There are three key steps to solving this problem:
1. Sammy started with x flavors and threw out y flavors, so he now has x - y flavors remaining.
2. To create combinations, we use the formula nCr = n!/(r!(n-r)!). Where n is the number of total flavors and r is the number we are combining.
3. Since we want 10-flavor combinations, and we know the number of flavors remaining is x - y, we can plug this value into the formula in place of n.
The key information needed is the original number of flavors (x) and the number thrown out (y) to determine the number remaining for combinations.
Gmat quant topic 8 probability solutionsRushabh Vora
1. The document contains multiple probability word problems involving coin flips, exam questions, picking items from different groups, weather outcomes, and more.
2. Key aspects involve counting the total possible outcomes and favorable outcomes to calculate probabilities, and using combinations, permutations, and other counting methods.
3. The correct solutions utilize precise calculations and reasoning to determine the probabilities requested in each problem.
Mathematics high school level quiz - Part IITfC-Edu-Team
The document outlines the format and questions for a mathematics quiz with multiple rounds. It begins with a two-part quiz where groups are given problem cards to solve. The subsequent rounds include warm-up questions testing concepts like geometry, averages, and number puzzles, as well as "real math" and logic rounds. Later rounds involve problem-solving, model-making to demonstrate algebraic identities, and a final written work discussion period.
This document contains a summary of questions and answers from multiple rounds of a math exam for Class 9 and 10 students. In the first round, it provides 5 multiple choice questions on topics of numbers, algebra, and trigonometry, along with the answers. The second round contains 7 additional multiple choice questions covering geometry, trigonometry, and algebra. The third round includes 4 short answer questions on volumes, areas, and ratios. Finally, the fourth round lists 8 algebra questions ranging from basic operations to solving equations to polynomial factorization.
This document contains a series of math quizzes covering various topics like multiplication, factoring, exponents, algebra, angles, and more. It includes over 50 questions testing skills like finding products, factoring numbers, solving equations, calculating angles of circles, and evaluating expressions. The quizzes start with simpler questions and increase in difficulty, containing up to 6 possible answer choices. After each short quiz, the number of correct answers is reported out of the total number of questions. At the end of the document, the participant is congratulated for getting all questions correct across every quiz.
This document contains information on various math topics organized under different subheadings:
1. The topics covered include algebra, shapes, data, numbers, adding/subtracting decimals and fractions, direct proportion, BIDMAS rules, factors and HCF, factor trees, directed numbers, equivalent fractions, ratio, finding percentages, HCF and LCM, indirect proportion, fractions/decimals/percentages, limits, multiples, multiplying/dividing fractions and decimals, ordering fractions and decimals, percentage increase/decrease, and rounding to significant figures.
2. There are examples and multi-step problems provided under each topic for practice.
3. The document serves as a review of various math concepts and skills for
This puzzle involves three friends who each think they are the smartest. To determine who is correct, a fourth friend secretly places a black or white spot on each of their foreheads without them knowing. When brought together in a room where each can see the others' foreheads, all three see two black spots and raise their hands. One then correctly states that he has a black spot on his own forehead. He was able to deduce this because if any had a white spot, not all three could see two black spots, so he must have a black spot as well.
Sam and Appu are mathematicians who each have two prime numbers written on their foreheads. They take turns stating whether they can deduce the numbers on their own forehead. After two turns each saying "don't know", it is now Sam's third turn. The numbers on Sam's forehead must be 13 and 17, as this is the only combination that satisfies the conditions given and allows both Sam and Appu to have correctly said "don't know" on their first two turns.
This document discusses counting combinations using the combination formula. It provides examples of counting the number of possible 3-topping pizzas that can be made from 8 toppings, forming a committee of 3 people from 10 club members. It also discusses distinguishing between combination and permutation problems, and how to solve problems involving multiple pools or scenarios by multiplying or adding the relevant combination terms.
Gmat quant topic 8 probability solutionsRushabh Vora
1. The document contains multiple probability word problems involving coin flips, exam questions, picking items from different groups, weather outcomes, and more.
2. Key aspects involve counting the total possible outcomes and favorable outcomes to calculate probabilities, and using combinations, permutations, and other counting methods.
3. The correct solutions utilize precise calculations and reasoning to determine the probabilities requested in each problem.
Mathematics high school level quiz - Part IITfC-Edu-Team
The document outlines the format and questions for a mathematics quiz with multiple rounds. It begins with a two-part quiz where groups are given problem cards to solve. The subsequent rounds include warm-up questions testing concepts like geometry, averages, and number puzzles, as well as "real math" and logic rounds. Later rounds involve problem-solving, model-making to demonstrate algebraic identities, and a final written work discussion period.
This document contains a summary of questions and answers from multiple rounds of a math exam for Class 9 and 10 students. In the first round, it provides 5 multiple choice questions on topics of numbers, algebra, and trigonometry, along with the answers. The second round contains 7 additional multiple choice questions covering geometry, trigonometry, and algebra. The third round includes 4 short answer questions on volumes, areas, and ratios. Finally, the fourth round lists 8 algebra questions ranging from basic operations to solving equations to polynomial factorization.
This document contains a series of math quizzes covering various topics like multiplication, factoring, exponents, algebra, angles, and more. It includes over 50 questions testing skills like finding products, factoring numbers, solving equations, calculating angles of circles, and evaluating expressions. The quizzes start with simpler questions and increase in difficulty, containing up to 6 possible answer choices. After each short quiz, the number of correct answers is reported out of the total number of questions. At the end of the document, the participant is congratulated for getting all questions correct across every quiz.
This document contains information on various math topics organized under different subheadings:
1. The topics covered include algebra, shapes, data, numbers, adding/subtracting decimals and fractions, direct proportion, BIDMAS rules, factors and HCF, factor trees, directed numbers, equivalent fractions, ratio, finding percentages, HCF and LCM, indirect proportion, fractions/decimals/percentages, limits, multiples, multiplying/dividing fractions and decimals, ordering fractions and decimals, percentage increase/decrease, and rounding to significant figures.
2. There are examples and multi-step problems provided under each topic for practice.
3. The document serves as a review of various math concepts and skills for
This puzzle involves three friends who each think they are the smartest. To determine who is correct, a fourth friend secretly places a black or white spot on each of their foreheads without them knowing. When brought together in a room where each can see the others' foreheads, all three see two black spots and raise their hands. One then correctly states that he has a black spot on his own forehead. He was able to deduce this because if any had a white spot, not all three could see two black spots, so he must have a black spot as well.
Sam and Appu are mathematicians who each have two prime numbers written on their foreheads. They take turns stating whether they can deduce the numbers on their own forehead. After two turns each saying "don't know", it is now Sam's third turn. The numbers on Sam's forehead must be 13 and 17, as this is the only combination that satisfies the conditions given and allows both Sam and Appu to have correctly said "don't know" on their first two turns.
This document discusses counting combinations using the combination formula. It provides examples of counting the number of possible 3-topping pizzas that can be made from 8 toppings, forming a committee of 3 people from 10 club members. It also discusses distinguishing between combination and permutation problems, and how to solve problems involving multiple pools or scenarios by multiplying or adding the relevant combination terms.
The document describes the rules for an inter-school mathematics quiz being organized by the Aryabhatta Mathematics Club. There will be 24 multiple choice questions asked in the quiz, with each correct answer earning 10 marks and no negative marking for incorrect answers. Each question must be answered within 1 minute and there are no transferable questions. The quiz then provides sample questions following this format to illustrate the types of math problems that will be asked.
The document discusses combinations and the binomial theorem. It explains that a combination is a selection of objects from a group where order does not matter. It also covers Pascal's triangle and the binomial expansion formula for (a + b)^n, which uses coefficients that are the combinations of n objects taken r at a time.
Important questions for class 10 maths chapter 15 probability with solionsExpertClass
expert's class ahmedabad by dhruv thakar
important notes of mathematics class 10
ncert solution also available ,mathematics adda
top institute in ahmedabad for mathematics
school and And graduation level competitive exam mathematics
dhruv sir mobile number-8320495706 ,bhai ka mathematics
Important questions for class 10 maths chapter 15 probability with solutionsExpertClass
1) The document provides solutions to 10 probability questions involving drawing cards from boxes or bags, rolling dice, tossing coins, and spinning arrows.
2) It includes the total possible outcomes, favorable outcomes, and probability calculation for each question.
3) The questions range from very short answer to long answer types and cover topics like simple probability, conditional probability, and independent events.
المراجعة النهائية فى الماث للشهادة الابتدائية الترم الأول لمدرسة الكمال أمنية وجدى
This document contains notes from a math student's copy book covering ratios and fractions. It includes examples and exercises on simplifying fractions, writing numbers divisible by certain numbers, finding ratios between quantities, and properties of ratios. Key concepts covered are ratios comparing two quantities, reducing ratios to simplest form, ratio properties such as keeping the same units and order, and using ratios to solve word problems involving quantities in a given ratio. Worked examples demonstrate finding ratios and using them to determine unknown values. Exercises provide practice applying these ratio and fraction skills.
1. Let x = Archibald's marbles. Then Muriel has 2x marbles and Oswald has 95 - x - 2x marbles.
2. Set up an equation where their total marbles equals 95: x + 2x + (95 - x - 2x) = 95
3. Solve the equation for x. Archibald has 25 marbles.
4. Check that with x = 25, their individual amounts add to 95.
This document summarizes a quiz with 3 rounds - MCQ, riddles, and visual. The MCQ round will eliminate incorrect answers. The riddles round is for fun and to refresh the mind. Solving the quiz successfully means becoming a mathematician. All questions are from class 9 syllabus except riddles. It then provides 6 sample MCQ questions and their answers related to areas of triangles, decimal expansions, polynomial degrees, lines/angles, and more. Finally, it lists 12 riddles/visual questions about mathematicians, circle properties, angle relationships, and more.
The document shows the step-by-step work to solve the linear equation 3x^2 + 4 = 13. It begins by multiplying both sides by 2 to eliminate the fraction. It then distributes and combines like terms before subtracting and dividing to isolate x. The final solution is x = 6.
This document contains a math lesson for first grade students covering topics like classifying objects into sets, comparing sets, numbers 1-9, ordinal numbers, and basic positional words like in front of, behind, on, under, left, and right. The lesson is broken into multiple parts with instructions and exercises for students to practice the concepts through activities like circling, coloring, comparing and arranging numbers.
What is four times three? 12 you might say, but no longer! In a new type of math— intersection math—
we will see that four times three is 18, two times two is 1, and that two times five is 10 (Hang on! That’s
not new!) Let’s spend some time together exploring this new math and answering the question: What is
1001 times 492?
This document contains topics related to algebra, shape, data, numbers, decimals, fractions, proportions, BIDMAS rules, factors and HCF, directed numbers, ratios, percentages, fractions and decimals, limits, multiples, and fractions. It includes examples and multi-step problems involving these mathematical concepts.
This document discusses factors, multiples, least common multiples (LCM), and highest common factors (HCF) of numbers. It provides examples and methods to find the LCM and HCF of sets of numbers. The LCM is the smallest number that is a multiple of all numbers, while the HCF is the largest number that divides all numbers. Methods include factorizing into prime factors or using the division method. Practice problems with solutions are provided to help understand these concepts.
This document provides information about square roots and real numbers. It includes:
1) Examples of finding square roots of perfect squares and estimating other square roots.
2) Classifying different types of real numbers such as rational numbers, irrational numbers, integers, natural numbers, and more.
3) Examples of classifying given numbers as rational, irrational, integers, and other categories.
This document contains instructions and questions for a mathematics exam. It begins by providing spaces for the student to write their name, centre number, and candidate number. It then lists the total marks, time allowed, and materials permitted. The document contains 19 multiple choice and free response questions testing a variety of math skills like arithmetic, algebra, geometry, statistics, and graphing. It provides formulae for reference.
1) The document describes a mental ability test consisting of 12 multiple choice questions arranged in a 4x3 grid.
2) Teams will take turns choosing a question from the grid and will have a time limit to answer based on difficulty. Correct answers earn 10 points.
3) A sample question is provided to illustrate the multiple choice format. The questions cover a range of math, logic, and reasoning skills.
The document provides information about the structure of a quiz competition between teams. It states that each team will be given one question carrying 10 marks and they will have two turns. It provides examples of questions asked in previous rounds along with the correct answers. These include maths, word and logic questions. It then presents a new set of questions that will be asked in the upcoming round.
The document contains solutions to 18 math and probability problems. Some key details:
- Problem 1 involves finding an odd number n such that the sum of even numbers between 1 and n equals 79*80.
- Problem 2 calculates the price at which a bushel of corn costs the same as a peck of wheat, given changing prices.
- Problem 3 determines the minimum number of people needed to have over a 50% chance that one was born in a leap year.
1. The document contains 10 math problems with solutions. The problems cover topics like arithmetic progressions, rates of change, probability, and geometry.
2. One problem involves finding the value of n given that the sum of even numbers between 1 and n is a specific value. The solution uses the formula for the sum of an arithmetic progression.
3. Another problem asks what fraction of a solution must be replaced if the original solution was 40% and replaced with 25% solution to get a final concentration of 35%. The solution sets up an equation to solve for the fraction replaced.
The document provides information about Vedic mathematics and FastMaths techniques. It begins by introducing FastMaths as a system of mathematical reasoning based on ancient Indian Vedic teachings. It is described as fast, efficient, and easy to learn. The document then provides examples of how FastMaths techniques can be used to quickly solve problems like multiplication and division. It further explains the concepts of place value, the 9-point number circle, factors, and finding the highest common factor in Vedic mathematics.
This document contains formulas for arithmetic, algebra, real numbers, fractions, decimals, percentages, probability, and inequalities. Some key formulas include:
1) The formula for combinations is nCk = n!/((n-k)!k!).
2) Simple interest is calculated as Interest = Principal × Rate × Time. Interest paid annually uses the formula P(1 + R)n, while interest paid monthly uses P(1 + R/12)12×t.
3) The probability of an event E occurring is calculated as P(E) = Number of outcomes for E / Total number of possible outcomes. The probability of events E and F both occurring is P(E and F)
The document describes the rules for an inter-school mathematics quiz being organized by the Aryabhatta Mathematics Club. There will be 24 multiple choice questions asked in the quiz, with each correct answer earning 10 marks and no negative marking for incorrect answers. Each question must be answered within 1 minute and there are no transferable questions. The quiz then provides sample questions following this format to illustrate the types of math problems that will be asked.
The document discusses combinations and the binomial theorem. It explains that a combination is a selection of objects from a group where order does not matter. It also covers Pascal's triangle and the binomial expansion formula for (a + b)^n, which uses coefficients that are the combinations of n objects taken r at a time.
Important questions for class 10 maths chapter 15 probability with solionsExpertClass
expert's class ahmedabad by dhruv thakar
important notes of mathematics class 10
ncert solution also available ,mathematics adda
top institute in ahmedabad for mathematics
school and And graduation level competitive exam mathematics
dhruv sir mobile number-8320495706 ,bhai ka mathematics
Important questions for class 10 maths chapter 15 probability with solutionsExpertClass
1) The document provides solutions to 10 probability questions involving drawing cards from boxes or bags, rolling dice, tossing coins, and spinning arrows.
2) It includes the total possible outcomes, favorable outcomes, and probability calculation for each question.
3) The questions range from very short answer to long answer types and cover topics like simple probability, conditional probability, and independent events.
المراجعة النهائية فى الماث للشهادة الابتدائية الترم الأول لمدرسة الكمال أمنية وجدى
This document contains notes from a math student's copy book covering ratios and fractions. It includes examples and exercises on simplifying fractions, writing numbers divisible by certain numbers, finding ratios between quantities, and properties of ratios. Key concepts covered are ratios comparing two quantities, reducing ratios to simplest form, ratio properties such as keeping the same units and order, and using ratios to solve word problems involving quantities in a given ratio. Worked examples demonstrate finding ratios and using them to determine unknown values. Exercises provide practice applying these ratio and fraction skills.
1. Let x = Archibald's marbles. Then Muriel has 2x marbles and Oswald has 95 - x - 2x marbles.
2. Set up an equation where their total marbles equals 95: x + 2x + (95 - x - 2x) = 95
3. Solve the equation for x. Archibald has 25 marbles.
4. Check that with x = 25, their individual amounts add to 95.
This document summarizes a quiz with 3 rounds - MCQ, riddles, and visual. The MCQ round will eliminate incorrect answers. The riddles round is for fun and to refresh the mind. Solving the quiz successfully means becoming a mathematician. All questions are from class 9 syllabus except riddles. It then provides 6 sample MCQ questions and their answers related to areas of triangles, decimal expansions, polynomial degrees, lines/angles, and more. Finally, it lists 12 riddles/visual questions about mathematicians, circle properties, angle relationships, and more.
The document shows the step-by-step work to solve the linear equation 3x^2 + 4 = 13. It begins by multiplying both sides by 2 to eliminate the fraction. It then distributes and combines like terms before subtracting and dividing to isolate x. The final solution is x = 6.
This document contains a math lesson for first grade students covering topics like classifying objects into sets, comparing sets, numbers 1-9, ordinal numbers, and basic positional words like in front of, behind, on, under, left, and right. The lesson is broken into multiple parts with instructions and exercises for students to practice the concepts through activities like circling, coloring, comparing and arranging numbers.
What is four times three? 12 you might say, but no longer! In a new type of math— intersection math—
we will see that four times three is 18, two times two is 1, and that two times five is 10 (Hang on! That’s
not new!) Let’s spend some time together exploring this new math and answering the question: What is
1001 times 492?
This document contains topics related to algebra, shape, data, numbers, decimals, fractions, proportions, BIDMAS rules, factors and HCF, directed numbers, ratios, percentages, fractions and decimals, limits, multiples, and fractions. It includes examples and multi-step problems involving these mathematical concepts.
This document discusses factors, multiples, least common multiples (LCM), and highest common factors (HCF) of numbers. It provides examples and methods to find the LCM and HCF of sets of numbers. The LCM is the smallest number that is a multiple of all numbers, while the HCF is the largest number that divides all numbers. Methods include factorizing into prime factors or using the division method. Practice problems with solutions are provided to help understand these concepts.
This document provides information about square roots and real numbers. It includes:
1) Examples of finding square roots of perfect squares and estimating other square roots.
2) Classifying different types of real numbers such as rational numbers, irrational numbers, integers, natural numbers, and more.
3) Examples of classifying given numbers as rational, irrational, integers, and other categories.
This document contains instructions and questions for a mathematics exam. It begins by providing spaces for the student to write their name, centre number, and candidate number. It then lists the total marks, time allowed, and materials permitted. The document contains 19 multiple choice and free response questions testing a variety of math skills like arithmetic, algebra, geometry, statistics, and graphing. It provides formulae for reference.
1) The document describes a mental ability test consisting of 12 multiple choice questions arranged in a 4x3 grid.
2) Teams will take turns choosing a question from the grid and will have a time limit to answer based on difficulty. Correct answers earn 10 points.
3) A sample question is provided to illustrate the multiple choice format. The questions cover a range of math, logic, and reasoning skills.
The document provides information about the structure of a quiz competition between teams. It states that each team will be given one question carrying 10 marks and they will have two turns. It provides examples of questions asked in previous rounds along with the correct answers. These include maths, word and logic questions. It then presents a new set of questions that will be asked in the upcoming round.
The document contains solutions to 18 math and probability problems. Some key details:
- Problem 1 involves finding an odd number n such that the sum of even numbers between 1 and n equals 79*80.
- Problem 2 calculates the price at which a bushel of corn costs the same as a peck of wheat, given changing prices.
- Problem 3 determines the minimum number of people needed to have over a 50% chance that one was born in a leap year.
1. The document contains 10 math problems with solutions. The problems cover topics like arithmetic progressions, rates of change, probability, and geometry.
2. One problem involves finding the value of n given that the sum of even numbers between 1 and n is a specific value. The solution uses the formula for the sum of an arithmetic progression.
3. Another problem asks what fraction of a solution must be replaced if the original solution was 40% and replaced with 25% solution to get a final concentration of 35%. The solution sets up an equation to solve for the fraction replaced.
The document provides information about Vedic mathematics and FastMaths techniques. It begins by introducing FastMaths as a system of mathematical reasoning based on ancient Indian Vedic teachings. It is described as fast, efficient, and easy to learn. The document then provides examples of how FastMaths techniques can be used to quickly solve problems like multiplication and division. It further explains the concepts of place value, the 9-point number circle, factors, and finding the highest common factor in Vedic mathematics.
This document contains formulas for arithmetic, algebra, real numbers, fractions, decimals, percentages, probability, and inequalities. Some key formulas include:
1) The formula for combinations is nCk = n!/((n-k)!k!).
2) Simple interest is calculated as Interest = Principal × Rate × Time. Interest paid annually uses the formula P(1 + R)n, while interest paid monthly uses P(1 + R/12)12×t.
3) The probability of an event E occurring is calculated as P(E) = Number of outcomes for E / Total number of possible outcomes. The probability of events E and F both occurring is P(E and F)
This document contains 52 probability and combinatorics word problems with multiple choice answers. The problems cover a wide range of topics including counting arrangements, combinations, probability calculations involving drawing balls from jars or flipping coins, and more. The correct answers to each problem are provided as multiple choice options.
The document contains 68 math and probability questions involving topics like probability, combinations, sequences, functions, and other quantitative reasoning concepts. The questions range in complexity and cover a wide variety of mathematical scenarios and problem types.
The document provides information about the quantitative section of the Graduate Management Admission Test (GMAT). It details that the section contains 37 questions to be completed in 75 minutes, including both problem solving and data sufficiency questions. Problem solving questions require solving the problem and choosing the best answer, while data sufficiency questions involve determining whether one, both, or neither of the given statement(s) are sufficient to answer the question. The document provides examples of several questions and explains the relevant terms and concepts.
Gmat Cretical Reasoning 700 to 800 club solutionsRushabh Vora
The document summarizes the conclusions that can be validly made based on the information provided in several arguments. It emphasizes that conclusions must only rely on the stated facts and avoid making unintended assumptions. Additionally, it examines several response options and determines which conclusions necessarily follow from the facts given without needing any additional information.
- The document is a study guide for the GMAT quant section containing 27 practice questions and their answer choices.
- It provides the questions, possible answer choices for each question, and brief explanations or statements of facts for some questions.
- The questions cover a range of math and logic topics including sets, ratios, percentages, probability, and word problems involving numbers.
The document discusses the benefits of exercise for mental health. Regular physical activity can help reduce anxiety and depression and improve mood and cognitive function. Exercise causes chemical changes in the brain that may help protect against mental illness and improve symptoms.
This document provides an overview of the structure of the GMAT exam. It discusses that the GMAT is a standardized test required for admission to most MBA programs. It is created and administered by GMAC and costs $250 to take. The test takes place on a computer at an official testing center and consists of an integrated reasoning section, quantitative section, verbal section, and analytical writing assessment, with the questions adapting in difficulty based on the test-taker's performance.
1. The point of inflection of the function y = x3 − 3x2 + 6x + 2000 is x = 1. At this point, the slope is 3.
2. The optimal length for Karen to climb is when x = √3/3, which is the solution to the equation 1 − 3x2 = 0 derived from finding the critical points of her change in position, x − x3.
3. By applying the Fundamental Theorem of Calculus, the constant C in the solution to the given integral 1/4π ∫0 sin x + C dx is equal to 2√2 − 4/π.
This document contains 10 math problems with solutions. The problems cover topics like probability, combinatorics, geometry, and complex analysis. The solutions provide detailed step-by-step workings to arrive at the final answers.
The document contains multiple choice questions related to probability and statistics. It includes 10 probability questions related to events like drawing balls from bags, rolling dice, tossing coins, etc. It also includes 5 statistics questions related to mean, median, standard deviation, distributions, etc. The document tests the reader's understanding of key probability and statistics concepts through multiple choice questions and answers.
The document discusses permutation and combination. It provides examples of calculating the number of permutations and combinations in different scenarios like arranging letters to form words or selecting members for a committee. It also solves problems involving permutations and combinations, like calculating the number of 5-digit numbers that can be formed from given digits that are divisible by 4, with and without repetition. The key concepts covered are the formulas for permutation (nPr) and combination (nCr), and knowing when to apply multiplication or addition based on the nature of the problem.
ISI CMI IMPORTANT AND COMPLICATED QUESTION WITH SOLUTION BY SOURAV SIR'S CLAS...SOURAV DAS
The document presents a multi-step solution to determining the value of k if a system of 3 equations has no solution. It is shown that if one of the equations can be generated from the other two, but with a different constant term, then the system will have no solutions. Through algebraic manipulation, it is determined that if k = 7, then one of the equations can be generated from the others with a different constant term, so the system would have no solutions when k = 7.
The document then presents solutions to 5 additional problems involving permutations, combinations, ratios, and determining the minimum number of objects based on given conditions. Finally, it provides a detailed step-by-step solution to counting the number of arrangements
The document discusses questions related to decimals, fractions, and quantitative aptitude tests. It provides examples of questions on topics like square roots, scientific notation, and permutations and combinations. It also provides links to download presentations on related topics like profit and loss, ratios, and business mathematics.
I am Stacy W. I am a Probability Homework Expert at statisticshomeworkhelper.com. I hold a Masters in Statistics from, the University of McGill, Canada
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This document provides a summary of key concepts related to normal distribution for statistical analysis, including definitions of frequency distribution and important probability distributions like normal, binomial, chi square, student T, and F distribution. It also lists the assumptions of normal and binomial distributions and provides examples of probability calculations for scenarios involving dice, cards, samples, and more.
This document provides a summary of key concepts related to normal distribution for statistical analysis, including definitions of frequency distribution and important probability distributions like normal, binomial, chi square, student T, and F distribution. It also lists the assumptions of normal and binomial distributions and provides examples of probability calculations for scenarios involving dice, cards, samples, and more.
1. The document discusses rules and principles for working with negative numbers and algebraic expressions involving negative numbers.
2. Key ideas include defining negative multiplication, such as -3 × 2, as -6; establishing that the order of factors does not matter in negative multiplication, similar to positive numbers; and simplifying algebraic expressions by using properties such as x - (y - z) = x - y + z.
3. General principles are stated for adding and subtracting negative numbers, and techniques are described for simplifying algebraic expressions involving negative terms.
This document contains examples of probability problems and their solutions. Example 6 calculates the probability of two friends having the same or different birthdays. Example 7 finds the probability of selecting a girl or boy as the class representative. Example 8 determines the probability of drawing different colored marbles from a box. The remaining examples involve calculating probabilities of outcomes when rolling dice, selecting shirts from a carton, and the sum of numbers rolled.
1. The document discusses probability and counting methods for computing probabilities of events related to gambling and drawing cards from a deck. It introduces concepts like theoretical probability, empirical probability, permutations, combinations, and using factorials and probabilities to solve counting problems.
2. As an example, it calculates the probability of drawing a pair of the same color cards from a deck and determines it is 49% based on counting the possible combinations.
3. It also discusses using probabilities to determine rational betting strategies, like folding unless holding a pair of the same color or suit when playing against others who each drew 2 cards.
The document is the questions and solutions from the 2003 Western Australian Junior Mathematics Olympiad individual and team competitions. It includes 10 individual questions testing various math skills like algebra, number theory, and geometry, along with the solutions. It also includes 5 team problems involving cutting strings into pieces and calculating the resulting number and sizes of pieces. The team problems can be solved using powers of 2 and 3 and formulas are provided for the longest and shortest starting strings that result in a given number of pieces.
This material is a part of PGPSE / CSE study material for the students of PGPSE / CSE students. PGPSE is a free online programme for all those who want to be social entrepreneurs / entrepreneurs
The document provides explanations and examples for concepts related to differentiation and integration including:
- The differentiation of various functions like 3x^2, e^x, x^5, etc. and explanations of the differentiation process
- The integration of functions like 3x^3 and what integration undoes from differentiation
- Examples of calculating permutations, combinations, and arrangements of different objects
This document contains questions and answers related to differentiation and integration. Some key questions addressed include:
- What is the differentiation of 3x^2 + 5x - 2?
- What is the integration of 12x^3?
- How many different 4 digit numbers can be made out of 1,2,3,4,0?
- In how many ways can 8 person sit around a round table?
The document provides worked examples and explanations for calculating derivatives and integrals of various functions.
This document provides examples and explanations of patterns, inductive reasoning, and geometry concepts. It includes a list of counting numbers and their properties, perfect squares up to 10, and examples of making conjectures based on patterns. Students are asked to identify patterns, extend patterns, test conjectures, solve equations, name points and planes, and describe intersections of lines and planes. The document covers fundamental geometry, algebra, and logic skills.
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1) The value of the expression 44 + 4 × 4 − 4 is 268.
2) Two of the four statements about sequences and patterns are true - that it follows the Fibonacci sequence and is an arithmetic sequence with a common difference of 4.
3) The area of the gray region overlapping two rectangles is 32 cm^2.
The document contains 35 math and word problems. The assistant provides concise 3-sentence summaries for each problem:
1) A man discusses promoting entrepreneurship and free education for all. He needs support from others to achieve these goals.
2) The problem asks to rearrange the letters in "PROBLEM" to make 7-letter words without repetition, with the answer being 5040 possible arrangements.
3) The problem asks the probability that the third coin tossed will be heads if 10 coins are tossed simultaneously, with the answer being 1/2 or 512 possible outcomes out of 1024 total.
4) The problem asks to count the ways to post 5 letters into 3 post boxes with any number allowed
This document provides an overview and table of contents for a grammar guidebook called "Ultimate GMAT Grammar". It was created by the website GMAT Club to help users prepare for the GMAT exam. The guidebook is divided into three parts that progress from basic to intermediate to advanced English grammar concepts. It includes rules, exercises, and practice tests covering topics like verbs, nouns, pronouns, questions, comparisons, conditionals, and more. The document encourages users to ask questions in the GMAT Club forum and wishes them good luck on the GMAT. It also advertises grammar preparation courses and discounts available through partner organizations.
This document is the introduction to the 4th edition of the Turbocharge Your GMAT Sentence Correction Study Guide published in 2012. It provides an overview of the guide's contents which include a GMAT idiom list, common grammar errors to avoid, detailed explanations of tested grammar topics, tips and strategies, a grammar review covering parts of speech and concepts, and 250 practice questions. It also includes brief descriptions of other books and materials in the Turbocharge Your GMAT series and information about Manhattan Review, the publisher, and its founder.
The document provides an overview of GMAT scoring, including:
1) The total score ranges from 200 to 800 and is the primary score considered by business schools, with most students scoring around 550 and top programs requiring over 718.
2) Section scores range from 11 to 51 on the quantitative and verbal scales, though percentiles are a also considered.
3) The AWA is scored separately on a scale of 0 to 6, with a 4 considered satisfactory.
This document provides a comprehensive guide to Sentence Correction errors seen on the GMAT. It begins with an overview of the types of errors tested, including structure/meaning/concision, subject-verb agreement, pronouns, modifiers, parallelism, comparisons, idiomatic constructions, and tense. Each error type is then described in more detail with rules and examples. The guide explains how to approach SC questions step-by-step and provides strategies for identifying common error patterns. Key aspects like subject-verb agreement, pronoun reference, modifier placement, and parallel structure are emphasized.
1. The document contains 13 multiple choice questions regarding inequalities and absolute values.
2. The questions require evaluating logical statements and determining if they are sufficient to answer whether given inequalities are true.
3. Detailed step-by-step working is shown for each question, identifying the reasoning for selecting the answer. Common mistakes are discussed to help understand the concepts.
The document provides an overview of combinatorics problems that may appear on the GMAT. It discusses the different levels of difficulty, from easy one-action problems to harder problems involving calculations or multiple limitations. The key is to identify the correct solution method rather than getting bogged down in calculations. Several examples of combinatorics problems are provided and solved using logical reasoning rather than formulas. Common formulas for permutations and combinations are also listed.
The document discusses permutations and combinations. It begins by introducing permutations as arrangements of objects in a definite order. It then presents the fundamental principle of counting, which states that if one event can occur in m ways and is followed by another that can occur in n ways, the total number of occurrences is m×n. It provides examples of using this principle to count outcomes. The document then defines permutations more specifically as arrangements of a number of objects taken some or all at a time. It presents a theorem stating the formula to calculate the number of permutations of n distinct objects taken r at a time is nPr = n(n-1)(n-2)...(n-r+1).
This document contains formulas for calculating properties of common geometric shapes such as triangles, rectangles, circles, spheres, and coordinate geometry. It includes formulas for:
- Calculating the perimeter, area, and properties of triangles like the Pythagorean theorem
- Calculating the perimeter, area, and properties of quadrilaterals like squares, rectangles, parallelograms, and trapezoids
- Calculating circumference, arc length, sectors, and properties of circles
- Calculating surface area and volume of three-dimensional shapes like cubes, cylinders, cones, and spheres
- Calculating distance, midpoints, slopes and the general form of lines in coordinate geometry
This document contains a series of quantitative questions related to topics like fractions, ratios, percentages, integers, algebra, geometry, counting, probability, and coordinate geometry. Each question is followed by multiple choice answers. The questions range in difficulty from introductory to advanced. This appears to be a practice test or set of example questions for the GMAT quantitative section.
The document contains multiple word problems involving fractions, percentages, ratios, and algebraic expressions. Summaries of three sample problems are provided below:
1) The first problem determines the greatest number of contributors to a fundraising campaign based on the minimum $50 contribution amount and total funds raised of $1,749. The greatest number of contributors is 34.
2) The second problem involves the ages of four people as consecutive integers that can be expressed in terms of one variable, Lillian's age. The sum of their ages is a number two more than a multiple of 4, making the correct answer 54.
3) The third problem is a word problem translated into algebraic expressions to relate the current and future ages
Gmat quant topic 6 co ordinate geometry solutionsRushabh Vora
1) The document contains multiple math word problems and their solutions.
2) One problem involves finding the area of an equilateral triangle given the length of its height is 13 units. Using properties of 30-60-90 triangles, the base is found to be 26/√3 and the area is 26√3/4.
3) Another problem asks which quadrant a line with slope -3/4 and y-intercept of 2 passes through. Placing points on the line shows it passes through Quadrants I, II, and IV.
1. The angles labeled (2a)° and (5a + 5)° are supplementary and add up to 180°. Solving for a gives a = 25. Similarly, the angles labeled (4b +10)° and (2b – 10)° are supplementary and add up to 180°. Solving for b gives b = 30. Therefore, a + b = 25 + 30 = 55.
2. Two parallel lines intersected by a transversal form eight angles. The acute angles are equal and the obtuse angles are equal. The acute angles are supplementary to the obtuse angles. Solving the equation relating the angles gives the answer.
3. The relationship between angles formed when
This document discusses different types of numbers and problems involving numbers. It provides examples of data sufficiency problems and their step-by-step solutions. The problems involve topics like divisibility, primes, consecutive integers, and algebraic expressions with variables representing integers. The document aims to explain the thought processes and approaches used to determine whether statements provide sufficient information to answer yes/no questions.
The document discusses weighted averages and how to calculate them. It provides an example of calculating the overall average weight of a group made up of two subgroups, men and women. Statement 2 alone provides enough information to determine the ratio of women to men in the group, but Statement 1 does not provide enough information on its own.
1. The document describes how to use a double set matrix to solve overlapping sets problems. It provides examples of full double set matrices with values filled in to solve 4 different overlapping sets problems.
2. The key aspects are that a double set matrix organizes the sets into rows and columns and the totals must sum correctly. Values can be derived using relationships between the given and total values.
3. The examples walk through setting up the double set matrix, filling in known values, deriving other values, and extracting the answer from the fully populated matrix.
1. There are two characteristics that dictate an exponential behavior: the base x must be a decimal with an absolute value less than 1, and it must be a negative number.
2. The document provides examples and explanations for why certain exponential expressions with a negative decimal base will always be positive or negative.
3. It evaluates various inequalities involving exponential expressions with a negative decimal base to determine relationships between the expressions.
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...Dr. Vinod Kumar Kanvaria
Exploiting Artificial Intelligence for Empowering Researchers and Faculty,
International FDP on Fundamentals of Research in Social Sciences
at Integral University, Lucknow, 06.06.2024
By Dr. Vinod Kumar Kanvaria
বাংলাদেশের অর্থনৈতিক সমীক্ষা ২০২৪ [Bangladesh Economic Review 2024 Bangla.pdf] কম্পিউটার , ট্যাব ও স্মার্ট ফোন ভার্সন সহ সম্পূর্ণ বাংলা ই-বুক বা pdf বই " সুচিপত্র ...বুকমার্ক মেনু 🔖 ও হাইপার লিংক মেনু 📝👆 যুক্ত ..
আমাদের সবার জন্য খুব খুব গুরুত্বপূর্ণ একটি বই ..বিসিএস, ব্যাংক, ইউনিভার্সিটি ভর্তি ও যে কোন প্রতিযোগিতা মূলক পরীক্ষার জন্য এর খুব ইম্পরট্যান্ট একটি বিষয় ...তাছাড়া বাংলাদেশের সাম্প্রতিক যে কোন ডাটা বা তথ্য এই বইতে পাবেন ...
তাই একজন নাগরিক হিসাবে এই তথ্য গুলো আপনার জানা প্রয়োজন ...।
বিসিএস ও ব্যাংক এর লিখিত পরীক্ষা ...+এছাড়া মাধ্যমিক ও উচ্চমাধ্যমিকের স্টুডেন্টদের জন্য অনেক কাজে আসবে ...
How to Build a Module in Odoo 17 Using the Scaffold MethodCeline George
Odoo provides an option for creating a module by using a single line command. By using this command the user can make a whole structure of a module. It is very easy for a beginner to make a module. There is no need to make each file manually. This slide will show how to create a module using the scaffold method.
How to Make a Field Mandatory in Odoo 17Celine George
In Odoo, making a field required can be done through both Python code and XML views. When you set the required attribute to True in Python code, it makes the field required across all views where it's used. Conversely, when you set the required attribute in XML views, it makes the field required only in the context of that particular view.
How to Fix the Import Error in the Odoo 17Celine George
An import error occurs when a program fails to import a module or library, disrupting its execution. In languages like Python, this issue arises when the specified module cannot be found or accessed, hindering the program's functionality. Resolving import errors is crucial for maintaining smooth software operation and uninterrupted development processes.
A workshop hosted by the South African Journal of Science aimed at postgraduate students and early career researchers with little or no experience in writing and publishing journal articles.
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UPRAHUL
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
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2. (1) SUFFICIENT: In analyzing statement (1), consider how many individuals would have to be
available to create 126 different 5 person teams. We don't actually have to figure this out as long as we
know that we could figure this out. Certainly by testing some values, we could figure this out. It turns
out that if there are 9 available individuals, then we could create exactly 126 different 5-person teams
(since 9! ÷ [(5!)(4!)] = 126). This value (9) represents x + 2. Thus x would equal 7.
(2) SUFFICIENT: The same logic applies to statement (2). Consider how many individuals would have
to be available to create 56 different 3-person teams. Again, we don't actually have to figure this out as
long as we know that we could figure this out. It turns out that if there are 8 available individuals, then
we could create exactly 56 different 3-person teams (since 8! ÷ [(5!)(3!)] = 56). This value (8) represents
x + 1. Thus x would equal 7. Statement (2) alone IS sufficient.
The correct answer is D.
38. This question is simply asking us to come up with the number of permutations that can be formed
when x people are seated in y chairs. It would seem that all we require is the values of x and y. Let's keep
in mind that the question stem adds that x and y must be prime integers.
(1) SUFFICIENT: If x and y are prime numbers and add up to 12, x and y must be either 7 and 5 or 5
and 7. Would the number of permutations be the same for both sets of values? Let's start with x = 7, y =
5. The number of ways to seat 7 people in 5 positions (chairs) is 7!/2!. We divide by 2! because 2 of the
people are not selected in each seating arrangement and the order among those two people is therefore
not significant. An anagram grid for this permutation would look like this:
A B C D E F G
1 2 3 4 5 N N
But what if x = 5 and y = 7? How many ways are there to position five people in 7 chairs? It turns out
the number of permutations is the same. One way to think of this is to consider that in addition to the
five people (A,B,C,D,E), you are seating two ghosts (X,X). The number of ways to seat A,B,C,D,E,X,X
would be 7!/2!. We divide by 2! to eliminate order from the identical X's.
Another way to look at this is by focusing on the chairs as the pool from which you are choosing. It's as
if we are fixing the people in place and counting the number of ways that different chair positions can be
assigned to those people. The same anagram grid as above would apply, but now the letters would
correspond to the 7 chairs being assigned to each of the five fixed people. Two of the chairs would be
unassigned, and thus we still divide by 2! to eliminate order between those two chairs.
(2) INSUFFICIENT: This statement does not tell us anything about the values of x and y, other than y >
x. The temptation in this problem is to think that you need statement 2 in conjunction with statement 1 to
distinguish between the x = 5, y = 7 and the x = 7, y = 5 scenarios.
The correct answer is A: Statement (1) ALONE is sufficient to answer the question, but statement (2)
alone is not.
39. Let W be the number of wins and L be the number of losses. Since the total number of hands equals 12
and the net winnings equal $210, we can construct and solve the following simultaneous equations: w +
l = 12, 100 w – 10 l = 210. so l = 9, w = 3. So we know that the gambler won 3 hands and lost 9. We do
not know where in the sequence of 12 hands the 3 wins appear. So when counting the possible outcomes
for the first 5 hands, we must consider these possible scenarios:
1) Three wins and two losses 2) Two wins and three losses 3) One win and four losses 4) No wins
and five losses
In the first scenario, we have WWWLL. We need to know in how many different ways we can arrange
these five letters:
GMAT / Quant / P&C / Page 2 of 4
3. 5!/2!3! = 10. So there are 10 possible arrangements of 3 wins and 2 losses. The second scenario --
WWLLL -- will yield the same result: 10. The third scenario -- WLLLL -- will yield 5 possible
arrangements, since the one win has only 5 possible positions in the sequence. The fourth scenario --
LLLLL -- will yield only 1 possible arrangement, since rearranging these letters always yields the same
sequence. Altogether, then, there are 10 + 10 + 5 + 1 = 26 possible outcomes for the gambler's first five
hands. The correct answer is C.
GMAT / Quant / P&C / Page 3 of 4
4. 40.
First, let's consider the different medal combinations that can be awarded to the 3 winners: (1) If there are NO
TIES then the three medals awarded are: GOLD, SILVER, BRONZE. (2) What if there is a 2-WAY tie? --If
there is a 2-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, SILVER. --If there is a 2-WAY
tie for SECOND, then the medals awarded are: GOLD, SILVER, SILVER. --There cannot be a 2-WAY tie for
THIRD (because exactly three medals are awarded in total). (3) What if there is a 3-WAY tie? --If there is a 3-
WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, GOLD. --There are no other possible 3-
WAY ties. Thus, there are 4 possible medal combinations: (1) G, S, B (2) G, G, S (3) G, S, S (4) G, G, G. Now
let's determine how many different ways each combination can be distributed. We'll do this by considering four
runners: Albert, Bob, Cami, and Dora.
COMBINATION 1: Gold, Silver, Bronze
Gold Medal Silver Medal Bronze Medal
Any of the 4
runners can
receive the gold
medal.
There are only 3 runners who
can receive the silver medal.
Why? One of the runners has
already been awarded the Gold
Medal.
There are only 2 runners who can
receive the bronze medal. Why?
Two of the runners have already
been awarded the Gold and Silver
medals.
4 possibilities 3 possibilities 2 possibilities
Therefore, there are different victory circles that will contain 1 GOLD, 1 SILVER, and 1
BRONZE medalist.
COMBINATION 2: Gold, Gold, Silver.
Using the same reasoning as for Combination 1, we see that there are 24 different victory circles that will
contain 2 GOLD medalists and 1 SILVER medalist. However, it is important to realize that these 24 victory
circles must be reduced due to "overcounting." To illustrate this, consider one of the 24 possible Gold-Gold-
Silver victory circles: Albert is awarded a GOLD. Bob is awarded a GOLD. Cami is awarded a SILVER. Notice
that this is the exact same victory circle as the following: Bob is awarded a GOLD. Albert is awarded a GOLD.
Cami is awarded a SILVER. Each victory circle has been "overcounted" because we have been counting each
different ordering of the two gold medals as a unique victory circle, when, in reality, the two different orderings
consist of the exact same victory circle. Thus, the 24 victory circles must be cut in half; there are actually only
12 unique victory circles that will contain 2 GOLD medalists and 1 SILVER medalist. (Note that we did not
have to worry about "overcounting" in Combination 1, because each of those 24 possibilities was unique.)
COMBINATION 3: Gold, Silver, Silver.
Using the same reasoning as for Combination 2, we see that there are 24 possible victory circles, but only 12
unique victory circles that contain 1 GOLD medalist and 2 SILVER medalists.
COMBINATION 4: Gold, Gold, Gold.
Here, once again, there are 24 possible victory circles. However, because all three winners are gold medalists,
there has been a lot of "overcounting!" How much overcounting? Let's consider one of the 24 possible Gold-
GMAT / Quant / P&C / Page 4 of 4
5. Gold-Gold victory circles: Albert is awarded a GOLD. Bob is awarded a GOLD. Cami is awarded a GOLD.
Notice that this victory circle is exactly the same as the following victory circles: Albert-GOLD, Cami-GOLD,
Bob-GOLD. Bob-GOLD, Albert-GOLD, Cami-GOLD. Bob-GOLD, Cami-GOLD, Albert-GOLD. Cami-
GOLD, Albert-GOLD, Bob-GOLD. Cami-GOLD, Bob-GOLD, Albert-GOLD. Each unique victory circle has
actually been counted 6 times! Thus we must divide 24 by 6 to find the number of unique victory circles. There
are actually only unique victory circles that contain 3 GOLD medalists. FINALLY, then, we have
the following:
(Combination 1) 24 unique GOLD-SILVER-BRONZE victory circles. (Combination 2) 12 unique GOLD-
GOLD-SILVER victory circles. (Combination 3) 12 unique GOLD-SILVER-SILVER victory circles.
(Combination 4) 4 unique GOLD-GOLD-GOLD victory circles. Thus, there are unique
victory circles. The correct answer is B.
41.
This question is not as complicated as it may initially seem. The trick is to recognize a recurring pattern
in the assignment of the guards. First, we have five guards (let's call them a, b, c, d, and e) and we
have to break them down into pairs. So how many pairs are possible in a group of five distinct entities?
We could use the combinations formula: nCr. where n is the number of items you are selecting from (the
pool) and k is the number of items you are selecting (the subgroup). Here we would get 5C2 = 10. So
there are 10 different pairs in a group of 5 individuals. However, in this particular case, it is actually
more helpful to write them out (since there are only 5 guards and 10 pairs, it is not so onerous): ab, ac,
ad, ae, bc, bd, be, cd, ce, de. Now, on the first night (Monday), any one of the ten pairs may be assigned,
since no one has worked yet. Let's say that pair ab is assigned to work the first night. That means no pair
containing either a or b may be assigned on Tuesday night. That rules out 7 of the 10 pairs, leaving only
cd, ce, and de available for assignment. If, say, cd were assigned on Tuesday, then on Wednesday no
pair containing either c or d could be assigned. This leaves only 3 pairs available for Wednesday: ab, ae,
and be. At this point the savvy test taker will realize that on any given night after the first, including
Saturday, only 3 pairs will be available for assignment. Those test takers who are really on the ball may
have realized right away that the assignment of any two guards on any night necessarily rules out 7 of
the 10 pairs for the next night, leaving only 3 pairs available on all nights after Monday. The correct
answer is Choice D; 3 different pairs will be available to patrol the grounds on Saturday night.
42. The key to this problem is to avoid listing all the possibilities. Instead, think of an arrangement of five
donuts and two dividers. The placement of the dividers determines which man is allotted which donuts,
as pictured below:
In this example, the first man receives one donut, the second man receives three donuts, and the third
man receives one donut. Remember that it is possible for either one or two of the men to be allotted no
donuts at all. This situation would be modeled with the arrangement below:
GMAT / Quant / P&C / Page 5 of 4
6. Here, the second man receives no donuts. Now all that remains is to calculate the number of ways in
which the donuts and dividers can be arranged: There are 7 objects. The number of ways in which 7
objects can be arranged can be computed by taking 7!: However, the two
dividers are identical, and the five donuts are identical. Therefore, we must divide 7! by 2! and by 5!:
The correct answer choice is A.
43. There are two possibilities in this problem. Either Kim and Deborah will both get chocolate chip cookies
or Kim and Deborah will both get oatmeal cookies. If Kim and Deborah both get chocolate chip cookies,
then there are 3 oatmeal cookies and 2 chocolate chip cookies left for the remaining four children. There
are 5!/3!2! = 10 ways for these 5 remaining cookies to be distributed--four of the cookies will go to the
children, one to the dog. (There are 5! ways to arrange 5 objects but the three oatmeal cookies are
identical so we divide by 3!, and the two chocolate chip cookies are identical so we divide by 2!.) If Kim
and Deborah both get oatmeal cookies, there are 4 chocolate chip cookies and 1 oatmeal cookie left for
the remaining four children. There are 5!/4! = 5 ways for these 5 remaining cookies to be distributed--
four of the cookies will go to the children, one to the dog. (There are 5! ways to arrange 5 objects but the
four chocolate chip cookies are identical so we divide by 4!.) Accounting for both possibilities, there are
10 + 5 = 15 ways for the cookies to be distributed. The correct answer is D.
44. In order to determine how many 10-flavor combinations Sammy can create, we simply need to know
how many different flavors Sammy now has. If Sammy had x flavors to start with and then threw out y
flavors, he now has x – y flavors. Therefore, we can rephrase this questions as: What is x – y ?
According to statement (1), if Sammy had x – y – 2 flavors, he could have made exactly 3,003 different
10-flavor bags. We could use the combination formula below to determine the value of x – y – 2, which
is equal to n in the equation below:
Solving this equation would require some time and more familiarity with factorials than is really
necessary for the GMAT. However, keep in mind that you do not need to solve this equation; you
merely need to be certain that the equation is solvable. (Note, if you begin testing values for n, you will
soon find that n = 15.) Once we know the value of n, we can easily determine the value of x – y, which
is simply 2 more than n. Thus, we know how many different flavors Sammy has, and could determine
how many different 10-flavor combinations he could make. Statement (2) is tells us that x = y + 17.
Subtracting y from both sides of the equation yields the equation x – y = 17. Thus, Sammy has 17
different flavors. This information is sufficient to determine the number of different 10-flavor
combinations he could make. The correct answer is D: Each statement ALONE is sufficient.
45. The three-dice combinations fall into 3 categories of outcomes:
1) All three dice have the same number 2) Two of the dice have the same number with the third having a
different number than the pair 3) All three dice have different numbers By calculating the number of
combinations in each category, we can determine the total number of different possible outcomes by
summing the number of possible outcomes in each category.
First, let’s calculate how many combinations can be made if all 3 dice have the same number. Since
there are only 6 numbers, there are only 6 ways for all three dice to have the same number (i.e., all 1’s,
or all 2’s, etc.). Second, we can determine how many combinations can occur if only 2 of the dice have
the same number. There are 6 different ways that 2 dice can be paired (i.e., two 1’s, or two 2’s or two
3’s, etc.). For each given pair of 2 dice, the third die can be one of the five other numbers. (For example
if two of the dice are 1’s, then the third die must have one of the 5 other numbers: 2, 3, 4, 5, or 6.)
Therefore there are 6 x 5 = 30 combinations of outcomes that involve 2 dice with the same number.
Third, determine how many combinations can occur if all three dice have different numbers. Think of
GMAT / Quant / P&C / Page 6 of 4
7. choosing three of the 6 items (each of the numbers) to fill three "slots." For the first slot, we can choose
from 6 possible items. For the second slot, we can choose from the 5 remaining items. For the third slot,
we can choose from the 4 remaining items. Hence, there are 6 x 5 x 4 = 120 ways to fill the three slots.
However, we do not care about the order of the items, so permutations like {1,2,5}, {5, 2, 1}, {2, 5, 1},
{2, 1, 5}, {5, 1, 2}, and {1, 5, 2} are all considered to be the same result. There are 3! = 6 ways that each
group of three numbers can be ordered, so we must divide 120 by 6 in order to obtain the number of
combinations where order does not matter (every 1 combination has 6 equivalent permutations). Thus
there are combinations where all three dice have different numbers. The total number of
combinations is the sum of those in each category or 6 + 30 + 20 = 56. The correct answer is C.
46. There are two different approaches to solving this problem. The first employs a purely algebraic
approach, as follows: Let us assume there are n teams in a double-elimination tournament. In order to
crown a champion, n – 1 teams must be eliminated, each losing exactly two games. Thus, the minimum
number of games played in order to eliminate all but one of the teams is 2(n – 1). At the time when the
(n – 1)th team is eliminated, the surviving team (the division champion) either has one loss or no losses,
adding at most one more game to the total played. Thus, the maximum number of games that can be
played in an n-team double-elimination tournament is 2(n – 1) + 1. There were four divisions with 9, 10,
11, and 12 teams each. The maximum number of games that could have been played in order to
determine the four division champions was (2(8) + 1) + (2(9) + 1) + (2(10) + 1) + (2(11) + 1) = 17 + 19
+ 21 + 23 = 80. The four division champions then played in a single-elimination tournament. Since each
team that was eliminated lost exactly one game, the elimination of three teams required exactly three
more games. Thus, the maximum number of games that could have been played in order to crown a
league champion was 80 + 3 = 83. The correct answer choice is (B). Another way to approach this
problem is to use one division as a concrete starting point. Let's think first about the 9-team division.
After 9 games, there are 9 losses. Assuming that no team loses twice (thereby maximizing the number of
games played), all 9 teams remain in the tournament. After 8 additional games, only 1 team remains
and is declared the division winner. Therefore, 9 + 8 = 17 games is the maximum # of games than can be
played in this tournament. We can generalize this information and apply it to the other divisions. To
maximize the # of games in the 10-team division, 10 + 9 = 19 games are played. To maximize the # of
games in the 11-team division, 11 + 10 = 21 games are played. To maximize the # of games in the 12-
team division, 12 + 11 = 23 games are played. Thus, the maximum number of games that could have
been played in order to determine the four division champions was 17 + 19 + 21 + 23 = 80. After 3
games in the single elimination tournament, there will be 3 losses, thereby eliminating all but the one
championship team. Thus, the maximum number of games that could have been played in order to
crown a league champion was 80 + 3 = 83. Once again, we see that the correct answer choice is (B).
47. The simplest way to solve this problem is to analyze one row at a time, and one square at a time in each
row if necessary. Let’s begin with the top row. First, let’s place a letter in left box; we have a choice of 3
different letters for this box: X, Y, or Z. Next, we place a letter in the top center box. Now we have only
2 options so as not to match the letter we placed in the left box. Finally, we only have 1 letter to choose
for the right box so as not to match either of the letters in the first two boxes. Thus, we have 3 x 2 x 1 or
6 ways to fill in the top row without duplicating a letter across it. Now let’s analyze the middle row by
assuming that we already have a particular arrangement of the top row, say the one given in the example
above (XYZ).
X Y Z
Given
Arrangement of
Top Row
Middle Row
LEFT Options
Middle Row
CENTER Options
Middle Row
RIGHT Options
GMAT / Quant / P&C / Page 7 of 4
8. Y X Z
Not Allowed: Z
is in Right
column twice
Z X Permissible
Z X Y Permissible
Y X
Not Allowed: Y
is in Center
column twice
Now let’s analyze the bottom row by assuming that we already have a particular arrangement of the top
and middle rows. Again, let’s use top and middle row arrangements given in the example above.
X Y Z
Given Arrangement
of Top Row
Y Z X
Give Arrangment
of Middle Row
Bottom Row
LEFT
OPtions
Bottom Row
CENTER
Options
Bottom Row
RIGHT
Options
Z X Y
Only this option is
permissible.
We can see that given fixed top and middle rows, there is only 1 possible bottom row that will work. (In
other words, the 3rd row is completely determined by the arrangement of the 1st and 2nd rows).
By combining the information about each row, we calculate the solution as follows: 6 possible top rows
× 2 possible middle rows × 1 possible bottom row = 12 possible grids. The correct answer is D.
48. This is a counting problem that is best solved using logic. First, let’s represent the line of women as
follows:
0000
0000
where the heights go from 1 to 8 in increasing order and the unknowns are designated 0s. Since the
women are arranged by their heights in increasing order from left to right and front to back, we know
that at a minimum, the lineup must conform to this:
0008
1000
Let’s further designate the arrangement by labeling the other individuals in the top row as X, Y and Z,
and the individuals in the bottom row as A, B, and C.
XYZ8
1ABC
Note that Z must be greater than at least 5 numbers (X, Y, B, A, and 1) and less than at least 1 number
(8). This means that Z can only be a 6 or a 7. Note that Y must be greater than at least 3 numbers (X, A
and 1) and less than at least 2 numbers (8 and Z). This means that Y can only be 4, 5, or 6. Note that X
must be greater than at least 1 number (1) and less than at least 3 numbers (8, Z and Y), This means that
X must be 2, 3, 4, or 5. This is enough information to start counting the total number of possibilities for
the top row. It will be easiest to use the middle unknown value Y as our starting point. As we
determined above, Y can only be 4, 5, or 6. Let’s check each case, making our conclusions logically: If
Y is 4, Z has 2 options (6 or 7) and X has 2 options (2 or 3). This yields 2 x 2 = 4 possibilities. If Y is 5,
Z has 2 options (6 or 7), and X has 3 options (2, 3, or 4). This yields 2 x 3 = 6 possibilities. If Y is 6, Z
GMAT / Quant / P&C / Page 8 of 4
9. has 1 option (7), and X has 4 options (2, 3, 4, or 5). This yields 1 x 4 = 4 possibilities. For each of the
possibilities above, the bottom row is completely determined because we have 3 numbers left, all of
which must be in placed increasing order. Hence, there are 4 + 6 + 4 = 14 ways for the women to pose.
The correct answer is (B).
49. One way to approach this problem is to pick an actual number to represent the variable n. This helps to
make the problem less abstract. Let's assume that n = 6. Since each of the regional offices must be
represented by exactly one candidate on the committee, the committee must consist of 6 members.
Further, because the committee must have an equal number of male and female employees, it must
include 3 men and 3 women. First, let's form the female group of the committee. There are 3 women to
be selected from 6 female candidates (one per region). One possible team selection can be represented as
follows, where A, B, C, D, E, & F represent the 6 female candidates:
A B C D E F
Yes Yes Yes No No No
In the representation above, women A, B, and C are on the committee, while women D, E, and F are not.
There are many other possible 3 women teams. Using the combination formula, the number of different
combinations of three female committee members is 6! / (3! × 3!) = 720/36 = 20. To ensure that each
region is represented by exactly one candidate, the group of men must be selected from the remaining
three regions that are not represented by female employees. In other words, three of the regions have
been “used up” in our selection of the female candidates. Since we have only 3 male candidates
remaining (one for each of the three remaining regions), there is only one possible combination of 3
male employees for the committee. Thus, we have 20 possible groups of three females and 1 possible
group of three males for a total of 20 × 1 = 20 possible groups of six committee members. Now, we
can plug 6 in for the variable n in each of the five answer choices. The answer choice that yields
the solution 20 is the correct expression. Therefore, B is the correct answer since plugging 6 in for n,
yields the following:
50. This is a relatively simple problem that can be fiendishly difficult unless you have a good approach to
solving it and a solid understanding of how to count. We will present two different strategies here.
Strategy 1: This problem seems difficult, because you need to figure out how many distinct orientations
the cube has relative to its other sides. Given that you can rotate the cube in an unlimited number of
ways, it is very difficult to keep track of what is going on – unless you have a system. Big hint: In order
to analyze how multiple things behave or compare or are arranged relative to each other, the first thing
one should do is pick a reference point and fix it. Here is a simple example. Let’s say you have a round
table with four seat positions and you want to know how many distinct ways you can orient 4 people
around it relative to each other (i.e., any two orientations where all 4 people have the same person to
their left and to their right are considered equivalent). Let’s pick person A as our reference point and
anchor her to the North position. Think about this next statement and convince yourself that it is true: By
choosing A as a fixed reference, all distinct arrangements of the other 3 people relative to A will
constitute the complete set of distinct arrangements of all 4 people relative to each other. Hence, fixing
the location of one person makes it significantly easier to keep track of what is going on. Given A is
fixed at the North, the 3 other people can be arranged in the 3 remaining seats in 3! = 6 ways, so there
are 6 distinct orientations of 4 people sitting around a circular table. Using the same principle, we can
conclude that, in general, if there are N people in a circular arrangement, after fixing one person at a
reference point, we have (N-1)! distinct arrangements relative to each other. Now let’s solve the
problem. Assume the six sides are: Top (or T), Bottom (or B), N, S, E, and W, and the six colors are
designated 1, 2, 3, 4, 5, and 6. Following the first strategy, let’s pick color #1 and fix it on the Top side
of the cube. If #1 is at the Top position, then one of the other 5 colors must be at the Bottom position
GMAT / Quant / P&C / Page 9 of 4
10. and each of those colors would represent a distinct set of arrangements. Hence, since there are exactly 5
possible choices for the color of the Bottom side, the number of unique arrangements relative to #1 in
the Top position is a multiple of 5. For each of the 5 colors paired with #1, we need to arrange the other
4 colors in the N, S, E, and W positions in distinct arrangements. Well, this is exactly like arranging 4
people around a circular table, and we have already determined that there are (n-1)! or 3! ways to do
that. Hence, the number of distinct patterns of painting the cube is simply 5 x 3! = 30. The correct
answer is B.
Strategy 2: There is another way to solve this kind of problem. Given one distinct arrangement or
pattern, you can try to determine how many equivalent ways there are to represent that one particular
arrangement or pattern within the set of total permutations, then divide the total number of permutations
by that number to get the number of distinct arrangements. This is best illustrated by example so let’s go
back to the 4 people arranged around a circular table. Assume A is in the North position, then going
clockwise we get B, then C, then D. Rotate the table 1/4 turn clockwise. Now we have a different
arrangement where D is at the North position, followed clockwise by A, then B, then C. BUT, this is
merely a rotation of the distinct relative position of the 4 people (i.e., everyone still has the same person
to his right and to his left) so they are actually the same arrangement. We can quickly conclude that
there are 4 equivalent or non-distinct arrangements for every distinct relative positioning of the 4 people.
We can arrange 4 people in a total of 4! = 24 ways. However, each DISTINCT arrangement has 4
equivalents, so in order to find the number of distinct arrangements, we need to divide 4! by 4, which
yields 3! or 6 distinct ways to arrange 4 people around a circular table, the same result we got using the
“fixed reference” method in Strategy 1. Generalizing, if there are N! ways to arrange N people around a
table, each distinct relative rotation can be represented in N ways (each 1/Nth rotation around the table)
so the number of distinct arrangements is N!/N = (N-1)! Now let’s use Strategy #2. Consider a cube that
is already painted in a particular way. Imagine putting the cube on the table, with color #1 on the top
side. Note, that by rotating the cube, we have 4 different orientations of this particular cube given color
#1 is on top. Using symmetry, we can repeat this analysis when #1 is facing any of the other 5
directions. Hence, for each of the six directions that the side painted with #1 can face, there are 4 ways
to orient the cube. Consequently, there are 6 x 4 = 24 total orientations of any one cube painted in a
particular manner. Since there are 6 sides and 6 colors, there are 6! or 720 ways to color the six sides
each with one color. However, we have just calculated that each DISTINCT pattern has 24 equivalent
orientations, so 720 must be divided by 24 to get the number of distinct patterns. This yields 720/24 =
30, confirming the answer found using Strategy #1. Again, the correct answer is B.
51. The first thing to recognize here is that this is a permutation with restrictions question. In such questions
it is always easiest to tackle the restricted scenario(s) first. The restricted case here is when all of the
men actually sit together in three adjacent seats. Restrictions can often be dealt with by considering the
limited individuals as one unit. In this case we have four women (w1, w2, w3, and w4) and three men (m1,
m2, and m3). We can consider the men as one unit, since we can think of the 3 adjacent seats as simply 1
seat. If the men are one unit (m), we are really looking at seating 5 individuals (w1, w2, w3, w4, and m) in
5 seats. There are 5! ways of arranging 5 individuals in a row. This means that our group of three men is
sitting in any of the “five” seats. Now, imagine that the one seat that holds the three men magically splits
into three seats. How many different ways can the men arrange themselves in those three seats? 3!. To
calculate the total number of ways that the men and women can be arranged in 7 seats such that the men
ARE sitting together, we must multiply these two values: 5!3!. However this problem asks for the
number of ways the theatre-goers can be seated such that the men are NOT seated three in a
row. Logically, this must be equivalent to the following: (Total number of all seat arrangements) –
(Number of arrangements with 3 men in a row). The total number of all seat arrangements is simply 7!
so the final calculation is 7! – 5!3!. The correct answer is C.
GMAT / Quant / P&C / Page 10 of 4
11. 52. It is important to first note that our point of reference in this question is all the possible subcommittees
that include Michael. We are asked to find what percent of these subcommittees also include Anthony.
Let's first find out how many possible subcommittees there are that must include Michael. If Michael
must be on each of the three-person committees that we are considering, we are essentially choosing
people to fill the two remaining spots of the committee. Therefore, the number of possible committees
can be found by considering the number of different two-people groups that can be formed from a pool
of 5 candidates (not 6 since Michael was already chosen). Using the anagram method to solve this
combinations question, we assign 5 letters to the various board members in the first row. In the second
row, two of the board members get assigned a Y to signify that they were chosen and the remaining 3
get an N, to signify that they were not chosen:
A B C D E
Y Y N N N
The number of different combinations of two-person committees from a group of 5 board members
would be the number of possible anagrams that could be formed from the word YYNNN = 5! / (3!2!) =
10. Therefore there are 10 possible committees that include Michael. Out of these 10 possible
committees, of how many will Anthony also be a member? If we assume that Anthony and Michael
must be a member of the three-person committee, there is only one remaining place to fill. Since there
are four other board members, there are four possible three-person committees with both Anthony and
Michael. Of the 10 committees that include Michael, 4/10 or 40% also include Anthony. The correct
answer is C. As an alternate method, imagine splitting the original six-person board into two equal
groups of three. Michael is automatically in one of those groups of three. Now, Anthony could occupy
any one of the other 5 positions -- the 2 on Michael's committee and the 3 on the other committee. Since
Anthony has an equal chance of winding up in any of those positions, his chance of landing on Michael's
committee is 2 out of 5, or 2/5 = 40%. Since that probability must correspond to the ratio of committees
asked for in the problem, the answer is achieved. Answer choice C is correct.
53. The easiest way to solve this question is to consider the restrictions separately. Let’s start by considering
the restriction that one of the parents must drive, temporarily ignoring the restriction that the two sisters
won't sit next to each other. This means that…
2 people (mother or father) could sit in the driver’s seat
4 people (remaining parent or one of the children) could sit in the front passenger seat
3 people 0could sit in the first back seat
2 people could sit in the second back seat
1 person could sit in the remaining back seat
The total number of possible seating arrangements would be the product of these various possibilities: 2 × 4 × 3
× 2 × 1 = 48. We must subtract from these 48 possible seating arrangements the number of seating
arrangements in which the daughters are sitting together. The only way for the daughters to sit next to each
other is if they are both sitting in the back. This means that…
2 people (mother or father) could sit in the driver’s seat
2 people (remaining parent or son) could sit in the front passenger seat
GMAT / Quant / P&C / Page 11 of 4
12. Now for the back three seats we will do something a little different. The back three seats must contain the two
daughters and the remaining person (son or parent). To find out the number of arrangements in which the
daughters are sitting adjacent, let’s consider the two daughters as one unit. The remaining person (son or parent)
is the other unit. Now, instead of three seats to fill, we only have two "seats," or units, to fill.
There are 2 × 1 = 2 ways to seat these two units.
However, the daughter-daughter unit could be d1d2 or d2d1
We must consider both of these possibilities so we multiply the 2 by 2! for a total of 4 seating possibilities in the
back. We could also have manually counted these possibilities: d1d2X, d2d1X, Xd1d2, Xd2d1
Now we must multiply these 4 back seat scenarios by the front seat scenarios we calculated earlier:
(2 × 2) × 4 = 16
front back
If we subtract these 16 "daughters-sitting-adjacent" scenarios from the total number of "parent-driving"
scenarios, we get: 48 – 16 = 32 The correct answer is B.
54.
Ignoring Frankie's requirement for a moment, observe that the six mobsters can be arranged 6! or 6 x 5 x 4 x 3 x
2 x 1 = 720 different ways in the concession stand line. In each of those 720 arrangements, Frankie must be
either ahead of or behind Joey. Logically, since the combinations favor neither Frankie nor Joey, each would be
behind the other in precisely half of the arrangements. Therefore, in order to satisfy Frankie's requirement, the
six mobsters could be arranged in 720/2 = 360 different ways. The correct answer is D.
55.
We can imagine that the admissions committee will choose a “team” of students to receive scholarships.
However, because there are three different levels of scholarships, it will not suffice to simply count the number
of possible “scholarship teams.” In other words, the committee must also place the “scholarship team” members
according to scholarship level. So, order matters. If we knew the number of scholarships to be granted at each of
the three scholarship levels, we could use the anagram method to count the number of ways the scholarships
could be doled out among the 10 applicants. To show that this is true, let’s invent a hypothetical case for which
there are 3 scholarships to be granted at each of the three levels. If we assign letters to the ten applicants from A
to J, and if T represents a $10,000 scholarship, F represents a $5,000 scholarship, O represents a $1,000
scholarship, and N represents no scholarship, then the anagram grid would look like this:
A B C D E F G H I
J
T T T F F F O O O
N
Our “word” is TTTFFFOOON. To calculate the number of different “spellings” of this “word,” we use the
following shortcut:
10!
(3!)(3!)(3!)(1!)
The 10 represents the 10 total applicants, the 3’s represent the 3 T’s, 3 F’s, and 3 O’s, and the 1
represents the 1 N. Simplifying this expression would yield the number of ways to distribute the
scholarships among the 10 applicants. So, knowing the number of scholarships to be granted at each of
the three scholarship levels allows us to calculate the answer to the question. Therefore, the rephrased
question is: "How many scholarships are to be granted at each of the three scholarship levels?" (1)
INSUFFICIENT: While this tells us the total number of scholarships to be granted, we still don’t know
how many from each level will be granted. (2) INSUFFICIENT: While this tells us that the number of
scholarships from each level will be equal, we still don’t know how many from each level will be
GMAT / Quant / P&C / Page 12 of 4
13. granted. (1) AND (2) SUFFICIENT: If there are 6 total scholarships to be granted and the same number
from each level will be granted, there will be two $10,000 scholarships, two $5,000 scholarships, and
two $1,000 scholarships granted. The correct answer is C.
56.
In order to know how many panels we can form when choosing three women and two men, we need to
know how many women and men we have to choose from. In this case, we need to know the value of x
(the number of women to choose from) and the value of y (the number of men to choose from). The
number of panels will be equal to the number of groups of three that could be chosen from x women
multiplied by the number of groups of two that could be chosen from y men. (1) INSUFFICIENT: This
statement tells us that choosing 3 from x + 2 would yield 56 groups. One concept that you need to know
for the exam is that when dealing with combinations and permutations, each result corresponds to a
unique set of circumstances. For example, if you have z people and know that choosing two of them
would result in 15 different possible groups of two, it must be true that z = 6. No other value of z would
yield exactly 15 different groups of two. So if you know how many subgroups of a certain size you can
choose from an unknown original larger group, you can deduce the size of the larger group. In the
present case, we know that choosing three women from x + 2 women would yield 56 groups of 3. These
numbers must correspond to a specific value of x. Do not worry if you do not know what value of x
would yield these results (in this case, x must equal 6, because the only way to obtain 56 groups
if choosing 3 is to choose from a group of 8. Since the statement tells us that 8 is 2 more than the value
of x, x must be 6). The GMAT does not expect you to memorize all possible results. It is enough to
understand the underlying concept: if you know the number of groups yielded (in this case 56), then you
know that there is only one possible value of x. (2) INSUFFICIENT: Knowing only that x = y + 1 tells
us nothing specific about the values of x and y. Infinitely many values of x and y satisfy this equation,
thus yielding infinitely many answers to the question. (1) AND (2) SUFFICIENT: Taking the
statements together, we know that (1) gives us the value of x and that (2) allows us to use that value of x
to determine the value of y. Remember that, with data sufficiency, we do not actually need to calculate
the values for x and y; it is enough to know that we can calculate them. The correct answer is C.
57.
To find the total number of possible committees, we need to determine the number of different five-
person groups that can be formed from a pool of 8 candidates. We will use the anagram method to solve
this combinations question. First, let's create an anagram grid and assign 8 letters in the first row, with
each letter representing one of the candidates. In the second row, 5 of the candidates get assigned a Y to
signify that they were chosen for a committee; the remaining 3 candidates get an N, to signify that they
were not chosen:
A B C D E F G H
Y Y Y Y Y N N N
The total number of possible five-person committees that can be created from a group of 8 candidates will be
equal to the number of possible anagrams that can be formed from the word YYYYYNNN = 8! / (5!3!) = 56.
Therefore, there are a total of 56 possible committees. The correct answer is D.
58. Using the anagram method to solve this combinations question, we assign 10 letters to the 10 teams in
the first row. In the second row, three of the teams are assigned numbers (1,2,3) representing gold, silver
and bronze medals. The remaining seven teams get an N, to signify that they do NOT receive a medal.
GMAT / Quant / P&C / Page 13 of 4
14. A B C D E F G H I J
1 2 3 N N N N N N N
The above anagram represents ONE possible way to assign the medals. The number of different possible ways
to assign the three medals to three of the 10 competing teams is equal to the number of possible anagrams
(arrangements of letters) that can be formed from the word 123NNNNNNN. Since there are 10 letters and 7
repeats, this equals 10! / 7! . Ans A.
59.
This problem cannot be solved through formula. Given that the drawer contains at least three socks of each color, we
know that at least one matched pair of each color can be removed. From the first nine socks, we can therefore make three
pairs, leaving three ‘orphans.’ To think through the problem, it is useful to conceptualize removing those nine socks from
the drawer. We will need additional information about any socks left in the drawer to solve the problem.
(1) INSUFFICIENT: Once those first nine socks have been removed, only two socks remain, but we do not have sufficient
information about the color of the two socks to solve the problem. If the two remaining socks are a matched pair, we can
add this final pair to the first three. This scenario results in four pairs and three orphans. However, if the final two socks
are mismatched, each will make a new pair with one of the original three orphans, resulting in five pairs and one orphan.
(2) INSUFFICIENT: This statement gives no information about how many socks are in the drawer.
(1) AND (2) INSUFFICIENT: Given that the drawer contains 11 socks and that there are an equal number of black and
gray socks, there are two possible scenarios. Three black, three gray, and five blue socks would yield four pairs total.
Four black, four gray, and three blue socks would yield five pairs total.
The correct answer is E.
60.
There are 3 × 2 × 4 = 24 possible different shirt-sweater-hat combinations that Kramer can wear. He wears the first one
on a Wednesday. The following Wednesday he will wear the 8th combination. The next Wednesday after that he will wear
the 15th combination. The next Wednesday after that he will wear the 22nd combination. On Thursday, he will wear the
23rd combination and on Friday he will wear the 24th combination.
Thus, the first day on which it will no longer be possible to wear a new combination is Saturday. The correct answer is E.
61.
With one letter, 26 stocks can be designated.
GMAT / Quant / P&C / Page 14 of 4
15. With two letters, 26*26 stocks can be designated.
With three letters, 26*26*26 stocks can be designated.
So, 26+26
2
+26
3
, the units digit is 8.
Answer is E
62.
C(4,1)C(6,2)+C(4,2)C(6,1)+C(4,3)=60+36+4=100
63.
C(9,1)C(9,1)C(8,1)C(7,1)=4536
64.
Answer: 10
65.
A derangement is a permutation in which none of the objects appear in their "natural" (i.e., ordered) place. For example,
the only derangements of (1, 2, 3) are (2, 3, 1) and (3, 1, 2), so !3 = 2. The function giving the number of distinct
derangements on n elements is called the sub-factorial !n and is equal to
If some, but not necessarily all, of the items are not in their original ordered positions, the configuration can be referred to
as a partial derangement. Among the n! possible permutations of n distinct items, examine the number R(n, k) of these
permutations in which exactly k items are in their original ordered positions. Then
R(n, n) = 1
R (n, n-1) = 0
, where denotes
n
Ck and !(n – k) is the sub-factorial.
In the given question, n = 4, k = 1, R (4, 1) =
4
C1 !(4 – 1) = 4 2 = 8. Total number of arrangements =
4
P4 = 4! = 24.
Answer = 8/24 = 1/3.
GMAT / Quant / P&C / Page 15 of 4
16. 66.
When solve such circular permutation questions, we just count one element less than the line permutation question.
The answer is P4,4=24
GMAT / Quant / P&C / Page 16 of 4