Let's consider first the girls, we got 4 3 - (B)

Counting handshakes is the same as picking groups of 2 out of something, and order does not matter, if Mike shakes Peter's hand that is the same as when Peter shakes Mike's. So we go

! !

= 56. Problems 1, 2, and 3 belong to the

same type of problem.

4 - (D) Again, order does not matter here, so because Peter already took one place in the three-people group, we got two spots available and only six people left, so the question in fact is asking: how many groups of 2 can we pick out of 6? So we go

! !

= 15. Problems 4, 5, and 6

belong to the same type of problem.

5 - (A) As in problem 4, order does not matter and the restriction is whom we pick and whom we don't. Here one spot is already taken by Mary so we got only 4 spots left. Patricia cannot be in the group, so we got in all 5 people to pick from (because Mary and Patricia are not eligible

girls and 2 spots for them, so we go

! !

ways of picking the girls. Now with the guys, we got 5 boys and 2 spots for them, so we go

! !

= 10 ways of picking the boys. What do we

do now? We multiply one by the other. Why? Because per every possible group of girls we got so many groups of boys, so we go (6)(10) = 60. Problems 7, 8, and 9 belong to the same type of problem.

8 - (C)

As in our problem before, we got 2

spots for our 6 biologists, so we go

! !

= 15 .

For our 5 engineers we got 3 spots, so we go

! ! !

= 10. Multiply one by the other (see

problem 7) and we get (15)(10) = 150. Problems 7, 8, and 9 belong to the same type of problem.

9 - (D) As in problems 7 and 8, we got 3 spots for our 5 ladies, so we go

! ! !

= 10, and 3 spots !

for our 6 gentlemen, so we go ! ! = 20, we then

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horacio.quiroga@gmail.com multiply one by the other and (10)(20) = 200. Problems 7, 8, and 9 belong to the same type of problem.

10 - (D)

Here "at least one" is our big clue. The point is, out of all the possible groups of 3 that we can form regardless of nationality, the only groups which are NOT good for us are the ones made up exclusively of Canadians. Regardless of nationality we got

! !'!

= 220

possible groups. Canadian-only groups will be ! ! !

= 20. Because the Canadian-only groups are

the ones which are not good fo us, se subtract them from the total so 220 - 20 = 200. Problems 10,11, and 12 belong to the same type of problem.

11 - (C)

As in problem 10 "at least one" is our clue here. From all the groups that we can form regardless of gender we are going to subtract the groups which are not good for us, that is the groups made up of girls only. So we go

! ! !

= 56, or all the groups regardless of

gender, and

! ! !

= 10, or the girl-only groups.

Subtract the latter from the former and we get 56 - 46 = 10. Problems 10,11, and 12 belong to the same type of problem.

12 - (B) Again, "at least one" is our clue here,

0(09 )

. In our case we get

( 9 )

= 14, problem

solved. Another way is using combinations, and we include it here because some GMAT problems are "hidden" diagonals of a polygon problem, such as problems 14 and 15 below. See that if our polygon has 7 sides, it has 7 vertices, and all the way of joining any two of our vertices are

! !

= 21. By the way, why do we

use the combinations formula? Because what we are doing is picking groups of 2 vertices out of 7, and order does NOT matter, meaning that line segment AB is the same as line segment BA. Now, of course by using our formula we got 21, which is more than 14, so in order to get to the answer we subtract the 7 sides of the polygon from 21 (because the sides are NOT diagonals) and we get 21 - 7 = 14. Problems 13,14, and 15 belong to the same type of problem.

14 - (D) First of all, as we saw in problem 3, handshakes are ways of picking groups of 2 out of a group of people and order does not matter, because Maria shaking Jane's hand would be the same as Jane shaking Maria's. But here we have the restriction that a girl cannot shake hands with the two girls standing next to her. So the problem could be rephrased as how many diagonals does a hexagon have? And so we go !

! !

= 15 and we subtract the 6 sides of the

and we will subtract from all the possible groups regardless of qualifications those groups made up production-line workers only. So we go

polygon (or the handshakes between any two girls standing next to each other), so we go 15 6= 9. Problems 13,14, and 15 belong to the same type of problem.

and 12 belong to the same type of problem.

15 - (E)

! − ! ! ! !

= 165 − 10 = 55. Problems 10,11,

13- (E)This

one looks like a geometry problem....because it is a geometry problem! But we include here because it can be looked at as a combinations problem too. One way of solving it is knowing by heart the formula that gives you the number of diagonals of an x-sided polygon:

We have a total of (3)(5) = 15 players. Imagine the 15 players standing on a circle with the members of each 3-player team standing next to each other. Now the problem became very much like problem 14 (handshakes are similar to chess games, or any 2-people game for that matter, because we pick 2 out of something and order does NOT matter), or a

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horacio.quiroga@gmail.com "hidden" number of diagonals of polygon problem. In this case we are being asked the number of diagonals of a 15-sided polygon and we could go

( )( )

= 95 . We can also solve

the problem using combinations and we go !

! !

= 105. From this we subtract the games

between members of the same team, which is the same as picking 2 people out of 3, or

! !

games per team, but we have 5 teams so we have (3)(5) = 15 "internal" games. In the end we do 105 - 15 = 95. Problems 13,14, and 15 belong to the same type of problem. In order to solve this problem you have to be very familiar with the way of solving problems 7, 8, and 9. Call F the number of ladies in the group and the equation we get is ! ! !

! ∗ ! !

! !( 9 )!

= 140. Because

! !( 9 )!

= 4 , and multiply both sides of the

= 35, we divide

both sides of the equation by 35 and get that

equation by 3! and we get that

! ( 9 )!

= 24. At

this point using algebra to solve the problem would be more complicated than simply plugging in the answer choices, one of which is &!

the value of F. Start from choice C and ( &9 )! =

720, much larger than 24, so we need a smaller value for F and try choice B, which will not work either. But try choice A and you get = 4! = 24,

which

course

works.

Problems 16,17, and 18 belong to the same type of problem.

17 - (C)

In order to solve this problem you have to be very familiar with the way of solving problems 7, 8, and 9. The equation we get is ! 0! ∗

! ! !(09 )!

and

!( 9 )!

= 10, and we hit the jackpot at our

first test, so C is our answer. Problems 16,17, and 18 belong to the same type of problem. 18 - (B) In order to solve this problem you have to be very familiar with the way of solving problems 7, 8, and 9. Call B the number of belts and the equation we get is Because

! ! !

= 60. Because

! !

= 6, we divide

! ! ∗ ! ! !( 9 )!

= 24.

= 4, we divide both sides of the

equation by 4 and get that

16 - (A)

! ( 9 )!

solve the problem would be more complicated than simply plugging in the answer choices, one of which is the value of x. Start from choice C

!( 9 )!

= 6. At this

point using algebra to solve the problem would be more complicated than simply plugging in the answer choices, one of which is the value of B. Start from choice C and

!( 9 )!

= 10, which is

not the value we are looking for, and because we need a smaller number we pick choice (B) and we get

!( 9 )!

= 6, which is the number we

were looking for. Problems 16,17, and 18 belong to the same type of problem. `19

- (D)

Call B the number boys, if you

exclude Peter you got that ( 9 )! = 35, !( 9 )! ( 9 )! = 210. ( 9 )!

( 9 )! !( 9 9 )!

= 35, or

multiply both sides by 3! and At this point using algebra to

solve the problem would be more complicated than simply plugging in the answer choices, one of which is the value of B. Start from choice C

as usual and we get

( 9 )! ( 9 )!

( )! ( )!

= 120, which

is less than 210 so we try a greater number for B and we go

( 9 )! ( 9 )!

( )! ( )!

= 210, which was the

number we were looking for. Problems 19 and 20 belong to the same type of problem.

both sides of the equation by 6 and get that 0!

!(09 )!

= 10. At this point using algebra to

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20 - (C)

The number of cars is C, so the

equation we get

(#9 )! !(#9 9 )!

= 28, or

(#9 )! !(#9 )!

28. At this point using algebra to solve the problem would be more complicated than simply plugging in the answer choices, one of which is the value of C. Start with choice C and ('9 )! !('9 )!

( )! = !( )!

factorials in the original expression and then simplify but to flip the bottom one and then simplify, arriving to the solution in a much faster way. Problems 21 and 22 belong to the same type of problem.

28, and we were lucky

22 - (B) This is not a difficult problem, really,

with our first choice. Problems 19 and 20 belong to the same type of problem.

I just include it here to teach a fast calculation trick, the ratio of the number of different fourtoys groups that can be formed to the number of different five-toys groups that can be formed is

we get

21 - (A) This is not a difficult problem, really, I just include it here to teach a fast calculation trick, the ratio of the number of different fivemember groups that can be formed to the number of different six-member groups that can be formed is

20! 5!15! 20! 6!14!

20!

5!15!

∗

6!14! 20!

2 5

, the

point here is not to make the calculation of the

15! 4!11! 15! 5!10!

15!

4!11!

∗

5!10! 15!

the point here is not to

make the calculation of the factorials in the original expression and then simplify but to flip the bottom one and then simplify, arriving to the solution in a much faster way. Problems 21 and 22 belong to the same type of problem.

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Explanations to Section 1.2 Combinatorics - Slots

23 (D) We got 3 ways of picking the first

30 (D) We subtract from all the possible ways

letter, three of picking the second, and three of picking the third one, so we go (3)(3)(3)=27. If you chose (A) you wrongly assumed that letters could not be repeated and so did (3)(2)(1)=6, but nowhere in the problem does it say that you cannot repeat your letter, in other words, AAA, AAB, BBB, or ABC are all possible codes.

of 5 women standing in line without restrictions (see problem 28) the ways in which actually Mary stands on the left spot. So we go 120 - 24 = 96.

31 (D) If Jamal stands on the left we got

24 (A) In this case, letters CANNOT be

(1)(5)(4)(3)(2)(1) = 120 ways, same thing for Anthony. Because it is Jamal or Anthony, we add both ways 120 + 120 = 240.

repeated (contrast this with problem 23) so we go (3)(2)(1)=6.

32 (C) We got 2 ways for the first spot and 1

25 (A) We got 5 ways of picking the first

way for the fifth. Once we take care of the boys we go for the girls, and in the end we got (2)(3)(2)(1)(1) = 12.

number, 1 one way of picking the second 1 (it has to be the same as the second one, and 5 ways of picking the third one, so we go (5)(1)(5) = 25 (there are no restrictions for the third digit, for instance, 555 is a valid code).

26 (D) In this case the third number MUST be different from the other two (contrast this with problem 21), so we go (5)(1)(4) = 20.

27 (C) We got 5 ways of picking the first letter, 21 o picking the second one, and 4 of picking the third one because it must be different from the first, so (5)(21)(4) = 420.

28 (E) Whenever you got items lined up, be it people, trees, cars, or Halloween pumpkins, the formula is the factorial of the number of items, in our case 5! = 120.

29 (C) We got only one option for the first spot that is Mary, 4 for the second, 3 for the third, 2 for the fourth, and 1 for the fifth, so we go (1)(4)(3)(2)(1) = 24.

33 (C) The two boys can take the first and second spots, or the second and third, or the third and fourth, or the fourth and fifth, or the fifth and sixth. So we got five different ways of the boys "taking their spots", but we can switch the boys (Peter on the left and Paul on the right or Paul on the left and Peter on the right), so we got (5)(2) = 10 ways of sitting the boys. Now, assume that the boys took spots 1 and 2, we then got (4)(3)(2)(1)=24 of sitting the girls, because we got 10 ways of sitting the boys and the same is going to be true no matter what spots the boys take. So because we got 10 ways of placing the boys and 24 ways of sitting the girls we go (10)(24) or (2)(5)(4!) = 240. 34 (A) It would be too difficult and time consuming to count all the possible ways of getting the six people in line with the two boys not standing next to each other. So what we do is we subtract from all the possible ways, or 6!, the ways in which the boys are actually sitting next to each other (see problem 33 above). In other

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horacio.quiroga@gmail.com words, we go 6! - (5)(4!)(2) = 6! - (5!)(2) = (6)(5!) - (5!)(2) = 5! (6 - 2) = (5!)(4).

35 (D) We got 3 ways of picking the middle digit (1, 3, or 5). And then because we CANNOT repeat digits we got 5 ways for the first, 4 for the second, and so on: (5)(4)(3)(2)(1) = 120. 36 (A) Our number will be a five-digit number, and numbers maybe repeated because we not

told otherwise. For our number to be odd the last digit must be odd and we got 3 options (1, 3, or 5) and for the first one only two options (4 or 5) because if not our number will be under 40,000. So we got (2)(6)(6)(6)(3) = 1296.

37 (B) Our number must be odd so it must end in 1, 3, or 5. Other than that we have no other restrictions (digits can be repeated), so we have (5)(5)(3) = 75

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Explanations to Section 1.3 - Combinatorics Miscellaneous

38 - (A)

43 - (C)

Circular arrangements with no fixed positions, so our solution is (6 - 1)! = 120.

First we calculate the number of two-letter codes, which is (8)(8) = 64, then we calculate the number o three-letter codes, which will be (8)(8)(8) = 512. Add them up (because it is one way or the other) and our answer is 512 + 64 = 576.

39 - (D) Circular arrangements with no fixed positions, so our solution is (5 - 1)! = 24. 40 - (E) This is a repeated letters problem and we solve it by dividing the number of letters in the word by the product of the factorials of the repeated (6)!

letters. So

(2)!(2)!

= 180.

41 - (B) This is a "hidden" repeatede letters problem that can be rephrased as: how many different 7-letter codes can be formed by rearranging the letters FFMMMLL? The answer is (7)!

(2)!(3)!(2)!

= 210.

42 - (C)

44 - (A) Now letters cannot be repeated. First we calculate the number of two-letter codes, which is (8)(7) = 56, then we calculate the number o three-letter codes, which will be (8)(7)(6) = 336. Add them up (because it is one way or the other) and our answer is 336 + 56 = 392. 45 - (D) The only restriction is that the thousandths digit must be odd (so we got 4 ways), therefore the number of codes we can get is (6)(5)(4)(4)(3)(2)(1) = 2,880. Notice that first we write our 4 ways in the thousandths digit and then we go 6, 5, 4, ... etc.

First calculate all the possible codes without 6!

restriction, this will be = 360, notice that we 2! divide by 2! because we got two 1s. Then we can calculate all the possible ways in which the two 1s can go together, which is 5 ways (firstsecond, second-third, third-fourth, fourth-fifth, fifth-sixth); now for each of the possible ways of putting the 1s together we can arrange the other 4 numbers in 4! says, or 24. So we got (5)(24)=120 ways of the two 1s going together. Subtract 120 from 360, and we get 240.

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Explanations to Section 2.1 - "Pure" Probability Problems

46 - (A)

In order to roll a 3 both times we need to roll a 3 first and a 3 second. Notice the boldface and italics of the "and", in probability jargon that means multiplication. So our scenario is:

3 and 3 = ) * ) * =

In order to roll 3 heads we need to roll a head the first time and a head the second time and a head the third time. Notice the boldface and italics of the "and", in probability jargon that means multiplication. So our scenario is:

48- (C) Now we got two roll a 3 exactly once; that means that we either roll a 3 the first roll and NOT a 3 in the second roll,or NOT a 3 in the first roll and a 3 in the second. Notice the boldface and italics of the "or", in probability jargon that means addition. So our scenarios are:

HTH = ) * ) * ) * = or

THH = ) * ) * ) * = Finally, because it is one way or the other or the other, we add our results and get

, ,

= .

Notice, and this is very important for more complicated binomial distribution problems, that the results of each scenario are identical, so instead of adding we could have multiplied

(3) ) * = .

50 - (C) Here your key expression is at least

subtract this from one and get 1 −

Finally, because it is one way or the other, we

NOT rolling a 4 in either roll is) * ) * =

not 3 and 3 = ) * ) * =

add our results and get

one, which means that we will subtract from 1 (or absolute certainty that something will happen) the only scenario which is not good for us, that is never rolling a 4. The probability of

3 and not 3 = ) * ) * =

H and H and H = ) * ) * ) * = .

HHT = ) * ) * ) * =

47 - (D)

prove useful in your MBA, taught in Stats class because it is very useful for quality control, for instance. Back to our problem, we have to get exactly two heads, so our possible scenarios are:

= = .

Contrast this way with the long (not recommended) way of solving the problem, which would be dealing with three separate scenarios and then adding them:

49 - (D)

This is a binomial distribution problem, and we will go into deeper explanations as our problems get more complicated. By the way, binomial distribution is one of the few things of the GMAT that will

4 & X = ) * ) * = or

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X&4=

) * ) *

54 - (A) Now we got a serious binomial distribution problem, see that we are looking at picking exactly two red marbles, and of course the other two ones must be blue, so one possible

scenario

4 &4 =) * ) * =

, ,

same

result as before but of course this way takes a much longer time, and when working with the GMAT you are basically in the business of shaving seconds.

51 - (E) Again, here your key expression is at least one, which means that we will subtract from 1 (or absolute certainty that something will happen) the only scenario which is not good for us, that is never getting a red marble. The probability of NOT getting a red marble (or the

probability of getting two blue ones is ) * ) * =

, subtract this from one and get 1 − =

notice that we do not care about the first roll, any number we roll is good for us! Then on the second roll we got one possible number, which must be equal to the first one, and finally in the third roll we have to pick a number different from the two other ones, so we go

getting

) &* )'* ) * ) * = ' '

If you try other scenarios, such as RBRB or BBRR, you will notice that you always end up . Now, the question is how many

possible scenarios we have, and to do so we got to go back to our combinatorics repeated letter problem and answer the question: how many possible four letter codes can we form by rearranging the letters RRBB? And the answer is !

! !

= 6. So we got 6 ways for our scenario to

) * ) * ) * = .

53 - (D)

Again, the key to solving this problem is to notice that we do not care about the first roll, any number we roll is good for us! Then on the second roll we got one possible number, which must be equal to the first one, and finally in the third roll we have to again pick

the same number, so we go ) * ) * ) * = .

happen and the answer then is (6) ) * = = .

55 - (A)

Yet another serious binomial distribution problem, see that we are looking at picking exactly two female students, and of course the other two ones must be male, so one

52 - (B) The key to solving this problem is to

RRBB

) *) *) *) * = ) *) *) * = ) *) * =

Now we add them up and get

is:

possible scenario is: FFMM = )'* ) * ) * ) * =

)'* ) * ) * ) * = ) *.

you

try

other

scenarios, such as FMFM or MFMF, you will notice that you always end up getting

. Now,

the question is how many possible scenarios we have, and to do so we got to go back to our combinatorics repeated letter problem and answer the question: how many possible four letter codes can we form by re-arranging the letters FFMM? And the answer is

! !

= 6. So

we got 6 ways for our scenario to happen and

the answer then is (6) ) * = = .

56 - (C)

Yet another serious binomial distribution problem, see that we are looking at picking exactly two defective computers, and of course the other three ones must be non-

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horacio.quiroga@gmail.com defective, so one possible scenario is: DDKKK

= ) *) *) *) *) * =

) * ) * .

If you try

other scenarios, such as DKDKK or KKKDD, you will notice that you always end up getting ) * ) *

. Now, the question is how many

possible scenarios we have, and to do so we got to go back to our combinatorics repeated letter problem and answer the question: how many possible five letter codes can we form by rearranging the letters DDKKK? And the answer

is ! ! = 10. So we got 10 ways for our scenario

to happen and the answer then is (10) ) * ) * .

57 - (E)To do this problem you have to realize one very important thing, the two dice will NOT land at the same time, one will land a fraction of a second after the other. Now, the time elapsing from the instant the first die shows its face to the instant the other die does so is immaterial to us, it could be 0.1 seconds, 1 second, or one day. So imagine the first roll is today, and the second one tomorrow, today the numbers which are good for us to add up to a 7 between the two dice are all the numbers from 1 through 6, inclusive, it all depends on what shows up on the second die. Say we roll the first die, then, and we get a 3; well, the next day ONLY a 4 will

do. In other ways, no matter what you roll the first time, the second time only one number will be good for you in order to get the desired sum.

So we got for the first die and for the second

die, or ) * ) * = .

58 - (C)

Following the same reasoning in problem 57, for the first die only a 1 or a 2 are good for us (because if we get a 3 or above, there is no way to add up to three), now depending on what we get on the first roll, only one number will be good on the second roll, so

we got ) * ) * = , and that is the probability

of getting a 3. To get an 8, only the numbers 2 through 6, inclusive, are good for us on the first roll and again, only one number will be good on

the second roll, so we got ) * ) * = , and that

is the probability of getting an 8. Now, because it is a 3 or an 8 we add both cases and get

= .

59 - (D) Two get a sum of 9 our possible rolls are 3-6, 4-5, 5-4, 6-3, see that we got 8 "turnups", two of which are 3s, so my probability is 2 8

= . 4

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Explanations to Section 2.2 - Probability Meets Comb Problems

60 - (B) The combinations way of going about

possible outcomes, which are all the groups of 4

this problem is by dividing all the possible groups of 1 boy and 2 girls that we can pick out of 4 boys and 4 girls, respectively (our favorable outcomes), by all the possible groups of 3 people that we can pick out of 8 people (all possible outcomes). Our expression will thus look like:

people out of 9, or

)

R! R! *) * S!T! ;!;! U! T!V!

'! . ! !

A good trick when the

calculation is not that easy is instead of calculating both combinations and dividing we flip the expressions and multiply

! ! ! '!

( )( )

= (')( ) =

(see

! ! ! ! ! '!

explanations

problems 21 and 22). =

= .

63 - (D)

61 - (E)

First of all let's calculate our favorable outcomes, which are all the possible groups of 3 people out of 7 that include both Peter and Michael. Notice that two spots out of 3 are already taken by Peter and Michael, so in the end our favorable outcomes will be all the ways in which we can fill up the spot that we still have free with any of the 5 people still not picked, so we are picking 1 person out of 5 or

! ! !

First of all let's calculate our favorable outcomes, which are all the possible groups of 4 balls out of 8 that include neither the blue nor the red ball. Notice that the 4 spots are free because we have not placed any ball yet, so in the end our favorable outcomes will be all the ways in which we can fill up the four spots that we still have free with any of the 6 balls still not picked (the red and blue ball are already out of the bag), so we are picking 4 balls out of 6 or ! ! !

= 15 ways. Then we divide that by all the

ways. Then we divide that by all the possible outcomes, which are all the groups of 3 people

possible outcomes, which are all the groups of 4

out of 7, or

calculation is not that easy is instead of calculating both combinations and dividing we

= .

! ! !

= 35. So in the end we get

62 - (A)

First of all let's calculate our favorable outcomes, which are all the possible groups of 4 people out of 9 that include Jamal but not Maria. Notice that one spot out of 4 is already taken by Maria, so in the end our favorable outcomes will be all the ways in which we can fill up the three spots that we still have free with any of the 7 people still not picked (we told Jamal to pack and go after Maria took her spot), so we are picking 3 people out of 7 or ! ! !

balls out of 8, or

! . ! !

A good trick when the

flip the expressions and multiply ! !

! !

( )( ) ( )( )

(see

! ! ! ! ! !

explanations

problems 21 and 22).

64 - (C) All the outcomes here will be all the groups of 3 marbles that we can pick out of 9, or 9!

. For the favorable outcomes, the key

3!6!

expression is "at least one" (see problems 10, 11, and 12) we subtract from all the groups, or

= 35 ways. Then we divide that by all the

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3!6!

horacio.quiroga@gmail.com the groups that contain green marbles only, or 5!

. So in the end we get

3!2!

9! 5! − 3!6! 3!2! 9! 3!6!

74 84

66 - (C)

groups of 3 students that we can pick out of 11, 11!

. For the favorable outcomes, the key

3!8!

possible groups of 2 Party A senators, or 5!

9). So in the end we get

the groups that contain boys only, or the end we get

11! 5! − 3!8! 3!2! 11! 3!8!

155 165

. So in

3!2!

2!4!

(see the explanations to problems 7, 8, and

and 12) we subtract from all the groups, or

3!8!

all the possible groups of 2 Party B senators, or

expression is "at least one" (see problems 10, 11, 11!

4!7!

For the favorable outcomes we multiply all the

2!3!

11!

groups of 4 senators out of a total of 11, or

65 - (A) All the outcomes here will be all the or

All the outcomes here will be the

)

6! 5! *) * 2!4! 2!3! 11! 4!7!

150 330

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15 33

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Explanations to Chapter 3 - Mixtures Problems

67 - (B) To get the ratio of the weight of the ham Jamal bought to the cheese he bought we use the method we learned above. The distance from the weighted average to the price of the ham is 3, and the the one from the weighted average to the cheese is 1, so the ratio will be the flipside of that or

W X

1 3

. Now our problem becomes a

ratios problem and 1= + 3= = 2, solve for x and x is .5, since he bought 3 parts of cheese we know that he bought 3(.5) = 1.5 pounds of cheese.

The distance from Jamal to the average is 4, and the distance Mary to the average is 6, so the ratio is

[

6 4

3 2

. Now say Jamal read 3 books and

Mary read 2 books, for a total of 5 books, Mary read 2 out of 5, or 40%. 72 - (C) The distance from the women to the average is .2, and the distance from the men to the average is .8, so the ratio is

$ _

= . = . Now we set our

ratios problem and 1= + 4= = 180, so x = 36 and since we got 1 part of men 1(36) = 36.

68 - (E)

73 - (E)

The distance from the nickels (the 5-cent coins) to the average is 5, and the distance from the quarters (the 25-cent coins) to the average is 15,

Since the weighted average is given in terms of the white t-shirts we will concentrate on them and ignore the red and the green ones. The distance from school A to the average is 2, and the distance from school B to the average is 8, so the ratio of students from A to be is 8:2, or 4:1. Now say 4 students came from A and 1 student

so the ratio is

Y Z

15 5

= . Now we set our ratios 1

problem and 3= + 1= = 8, so x = 2 and since we got 3 parts of nickels 3(2) = $6.

came from B, for a total of 5 students, so of the

69 - (B) The distance from the workers to the average is 2, and the distance from the managers to the average is 4, so the ratio is

[

2 4

= . Now we 2

set our ratios problem and 1= + 2= = 90, so x = 30 and since we got 1 part of managers 30 (1) = 30. 70 - (A) The distance from the female attendants to the average is 300, and the distance from the male attendants to the average is 100, so the ratio is ]

[

100 300

= . 3

71 - (B)

students came from A, or 80%.

74- (D) We will use the savings account average to find the ratio of men to women in Springville. Since the distance from men to the average is 1 and the distance from women to the average is 2, we know that the ratio of men to women is 2:1. Now imagine you got exactly 2 guys and 1 lady, the weighted average for the checking accounts will be

2(800)+1(1400) 2+1

=1,000.

75 - (B) Statement 1) is not sufficient because we have no idea about the number of apples.

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horacio.quiroga@gmail.com Statement 2) is sufficient because it gives us the ratio, the "weight", of the apples and oranges sold. Put it this way, imagine she sold exactly 3 oranges and 1 apple, we can calculate a weighted average:

3(60)+1(80) 3+1

= 65. But we

need not do this calculation, because the big principle to calculating weighted averages is to know the "weight". 76 - (A) Statement 1) is sufficient because it gives me the "weight", see problem 75. Statement 2) is not sufficient because I have no idea about the number of female students.

From the stem of the problem we can calculate using our method that the ratio of a to b is 3:2. Statement 1) sufficient because we can do a 3

direct proportion: = 2

12 5

and we solve for b.

Statement 2) is also sufficient because we know 3

that = and a + b = 20 (remember that a + b = 5

Statement 2) is not sufficient either because we get the ratio of the assets under management but not what percentage of each is invested in biotechs. Together the two statements are sufficient because, say, Aggresinv manages $2 and Conservinv manages $1 we can do a weighted average equation:

77 - (D)

Statement 1) is not sufficient because we have no idea about the assets under management by each fund. Say Aggresinv is a very small fund and Conservinv is a very large one, that 70% of Aggresinv will not weigh that much. The opposite would be true if Conservinv were very small and Aggresinv very large.

c), so we got two equations with two unknowns and we can solve for b. 78 - (A) Statement 1) is sufficient because it gives us the weighted average and so we can calculate the ratio of men to women

\ [

= , so if the club has 1

3 ladies and 1 guy then 3 out of 4 people are women, or 75%.

2(.7)+1(.4) 2+1

= .60.

80 - (A) Statement 1) is sufficient because if you think about it, the weighted average of people who will apply for college will be anywhere between 76 and 80, inclusive. Suppose our school were a girl-only school then 80 percent of the students will apply for college, but no way a larger percentage will do so. Statement 2) is not sufficient because there is no way of telling how many students are female and how many are male. 81 - (B) Statement 1) is not sufficient because we do not know how many people live in town A.

Statement 2) is not sufficient because we have no idea about the number of men in the club.

Statement 2) is sufficient because we know that 60% of the people make an average of 36k and therefore 40% make an average of 55k, so we can calculate our weighted average.

79 - (C)

82 - (C)

Notice that we are asked what percent of the combined assets of the two funds are invested in biotech stocks.

Statement 1) is not sufficient because we do not know anything about the income of people who live in town B.

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horacio.quiroga@gmail.com Statement 2) is not sufficient because we do not know anything about the income of people who live in town A. Together we know that the distance from A to the average is 3 and the distance from B to the

average is 2, so flipping our distances we know that the ratio of the populations of A to B will be 2:3. Say 2 people live in A and 3 people live in B, then 2 out of 5 people live in A, or 40%.

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Explanations to Chapter 4 - Rate (work)

83 - (B) Simply use our formula

( )( )

= = 1 . Remember, to find the time it

will take two items to do a job together multiply the times in the numerator and add them in the denominator.

84 – (C)Simply use our formula '

( )( ) ,

= 2 . Remember, to find the time it will

take two items to do a job together multiply the times in the numerator and add them in the denominator.

85 – (B) First of all, we do not care about the 500 widgets, it could 500 or 1,000, or a billion,it would not change anything. For us the 500 widgets are one job. So statement (1) is not sufficient because we get infinite values for x and y, which will yield different results when plugged into our formula. State (2) is sufficient because it is precisely applying our formula and it tells us the time 01

1.2 hours so we work with nicer numbers. ,0

= 1.2, multiply

both sides by (3 + x)and distribute the right side and get that 3x = 3.6 + 1.2x , send the 1.2x to the left and 1.8x = 3.6, and x = 2.

87 – (B) From the stem of the problem we 01

(0)( ) 0,

= 3 , and

8x = 3x + 24, and of course we can solve for x.

88 – (A) Call J the time it takes Jamal to do the job, so Pedro will do it in 2J hours (it takes Pedro twice the time) and that is the question we are trying to answer. The equation we get is

( `)(`)

`,`

= 6, and 2JJ =

(3J)(6), divide both sides by J and 2J = 18, which is what we are looking for, because the question is how long it will take Pedro to do the job alone.

89 – (D) This is a typical case of reverse proportion because more workers will take a shorter time and fewer workers a longer time, so we multiply corresponding tings (8)(3) = (x)(2), solve for x and x = 12.

90 – (D) Again, a typical case of reverse

86 – (A)First of all, transform the 1 into ( )(0)

because our equation now is

it will take to do the job 0,1 = .

Using our equation

sufficient because we get a value for our denominator and so xy = (12.8)(3), but see that there are infinite values of x that we can get from here. Statement (2) is sufficient

get that 0,1 = 3. Statement (1) is not

proportion because more combines will take a shorter time and fewer combines a longer time, so we multiply corresponding tings (5)(8) = (4)(x), solve for x and x = 10.

91 – (C)Again, a typical case of reverse proportion because more pipes will take a shorter time and fewer pipes a longer time, so we multiply corresponding tings (3)(6) = (x)(2), solve for x and x = 9. Now, heads up, we are being asked how many more pipes, not how many pipes, so our job is still not

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done and we subtract 3 from the new number of pipes, so 9 – 3 = 6.

92 – (A) And yet another example of reverse proportion because more workers will take a shorter time and fewer workers a longer time, so we multiply corresponding tings (4)(9) = (3)(x), solve for x and x = 12. Now, heads up, we are being asked how many more hours, not how many hours, so our job is still not done and we subtract 9 from the new number of worker, so 12 – 9= 3.

of an hour, or 20 minutes to do

of the job

and therefore complete the task at hand.

95 - (A) They can do the job together in ( )( ) ,

hours. So their joint rate is

remember that the rate is always the flipside, or reciprocal in math parlance, of the time it will take to do the job, so in 1.5 hours, using RT = D, they will do

) * (1.5) =

of the job, therefore

1 − = of the job remains to be done.

Because Maria can do the full job in 5 hours,

93 – (C) Here we have to use the RT = D equation. See that if Sandra can do the full job in 6 hours then she does one hour, so

per hour is her rate. If she

) * (2) = of the job while working alone,

therefore 1 − = of the job remains to be

done. Now, both ladies can do the full job together in

( )( ) ,

hours. So to do

of the job they will take hours.

hour to

complete the job.

of the job in

worked alone for two hours she did of

it will take her (5) ) * = 1

96 - (B) Here the secret is very simple, the ratio of the fraction of the job done by one item to the fraction of the job done by the other item is the reciprocal of the time it takes them alone to do the job, so if the ratio of their times is 5:4 then the ratio of the parts of the job done is 4:5. See that we do not care about the total time it took them to do the job, totally useless info.

97 - (E) As in problem 96 the secret is very 94 – (C) Together they can do the job in ( )( ) ,

hours. So their joint rate is ,

remember that the rate is always the flipside, or reciprocal in math parlance, of the time it will take to do the job, so in one

hour, using RT = D, they will do ) * (1) =

of the job, therefore 1 − =

of the job

remains to be done. Because pipe can do the

full job in 2 hours, it will take it (2) ) * =

simple, the ratio of the fraction of the job done by one person to the fraction of the job done by the other person is the reciprocal of the time it takes them alone to do the job, so if the ratio of their times is 20:30 then the ratio of the parts of the job done is 30:20, or 3:2.

98 - (D) As in problems 96 and 97, the ratio of the fraction of the job done by one person to the fraction of the job done by the other person is the reciprocal of the time it

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takes them alone to do the job, so if the ratio of their times is 4:7, the ratio of the parts of the job done will be 7:4. In this problem the question is what fraction of the whole job was done by Lisa; well, imagine that Lisa painted 7 square feet and Maria 4 square feet, together they painted 11 square feet, so Lisa painted

of the wall.

ratio of the fraction of the job done by one machine to the fraction of the job done by the other machine is the reciprocal of the time it takes them alone to do the job, so if the ratio of their times is 3:5, the ratio of the parts of the job done will be 5:3. In this problem the question is what fraction of the whole job was done by Machine B; well, imagine that Machine A manufactured 5 widgets and Machine B manufactured 3 widgets, together they manufactured 8 widgets, so Machine B machine

all

the

:5, then the ratio of the part of the job

done by Paul and Jose to that done by Mary will be, yes, you guessed it right, the

reciprocal of the times, or 5: . Now, if John plus Paul painted 5 square feet or feet, and Mary painted

widgets

manufactured, or 37.5%.

square feet, then ,

the three of them together painted

99 - (A) As in problems 96, 97 and 98, the

manufactured

square feet, so Mary painted

S; a Ra a

= =

101 - (A) As in problems 96, 97, 98, and 99 the ratio of the fraction of the job done by one pipe to the fraction of the job done by the other pipe is the reciprocal of the time it takes them alone to do the job. Of course, in this problem things are not so simple because we got three pipes working together. In order to make it fit our previous model, we consider Pipe A and Pipe B one pipe, and using our formula, Pipes A and B will do the whole job in

( )( ) ,

hours.

Now, if the ratio of time by Pipes A and B

100 - (D)As in problems 96, 97, 98, and 99 the ratio of the fraction of the job done by one person to the fraction of the job done by the other person is the reciprocal of the time it takes them alone to do the job. Of course, in this problem things are not so simple because we got three people working. In order to make it fit our previous model, we consider Paul and Jose one person, and using our formula, they (Paul and Jose together) will do the whole job in ( )( ) ,

hours. Now, if the ratio of time

together to that of Pipe C is

:8, then the

ratio of the part of the job done by Pipes A and B together to that done by Pipe C will

be, or 8: . Now, if Pipe A plus Pipe B

poured, say, 8 gallons of water into the tank or

gallons, and Pipe C poured

gallons,

then the three of them together poured ,

SV U ab U

gallons, so Pipe C poured

= '.

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Explanations to Chapter 5 - Rate (motion)

102 - (C) This is a "encounter" problem, so we add the rates and 60 + 50 = 110 miles per hour, that is the rate at which our two trains move towards each other. Using the RT = D formula, we get that (110)(T) = 220 miles, solve for T, and T = 2 hours.

103 - (E) Again, we have an encounter but with a twist, train Y started traveling one hour later than train X, so train X had already covered 70 miles (it moves at a rate of 70 mph) when train Y started moving. Therefore, the distance they will have to cover since Y started its trip will be 550 - 70 = 480. Since their "relative" speed is 70 + 50 = 120 miles per hour, using RT = D we get that (120)(T) = 480, and T = 4. Now, heads up!!! Four hours is the time since train Y started traveling, and we are being asked for the distance covered by X. So train X's time will be 5 hours, and again, using RT = D, we get that (70)(5) = 350.

104 - (B) Again an encounter problem. Car B starts traveling towards car A half an hour after car A started its trip. Because car A travels at a rate of 60 mph, it will have covered 30 miles in half an hour, so the distance they will have to cover together will be 330 - 30 = 300 miles. Since they travel at a relative speed of 60 + 40 = 100 mph, using RT = D equals (100)(T) = (300), so T = 3 hours. Now, we got 3 hours since car B started moving, that is at 3:30 pm (that is half an hour after car A did), so we add 3

hours to 3:30 pm and they will pass each other at 6:30 pm.

105 - (C) In this case we got a separation problem, our two cars are getting away from each other at a relative rate of 80 + 90 = 170 mph. Using RT = D it will take them (170)(T) = (510), solve for T and T = 3 hours.

106 - (D) Car A started traveling one hour before car B, so it had already covered 50 miles when car B started moving. After B starts moving they are getting away from each other at a relative rate of 50 + 70 = 120 mph. Now, because car A had already covered 50 miles while B was not moving, they will cover together a total of 410 - 50 = 360 miles. Using RT = D, (120)(T) = (360), solve for T and T = 3 hours. Now, do not forget that car A started one hour earlier, so its time is the 3 hours it took the two cars to get 360 miles apart plus the one hour the A covered alone. So A's time is 4 hours, and at a speed of 50 mph, it covered (50)(4) = 200 miles. 107 - (C) Train A started traveling half an hour before train B, so it had already covered 30 miles (it is travelling at 60 mph so in half an hour it does 30 miles) when train B started moving. After B starts moving they are getting away from each other at a relative rate of 60 + 80 = 140 mph. Now, because train A had already covered 30 miles while B was not moving, they will

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cover together a total of 310 - 30 = 280 miles. Using RT = D, (140)(T) = (280), solve for T and T = 2 hours. Now, those 2 hours are the time that train B travelled until they were 310 miles apart, so train B travelled (80)(2) = (160).

108 – (A) This is a catch up case, because A is travelling at 60 mph and B at 40 mph, A is catching up at a rate of 60 – 40 = 20 mph, since A has to catch up 100 miles, using RT = D we get (20)(T) = 100, so T = 5 hours.

109 – (B) Again, a catch up case, A left half an hour later, so B had already travelled 30 miles when A started chasing it. So A has to catch up 30 miles, and we know A’s catch up rate is 80 – 60 = 20 mph. Using RT = D, (20)(T) = 30, and T = 1.5 hours.

110 – (D) If Horacio evacuated 2 hours before Ricardo did, then Horacio was 120 miles ahead of Ricardo when Ricardo started driving towards Houston; so Ricardo has to catch up 120 miles and he is doing so at a rate of 90 – 60 = 30 mph. Using RT = D, (30)(T) = 120, solving for T it will take Ricardo 4 hours to catch up with Horacio, since Ricardo is driving at 90 mph he will be (90)(4) = 360 miles from New Orleans when he catches up with Horacio.

111 – (C) Ricardo is catching up at a rate of 90 – 60 = 30 mph. If Horacio is 30 miles ahead of Ricardo, then using RT = D we got (30)(T) = 30, solve for T and T = 1. Now, not only does Ricardo have to catch up but also gain 60 miles on Horacio, which will take Ricardo (30)(T) = 60, solve for T, and

T = 2. So Ricardo will have travelled for 1 + 2 = 3 hours until he is 60 miles ahead of Horacio, so he will have covered (90)(3) = 270 miles, add the 30 miles he already was away from New Orleans when our story started, he will have travelled 270 + 30 = 300 miles from New Orleans when he is 60 miles ahead from Horacio.

112 – (B) If it will take Mike 2 hours to complete the 12 miles and Amanda 3 hours to walk the same distance, meaning that the ratio of their times to do a certain distance is 2:3, then the ratio of the distances covered when they pass each other will be the flipside of that, or 3:2 (take a look at the explanations of problems 96 to 101, inclusive, what is true of time it takes to do a job and fraction of work done is true also of time it takes to cover a certain distance and distance walked). Now our problem becomes a straight ratio problem and 3x + 2x = 12 miles, solve for x and x = 2.4. So Mike will have walked 3 parts worth 2.4 each, so he will be (3)(2.4) = 7.2 miles from his starting point when he passes Amanda.

113 – (A) If Bill drove for half an hour at 80 mph before his car broke down, then his car broke down 40 miles from X During the same half hour Horacio drove 30 miles (he is driving at 60 mph), so he will be 40 – 30 = 10 miles behind Rahul when Rahul car breaks down. The question in the end is how long it will take Horacio to drive 10 miles if he is driving at 60 mph. Using RT = D, (60)(T) = 10, solve for T and T = an hour, or 10 minutes.

= of &

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114 – (B) If Cindy walked for half an hour at 6 mph before stopping to rest, then she stopped 3 miles from her home. Vanessa started walking exactly when Cindy stopped, so the question in the end is how long it will take Vanessa to walk 3 miles if she is walking at 4 mph. Using RT = D, (4)(T) = 3, solve for T and T = minutes.

of an hour, or 45

use the harmonic average formula and we ( )( )( ) ( ),( )

= 4.8. See that by using the

( )( )(0)

= 3. Multiply both sides by 2 +

( ),(0)

x and we get that 4x = 6 + 3x, so x = 6 and. See that by using the harmonic average formula we do not pay attention at all at the distance Jose walks.

119 – (A) Because this is a round trip we use the harmonic average formula and we get

115 – (C) Because this is a round trip we get

get

( )(2)(3) (2),(3)

. See that by using the harmonic

average formula we do not pay attention at all at the distance Maria walks.

120 – (B) Because this is a round trip we

harmonic average formula we do not pay attention at all at the distance she walks.

use the harmonic average formula and so we

116 – (A) Because this is a half and half

using the harmonic average formula we do not pay attention at all at the distance Paz walks.

using the harmonic average formula we do not pay attention at all at the distance Elida walks, so statement 1) is insufficient because it gives us the distance but no clue about x. On the other hadn, statement 2) if sufficient because it gives us a value for x and we can answer our question.

117 – (E) Because this is a round trip we

121 – (D) Peter’s RT = D equation

use the harmonic average formula and we

(Rp)(90) = D. Keisha’s time is 60 minutes because she started walking 30 minutes after Peter did, so her RT = D equation is (Rk)(60) = D. Notice that the distance is the same because they meet exactly half of the way between their starting points. So we set our equations equal to each other and (Rp)(90) = (Rk)(60) cross divide and we get

trip we use the harmonic average formula and we get

get

( )( &)(0) ( &),(0)

( )( &)('&) ( &),('&)

= 72. See that by

= 48. Multiply both sides by

40 + x and we get that 80x = 1920 + 48x, so 32x = 1920 and x = 60. See that by using the harmonic average formula we do not pay attention at all at the distance Jamal walks.

118 – (A) Because this is a round trip we use the harmonic average formula and we

want to get a value for

c d

( )( )(0) ( ),(0)

. See that by

= '& = .

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Explanations to Section 6 - Overlapping Sets

122 - (D)

124 - (C)

123 - (A) 125 - (B) Notice the expression "everybody has a watch or a ring or both", this means that our "not watch not ring" box has a zero.

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horacio.quiroga@gmail.com 126 - (D)

128 - (D)

127 - (A)

And we get an equation 40 + .20(w) = w, so 40 = .80(w), and w = 50.

TIP: in these problems in which numbers and percentages are combined, it is better to do the percentage calculation and then calculate the number. See how we do our table with percentages:

129 - (C)

Very easily we found that 10% of 80 took a test but did not write a paper, so 8 people is our answer.

And we again get an equation: 20 + .40(i) = i, so

20 = .60(i), and i = 33 %

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horacio.quiroga@gmail.com 130 - (D)

132 - (A) Again, we work with percentages and then calculate the actual numbers:

And again we get an equation: 35 + .5(H) = H, so 35 = .50 (H), and H = 70. 131 - (D)

And 50% of 1200 is 600, or choice A.

Because percentages and number are mixed we work with percentages first:

133 - (D) Again, we work with percentages and then calculate the actual numbers:

And 50% of 240, or 120 is our answer. And 10% of 360 is 36, or choice D.

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horacio.quiroga@gmail.com 134 - (C)

And we got an equation: = + 160 − 2= = 150, solve for x and x is 10, but we are looking for 160 - 2x = 140. 135 - (D) Watch out!! We are told that 60 of the females have blue t-shirts!!

Again, we work with percentages and then calculate the actual numbers:

And we got an equation: = + 50 − 2= = 40, solve for x and x is 10, but we are looking for 50 - x = 40. Now, dont forget that we so far worked with percentages, and 40 is in fact 40% of 180, or 72. 137 - (A) This three set problem is arguably one of the most difficult overlapping sets problems found in the GMAT. Although infrequent, it is worth taking a look at.

We solve the equation: = + 2= − 20 = 220, solve for x and x = 80. We want the number of students that have blue t-shirts, which is 60 + 80 = 140.

Call S all the people who play exactly one game, D all the people who play exactly two games, and T all the people who play all three games. We are told that out of 65 people 12 do not play any game at all so 65 - 12 = 53 people play at least one game. Now, if we add the 28 people that play backgammon, the 23 who play chess, and the 24 who play bridge, we get 75, why is that? Because we are double counting some of the people and triple counting others. And we get these two equations: 75 = 3T + 2D + S 53 = T + D + S Subtract on from the other and we get

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horacio.quiroga@gmail.com 22 = 2T + D, and because we are told that the people who play the three games are two-thirds of those who play none, then T = 8, so 22 = 2(8) + D, and D = 6.

139 - (D) This is the info given on the problem stem.

138 - (D) This three set problem is arguably one of the most difficult overlapping sets problems found in the GMAT. Although infrequent, it is worth taking a look at. Call S all the people who speak exactly one language, D all the people who speak exactly two languages, and T all the people who speak all three languages. We are told that out of 73 people 15 do not speak any language at all so 73 - 15 = 58 people speak at least one language. Now, if we add the 26 people that speak German, the 27 who speak Italian, and the 34 who speak Spanish, we get 87, why is that? Because we are double counting some of the people and triple counting others. And we get these two equations:

The info given in statement 1), coupled with the stem info, is sufficient to answer the question, as we can see on the chart below:

87 = 3T + 2D + S 58 = T + D + S Subtract on from the other and we get 29 = 2T + D, and because we are told that the people who speak the three languages are onethird of those who speak none, then T = 5, so 29 = 2(5) + D, and D = 19, so 58 = 5 + 19 + S, so S = 34. The info given in statement 1), coupled with the stem info, is sufficient to answer the question, as we can see on the chart below:

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140 - (E) No info is given in the stem question. Statement 1) is not sufficient, because although we learn that 25% of the women have a car, we do not know what percent of the population are women, as you can see in the chart below.

Even if we put the info given in both statements together we still cannot answer the question because we do not know what percentage of the population are women and what percentage are men.

Statement 2) is not sufficient, because although we learn that 29% of the men have a car, we do not know what percent of the population are men, as you can see in the chart below.

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horacio.quiroga@gmail.com 141 - (A) The chart below shows the info given in the stem. Notice that as usual when the problem mixes numbers with percentages, we work with percentages first.

Statement 1) is sufficient, because we learn that 15 percent of 55 percent of the coins are Russian coins with a value of more than 5 rubles, so 35 percent of 55 percent, or 38.5 of all the coins are Russian coins with a value not higher than 5 rubles, and by multiplying 38.5% by 1,200 we get our answer. See the chart below:

Statement 2) tells us that a certain percentage of the coins are gold coins, but that does nothing to do with the question. 142 - (D) The chart shows the info we get from the stem:

Statement 1) is sufficient because we can plug in a 100 in the n partners - at most 8 yrs box and therefore calculate the rest, as shown on the chart below:

Statement 2) is sufficient because we can plug in a 200 in the partners - at most 8 yrs box and therefore calculate the rest, as shown on the chart below:

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horacio.quiroga@gmail.com 143 - (A) The chart shows the info we get from the stem:

Statement 1) is sufficient, as shown in the chart below:

We can write the equation: . 7\ + .4[ = .6\ + .6[ and . 1\ = .2[, eI

_ $

= .

Statement 2) is not sufficient because it tells nothing about the distribution between man and women.

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Explanations to Chapter 7 - Statistics 144 - (E)

147 - (E)

If 34 percent of the distribution lies at approximately 1 standard deviation of the mean, to the left of which lies exactly 50% of the distribution, then to the right of one standard deviation we have 50% + 34% = 84% of the distribution, therefore 16% will lie to the right of x + s.

I) Think of a smaller set, say -10, -5, 0, 5, 10 and 15. If you multiply al the terms by 3 then you get -30, -15, 0, 15, 30, and 45. See that the set got more spread around the mean, so the standard deviation definitely changed. The mean, which because this is an evenly spaced set will also be the median, will go from 2.5 to 7.5, so both mean and median will change. The same thing will happen to our original set. So this statement is true.

145 - (A) As we explained above in our introduction to the concept of standard deviation, the closer to the mean the elements of the set are, the less the standard deviation. Notice that choice (A) is the one that really introduces one element that is further away from the mean than all the other ones, that is -8, so (A) is our answer. Take a look at all the others and you will see that, for instance, (C) will not change the standard deviation much because 7 and 7 are exactly on the mean.

II) This is not true, because the negative terms will become positive and therefore the set will become closer to the mean and the standard deviation will become smaller. III) This is true. If we add a constant to all the elements in our set the standard deviation will not change; but because the elements will grow greater, then the median and mean will change. So our answer is choice (E).

146 - (E) Remember, if you add or subtract a constant to all the elements in a set, the standard deviations DOES NOT change, so choice (E) is our answer.

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horacio.quiroga@gmail.com 148 - (B)

151 - (A)

From the stem of the problem we can calculate the standard deviation of the heights of the players of team A. To compare both standard deviations and therefore answer our question we have to know the standard deviation of team B.

If the mean of the 7 positive integers is 25 then the sum is 175, if the median is 30, the first three elements from least to most could be 1, 2, and 3, and then we could tailor-make the three elements to the right of 30 in order to add up to 175. Example: 1, 2, 3, 30, 40, 49, 50.

Statement 1) is NOT sufficient, we know both sets have the same mean, but we have no idea about the distribution around the mean, which would give us the standard deviation. Statement 2) is sufficient, if we know the variance we know the standard deviation of team B. 149 - (E) Notice first that each person in the three, towns makes exactly $30,000, $35,000, and $40,000 and the weighted average is $37,500, which is to the right of $35,000. For that to happen, the population of town C must be greater than that of A in order to "pull" the weighted average to its side. 150 - (A) As we said in our introduction to standard deviation above, adding a constant to all the elements will NOT change the standard deviation, so our answer is (A).

152 - (A) Statement 1) is sufficient, because if the number of apples in each basket is the same then the standard deviation is 0. Statement 2) is NOT sufficient, the median tells us nothing about the distribution, which is essential to determining the standard deviation. 153 - (E) Statement 1) is insufficient because we learn nothing about the rest of the elements of set P. Statement 2) is insufficient because we learn nothing about the rest of the elements of set P. Even if put both statements together, we do not know anything about the rest of the elements in each set. For instance, one of the sets could have a quite large element and a very small one that could increase drastically the standard deviation. Conversely, the other set could have the rest of the elements quite close to the mean thus making the standard deviation quite smaller.

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