Basics of geometry including concepts of triangles, similar triangles, quadrilaterals, polygons and circles are oft repeated questions both in the GMAT Problems Solving section and in the GMAT Data Sufficiency section of the GMAT Math Section. A good grasp of the fundamentals of Geometry will help you crack questions from this topic with relative ease. 4GMAT's GMAT Math Lesson Book in this chapter covers the following concepts:
1. Definition of various terms used in geometry
2. Properties of concepts explained, with illustrations
3. Detailed explanation of the concepts in angles
4. Introduction into triangles and related properties with illustrative examples
5. Different types of triangles
6. Concepts and theorems, such as Apollonius Theorem and Pythagoras Theorem, associated with triangles with illustrative examples
7. Explanation of relationship between area of triangles, inradius and circumradius
8. Introduction of different polygons, meaning of quadrilaterals, squares, rectangles, rhombus, parallelogram, kite and trapezium
9. Circles and various associated concepts and properties, with illustrative examples
10. Different types of circles
11. 22 solved examples with illustrations
12. 13 exercise problems with the answer key and explanatory answers
Here is an example of a typical solved example in this chapter.
Sample Question
Question
In a TriangleABC. AB = 4.5, AC = 6 and BC = 7.5. The points D & E lie on the line segment BC with BD = 1.5 and EC = 3. What is the value of the /_DAE?
Explanatory Answer
Geometry Solved Example - Solution Image The triangle is right-angled (4.5, 6 and 7.5).
Therefore /_A is right-angled.
BD = 1.5, EC = 3 => BE = 4.5 and DC = 6
Hence AB = BE = 4.5 and AC = CD = 6 and
Therefore
/_BAE = /_BEA = /_DEA and /_DAC = /_ADE
In triangle ADE, /_ADE + /_DEA + /_DAE = 180o
=> /_DAC + /_BAE + /_DAE = 180o
=> (90o + /_DAE) + /_DAE = 180o
=> /_DAE = 45o
Sunday, August 26, 2007
Set Theory
Basics of set language theory such as union of sets, intersection of sets get tested in the GMAT problem solving and data sufficiency sections. With a good grasp of the basic formulae governing set language, one should be able to attempt these questions with ease. 4GMAT's GMAT Math Lesson Book in this chapter covers the following concepts
1. Introduction of the concept of sets and methods to represent sets.
2. Different types of sets explained with illustrative examples.
3. Cardinal number of sets, power sets, number of subsets for a set.
4. The De-Morgan's Law.
5. Using Venn Diagrams to solve set language questions explained with illustrative examples.
6. 7 solved examples, including Venn Diagram representation.
7. 5 Comprehensive exercise problems with answer key and explanatory answers.
8. An objective type speed test with around 35 questions along with explanatory answers and answer key are provided for the speed test.
Here is an example of a typical solved example in this chapter.
Sample Question
Question
In a room of 50 people whose dresses have either red or white color, 30 are wearing red dress, 16 are wearing a combination of red and white. How many are wearing dresses that have only white color?
Explanatory Answer
Venn Diagram Number of people wearing a red dress = 30
i.e., n(R) = 30
Number of people wearing a combination of red and white = 16
i.e., n (R Intersection W) = 16
The total number of people in the room = number of people who are wearing dresses that have either red or white color = n (R Union W) = 50.
We know,
n (R Union W) = n(R) + n(W) - n(R Intersection W)
50 = 30 + n(W) - 16
50 - 14 = n(W) - 16
n(W) = 36
i.e., the number of people who are wearing a white dress = 36.
Therefore, number of people who are wearing white dress only = n(W) - n(R Intersection W) = 36 - 16 = 20
Additional Practice Questions in Set Language
Looking for additional practice questions in Set Theory? A collection of some questions available at
1. Introduction of the concept of sets and methods to represent sets.
2. Different types of sets explained with illustrative examples.
3. Cardinal number of sets, power sets, number of subsets for a set.
4. The De-Morgan's Law.
5. Using Venn Diagrams to solve set language questions explained with illustrative examples.
6. 7 solved examples, including Venn Diagram representation.
7. 5 Comprehensive exercise problems with answer key and explanatory answers.
8. An objective type speed test with around 35 questions along with explanatory answers and answer key are provided for the speed test.
Here is an example of a typical solved example in this chapter.
Sample Question
Question
In a room of 50 people whose dresses have either red or white color, 30 are wearing red dress, 16 are wearing a combination of red and white. How many are wearing dresses that have only white color?
Explanatory Answer
Venn Diagram Number of people wearing a red dress = 30
i.e., n(R) = 30
Number of people wearing a combination of red and white = 16
i.e., n (R Intersection W) = 16
The total number of people in the room = number of people who are wearing dresses that have either red or white color = n (R Union W) = 50.
We know,
n (R Union W) = n(R) + n(W) - n(R Intersection W)
50 = 30 + n(W) - 16
50 - 14 = n(W) - 16
n(W) = 36
i.e., the number of people who are wearing a white dress = 36.
Therefore, number of people who are wearing white dress only = n(W) - n(R Intersection W) = 36 - 16 = 20
Additional Practice Questions in Set Language
Looking for additional practice questions in Set Theory? A collection of some questions available at
Probability
The concepts in this topic are essentially an extension of the fundamentals learnt in Permutation Combination. Many of the probability questions are restricted permutation questions. 4GMAT's GMAT Math Lesson Book in this chapter covers the following concepts:
1. Introduction to the concept of probability
2. Meaning of sample space and events along with illustrative examples to explain the same
3. Types of experiments in probability
4. Introduction to independent events, mutually exclusive events and probability of complimentary events.
5. Explanation of compound events
6. Method to evaluate the probability of an event with illustrative example and shortcut methods
7. 21 solved examples covering typical questions in tossing of coins, rolling of dice, picking cards from a pack of cards.
8. 15 exercise problems with the answer key and also explanatory answers
9. An objective type speed test with around 40 questions along with explanatory answers and answer key are provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question
A bag contains 5 yellow balls and 6 orange balls. When 4 balls are drawn at random simultaneously from the bag, what is the probability that not all of the balls drawn are orange?
Explanatory Answer
Sample Space (Denominator) : Four balls can be drawn from a bag containing 11 balls in 11C4 ways
Event (Numerator) : The number of ways in which all four balls drawn will be orange = 6C4.
Probability: The probability that all four balls drawn are orange
=
Therefore, the probability that not all of the balls drawn are orange = 1 - probability of all four being orange
=>
=>
1. Introduction to the concept of probability
2. Meaning of sample space and events along with illustrative examples to explain the same
3. Types of experiments in probability
4. Introduction to independent events, mutually exclusive events and probability of complimentary events.
5. Explanation of compound events
6. Method to evaluate the probability of an event with illustrative example and shortcut methods
7. 21 solved examples covering typical questions in tossing of coins, rolling of dice, picking cards from a pack of cards.
8. 15 exercise problems with the answer key and also explanatory answers
9. An objective type speed test with around 40 questions along with explanatory answers and answer key are provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question
A bag contains 5 yellow balls and 6 orange balls. When 4 balls are drawn at random simultaneously from the bag, what is the probability that not all of the balls drawn are orange?
Explanatory Answer
Sample Space (Denominator) : Four balls can be drawn from a bag containing 11 balls in 11C4 ways
Event (Numerator) : The number of ways in which all four balls drawn will be orange = 6C4.
Probability: The probability that all four balls drawn are orange
=
Therefore, the probability that not all of the balls drawn are orange = 1 - probability of all four being orange
=>
=>
Permutation and Combination
Permutation combination is considered to be a hard topic by many GMAT aspirants. It is widely believed that questions from this topic appear when you score in the higher percentile levels in the GMAT Math section. If you have good understanding of the basics of this topic, it will be quite easy to crack questions that appear from permutation combination. 4GMAT's Math Lesson Book in this chapter covers the following areas
1. Independent Events, Product Rule, Sampling with and without replacement, sampling with and without ordering (arrangement).
2. Introduction to Permutation, combination. Difference between permutation and combination. Formulae associated with permutation and combination.
3. Examples of sampling with replacements, r-sequence and r-multisets.
4. Solved examples involving permutation and combination concepts in listing numbers
5. Solved examples involving re-arranging letters of words and their ranks
6. Concepts and solved examples on tossing of coins
7. Concepts and solved examples on rolling of a die and multiple dice
8. Solved examples on drawing one or more cards from a pack of cards
9. Typical permutation problems such as circular permutation, arranging boys and girls in a line etc.,
10. Typical combination problems such as questions on making musical albums, chess boards etc.,
11. About 5 illustrative examples to explain concepts; over 50 solved examples (with shortcuts wherever applicable to hard and tough questions) to acquaint you with as many different questions as possible; around 25 exercise problems with answer key and explanatory answers to provide you with practice and an objective type speed test with 50 questions. Explanatory answers and answer key are provided for the speed test.
Here is an example of a typical solved example in this chapter.
Sample Question
Question:
Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same when looked at in a mirror. They are called symmetric letters. Other letters in the alphabet are asymmetric letters. How many three letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter?
Explanatory Answer
There are three possible cases that will satisfy the condition of forming three letter passwords with at least 1 symmteric letter.
Case 1: 1 symmetric and 2 asymmetric or
Case 2: 2 symmetric and 1 asymmetric or
Case 3: all 3 symmetric
= {(11C1 * 15C2) + (11C2 * 15C1) + 11C3} * 3!
= {11 * permutation + permutation * 15 + permutation} x 6
= {1155 + 825 + 165} * 6
= 2145 * 6 = 12870
1. Independent Events, Product Rule, Sampling with and without replacement, sampling with and without ordering (arrangement).
2. Introduction to Permutation, combination. Difference between permutation and combination. Formulae associated with permutation and combination.
3. Examples of sampling with replacements, r-sequence and r-multisets.
4. Solved examples involving permutation and combination concepts in listing numbers
5. Solved examples involving re-arranging letters of words and their ranks
6. Concepts and solved examples on tossing of coins
7. Concepts and solved examples on rolling of a die and multiple dice
8. Solved examples on drawing one or more cards from a pack of cards
9. Typical permutation problems such as circular permutation, arranging boys and girls in a line etc.,
10. Typical combination problems such as questions on making musical albums, chess boards etc.,
11. About 5 illustrative examples to explain concepts; over 50 solved examples (with shortcuts wherever applicable to hard and tough questions) to acquaint you with as many different questions as possible; around 25 exercise problems with answer key and explanatory answers to provide you with practice and an objective type speed test with 50 questions. Explanatory answers and answer key are provided for the speed test.
Here is an example of a typical solved example in this chapter.
Sample Question
Question:
Each of the 11 letters A, H, I, M, O, T, U, V, W, X and Z appears same when looked at in a mirror. They are called symmetric letters. Other letters in the alphabet are asymmetric letters. How many three letter computer passwords can be formed (no repetition allowed) with at least one symmetric letter?
Explanatory Answer
There are three possible cases that will satisfy the condition of forming three letter passwords with at least 1 symmteric letter.
Case 1: 1 symmetric and 2 asymmetric or
Case 2: 2 symmetric and 1 asymmetric or
Case 3: all 3 symmetric
= {(11C1 * 15C2) + (11C2 * 15C1) + 11C3} * 3!
= {11 * permutation + permutation * 15 + permutation} x 6
= {1155 + 825 + 165} * 6
= 2145 * 6 = 12870
Arithmetic and Multiplicative Progressions (AP, GP).
One could expect one or two questions based on the topic sequences and series. The typical series tested included arithmetic progression and geometric or multiplicative progression. 4GMAT's GMAT Math Lesson Book in this chapter covers the following concepts in the topic sequences and progressions.
1. Introduction to Arithmetic Progression
2. Explanation with formulae to find the nth term of an arithmetic progression and the sum of n terms of an arithmetic progression.
3. Illustrative and solved examples to find the value of the common difference, first term and the number of terms, given the sum of n terms or the nth term and first term of an AP.
4. Property changes when an AP is transformed by addition, multiplication of constant terms.
5. Introduction to Multiplicative Progression (Geometric Progression)
6. Formulae to find the nth term of a multiplicative progression and the sum up to n terms of a multiplicative progression.
7. Introduction to the concept of infinitely decreasing multiplicative progression and the formulae to find the sum of such a sequence.
8. Geometric Mean or the mean of the Multiplicative progression.
9. Transformation witnessed when a Multiplicative Progression is multiplied by a constant.
10. Introduction to Harmonic progression. Formulae to find the nth term of a Harmonic Progression.
11. Relation between Arithmetic mean, Multiplicative mean and Harmonic mean.
12. 2 illustrative examples to explain concepts; 25 solved examples (with shortcuts wherever applicable) to acquaint you with as many different questions as possible; around 20 exercise problems with answer key and explanatory answers to provide you with practice and an objective type speed test with 40+ questions. Explanatory answers and answer key are provided for the speed test.
Here is an example of a typical solved example in this chapter.
Sample Question
Question
There are 4 terms in an A.P. such that the sum of the two means is 21 and the product of the extremes is 54. What are the terms of the A.P?
Explanatory Answer
Let the four terms be a - 3d, a - d, a + d and a + 3d.
The sum of the two means = a - d + a + d = 2a = 21 or a = 10.5
The product of the two extremes = (a - 3d)(a + 3d) = a2 - 9d2 = 54 => 10.52 - 9d2 = 54
=> 9d2 = 110.25 - 54 = 56.25
=> 3d = 7.5 => d = 2.5.
The four terms are a - 3d = 10.5 - 7.5 = 3, a - d = 10.5 - 2.5 = 8; a + d = 10.5 + 2.5 = 13 and a + 3d = 10.5 + 7.5 = 18. ie. 3, 8, 13, 18
Note
In the above expression, the term 'a' is not the first term as is generally assumed and the common difference is not 'd'. There is actually no term as 'a' as part of this progression. The terms are a - 3d, a - d ... and the common difference is 2d.
1. Introduction to Arithmetic Progression
2. Explanation with formulae to find the nth term of an arithmetic progression and the sum of n terms of an arithmetic progression.
3. Illustrative and solved examples to find the value of the common difference, first term and the number of terms, given the sum of n terms or the nth term and first term of an AP.
4. Property changes when an AP is transformed by addition, multiplication of constant terms.
5. Introduction to Multiplicative Progression (Geometric Progression)
6. Formulae to find the nth term of a multiplicative progression and the sum up to n terms of a multiplicative progression.
7. Introduction to the concept of infinitely decreasing multiplicative progression and the formulae to find the sum of such a sequence.
8. Geometric Mean or the mean of the Multiplicative progression.
9. Transformation witnessed when a Multiplicative Progression is multiplied by a constant.
10. Introduction to Harmonic progression. Formulae to find the nth term of a Harmonic Progression.
11. Relation between Arithmetic mean, Multiplicative mean and Harmonic mean.
12. 2 illustrative examples to explain concepts; 25 solved examples (with shortcuts wherever applicable) to acquaint you with as many different questions as possible; around 20 exercise problems with answer key and explanatory answers to provide you with practice and an objective type speed test with 40+ questions. Explanatory answers and answer key are provided for the speed test.
Here is an example of a typical solved example in this chapter.
Sample Question
Question
There are 4 terms in an A.P. such that the sum of the two means is 21 and the product of the extremes is 54. What are the terms of the A.P?
Explanatory Answer
Let the four terms be a - 3d, a - d, a + d and a + 3d.
The sum of the two means = a - d + a + d = 2a = 21 or a = 10.5
The product of the two extremes = (a - 3d)(a + 3d) = a2 - 9d2 = 54 => 10.52 - 9d2 = 54
=> 9d2 = 110.25 - 54 = 56.25
=> 3d = 7.5 => d = 2.5.
The four terms are a - 3d = 10.5 - 7.5 = 3, a - d = 10.5 - 2.5 = 8; a + d = 10.5 + 2.5 = 13 and a + 3d = 10.5 + 7.5 = 18. ie. 3, 8, 13, 18
Note
In the above expression, the term 'a' is not the first term as is generally assumed and the common difference is not 'd'. There is actually no term as 'a' as part of this progression. The terms are a - 3d, a - d ... and the common difference is 2d.
Pipes, Cisterns and Work, Time
Concepts Covered The typical questions that appear from this topic are word problems. The basic concept tested in this chapter is analogous to the concepts tested in Speed Time Distance. So, if you have a good grasp of Speed Time Distance topic, then you will be able to understand and solve questions in this topic with a lot of ease. 4GMAT's GMAT Math Lesson Book in this chapter covers the following concepts:
1. The chapter begins with the rules governing the questions in pipes and cisterns and it is explained with illustrative examples
2. There are 17 solved examples in questions related to pipes and cisterns. Some of the tough math questions in this topic are featured here.
3. The concepts regarding work and time are explained with examples.
4. There are 13 solved examples for questions related to work and time. Shorcuts to few hard math questions in work time are provided.
5. 24 exercise problems in both pipes and cisterns as well as work and time have been provided with the answer key and explanatory answers to the questions
6. An objective type speed test with around 50 questions with explanatory answers and answer key provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question
Pipe A fills a tank in 30 minutes. Pipe B can fill the same tank 5 times as fast as pipe A. If both the pipes were kept open when the tank is empty, how much time will it take for the tank to overflow?
Explanatory Answer
Pipe B fills the tank 5 times as fast as pipe A.
Therefore, pipe B will fill the tank in one-fifth of the time that pipe A takes.
Pipe B will fill the tank in 30 / 5 = 6 minutes.
In 1 minute, pipe A will fill 1 / 30th of the tank and pipe B will fill 1 / 6th of the tank.
Therefore, together, the two pipes will fill (1/30 + 1/6) = (1+5 / 30) = 6/30 = 1/5th of the tank in a minute
Hence, the two pipes working together will take 5 minutes to fill the tank.
1. The chapter begins with the rules governing the questions in pipes and cisterns and it is explained with illustrative examples
2. There are 17 solved examples in questions related to pipes and cisterns. Some of the tough math questions in this topic are featured here.
3. The concepts regarding work and time are explained with examples.
4. There are 13 solved examples for questions related to work and time. Shorcuts to few hard math questions in work time are provided.
5. 24 exercise problems in both pipes and cisterns as well as work and time have been provided with the answer key and explanatory answers to the questions
6. An objective type speed test with around 50 questions with explanatory answers and answer key provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question
Pipe A fills a tank in 30 minutes. Pipe B can fill the same tank 5 times as fast as pipe A. If both the pipes were kept open when the tank is empty, how much time will it take for the tank to overflow?
Explanatory Answer
Pipe B fills the tank 5 times as fast as pipe A.
Therefore, pipe B will fill the tank in one-fifth of the time that pipe A takes.
Pipe B will fill the tank in 30 / 5 = 6 minutes.
In 1 minute, pipe A will fill 1 / 30th of the tank and pipe B will fill 1 / 6th of the tank.
Therefore, together, the two pipes will fill (1/30 + 1/6) = (1+5 / 30) = 6/30 = 1/5th of the tank in a minute
Hence, the two pipes working together will take 5 minutes to fill the tank.
Circular and Straight Races
The chapter on Races is essentially an extention to the concepts covered in the Speed Time and Distance Chapter and Ratio Proportion chapter. The lesson book has questions on two types of races - Straight Races and Circular Races. 4GMAT's GMAT Math Lesson Book in this chapter covers the following concepts:
1. Definition of the term race
2. Explanation of the concept of races along a straight track
3. Illustrative Examples are given to explain the above mentioned concept
4. 9 solved examples are provided with detailed explanation for races along a straight track
5. The concept of race in a circular path is explained along with examples
6. 4 solved examples are given with detailed explanation
7. 13 exercise problem of both types of races with the answer key and explanatory answers
8. An objective type speed test with around 40 questions with explanatory answers and answer key provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question:
P and Q run around a circular track of radius 49 meters starting from the same point at the same time and in the same direction. If 'P' runs at a speed of 15m/min and 'Q' runs at a speed of 10m/min, when will they meet again for the first time?
Explanatory Answer
As P and Q run in the same direction, their relative speed = 15 - 10 = 5 m/min
Circumference of the circular track is the length of the circular track = 2 * pi * r = 2 * (22/7) * 49 = 308 m.
Time taken for the two of them to meet for the first time after they have started a race along a circular track, starting simultaneously from the same point is given by
(length of the circular track / relative speed) = 308 / 5 = 61.6 minutes
Therefore, they will meet for the first time after 1 hour, 1 minute and 36 seconds.
1. Definition of the term race
2. Explanation of the concept of races along a straight track
3. Illustrative Examples are given to explain the above mentioned concept
4. 9 solved examples are provided with detailed explanation for races along a straight track
5. The concept of race in a circular path is explained along with examples
6. 4 solved examples are given with detailed explanation
7. 13 exercise problem of both types of races with the answer key and explanatory answers
8. An objective type speed test with around 40 questions with explanatory answers and answer key provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question:
P and Q run around a circular track of radius 49 meters starting from the same point at the same time and in the same direction. If 'P' runs at a speed of 15m/min and 'Q' runs at a speed of 10m/min, when will they meet again for the first time?
Explanatory Answer
As P and Q run in the same direction, their relative speed = 15 - 10 = 5 m/min
Circumference of the circular track is the length of the circular track = 2 * pi * r = 2 * (22/7) * 49 = 308 m.
Time taken for the two of them to meet for the first time after they have started a race along a circular track, starting simultaneously from the same point is given by
(length of the circular track / relative speed) = 308 / 5 = 61.6 minutes
Therefore, they will meet for the first time after 1 hour, 1 minute and 36 seconds.
Speed, Time and Distance
It is an important topic and includes relationship between speed, time, distance. Different units of measurement of these quantities. Also included are concepts on average speed and relative speed. 4GMAT's GMAT Math Lesson Book in this chapter covers the following concepts:
1. The meaning of distance, time and speed has been given and their respective units of measurement have also been given.
2. The relationship between the three has also been highlighted and for the better understanding of the concept illustrative examples have been provided.
3. Simple conversion factors for different units of speed, time and distance such as conversion between mph to kmph or m/sec.
4. Method to calculate average speed under three different circumstances has been explained along with examples
5. Definition of the term relative speed and the method of calculation when traveling in the same direction and different directions are explained
6. 23 solved examples of the above mentioned concepts are given with detailed explanation and the short cuts that can be used to arrive at the result quicker have been mentioned wherever applicable
7. In addition, boats in streams (Upstream and downstream) have been explained and to aid the understanding process solved examples have also been worked out
8. 24 exercise problems have been asked and the answer to the questions has also been provided along with explanatory answers
9. An objective type speed test with around 40 questions with explanatory answers and answer key provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question
A boat travels upstream at a speed of 30 mph and then downstream at a speed of 45 mph. What is the speed of the boat in still water and what is the speed of the stream?
Explanatory Answer
When the downstream speed (D) and upstream (U) of a boat are given:
The speed of boat in still water, B = (D + U) / 2 = (45 + 30) / 2 = 37.5 mph
Speed of the stream S = ((D - U) / 2) = ((45 - 30) / 2) = 7.5 mph
1. The meaning of distance, time and speed has been given and their respective units of measurement have also been given.
2. The relationship between the three has also been highlighted and for the better understanding of the concept illustrative examples have been provided.
3. Simple conversion factors for different units of speed, time and distance such as conversion between mph to kmph or m/sec.
4. Method to calculate average speed under three different circumstances has been explained along with examples
5. Definition of the term relative speed and the method of calculation when traveling in the same direction and different directions are explained
6. 23 solved examples of the above mentioned concepts are given with detailed explanation and the short cuts that can be used to arrive at the result quicker have been mentioned wherever applicable
7. In addition, boats in streams (Upstream and downstream) have been explained and to aid the understanding process solved examples have also been worked out
8. 24 exercise problems have been asked and the answer to the questions has also been provided along with explanatory answers
9. An objective type speed test with around 40 questions with explanatory answers and answer key provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question
A boat travels upstream at a speed of 30 mph and then downstream at a speed of 45 mph. What is the speed of the boat in still water and what is the speed of the stream?
Explanatory Answer
When the downstream speed (D) and upstream (U) of a boat are given:
The speed of boat in still water, B = (D + U) / 2 = (45 + 30) / 2 = 37.5 mph
Speed of the stream S = ((D - U) / 2) = ((45 - 30) / 2) = 7.5 mph
9. Inequalities
Inequalities is an important topic in the GMAT problem solving and GMAT data sufficiency sections. One can expect quite a few questions from this topic. Once the basics are understood and mastered, solving these questions is quite easy. 4GMAT's GMAT Math Lesson Book in this chapter covers the following concepts
1. Explanation to the basic concept of inequalities.
2. Various basic rules governing inequalities such as multiplying both sides of an inequality with a negative number, inequalities with algebraic equations such as quadratic expressions, inequalities with exponents, inequalities with modulus.
3. 16 solved examples, which also contain additional notes that help solve questions in inequalities quickly.
4. 17 exercise problems with the answer key and also explanatory answers.
5. An objective type speed test with around 35 questions along with explanatory answers and answer key are provided for the speed test.
Here is an example of a typical solved example in this chapter.
Sample Question
Question
Find the range of values for x for which 1 / (x - 2) > -2?
Explanatory Answer
As there is a linear term in 'x' in the denominator, we can eliminate the x terms in the denominator by multiplying both sides of the inequality by (x - 2)2 in the numerator
(x - 2) > -2(x - 2)2
=> (x - 2) > -2(x2 - 4x + 4)
=> (x - 2) > -2(2x2 + 8x - 8)
=> 0 > -2x2 + 7x - 6
=> 2x2 - 4x - 3x + 6 > 0
=> (2x - 3) (x - 2) > 0
The range of values of x that satisfy the above inequality are
In other words, x does not lie between 3 / 2 and 2.
1. Explanation to the basic concept of inequalities.
2. Various basic rules governing inequalities such as multiplying both sides of an inequality with a negative number, inequalities with algebraic equations such as quadratic expressions, inequalities with exponents, inequalities with modulus.
3. 16 solved examples, which also contain additional notes that help solve questions in inequalities quickly.
4. 17 exercise problems with the answer key and also explanatory answers.
5. An objective type speed test with around 35 questions along with explanatory answers and answer key are provided for the speed test.
Here is an example of a typical solved example in this chapter.
Sample Question
Question
Find the range of values for x for which 1 / (x - 2) > -2?
Explanatory Answer
As there is a linear term in 'x' in the denominator, we can eliminate the x terms in the denominator by multiplying both sides of the inequality by (x - 2)2 in the numerator
(x - 2) > -2(x - 2)2
=> (x - 2) > -2(x2 - 4x + 4)
=> (x - 2) > -2(2x2 + 8x - 8)
=> 0 > -2x2 + 7x - 6
=> 2x2 - 4x - 3x + 6 > 0
=> (2x - 3) (x - 2) > 0
The range of values of x that satisfy the above inequality are
In other words, x does not lie between 3 / 2 and 2.
7. Ratio, Proportion and Variation
Ratio Proportion Variation is a fundamental topic in GMAT Math. Many word problems in the problem solving section of GMAT Math section are essentially ratio proportion questions. 4GMAT's GMAT Math Review Book in problem solving covers the following concepts in this chapter:
1. The meaning of the term ratio and an illustrative example to explain the same
2. The rules governing ratios and fractions
3. Definition and explanation of the term proportion, continued proportion, direct variation and inverse variation
4. 35 solved examples with explanation that helps the process of understanding and also the essential short cuts that would save time in arriving at the solution as and when applicable
5. 21 exercise problems with the answer key and explanatory answers
6. An objective type speed test with around 50 questions with explanatory answers and answer key provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question
What is the value of 'x' if it is the mean proportional of x - 4 and x + 8?
Explanatory Answer
As x is the mean proportional of x - 4 and x + 8, we have x - 4 : x :: x : x + 8
Therefore,
Or x2 = (x-4)(x+8) => x2 = x2 + 4x - 32
Therefore, x = 8
1. The meaning of the term ratio and an illustrative example to explain the same
2. The rules governing ratios and fractions
3. Definition and explanation of the term proportion, continued proportion, direct variation and inverse variation
4. 35 solved examples with explanation that helps the process of understanding and also the essential short cuts that would save time in arriving at the solution as and when applicable
5. 21 exercise problems with the answer key and explanatory answers
6. An objective type speed test with around 50 questions with explanatory answers and answer key provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question
What is the value of 'x' if it is the mean proportional of x - 4 and x + 8?
Explanatory Answer
As x is the mean proportional of x - 4 and x + 8, we have x - 4 : x :: x : x + 8
Therefore,
Or x2 = (x-4)(x+8) => x2 = x2 + 4x - 32
Therefore, x = 8
5. Profit and Loss
Percentages, Fractions are basic concepts that appear frequently as part of the GMAT Problem Solving section. It is advised that you get a good grasp of converting percentages to equivalent fractions and fractions to equivalent percentages to able to solve questions in this topic with ease. 4GMAT's GMAT Math Lesson Book in this chapter covers the following concepts:
1. The meaning of the term percent and the need for using percentages
2. The relation between percents, fractions and decimals and the rules to convert from one form to another with illustrative examples
3. Explanation of the percentage change in values along with illustrative examples
4. 23 solved examples with detailed explanation and also the special mention of shortcuts that can be used as and when necessary
5. 29 exercise problems with answer key and explanatory answers to enhance understanding
6. An objective type speed test with around 40 questions with explanatory answers and answer key provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question
In an examination a candidate who secured 20% of the maximum marks failed by 10 marks. Another candidate who secured 30% of the maximum marks got 20 marks more than the pass mark. What is the pass mark in this examination?
Explanatory Answer
Let the pass mark in the examination be P and the maximum marks be M.
Candidate 1 secured 20% of maximum marks that was 10 marks less than pass mark
i.e., P = 20% of M + 10
Candidate 2 secured 30% of maximum marks that was 20 marks more than pass mark.
i.e., P= 30% of M - 20
Equating the two relations, we get 20% of M + 10 = 30% of M - 20 or 10% of M = 30
If 10% of the maximum marks is 30, then maximum marks, M= 300
Therefore, the pass mark is obtained by substituting for M in any one of the equations.
i.e., P = 30% of 300 - 20 = 90 - 20 = 70. Pass mark, P = 70
1. The meaning of the term percent and the need for using percentages
2. The relation between percents, fractions and decimals and the rules to convert from one form to another with illustrative examples
3. Explanation of the percentage change in values along with illustrative examples
4. 23 solved examples with detailed explanation and also the special mention of shortcuts that can be used as and when necessary
5. 29 exercise problems with answer key and explanatory answers to enhance understanding
6. An objective type speed test with around 40 questions with explanatory answers and answer key provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question
In an examination a candidate who secured 20% of the maximum marks failed by 10 marks. Another candidate who secured 30% of the maximum marks got 20 marks more than the pass mark. What is the pass mark in this examination?
Explanatory Answer
Let the pass mark in the examination be P and the maximum marks be M.
Candidate 1 secured 20% of maximum marks that was 10 marks less than pass mark
i.e., P = 20% of M + 10
Candidate 2 secured 30% of maximum marks that was 20 marks more than pass mark.
i.e., P= 30% of M - 20
Equating the two relations, we get 20% of M + 10 = 30% of M - 20 or 10% of M = 30
If 10% of the maximum marks is 30, then maximum marks, M= 300
Therefore, the pass mark is obtained by substituting for M in any one of the equations.
i.e., P = 30% of 300 - 20 = 90 - 20 = 70. Pass mark, P = 70
4. Percentages
Percentages, Fractions are basic concepts that appear frequently as part of the GMAT Problem Solving section. It is advised that you get a good grasp of converting percentages to equivalent fractions and fractions to equivalent percentages to able to solve questions in this topic with ease. 4GMAT's GMAT Math Lesson Book in this chapter covers the following concepts:
1. The meaning of the term percent and the need for using percentages
2. The relation between percents, fractions and decimals and the rules to convert from one form to another with illustrative examples
3. Explanation of the percentage change in values along with illustrative examples
4. 23 solved examples with detailed explanation and also the special mention of shortcuts that can be used as and when necessary
5. 29 exercise problems with answer key and explanatory answers to enhance understanding
6. An objective type speed test with around 40 questions with explanatory answers and answer key provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question
In an examination a candidate who secured 20% of the maximum marks failed by 10 marks. Another candidate who secured 30% of the maximum marks got 20 marks more than the pass mark. What is the pass mark in this examination?
Explanatory Answer
Let the pass mark in the examination be P and the maximum marks be M.
Candidate 1 secured 20% of maximum marks that was 10 marks less than pass mark
i.e., P = 20% of M + 10
Candidate 2 secured 30% of maximum marks that was 20 marks more than pass mark.
i.e., P= 30% of M - 20
Equating the two relations, we get 20% of M + 10 = 30% of M - 20 or 10% of M = 30
If 10% of the maximum marks is 30, then maximum marks, M= 300
Therefore, the pass mark is obtained by substituting for M in any one of the equations.
i.e., P = 30% of 300 - 20 = 90 - 20 = 70. Pass mark, P = 70
1. The meaning of the term percent and the need for using percentages
2. The relation between percents, fractions and decimals and the rules to convert from one form to another with illustrative examples
3. Explanation of the percentage change in values along with illustrative examples
4. 23 solved examples with detailed explanation and also the special mention of shortcuts that can be used as and when necessary
5. 29 exercise problems with answer key and explanatory answers to enhance understanding
6. An objective type speed test with around 40 questions with explanatory answers and answer key provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question
In an examination a candidate who secured 20% of the maximum marks failed by 10 marks. Another candidate who secured 30% of the maximum marks got 20 marks more than the pass mark. What is the pass mark in this examination?
Explanatory Answer
Let the pass mark in the examination be P and the maximum marks be M.
Candidate 1 secured 20% of maximum marks that was 10 marks less than pass mark
i.e., P = 20% of M + 10
Candidate 2 secured 30% of maximum marks that was 20 marks more than pass mark.
i.e., P= 30% of M - 20
Equating the two relations, we get 20% of M + 10 = 30% of M - 20 or 10% of M = 30
If 10% of the maximum marks is 30, then maximum marks, M= 300
Therefore, the pass mark is obtained by substituting for M in any one of the equations.
i.e., P = 30% of 300 - 20 = 90 - 20 = 70. Pass mark, P = 70
3. Quadratic Equations
Quadratic Equations or equations of the second order where the variables have a power of 2 is tested. Solutions, roots, types of roots of quadratic equations are tested. Factorizing a quadratic equation to find its solutions is also tested in GMAT. 4GMAT's GMAT Math Lesson Book in this chapter covers the following concepts:
1. Introduction to quadratic equations and roots of quadratic equations
2. Concept of discriminant of quadratic equation
3. Method to calculate the sum, difference and product of roots along with illustrative example
4. Method to form a quadratic equation, given its roots; explained with illustrative examples
5. Definition of conjugate roots
6. Concepts in quadratic equations such as common roots, minimum and maximum value of quadratic equations
7. Method to determine nature of roots (real, rational, equal, imaginary), using value of the discriminant
8. Method to determine the value and sign of roots using value and sign of the coefficients of the quadratic equation.
9. 15 solved examples to explain above mentioned concepts
10. 18 exercise problems with the answer key and also explanatory answers
11. An objective type speed test with around 40 questions along with explanatory answers and answer key are provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question
If one root of the quadratic equation 14x2 + 5x - 1 = 0 is 2, what is the other root?
Explanatory Answer
Let a and b be the two roots of the quadratic equation given above.
The sum of the roots a + b = -b/a = -5/14
As one of the roots a is 2, and the sum of the roots is -5/14, the other root b = -5/14 - 2 = -33/14
1. Introduction to quadratic equations and roots of quadratic equations
2. Concept of discriminant of quadratic equation
3. Method to calculate the sum, difference and product of roots along with illustrative example
4. Method to form a quadratic equation, given its roots; explained with illustrative examples
5. Definition of conjugate roots
6. Concepts in quadratic equations such as common roots, minimum and maximum value of quadratic equations
7. Method to determine nature of roots (real, rational, equal, imaginary), using value of the discriminant
8. Method to determine the value and sign of roots using value and sign of the coefficients of the quadratic equation.
9. 15 solved examples to explain above mentioned concepts
10. 18 exercise problems with the answer key and also explanatory answers
11. An objective type speed test with around 40 questions along with explanatory answers and answer key are provided for the speed test
Here is an example of a typical solved example in this chapter.
Sample Question
Question
If one root of the quadratic equation 14x2 + 5x - 1 = 0 is 2, what is the other root?
Explanatory Answer
Let a and b be the two roots of the quadratic equation given above.
The sum of the roots a + b = -b/a = -5/14
As one of the roots a is 2, and the sum of the roots is -5/14, the other root b = -5/14 - 2 = -33/14
2. Linear Equations
Concepts Covered
Simple linear Equations, equations of the first order with one, two or three variables are tested in GMAT Problem Solving and Data Sufficiency sections of the GMAT CAT Test. 4GMAT's GMAT Math Lesson Book in this chapter covers the following concepts:
1. Definition of simple equation and degree of equation
2. Method to solve equations
3. Definition of linear equation
4. Illustrative examples to explain the above concepts
5. Definition of independent equation, dependent equation, quadratic equation, linear equation in two variables
6. Introduction to the concept of system of linear equations and the two methods to solve simultaneous linear equations, namely, solving by substitution and solving by elimination
7. Different types of solutions to a system of linear equations
8. Special cases in linear equations
9. 11 solved examples that would help understand the concepts explained in the chapter. Includes word problems.
10. 8 exercise problems with the answer key and also explanatory answers
Here is an example of a typical solved example in this chapter.
Sample Question
Question
Solve the following system of equations
5x + 4y -z = 26; 3x + y + z = 5; -2x + 5y - 2z = 19
Explanatory Answer
5x + 4y - z = 26 (1)
3x + y + z = 5 (2)
-2x + 5y - 2z = 19 (3)
Adding equations (1) and (2), we get 8x + 5y = 31 (4)
Multiplying equation (2) by 2 and adding with equation (3), we get ; 4x + 7y = 29 (5)
Multiplying equation (5) by 2 and deducting it from equation (4), we get y =3
Substituting y = 3 in equation (5), we get x = 2
Substituting values of x and y in equation (1), we get z = -4
Simple linear Equations, equations of the first order with one, two or three variables are tested in GMAT Problem Solving and Data Sufficiency sections of the GMAT CAT Test. 4GMAT's GMAT Math Lesson Book in this chapter covers the following concepts:
1. Definition of simple equation and degree of equation
2. Method to solve equations
3. Definition of linear equation
4. Illustrative examples to explain the above concepts
5. Definition of independent equation, dependent equation, quadratic equation, linear equation in two variables
6. Introduction to the concept of system of linear equations and the two methods to solve simultaneous linear equations, namely, solving by substitution and solving by elimination
7. Different types of solutions to a system of linear equations
8. Special cases in linear equations
9. 11 solved examples that would help understand the concepts explained in the chapter. Includes word problems.
10. 8 exercise problems with the answer key and also explanatory answers
Here is an example of a typical solved example in this chapter.
Sample Question
Question
Solve the following system of equations
5x + 4y -z = 26; 3x + y + z = 5; -2x + 5y - 2z = 19
Explanatory Answer
5x + 4y - z = 26 (1)
3x + y + z = 5 (2)
-2x + 5y - 2z = 19 (3)
Adding equations (1) and (2), we get 8x + 5y = 31 (4)
Multiplying equation (2) by 2 and adding with equation (3), we get ; 4x + 7y = 29 (5)
Multiplying equation (5) by 2 and deducting it from equation (4), we get y =3
Substituting y = 3 in equation (5), we get x = 2
Substituting values of x and y in equation (1), we get z = -4
1. Number Theory
Number theory questions cover a wide array of concepts that range from simple questions on LCM, HCF to questions on factors, remainders, number of factors etc. 4GMAT's GMAT Math Lesson Book in this hard math topic covers the following concepts in Number Theory.
1. Definition of types of numbers from Natural Number to Real numbers, complex numbers and their conjugates.
2. Irrational numbers, surds, binomial quadratic surds and their conjugates.
3. Method to determine if a given number is a prime number.
4. Concepts and examples of co-prime numbers.
5. Tests of divisibility including shortcuts for divisibility of numbers such as '7' and '13'.
6. Concepts and word problem examples to elucidate LCM and HCF.
7. Fractions - Proper fractions, improper fractions, LCM and HCF of fractions.
8. Factors and method to find the number of factors for a given integer. Method to find the sum of all positive factors of a natural number.
9. Properties of remainders and quotients of division of large numbers such as 371.
10. Remainder theorem and its application with examples, which is considered to be a difficult topic with tough questions is explained with examples.
11. Factorials and highest power of a number contained in a factorial.
12. Properties of unit digits of numbers raised to large powers. For example unit digit of 3231179.
13. Conversion of numbers from one base to another and back to decimal notation.
14. 19 illustrative examples to explain the concepts; 34 solved examples (with shortcuts wherever applicable) to acquaint you with as many different questions as possible; 30 exercise problems with answer key and explanatory answers to provide you with practice and an objective type speed test with around 60 questions. Explanatory answers and answer key are provided for the speed test.
Here is an example of a typical solved example in this chapter.
Sample Question
Question
A green light flashes thrice in a minute and a red light flashes 4 times in 2 minutes. If the two lights start flashing together, how often will they flash together in an hour?
Explanatory Answer
The green light flashes thrice in a minute.
Therefore, the green light will flash once every 20 seconds.
The red light flashes 4 times in 2 minutes
Therefore, the red light will flash once every 30 seconds.
The green light will flash at multiples of 20 seconds. i.e., 20, 40, 60, 80 ... seconds.
The red light will flash at multiples of 30 seconds. i.e., 30, 60, 90, 120 ... seconds.
Hence, the two lights will flash together at time intervals that are common multiples to both 20 and 30.
The Least Common Multiple (LCM) of 20 and 30 = 2*3*10 = 60.
i.e., the two lights will flash together every 60 seconds or 1 minute.
Therefore, they will flash together 60 times in an hour.
1. Definition of types of numbers from Natural Number to Real numbers, complex numbers and their conjugates.
2. Irrational numbers, surds, binomial quadratic surds and their conjugates.
3. Method to determine if a given number is a prime number.
4. Concepts and examples of co-prime numbers.
5. Tests of divisibility including shortcuts for divisibility of numbers such as '7' and '13'.
6. Concepts and word problem examples to elucidate LCM and HCF.
7. Fractions - Proper fractions, improper fractions, LCM and HCF of fractions.
8. Factors and method to find the number of factors for a given integer. Method to find the sum of all positive factors of a natural number.
9. Properties of remainders and quotients of division of large numbers such as 371.
10. Remainder theorem and its application with examples, which is considered to be a difficult topic with tough questions is explained with examples.
11. Factorials and highest power of a number contained in a factorial.
12. Properties of unit digits of numbers raised to large powers. For example unit digit of 3231179.
13. Conversion of numbers from one base to another and back to decimal notation.
14. 19 illustrative examples to explain the concepts; 34 solved examples (with shortcuts wherever applicable) to acquaint you with as many different questions as possible; 30 exercise problems with answer key and explanatory answers to provide you with practice and an objective type speed test with around 60 questions. Explanatory answers and answer key are provided for the speed test.
Here is an example of a typical solved example in this chapter.
Sample Question
Question
A green light flashes thrice in a minute and a red light flashes 4 times in 2 minutes. If the two lights start flashing together, how often will they flash together in an hour?
Explanatory Answer
The green light flashes thrice in a minute.
Therefore, the green light will flash once every 20 seconds.
The red light flashes 4 times in 2 minutes
Therefore, the red light will flash once every 30 seconds.
The green light will flash at multiples of 20 seconds. i.e., 20, 40, 60, 80 ... seconds.
The red light will flash at multiples of 30 seconds. i.e., 30, 60, 90, 120 ... seconds.
Hence, the two lights will flash together at time intervals that are common multiples to both 20 and 30.
The Least Common Multiple (LCM) of 20 and 30 = 2*3*10 = 60.
i.e., the two lights will flash together every 60 seconds or 1 minute.
Therefore, they will flash together 60 times in an hour.
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