Datasets:

j-js
/

gmat-quant-corpus

Error code:   DatasetGenerationError
Exception:    CastError
Message:      Couldn't cast
id: string
source_name: string
source_file: string
page: int64
topic_guess: string
question_text: string
options: struct<A: string, B: string, C: string, D: string, E: string>
  child 0, A: string
  child 1, B: string
  child 2, C: string
  child 3, D: string
  child 4, E: string
correct_answer: null
solution_text: null
page_start: int64
source_type: string
text: string
page_end: int64
to
{'id': Value('string'), 'source_name': Value('string'), 'source_file': Value('string'), 'source_type': Value('string'), 'page_start': Value('int64'), 'page_end': Value('int64'), 'topic_guess': Value('string'), 'text': Value('string')}
because column names don't match
Traceback:    Traceback (most recent call last):
                File "/usr/local/lib/python3.12/site-packages/datasets/builder.py", line 1872, in _prepare_split_single
                  for key, table in generator:
                                    ^^^^^^^^^
                File "/usr/local/lib/python3.12/site-packages/datasets/packaged_modules/json/json.py", line 260, in _generate_tables
                  self._cast_table(pa_table, json_field_paths=json_field_paths),
                  ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
                File "/usr/local/lib/python3.12/site-packages/datasets/packaged_modules/json/json.py", line 120, in _cast_table
                  pa_table = table_cast(pa_table, self.info.features.arrow_schema)
                             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
                File "/usr/local/lib/python3.12/site-packages/datasets/table.py", line 2272, in table_cast
                  return cast_table_to_schema(table, schema)
                         ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
                File "/usr/local/lib/python3.12/site-packages/datasets/table.py", line 2218, in cast_table_to_schema
                  raise CastError(
              datasets.table.CastError: Couldn't cast
              id: string
              source_name: string
              source_file: string
              page: int64
              topic_guess: string
              question_text: string
              options: struct<A: string, B: string, C: string, D: string, E: string>
                child 0, A: string
                child 1, B: string
                child 2, C: string
                child 3, D: string
                child 4, E: string
              correct_answer: null
              solution_text: null
              page_start: int64
              source_type: string
              text: string
              page_end: int64
              to
              {'id': Value('string'), 'source_name': Value('string'), 'source_file': Value('string'), 'source_type': Value('string'), 'page_start': Value('int64'), 'page_end': Value('int64'), 'topic_guess': Value('string'), 'text': Value('string')}
              because column names don't match
              
              The above exception was the direct cause of the following exception:
              
              Traceback (most recent call last):
                File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1347, in compute_config_parquet_and_info_response
                  parquet_operations = convert_to_parquet(builder)
                                       ^^^^^^^^^^^^^^^^^^^^^^^^^^^
                File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 980, in convert_to_parquet
                  builder.download_and_prepare(
                File "/usr/local/lib/python3.12/site-packages/datasets/builder.py", line 884, in download_and_prepare
                  self._download_and_prepare(
                File "/usr/local/lib/python3.12/site-packages/datasets/builder.py", line 947, in _download_and_prepare
                  self._prepare_split(split_generator, **prepare_split_kwargs)
                File "/usr/local/lib/python3.12/site-packages/datasets/builder.py", line 1739, in _prepare_split
                  for job_id, done, content in self._prepare_split_single(
                                               ^^^^^^^^^^^^^^^^^^^^^^^^^^^
                File "/usr/local/lib/python3.12/site-packages/datasets/builder.py", line 1922, in _prepare_split_single
                  raise DatasetGenerationError("An error occurred while generating the dataset") from e
              datasets.exceptions.DatasetGenerationError: An error occurred while generating the dataset

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id string	source_name string	source_file string	source_type string	page_start int64	page_end int64	topic_guess string	text string
GMAT Club Math Book 2024 v8_p1_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	1	1	general	GMAT Club Math Book 8th Edition May 2024
GMAT Club Math Book 2024 v8_p2_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	2	2	word_problems	Table of Contents: Number Theory ………………………………………………… 2 Remainders ..…………………………………………………….18 Algebra …………………………………………………………….25 Sequences and Progressions ……………………………… 33 Functions / Coordinate Geometry ……………………….. 39 Word Problems………………………………………….……….59 Work Problems………………………………………….……….64 Distance Rate Time Problems……………………………….68 Overlapping Sets ……………………………………………….77 Probability ……………………………………………………….. 83 Combinations …………………………………………………… 91 Standard Deviation ……………………………………………. 96 Thank you for using GMAT Club Math Book. This is not a beginner’s guide as this work focuses on more advanced topics. Significant effort was invested by the authors and long-time GMAT Club members into creating a detailed overview of some of the challenging and misunderstood topics. I hope this book serves you well and big thanks to Bunuel, Walker, Shrouded1, and sriharimurthy! -BB, Founder of GMAT Club May 4, 2024 GMAT Club Math Book 2
GMAT Club Math Book 2024 v8_p3_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	3	3	number_theory	GMAT Club Math Book NUMBER THEORY Definition Number Theory is concerned with the properties of numbers in general, and in particular integers. As this is a huge issue we decided to divide it into smaller topics. Below is the list of Number Theory topics. GMAT Number Types GMAT deals with only Real Numbers: Integers, Fractions and Irrational Numbers. INTEGERS Definition Integers are defined as: all negative natural numbers , zero , and positive natural numbers . Note that integers do not include decimals or fractions - just whole numbers. Even and Odd Numbers An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. An even number is an integer of the form , where is an integer. An odd number is an integer that is not evenly divisible by 2. An odd number is an integer of the form , where is an integer. Zero is an even number. Addition / Subtraction: even +/- even = even; even +/- odd = odd; odd +/- odd = even. Multiplication: even * even = even; even * odd = even; odd * odd = odd. Division of two integers can result into an even/odd integer or a fraction. ZERO: 1. 0 is an integer. {. . . , −4, −3, −2, −1} {0} {1, 2, 3, 4, . . . } n = 2 k k n = 2 k+ 1 k GMAT Club Math Book 3
GMAT Club Math Book 2024 v8_p4_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	4	4	fractions_decimals_percents	2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even. 3. 0 is neither positive nor negative integer (the only one of this kind). 4. 0 is divisible by EVERY integer. IRRATIONAL NUMBERS Fractions (also known as rational numbers) can be written as terminating (ending) or repeating decimals (such as 0.5, 0.76, or 0.333333....). On the other hand, all those numbers that can be written as non- terminating, non-repeating decimals are non-rational, so they are called the "irrationals". Examples would be ("the square root of two") or the number pi ( ~3.14159..., from geometry). The rationals and the irrationals are two totally separate number types: there is no overlap. Putting these two major classifications, the rationals and the irrationals, together in one set gives you the "real" numbers. POSITIVE AND NEGATIVE NUMBERS A positive number is a real number that is greater than zero. A negative number is a real number that is smaller than zero. Zero is not positive, nor negative. Multiplication: positive * positive = positive positive * negative = negative negative * negative = positive Division: positive / positive = positive positive / negative = negative negative / negative = positive 2√ π = GMAT Club Math Book 4
GMAT Club Math Book 2024 v8_p5_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	5	5	number_theory	Prime Numbers A Prime number is a natural number with exactly two distinct natural number divisors: 1 and itself. Otherwise a number is called a composite number. Therefore, 1 is not a prime, since it only has one divisor, namely 1. A number is prime if it cannot be written as a product of two factors and , both of which are greater than 1: n = ab. • The first twenty-six prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101 • Note: only positive numbers can be primes. • There are infinitely many prime numbers. • The only even prime number is 2, since any larger even number is divisible by 2. Also 2 is the smallest prime. • All prime numbers except 2 and 5 end in 1, 3, 7 or 9, since numbers ending in 0, 2, 4, 6 or 8 are multiples of 2 and numbers ending in 0 or 5 are multiples of 5. Similarly, all prime numbers above 3 are of the form or , because all other numbers are divisible by 2 or 3. • Any nonzero natural number can be factored into primes, written as a product of primes or powers of primes. Moreover, this factorization is unique except for a possible reordering of the factors. • Prime factorization: every positive integer greater than 1 can be written as a product of one or more prime integers in a way which is unique. For instance integer with three unique prime factors , , and can be expressed as , where , , and are powers of , , and , respectively and are . Example: . • Verifying the primality (checking whether the number is a prime) of a given number can be done by trial division, that is to say dividing by all integer numbers smaller than , thereby checking whether is a multiple of . Example: Verifying the primality of : is little less than , from integers from to , is divisible by , hence is not prime. Note that, it is only necessary to try dividing by prime numbers up to , since if n has any divisors at all (besides 1 and n), then it must have a prime divisor. • If is a positive integer greater than 1, then there is always a prime number with . n > 1 a b 6n− 1 6n+ 1 n n a b c n = ∗ ∗ap bq cr p q r a b c ≥ 1 4200 = ∗ 3 ∗ ∗ 723 52 n n n√ n m ≤ n√ 161 161−− −√ 13 2 13 161 7 161 n√ n p n < p < 2 n GMAT Club Math Book 5
GMAT Club Math Book 2024 v8_p6_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	6	6	number_theory	Factors A divisor of an integer , also called a factor of , is an integer which evenly divides without leaving a remainder. In general, it is said is a factor of , for non-zero integers and , if there exists an integer such that . • 1 (and -1) are divisors of every integer. • Every integer is a divisor of itself. • Every integer is a divisor of 0, except, by convention, 0 itself. • Numbers divisible by 2 are called even and numbers not divisible by 2 are called odd. • A positive divisor of n which is different from n is called a proper divisor. • An integer n > 1 whose only proper divisor is 1 is called a prime number. Equivalently, one would say that a prime number is one which has exactly two factors: 1 and itself. • Any positive divisor of n is a product of prime divisors of n raised to some power. • If a number equals the sum of its proper divisors, it is said to be a perfect number. Example: The proper divisors of 6 are 1, 2, and 3: 1+2+3=6, hence 6 is a perfect number. There are some elementary rules: • If is a factor of and is a factor of , then is a factor of . In fact, is a factor of for all integers and . • If is a factor of and is a factor of , then is a factor of . • If is a factor of and is a factor of , then or . • If is a factor of , and , then a is a factor of . • If is a prime number and is a factor of then is a factor of or is a factor of . Finding the Number of Factors of an Integer First make prime factorization of an integer , where , , and are prime factors of and , , and are their powers. The number of factors of will be expressed by the formula . NOTE: this will include 1 and n itself. Example: Finding the number of all factors of 450: Total number of factors of 450 including 1 and 450 itself is factors. Greatest Common Factor (Divisior) - GCF (GCD) The greatest common divisor (gcd), also known as the greatest common factor (gcf), or highest common factor (hcf), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder. To find the GCF, you will need to do prime-factorization. Then, multiply the common factors (pick the lowest n n n m n m n k n = km a b a c a (b+ c) a (mb+ nc) m n a b b c a c a b b a a = b a = − b a bc gcd(a, b) = 1 c p p ab p a p b n = ∗ ∗ap bq cr a b c n p q r n (p+ 1)(q+ 1)(r+ 1) 450 = ∗ ∗21 32 52 (1 + 1) ∗ (2 + 1) ∗ (2 + 1) = 2 ∗ 3 ∗ 3 = 18 GMAT Club Math Book 6
GMAT Club Math Book 2024 v8_p7_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	7	7	number_theory	power of the common factors). • Every common divisor of a and b is a divisor of gcd(a, b). • ab=gcd(a, b)lcm(a, b) Lowest Common Multiple - LCM The lowest common multiple or lowest common multiple (lcm) or smallest common multiple of two integers a and b is the smallest positive integer that is a multiple both of a and of b. Since it is a multiple, it can be divided by a and b without a remainder. If either a or b is 0, so that there is no such positive integer, then lcm(a, b) is defined to be zero. To find the LCM, you will need to do prime-factorization. Then multiply all the factors (pick the highest power of the common factors). Perfect Square A perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is an perfect square. There are some tips about the perfect square: • The number of distinct factors of a perfect square is ALWAYS ODD. • The sum of distinct factors of a perfect square is ALWAYS ODD. • A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. • Perfect square always has even number of powers of prime factors. Divisibility Rules 2 - If the last digit is even, the number is divisible by 2. 3 - If the sum of the digits is divisible by 3, the number is also. 4 - If the last two digits form a number divisible by 4, the number is also. 5 - If the last digit is a 5 or a 0, the number is divisible by 5. 6 - If the number is divisible by both 3 and 2, it is also divisible by 6. 7 - Take the last digit, double it, and subtract it from the rest of the number, if the answer is divisible by 7 (including 0), then the number is divisible by 7. 8 - If the last three digits of a number are divisible by 8, then so is the whole number. 9 - If the sum of the digits is divisible by 9, so is the number. 10 - If the number ends in 0, it is divisible by 10. 11 - If you sum every second digit and then subtract all other digits and the answer is: 0, or is divisible by 11, then the number is divisible by 11. Example: to see whether 9,488,699 is divisible by 11, sum every second digit: 4+8+9=21, then subtract the sum of other digits: 21-(9+8+6+9)=-11, -11 is divisible by 11, hence 9,488,699 is divisible by 11. 12 - If the number is divisible by both 3 and 4, it is also divisible by 12. 25 - Numbers ending with 00, 25, 50, or 75 represent numbers divisible by 25. GMAT Club Math Book 7
GMAT Club Math Book 2024 v8_p8_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	8	8	number_theory	• Note: 0!=1. • Note: factorial of negative numbers is undefined. Trailing zeros: Trailing z eros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow. 125000 has 3 trailing zeros; The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer , can be determined with this formula: , where k must be chosen such that . It's easier if you look at an example: How many zeros are in the end (after which no other digits follow) of ? (denominator must be less than 32, is less) Hence, there are 7 zeros in the end of 32! The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero. Finding the number of powers of a prime number , in the . The formula is: ... till What is the power of 2 in 25!? Finding the power of non-prime in n!: How many powers of 900 are in 50! Make the prime factorization of the number: , then find the powers of these prime numbers in the n!. Find the power of 2: = Find the power of 3: = Find the power of 5: n Factorials Factorial of a non-negative integer, denoted by n!, is the product of all positive integers less than or equal to n. E.g. .5! = 1 ∗ 2 ∗ 3 ∗ 4 ∗ 5 n + + +. . . +n 5 n 52 n 53 n 5k ≤ n5k 32! + = 6 + 1 = 732 5 32 52 = 2552 p n! + +n p n p2 n p3 ≤ npx + + + = 12 + 6 + 3 + 1 = 2225 2 25 4 25 8 25 16 900 = ∗ ∗22 32 52 + + + +50 2 50 4 50 8 50 16 50 32 = 25 + 12 + 6 + 3 + 1 = 47 247 + + = 16 + 5 + 1 = 2250 3 50 9 50 27 322 GMAT Club Math Book 8
GMAT Club Math Book 2024 v8_p9_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	9	9	number_theory	= We need all the prime {2,3,5} to be represented twice in 900, 5 can provide us with only 6 pairs, thus there is 900 in the power of 6 in 50!. Consecutive Integers Consecutive integers are integers that follow one another, without skipping any integers. 7, 8, 9, and -2, -1, 0, 1, are consecutive integers. • Sum of consecutive integers equals the mean multiplied by the number of terms, . Given consecutive integers , , (mean equals to the average of the first and last terms), so the sum equals to . • If n is odd, the sum of consecutive integers is always divisible by n. Given , we have consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3. • If n is even, the sum of consecutive integers is never divisible by n. Given , we have consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4. • The product of consecutive integers is always divisible by . Given consecutive integers: . The product of 345*6 is 360, which is divisible by 4!=24. Evenly Spaced Set Evenly spaced set or an arithmetic progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. The set of integers is an example of evenly spaced set. Set of consecutive integers is also an example of evenly spaced set. • If the first term is and the common difference of successive members is , then the term of the sequence is given by: • In any evenly spaced set the arithmetic mean (average) is equal to the median and can be calculated by the formula , where is the first term and is the last term. Given the set , . • The sum of the elements in any evenly spaced set is given by: , the mean multiplied by the number of terms. OR, • Special cases: Sum of n first positive integers: Sum of n first positive odd numbers: , where is the last, term and given by: . Given first odd positive integers, then their sum equals to . Sum of n first positive even numbers: , where is the last, term and given by: . Given first positive even integers, then their sum equals to . • If the evenly spaced set contains odd number of elements, the mean is the middle term, so the sum is + = 10 + 2 = 1250 5 50 25 512 n n {−3, −2, −1, 0, 1, 2} mean = = −−3+2 2 1 2 − ∗ 6 = −31 2 {9, 10, 11} n = 3 {9, 10, 11, 12} n = 4 n n! n = 4 {3, 4, 5, 6} {9, 13, 17, 21} a1 d nth = + d(n− 1)an a1 mean = median = +a1 an 2 a1 an {7, 11, 15, 19} mean = median = = 137+19 2 Sum = ∗ n +a1 an 2 Sum = ∗ n 2 + d(n−1)a1 2 1 + 2+. . . + n = ∗ n1+n 2 + +. . . + = 1 + 3+. . . + =a1 a2 an an n2 an nth = 2 n− 1an n = 5 1 + 3 + 5 + 7 + 9 = = 2552 + +. . . + = 2 + 4+. . . +a1 a2 an an= n(n+ 1) an nth = 2 nan n = 4 2 + 4 + 6 + 8 = 4(4 + 1) = 20 GMAT Club Math Book 9
GMAT Club Math Book 2024 v8_p10_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	10	10	fractions_decimals_percents	middle term multiplied by number of terms. There are five terms in the set {1, 7, 13, 19, 25}, middle term is 13, so the sum is 13*5 =65. FRACTIONS Definition Fractional numbers are ratios (divisions) of integers. In other words, a fraction is formed by dividing one integer by another integer. Set of Fraction is a subset of the set of Rational Numbers. Fraction can be expressed in two forms fractional representation and decimal representation . Fractional representation Fractional representation is a way to express numbers that fall in between integers (note that integers can also be expressed in fractional form). A fraction expresses a part-to-whole relationship in terms of a numerator (the part) and a denominator (the whole). • The number on top of the fraction is called numerator or nominator. The number on bottom of the fraction is called denominator. In the fraction, , 9 is the numerator and 7 is denominator. • Fractions that have a value between 0 and 1 are called proper fraction. The numerator is always smaller than the denominator. is a proper fraction. • Fractions that are greater than 1 are called improper fraction. Improper fraction can also be written as a mixed number. is improper fraction. • An integer combined with a proper fraction is called mixed number. is a mixed number. This can also be written as an improper fraction: Converting Improper Fractions • Converting Improper Fractions to Mixed Fractions: 1. Divide the numerator by the denominator 2. Write down the whole number answer 3. Then write down any remainder above the denominator Example #1: Convert to a mixed fraction. Solution: Divide with a remainder of . Write down the and then write down the remainder above the denominator , like this: • Converting Mixed Fractions to Improper Fractions: 1. Multiply the whole number part by the fraction's denominator 2. Add that to the numerator 3. Then write the result on top of the denominator Example #2: Convert to an improper fraction. Solution: Multiply the whole number by the denominator: . Add the numerator to that: . Then write that down above the denominator, like this: ( )m n (a. bcd) 9 7 1 3 5 2 4 3 5 23 5 11 4 = 211 4 3 2 3 4 2 3 4 3 2 5 3 ∗ 5 = 15 15 + 2 = 17 17 5 GMAT Club Math Book 10
GMAT Club Math Book 2024 v8_p11_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	11	11	fractions_decimals_percents	Reciprocal Reciprocal for a number , denoted by or , is a number which when multiplied by yields . The reciprocal of a fraction is . To get the reciprocal of a number, divide 1 by the number. For example reciprocal of is , reciprocal of is . Operation on Fractions • Adding/Subtracting fractions: To add/subtract fractions with the same denominator, add the numerators and place that sum over the common denominator. To add/subtract fractions with the different denominator, find the Least Common Denominator (LCD) of the fractions, rename the fractions to have the LCD and add/subtract the numerators of the fractions • Multiplying fractions: To multiply fractions just place the product of the numerators over the product of the denominators. • Dividing fractions: Change the divisor into its reciprocal and then multiply. Example #1: Example #2: Given , take the reciprocal of . The reciprocal is . Now multiply: . Decimal Representation The decimals has ten as its base. Decimals can be terminating (ending) (such as 0.78, 0.2) or repeating (recuring) decimals (such as 0.333333....). Reduced fraction (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only (denominator) is of the form , where and are non-negative integers. For example: is a terminating decimal , as (denominator) equals to . Fraction is also a terminating decimal, as and denominator . Converting Decimals to Fractions • To convert a terminating decimal to fraction: 1. Calculate the total numbers after decimal point 2. Remove the decimal point from the number 3. Put 1 under the denominator and annex it with "0" as many as the total in step 1 4. Reduce the fraction to its lowest terms Example: Convert to a fraction. 1: Total number after decimal point is 2. 2 and 3: . 4: Reducing it to lowest terms: • To convert a recurring decimal to fraction: 1. Separate the recurring number from the decimal fraction 2. Annex denominator with "9" as many times as the length of the recurring number 3. Reduce the fraction to its lowest terms Example #1: Convert to a fraction. x 1 x x−1 x 1 a b b a 3 1 3 5 6 6 5 + = + =3 7 2 3 9 21 14 21 23 21 3 5 2 2 1 2 ∗ =3 5 1 2 3 10 a b b 2n5m m n 7 250 0.028 250 2 ∗ 53 3 30 =3 30 1 10 10 = 2 ∗ 5 0.56 56 100 =56 100 14 25 0.393939... GMAT Club Math Book 11
GMAT Club Math Book 2024 v8_p12_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	12	12	fractions_decimals_percents	1: The recurring number is . 2: , the number is of length so we have added two nines. 3: Reducing it to lowest terms: . • To convert a mixed-recurring decimal to fraction: 1. Write down the number consisting with non-repeating digits and r epeating digits. 2. Subtract non-repeating number from above. 3. Divide 1-2 by the number with 9's and 0's: for every repeating digit write down a 9, and for every non- repeating digit write down a zero after 9's. Example #2: Convert to a fraction. 1. The number consisting with non-repeating digits and repeating digits is 2512; 2. Subtract 25 (non-repeating number) from above: 2512-25=2487; 3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300. Rounding Rounding is simplifying a number to a certain place value. To round the decimal drop the extra decimal places, and if the first dropped digit is 5 or greater, round up the last digit that you keep. If the first dropped digit is 4 or smaller, round down (keep the same) the last digit that you keep. Example: 5.3485 rounded to the nearest tenth = 5.3, since the dropped 4 is less than 5. 5.3485 rounded to the nearest hundredth = 5.35, since the dropped 8 is greater than 5. 5.3485 rounded to the nearest thousandth = 5.349, since the dropped 5 is equal to 5. Ratios and Proportions Given that , where a, b, c and d are non-zero real numbers, we can deduce other proportions by simple Algebra. These results are often referred to by the names mentioned along each of the properties obtained. - invertendo - alternendo - componendo - dividendo - componendo & dividendo 39 39 99 39 2 =39 99 13 33 0.2512(12) =a b c d =b a d c =a c b d =a+b b c+d d =a−b b c−d d =a+b a−b c+d c−d GMAT Club Math Book 12
GMAT Club Math Book 2024 v8_p13_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	13	13	word_problems	EXPONENTS Exponents are a "shortcut" method of showing a number that was multiplied by itself several times. For instance, number multiplied times can be written as , where represents the base, the number that is multiplied by itself times and represents the exponent. The exponent indicates how many times to multiple the base, , by itself. Exponents one and zero: Any nonzero number to the power of 0 is 1. For example: and • Note: the case of 0^0 is not tested on the GMAT. Any number to the power 1 is itself. Powers o f zero: If the exponent is positive, the power of zero is zero: , where . If the exponent is negative, the power of zero ( , where ) is undefined, because division by zero is implied. Powers of one: The integer powers of one are one. Negative powers: Powers of minus one: If n is an even integer, then . If n is an odd integer, then . Operations involving the same exponents: Keep the exponent, multiply or divide the bases and not Operations involving the same bases: Keep the base, add or subtract the exponent (add for multiplication, subtract for division) Fraction as power: Exponential Equations: When solving equations with even exponents, we must consider both positive and negative possibilities for the solutions. a n an a n n a = 1a0 = 150 (−3 = 1)0 = aa1 = 00n n > 0 0n n < 0 = 11n =a−n 1 an (−1 = 1)n (−1 = −1)n ∗ = ( aban bn )n = (an bn a b )n ( =am)n amn =amn a( )mn (am)n ∗ =an am an+m =an am an−m =a 1 n a√n =a m n am−− −√n GMAT Club Math Book 13
GMAT Club Math Book 2024 v8_p14_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	14	14	sequences_patterns	For instance , the two possible solutions are and . When solving equations with odd exponents, we'll have only one solution. For instance for , solution is and for , solution is . Exponents and divisibility: is ALWAYS divisible by . is divisible by if is even. is divisible by if is odd, and not divisible by a+b if n is even. LAST DIGIT OF A PRODUCT Last digits of a product of integers are last digits of the product of last digits of these integers. For instance last 2 digits of 8459512408613 would be the last 2 digits of 4512813=540104=404=160=60 Example: The last digit of 859458958307=597=45*7=35=5? LAST DIGIT OF A POWER Determining the last digit of : 1. Last digit of is the same as that of ; 2. Determine the cyclicity number of ; 3. Find the remainder when divided by the cyclisity; 4. When , then last digit of is the same as that of and when , then last digit of is the same as that of , where is the cyclisity number. • Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base. • Integers ending with 2, 3, 7 and 8 have a cyclicity of 4. • Integers ending with 4 (eg. ) have a cyclisity of 2. When n is odd will end with 4 and when n is even will end with 6. • Integers ending with 9 (eg. ) have a cyclisity of 2. When n is odd will end with 9 and when n is even will end with 1. Example: What is the last digit of ? Solution: Last digit of is the same as that of . Now we should determine the cyclisity of : 1. 7^1=7 (last digit is 7) 2. 7^2=9 (last digit is 9) 3. 7^3=3 (last digit is 3) 4. 7^4=1 (last digit is 1) 5. 7^5=7 (last digit is 7 again!) ... So, the cyclisity of 7 is 4. Now divide 39 (power) by 4 (cyclisity), remainder is 3.So, the last digit of is the same as that of the last digit of , is the same as that of the last digit of , which is . ROOTS Roots (or radicals) are the "opposite" operation of applying exponents. For instance x^2=16 and square root of 16=4. = 25a2 5 −5 = 8a3 a = 2 = −8a3 a = −2 −an bn a− b −an bn a+ b n +an bn a+ b n n n n (xyz)n (xyz)n zn c z r n r > 0 (xyz)n zr r = 0 (xyz)n zc c (xy4)n (xy4)n (xy4)n (xy9)n (xy9)n (xy9)n 12739 12739 739 7 12739 739 73 3 GMAT Club Math Book 14
GMAT Club Math Book 2024 v8_p15_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	15	15	fractions_decimals_percents	General rules: • and . • • • • • , when , then and when , then • When the GMAT provides the square root sign for an even root, such as or , then the only accepted answer is the positive root. That is, , NOT +5 or -5. In contrast, the equation has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT. • Odd roots will have the same sign as the base of the root. For example, and . • For GMAT it's good to memorize following values: PERCENTS A percentage is a way of expressing a number as a fraction of 100 (per cent meaning "per hundred"). It is often denoted using the percent sign, "%", or the abbreviation "pct". Since a percent is an amount per 100, percents can be represented as fractions with a denominator of 100. For example, 25% means 25 per 100, 25/100 and 350% means 350 per 100, 350/100. • A percent can be represented as a decimal. The following relationship characterizes how percents and decimals interact. Percent Form / 100 = Decimal Form For example: What is 2% represented as a decimal? Percent Form / 100 = Decimal Form: 2%/100=0.02 • Percent change General formula for percent increase or decrease, (percent change): Example: A company received $2 million in royalties on the first $10 million in sales and then $8 million in royalties on the next $100 million in sales. By what percent did the ratio of royalties to sales decrease from the first $10 million in sales to the next $100 million in sales? Solution: Percent decrease can be calculated by the formula above: =x√ y√ xy− −√ = x√ y√ x y − − √ ( =x√ )n xn− −√ =x 1 n x√n =x n m xn− −√m + ≠a√ b√ a+ b− − − −√ = \|x\|x2− −√ x ≤ 0 = − xx2− −√ x ≥ 0 = xx2− −√ x√ x√4 = 525− −√ = 25x2 = 5125−− −√3 = −4−64− − − −√3 ≈ 1.412√ ≈ 1.733√ ≈ 2.245√ ≈ 2.456√ ≈ 2.657√ ≈ 2.838√ ≈ 3.1610− −√ Percent = ∗ 100Change Original GMAT Club Math Book 15
GMAT Club Math Book 2024 v8_p16_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	16	16	word_problems	, so the royalties decreased by 60%. • Simple Interest Simple interest = principal * interest rate * time Example: If $15,000 is invested at 10% simple annual interest, how much interest is earned after 9 months? Solution: $15,0000.19/12 = $1125 • Compound Interest , where C = the number of times compounded annually. If C=1, meaning that interest is compounded once a year, then the formula will be: , where time is number of years. Example: If $20,000 is invested at 12% annual interest, compounded quarterly, what is the balance after 2 year? Solution: ORDER OF OPERATIONS - PEMDAS Perform the operations inside a Parenthesis first (absolute value signs also fall into this category), then Exponents, then Multiplication and Division, from left to right, then Addition and Subtraction, from left to right - PEMDAS. Special cases: • An exclamation mark indicates that one should compute the factorial of the term immediately to its left, before computing any of the lower-precedence operations, unless grouping symbols dictate otherwise. But means while ; a factorial in an exponent applies to the exponent, while a factorial not in the exponent applies to the entire power. • If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: and not
GMAT Club Math Book 2024 v8_p16_c2	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	16	16	word_problems	n the exponent applies to the entire power. • If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: and not Percent = ∗ 100 =Change Original = ∗ 100 = 60 −2 10 8 100 2 10 Balance(final) = principal∗ (1 + interest C )time∗C Balance(final) = principal∗ (1 + interest)time Balance = 20, 000 ∗ (1 + = 20, 000 ∗ (1.03 = $25, 335.40.12 4 )2∗4 )8 !32 ( )! = 9!32 =25! 2120 =amn a( )mn (am)n GMAT Club Math Book 16
GMAT Club Math Book 2024 v8_p17_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	17	17	number_theory	PRACTICE QUESTIONS Easy: 1. https://gmatclub.com/forum/if-n-is-a-pr ... 22490.html 2. https://gmatclub.com/forum/if-x-is-an-i ... 68233.html 3. https://gmatclub.com/forum/how-many-two ... 91802.html 4. https://gmatclub.com/forum/if-n-is-an-i ... 96084.html 5. https://gmatclub.com/forum/root-80-root-139868.html 6. https://gmatclub.com/forum/if-x-4-and-y ... 28707.html 7. https://gmatclub.com/forum/4th-root-of- ... 59975.html 8. https://gmatclub.com/forum/what-is-the- ... 03027.html 9. https://gmatclub.com/forum/if-n-is-the- ... 28102.html 10. https://gmatclub.com/forum/if-x-and-y-a ... 41958.html Medium: 1. https://gmatclub.com/forum/how-many-int ... 10744.html 2. https://gmatclub.com/forum/topic-143744.html 3. https://gmatclub.com/forum/if-the-sum-o ... 10512.html 4. https://gmatclub.com/forum/amy-s-grade- ... 98164.html 5. https://gmatclub.com/forum/if-3-4-2-3-5 ... l#p1526295 6. https://gmatclub.com/forum/what-is-63908.html 7. https://gmatclub.com/forum/the-sum-of-t ... 32773.html 8. https://gmatclub.com/forum/how-many-of- ... 00708.html 9. https://gmatclub.com/forum/which-of-the ... 25989.html 10. https://gmatclub.com/forum/what-is-the- ... 07096.html Hard: 1. https://gmatclub.com/forum/for-every-po ... 12521.html 2. https://gmatclub.com/forum/how-many-int ... 99697.html 3. https://gmatclub.com/forum/when-the-pos ... 72713.html 4. https://gmatclub.com/forum/if-x-a-and-b ... 01946.html 5. https://gmatclub.com/forum/the-function ... 03852.html 6. https://gmatclub.com/forum/for-prime-nu ... 98508.html 7. https://gmatclub.com/forum/how-many-pos ... 56052.html 8. https://gmatclub.com/forum/if-each-expr ... 87153.html 9. https://gmatclub.com/forum/what-is-the- ... 51159.html 10. https://gmatclub.com/forum/sqrt-7-sqrt- ... 68800.html
GMAT Club Math Book 2024 v8_p17_c2	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	17	17	general	forum/if-each-expr ... 87153.html 9. https://gmatclub.com/forum/what-is-the- ... 51159.html 10. https://gmatclub.com/forum/sqrt-7-sqrt- ... 68800.html GMAT Club Math Book 17
GMAT Club Math Book 2024 v8_p18_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	18	18	number_theory	GMAT Club Forum Remainders https://gmatclub.com/forum/remainders-144665.html REMAINDERS created by: Bunuel Definition If and are positive integers, there exist unique integers and , called the quotient and remainder, respectively, such that and . For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since . Notice that means that remainder is a non-negative integer and always less than divisor. This formula can also be written as . Properties When is divided by the remainder is 0 if is a multiple of . For example, 12 divided by 3 yields the remainder of 0 since 12 is a multiple of 3 and . When a smaller integer is divided by a larger integer, the quotient is 0 and the remainder is the smaller integer. For example, 7 divided by 11 has the quotient 0 and the remainder 7 since The possible remainders when positive integer is divided by positive integer can range from 0 to . For example, possible remainders when positive integer is divided by 5 can range from 0 (when y is a multiple of 5) to 4 (when y is one less than a multiple of 5). x y q r y = divisor∗ quotient+ remainder = xq+ r 0 ≤ r < x 15 = 6 ∗ 2 + 3 0 ≤ r < x = q+y x r x y x y x 12 = 3 ∗ 4 + 0 7 = 11 ∗ 0 + 7 y x x− 1 y GMAT Club Math Book 18
GMAT Club Math Book 2024 v8_p19_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	19	19	number_theory	If a number is divided by 10, its remainder is the last digit of that number. If it is divided by 100 then the remainder is the last two digits and so on. For example, 123 divided by 10 has the remainder 3 and 123 divided by 100 has the remainder of 23. Example #1 (easy) If the remainder is 7 when positive integer n is divided by 18, what is the remainder when n is divided by 6? A. 0 B. 1 C. 2 D. 3 E. 4 When positive integer n is dived by 18 the remainder is 7: . Now, since the first term (18q) is divisible by 6, then the remainder will only be from the second term, which is 7. 7 divided by 6 yields the remainder of 1. Answer: B. Discuss this question HERE. Example #2 (easy) If n is a prime number greater than 3, what is the remainder when n^2 is divided by 12 ? A. 0 B. 1 C. 2 D. 3 E. 5 There are several algebraic ways to solve this question, but the easiest way is as follows: since we cannot have two correct answers just pick a prime greater than 3, square it and see what would be the remainder upon division of it by 12. If , then . The remainder upon division 25 by 12 is 1. Answer: B. Discuss this question HERE. n = 18q + 7 n = 5 = 25n2 GMAT Club Math Book 19
GMAT Club Math Book 2024 v8_p20_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	20	20	statistics	Example #3 (easy) What is the tens digit of positive integer x ? (1) x divided by 100 has a remainder of 30. (2) x divided by 110 has a remainder of 30. (1) x divided by 100 has a remainder of 30. We have that : 30, 130, 230, ... as you can see every such number has 3 as the tens digit. Sufficient. (2) x divided by 110 has a remainder of 30. We have that : 30, 140, 250, 360, ... so, there are more than 1 value of the tens digit possible. Not sufficient. Answer: A. Discuss this question HERE. Example #4 (easy) What is the remainder when the positive integer n is divided by 6? (1) n is multiple of 5 (2) n is a multiple of 12 (1) n is multiple of 5. If n=5, then n yields the remainder of 5 when divided by 6 but if n=10, then n yields the remainder of 4 when divided by 6. We already have two different answers, which means that this statement is not sufficient. (2) n is a multiple of 12. Every multiple of 12 is also a multiple of 6, thus n divided by 6 yields the remainder of 0. Sufficient. Answer: B. Discuss this question HERE. Example #5 (medium) If s and t are positive integers such that s/t = 64.12, which of the following could be the remainder when s is divided by t ? A. 2 B. 4 C. 8 D. 20 E. 45 divided by yields the remainder of can always be expressed as: (which is the same as ), where is the quotient and is the remainder. Given that , so according to the above , which means that must be a multiple of 3. Only option E offers answer which is a multiple of 3 Answer. E. Discuss this question HERE. x = 100q + 30 x = 110p + 30 s t r = q+s t r t s = qt+ r q r = 64.12 = 64 = 64 = 64 +s t 12 100 3 25 3 25 =r t 3 25 r GMAT Club Math Book 20
GMAT Club Math Book 2024 v8_p21_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	21	21	number_theory	Example #6 (medium) Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30? A. 3 B. 12 C. 18 D. 22 E. 28 Positive integer n leaves a remainder of 4 after division by 6: . Thus n could be: 4, 10, 16, 22, 28, ... Positive integer n leaves a remainder of 3 after division by 5: . Thus n could be: 3, 8, 13, 18, 23, 28, ... There is a way to derive general formula for (of a type , where is a divisor and is a remainder) based on above two statements: Divisor would be the least common multiple of above two divisors 5 and 6, hence . Remainder would be the first common integer in above two patterns, hence . Therefore general formula based on both statements is . Hence the remainder when positive integer n is divided by 30 is 28. Answer. E. Discuss this question HERE. Example #7 (medium) If x^3 - x = n and x is a positive integer greater than 1, is n divisible by 8? (1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer. , notice that we have the product of three consecutive integers. Now, notice that if , then and are consecutive even integers, thus one of them will also be divisible by 4, which will make divisible by 24=8 (basically if then will be divisible by 83=24). (1) When 3x is divided by 2, there is a remainder. This implies that , which means that . Therefore is divisible by 8. Sufficient. (2) x = 4y + 1, where y is an integer. We have that , thus is divisible by 8. Sufficient. n = 6 p+ 4 n = 5 q+ 3 n n = mx+ r x r x x = 30 r r = 28 n = 30m + 28 − x = x( − 1) = ( x− 1)x(x+ 1)x3 x2 x = odd x− 1 x+ 1 (x− 1)(x+ 1) x = odd (x− 1)x(x+ 1) 3x = odd x = odd (x− 1)x(x+ 1) x = even+ odd = odd (x− 1)x(x+ 1) GMAT Club Math Book 21
GMAT Club Math Book 2024 v8_p22_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	22	22	algebra	Answer: D. Discuss this question HERE. Example #8 (hard) When 51^25 is divided by 13, the remainder obtained is: A. 12 B. 10 C. 2 D. 1 E. 0 , now if we expand this expression all terms but the last one will have in them, thus will leave no remainder upon division by 13, the last term will be . Thus the question becomes: what is the remainder upon division -1 by 13? The answer to this question is 12: . Answer: A. Discuss this question HERE. Example #9 (hard) When positive integer x is divided by 5, the remainder is 3; and when x is divided by 7, the remainder is 4. When positive integer y is divided by 5, the remainder is 3; and when y is divided by 7, the remainder is 4. If x > y, which of the following must be a factor of x - y? A. 12 B. 15 C. 20 D. 28 E. 35 When the positive integer x is divided by 5 and 7, the remainder is 3 and 4, respectively: (x could be 3, 8, 13, 18, 23, ...) and (x could be 4, 11, 18, 25, ...). We can derive general formula based on above two statements the same way as for the example above: Divisor will be the least common multiple of above two divisors 5 and 7, hence 35. Remainder will be the first common integer in above two patterns, hence 18. So, to satisfy both this conditions x must be of a type (18, 53, 88, ...); The same for y (as the same info is given about y): ; . Thus must be a multiple of 35. = (52 − 15125 )25 52 = 13 ∗ 4 (−1 = −1)25 −1 = 13 ∗ (−1) + 12 x = 5 q+ 3 x = 7 p+ 4 x = 35m + 18 y = 35n + 18 x− y = (35m + 18) − (35n + 18) = 35( m− n) x− y GMAT Club Math Book 22
GMAT Club Math Book 2024 v8_p23_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	23	23	number_theory	Answer: E. Discuss this question HERE. Example #10 (hard) If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4? (1) When p is divided by 8, the remainder is 5 (2) x – y = 3 (1) When p is divided by 8, the remainder is 5. This implies that . Since given that , then --> . So, . Now, if then and if then , so in any case --> --> in order to be multiple of 4 must be multiple of 16 but as we see it's not, so is not multiple of 4. Sufficient. (2) x – y = 3 --> --> but not sufficient to say whether it's multiple of 4. Answer: A. Discuss this question HERE. Example #11 (hard) and are positive integers. Is the remainder of bigger than the remainder of ? (1) . (2) The remainder of is 2 First of all any positive integer can yield only three remainders upon division by 3: 0, 1, or 2. Since, the sum of the digits of and is always 1 then the remainders of and are only dependent on the value of the number added to and . There are 3 cases: If the number added to them is: 0, 3, 6, 9, ... then the remainder will be 1 (as the sum of the digits of and will be 1 more than a multiple of 3); If the number added to them is: 1, 4, 7, 10, ... then the remainder will be 2 (as the sum of the digits of and will be 2 more than a multiple of 3); If the number added to them is: 2, 5, 8, 11, ... then the remainder will be 0 (as the sum of the digits of and will be a multiple of 3). (1) . Not sufficient. p = 8 q+ 5 = +x2 y2 y = odd = 2 k+ 1 8q+ 5 = + (2 k+ 1x2 )2 = 8 q+ 4 − 4 − 4 k = 4(2 q+ 1 − − k)x2 k2 k2 = 4(2 q+ 1 − − k)x2 k2 k = odd 2q+ 1 − − k = even+ odd− odd− odd = oddk2 k = even 2q+ 1 − − k = even+ odd− even− even = oddk2 2q+ 1 − − k = oddk2 = 4 ∗ oddx2 x x2 x x− odd = 3 x = even m n +n10m 3 +m10n 3 m > n n 3 10m 10n +n10m 3 +m10n 3 10m 10n 10m 10n 10m 10n 10m 10n m > n GMAT Club Math Book 23
GMAT Club Math Book 2024 v8_p24_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	24	24	number_theory	(2) The remainder of is --> is: 2, 5, 8, 11, ... so we have the third case. Which means that the remainder of is 0. Now, the question asks whether the remainder of , which is 0, greater than the reminder of , which is 0, 1, or 2. Obviously it cannot be greater, it can be less than or equal to. So, the answer to the question is NO. Sufficient. Answer: B. Discuss this question HERE. Check more PS questions on remainders here n 3 2 n +n10m 3 +n10m 3 +m10n 3 GMAT Club Math Book 24
GMAT Club Math Book 2024 v8_p25_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	25	25	algebra	GMAT Club Math Book Algebra https://gmatclub.com/forum/algebra-101576.html Algebra 101 Scope Manipulation of various algebraic expressions Equations in 1 & more variables Dealing with non-linear equations Algebraic identities Notation & Assumptions In this document, lower case roman alphabets will be used to denote variables such as a,b,c,x,y,z,w In general it is assumed that the GMAT will only deal with real numbers ( ) or subsets of such as Integers ( ), rational numbers ( ) etc Concept of variables A variable is a place holder, which can be used in mathematical expressions. They are most often used for two purposes : (a) In Algebraic Equations : To represent unknown quantities in known relationships. For eg : "Mary's age is 10 more than twice that of Jim's", we can represent the unknown "Mary's age" by x and "Jim's age" by y and then the known relationship is (b) In Algebraic Identities : These are generalized relationships such as , which says for any number, if you square it and take the root, you get the absolute value back. So the variable acts like a true placeholder, which may be replaced by any number. Basic rules of manipulation A. When switching terms from one side to the other in an algebraic expression + becomes - and vice versa. Eg. B. When switching terms from one side to the other in an algebraic expression * becomes / and vice versa. Eg. C. you can add/subtract/multiply/divide both sides by the same amount. Eg. D. you can take to the exponent or bring from the exponent as long as the base is the same. Eg 1. Eg 2. It is important to note that all the operations above are possible not just with constants but also with variables themselves. So you can "add x" or "multiply with y" on both sides while maintaining the expression. But what you need to be very careful about is when dividing both sides by a variable. When you divide both sides by a variable (or do operations like "canceling x on both sides") you implicitly assume that the variable cannot be equal to 0, as division by 0 is undefined. This is a concept shows up very often on GMAT R R Z Q x = 2 y+ 10 = \|x\|x2− −√ x+ y = 2 z ⇒ x = 2 z− y 4 ∗ x = ( y+ 1 ⇒ x =)2 (y+1)2 4 x+ y = 10z ⇒ =x+y 43 10z 43 + 2 = z ⇒ =x2 4 +2x2 4z = ⇒ = ⇒ 4 x = 3 y24x 8y 24x 23y GMAT Club Math Book 25
GMAT Club Math Book 2024 v8_p26_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	26	26	algebra	questions. Degree of an expression The degree of an algebraic expression is defined as the highest power of the variables present in the expression. Degree 1 : Linear Degree 2 : Quadratic Degree 3 : Cubic Degree 4 : Bi-quadratic Egs : the degree is 1 the degree is 3 the degree of x is 3, degree of z is 5, degree of the expression is 5 Solving equations of degree 1 : LINEAR Degree 1 equations or linear equations are equations in one or more variable such that degree of each variable is one. Let us consider some special cases of linear equations : One variable Such equations will always have a solution. General form is and solution is One equation in Two variables This is not enough to determine x and y uniquely. There can be infinitely many solutions. Two equations in Two variables If you have a linear equation in 2 variables, you need at least 2 equations to solve for both variables. The general form is : If then there are infinite solutions. Any point satisfying one equation will always satisfy the second If then there is no such x and y which will satisfy both equations. No solution In all other cases, solving the equations is straight forward, multiply eq (2) by a/d and subtract from (1). More than two equations in Two variables Pick any 2 equations and try to solve them : Case 1 : No solution --> Then there is no solution for bigger set Case 2 : Unique solution --> Substitute in other equations to see if the solution works for all others Case 3 : Infinite solutions --> Out of the 2 equations you picked, replace any one with an un-picked equation and repeat. More than 2 variables This is not a case that will be encountered often on the GMAT. But in general for n variables you will need at least n equations to get a unique solution. Sometimes you can assign unique values to a subset of variables using less than n equations using a small trick. For example consider the equations : In this case you can treat as a single variable to get : These can be solved to get x=0 and 2y+5z=20 There is a common misconception that you need n equations to solve n variables. This is not true. x+ y + x+ 2x3 +x3 z5 ax = b x = ( b/a) ax+ by = c dx+ ey = f (a/d) = ( b/e) = ( c/f) (a/d) = ( b/e) ≠ ( c/f) x+ 2y+ 5z = 20 x+ 4y+ 10z = 40 2y+ z x+ (2y+ 5z) = 20 x+ 2 ∗ (2 y+ 5z) = 40 GMAT Club Math Book 26
GMAT Club Math Book 2024 v8_p27_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	27	27	algebra	Solving equations of degree 2 : QUADRATIC The general form of a quadratic equation is The equation has no solution if The equation has exactly one solution if This equation has 2 solutions given by if The sum of roots is The product of roots is If the roots are and , the equation can be written as A quick way to solve a quadratic, without the above formula is to factorize it : Step 1> Divide throughout by coeff of x^2 to put it in the form Step 2> Sum of roots = -d and Product = e. Search for 2 numbers which satisfy this criteria, let them be f,g Step 3> The equation may be re-written as (x-f)(x-g)=0. And the solutions are f,g Eg. The sum is -11 and the product is 30. So numbers are -5,-6 Solvingequations with DEGREE>2 You will never be asked to solved higher degree equations, except in some cases where using simple tricks these equations can either be factorized or be reduced to a lower degree or both. What you need to note is that an equation of degree n has at most n unique solutions. Factorization This is the easiest approach to solving higher degree equations. Though there is no general rule to do this, generally a knowledge of algebraic identities helps. The basic idea is that if you can write an equation in the form : where each of A,B,C are algebraic expressions. Once this is done, the solution is obtained by equating each of A,B,C to 0 one by one. Eg. So the solution is x=0,-5,-6 Reducing to lower degree This is useful sometimes when it is easy to see that a simple variable substitution can reduce the degree. Eg. Here let a + bx+ c = 0x2 < 4 acb2 = 4 acb2 −b± −4acb2− − − − − −√ 2a > 4 acb2 −b a c a r1 r2 (x− )( x− ) = 0r1 r2 + dx+ e = 0x2 + 11x + 30 = 0x2 + 11x + 30 = + 5 x+ 6x+ 30 = x(x+ 5) + 6( x+ 5) = ( x+ 5)(x+ 6)x2 x2 A∗ B∗ C = 0 + 11 + 30x = 0x3 x2 x∗ ( + 11x + 30) = 0x2 x∗ (x+ 5) ∗ ( x+ 6) = 0 − 3 + 2 = 0x6 x3 y = x3 − 3y+ 2 = 0y2 (y− 2)(y− 1) = 0 GMAT Club Math Book 27
GMAT Club Math Book 2024 v8_p28_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	28	28	number_theory	So the solution is y=1 or 2 or x^3=1 or 2 or x=1 or Other tricks Sometimes we are given conditions such as the variables being integers which make the solutions much easier to find. When we know that the solutions are integral, often times solutions are easy to find using just brute force. Eg. and we know a,b are integers such that a<b We can solve this by testing values of a and checking if we can find b a=1 b=root(115) Not integer a=2 b=root(112) Not integer a=3 b=root(107) Not integer a=4 b=root(100)=10 a=5 b=root(91) Not integer a=6 b=root(80) Not integer a=7 b=root(67) Not integer a=8 b=root(52)<a So the answer is (4,10) Algebraic Identities These can be very useful in simplifying & solving a lot of questions : 2√3 + = 116a2 b2 (x+ y = + + 2 xy)2 x2 y2 (x− y = + − 2 xy)2 x2 y2 − = ( x+ y)(x− y)x2 y2 (x+ y − (x− y = 4 xy)2 )2 (x+ y+ z = + + + 2( xy+ yz+ zx))2 x2 y2 z2 + = ( x+ y)( + − xy)x3 y3 x2 y2 − = ( x− y)( + + xy)x3 y3 x2 y2 GMAT Club Math Book 28
GMAT Club Math Book 2024 v8_p29_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	29	29	algebra	ABSOLUTE VALUE (Modulus) Definition The absolute value (or modulus) of a real number x is x's numerical value without regard to its sign. For example, ; ; Graph: Important properties: How to approach equations with moduli It's not easy to manipulate with moduli in equations. There are two basic approaches that will help you out. Both of them are based on two ways of representing modulus as an algebraic expression. 1) . This approach might be helpful if an equation has × and /. 2) \|x\| equals x if x>=0 or -x if x<0. It looks a bit complicated but it's very powerful in dealing with moduli and the most popular approach too (see below). 3-steps approach: General approach to solving equalities and inequalities with absolute value: 1. Open modulus and set conditions. To solve/open a modulus, you need to consider 2 situations to find all roots: Positive (or rather non-negative) Negative \|x\| \|3\| = 3 \| − 12\| = 12 \| − 1.3\| = 1.3 \|x\| ≥ 0 \|0\| = 0 \| − x\| = \|x\| \|x\| + \|y\| ≥ \|x + y\| \|x\| = x2− −√ GMAT Club Math Book 29
GMAT Club Math Book 2024 v8_p30_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	30	30	algebra	For example, a) Positive: if , we can rewrite the equation as: b) Negative: if , we can rewrite the equation as: We can also think about conditions like graphics. is a key point in which the expression under modulus equals zero. All points right are the first condition and all points left are second condition . 2. Solve new equations: a) --> x=5 b) --> x=-3 3. Check conditions for each solution: a) has to satisfy initial condition . . It satisfies. Otherwise, we would have to reject x=5. b) has to satisfy initial condition . . It satisfies. Otherwise, we would have to reject x=-3. 3-steps approach for complex problems Let’s consider following examples, Example #1 Q.: . How many solutions does the equation have? Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions: a) . --> . We reject the solution because our condition is not satisfied (-1 is not less than -8) b) . --> . We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.) c) . --> . We reject the solution because our condition is not satisfied (9 is not within (-3,4) interval.) d) . --> . We reject the solution because our condition is not satisfied (-1 is not more than 4) (Optional) The following illustration may help you understand how to open modulus at different conditions. Answer: 0 Example #2 Q.: . What is x? Solution: There are 2 conditions: a) --> or . --> . x e { , } and both solutions satisfy the condition. \|x− 1\| = 4 (x− 1) ≥ 0 x− 1 = 4 (x− 1) < 0 −(x− 1) = 4 x = 1 (x > 1) (x < 1) x− 1 = 4 −x+ 1 = 4 x = 5 x− 1 >= 0 5 − 1 = 4 > 0 x = −3 x− 1 < 0 −3 − 1 = −4 < 0 \|x+ 3\| − \|4 − x\| = \|8 + x\| x < −8 −(x+ 3) − (4 − x) = −(8 + x) x = −1 −8 ≤ x < −3 −(x+ 3) − (4 − x) = (8 + x) x = −15 −3 ≤ x < 4 (x+ 3) − (4 − x) = (8 + x) x = 9 x ≥ 4 (x+ 3) + (4 − x) = (8 + x) x = −1 \| − 4\| = 1x2 ( − 4) ≥ 0x2 x ≤ −2 x ≥ 2 − 4 = 1x2 = 5x2 − 5√ 5√ GMAT Club Math Book 30
GMAT Club Math Book 2024 v8_p31_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	31	31	word_problems	b) --> . --> . x e { , } and both solutions satisfy the condition. (Optional) The following illustration may help you understand how to open modulus at different conditions. Answer: , , , Tip & Tricks The 3-steps method works in almost all cases. At the same time, often there are shortcuts and tricks that allow you to solve absolute value problems in 10-20 sec. I. Thinking of inequality with modulus as a segment at the number line. For example, Problem: 1<x<9. What inequality represents this condition? A. \|x\|<3 B. \|x+5\|<4 C. \|x-1\|<9 D. \|-5+x\|<4 E. \|3+x\|<5 Solution: 10sec. Traditional 3-steps method is too time-consume technique. First of all we find length (9-1)=8 and center (1+8/2=5) of the segment represented by 1<x<9. Now, let’s look at our options. Only B and D has 8/2=4 on the right side and D had left site 0 at x=5. Therefore, answer is D. II. Converting inequalities with modulus into a range expression. In many cases, especially in DS problems, it helps avoid silly mistakes. For example, \|x\|<5 is equal to x e (-5,5). \|x+3\|>3 is equal to x e (-inf,-6)&(0,+inf) III. Thinking about absolute values as the distance between points at the number line. For example, Problem: A<X<Y<B. Is \|A-X\| <\|X-B\|? 1) \|Y-A\|<\|B-Y\| Solution: We can think of absolute values here as the distance between points. Statement 1 means than the distance between Y and A is less than that between Y and B. Because X is between A and Y, \|X-A\| < \|Y-A\| and at the same time the distance between X and B will be larger than that between Y and B (\|B-Y\|<\|B-X\|). Therefore, statement 1 is sufficient. Pitfalls The most typical pitfall is ignoring the third step in opening modulus - always check whether your solution satisfies conditions. ( − 4) < 0x2 −2 < x < 2 −( − 4) = 1x2 = 3x2 − 3√ 3√ − 5√ − 3√ 3√ 5√ GMAT Club Math Book 31
GMAT Club Math Book 2024 v8_p32_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	32	32	coordinate_geometry	Practice Questions Easy: 1. https://gmatclub.com/forum/which-of-the ... 44267.html 2. https://gmatclub.com/forum/if-y-is-an-i ... 39867.html 3. https://gmatclub.com/forum/on-the-numbe ... 16027.html 4. https://gmatclub.com/forum/if-k-2-m-2-w ... 07128.html 5. https://gmatclub.com/forum/if-y-0-which ... 09621.html 6. https://gmatclub.com/forum/which-of-the ... 08204.html 7. https://gmatclub.com/forum/the-function ... 96171.html 8. https://gmatclub.com/forum/4-272003.html 9. https://gmatclub.com/forum/how-many-int ... 88429.html 10. https://gmatclub.com/forum/if-x-is-a-ne ... 64088.html Medium: 1. https://gmatclub.com/forum/what-is-the- ... 79892.html 2. https://gmatclub.com/forum/if-x-y-1-whi ... 22118.html 3. https://gmatclub.com/forum/if-x-y-0-whi ... 86959.html 4. https://gmatclub.com/forum/for-how-many ... 68675.html 5. https://gmatclub.com/forum/given-the-in ... 18973.html 6. https://gmatclub.com/forum/hot-competit ... 33254.html 7. https://gmatclub.com/forum/around-the-w ... 15784.html 8. https://gmatclub.com/forum/around-the-w ... 15984.html 9. https://gmatclub.com/forum/if-q-is-the- ... 31226.html 10. https://gmatclub.com/forum/what-is-the- ... 96176.html Hard: 1. https://gmatclub.com/forum/if-y-x-5-x-5 ... 73626.html 2. https://gmatclub.com/forum/if-x-0-and-x ... 43572.html 3. https://gmatclub.com/forum/if-x-3-4-and ... 10071.html 4. https://gmatclub.com/forum/what-is-the- ... 98596.html 5. https://gmatclub.com/forum/how-many-roo ... 79379.html 6. https://gmatclub.com/forum/which-of-the ... 54341.html 7. https://gmatclub.com/forum/if-x-2-x-5-0 ... 24552.html 8. https://gmatclub.com/forum/what-is-the- ... 48153.html 9. https://gmatclub.com/forum/what-is-the- ... 99579.html 10. https://gmatclub.com/forum/what-is-the- ... 20033.html GMAT Club Math Book 32
GMAT Club Math Book 2024 v8_p33_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	33	33	sequences_patterns	GMAT Club Math Book Algebra: Sequences and Progressions https://gmatclub.com/forum/algebra-sequences-and-progressions-101891.html Sequences & Progressions created by: shrouded1 Definition Sequence : It is an ordered list of objects. It can be finite or infinite. The elements may repeat themselves more than once in the sequence, and their ordering is important unlike a set Arithmetic Progressions Definition It is a special type of sequence in which the difference between successive terms is constant. General Term is the ith term is the common difference is the first term Defining Properties Each of the following is necessary & sufficient for a sequence to be an AP : Constant If you pick any 3 consecutive terms, the middle one is the mean of the other two For all i,j > k >= 1 : Summation The sum of an infinite AP can never be finite except if & The general sum of a n term AP with common difference d is given by The sum formula may be re-written as = + d = + ( n− 1)dan an−1 a1 ai d a1 − =ai ai−1 = −ai ak i−k −aj ak j−k = 0a1 d = 0 (2a+ (n− 1)d)n 2 n∗ Avg( , ) = ∗ ( FirstTerm + LastTerm)a1 an n 2 GMAT Club Math Book 33
GMAT Club Math Book 2024 v8_p34_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	34	34	sequences_patterns	Examples 1. All odd positive integers : {1,3,5,7,...} 2. All positive multiples of 23 : {23,46,69,92,...} 3. All negative reals with decimal part 0.1 : {-0.1,-1.1,-2.1,-3.1,...} Geometric Progressions Definition It is a special type of sequence in which the ratio of consequetive terms is constant General Term is the ith term is the common ratio is the first term Defining Properties Each of the following is necessary & sufficient for a sequence to be an GP : Constant If you pick any 3 consecutive terms, the middle one is the geometric mean of the other two For all i,j > k >= 1 : Summation The sum of an infinite GP will be finite if absolute value of r < 1 The general sum of a n term GP with common ratio r is given by If an infinite GP is summable (\|r\|<1) then the sum is Examples 1. All positive powers of 2 : {1,2,4,8,...} 2. All negative powers of 4 : {1/4,1/16,1/64,1/256,...} Harmonic Progressions Definition It is a special type of sequence in which if you take the inverse of every term, this new sequence forms an HP = 1, d = 2a1 = 23, d = 23a1 = −0.1, d = −1a1 = ∗ r = ∗bn bn−1 a1 rn−1 bi r b1 = bi bi−1 ( = ( bi bk )j−k bj bk )i−k ∗b1 −1rn r−1 b1 1−r = 1, r = 2b1 = 1/4, r = 1/4, sum = = (1/3)b1 1/4 (1−1/4) GMAT Club Math Book 34
GMAT Club Math Book 2024 v8_p35_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	35	35	sequences_patterns	Important Properties Of any three consecutive terms of a HP, the middle one is always the harmonic mean of the other two, where the harmonic mean (HM) is defined as : Or in other words : APs, GPs, HPs : Linkage Each progression provides us a definition of "mean" : Arithmetic Mean : OR Geometric Mean : OR Harmonic Mean : OR For all non-negative real numbers : AM >= GM >= HM In particular for 2 numbers : AM * HM = GM * GM Example : Let a=50 and b=2, then the AM = (50+2)0.5 = 26 ; the GM = sqrt(502) = 10 ; the HM = (2502)/(52) = 3.85 AM > GM > HM AMHM = 100 = GM^2 Misc Notes A subsequence (any set of consequutive terms) of an AP is an AP A subsequence (any set of consequutive terms) of a GP is a GP A subsequence (any set of consequutive terms) of a HP is a HP If given an AP, and I pick out a subsequence from that AP, consisting of the terms such that are in AP then the new subsequence will also be an AP For Example : Consider the AP with {1,3,5,7,9,11,...}, so a_n=1+2(n- 1)=2n-1 Pick out the subsequence of terms New sequence is {9,19,29,...} which is an AP with and If given a GP, and I pick out a subsequence from that GP, consisting of the terms such that are in AP then the new subsequence will also be a GP ∗ ( + ) =1 2 1 a 1 b 1 HM(a,b) HM(a, b) = 2ab a+b a+b 2 a1+..+an n ab− −√ (a1∗. . ∗an) 1n 2ab a+b n +..+1 a1 1 an , , , . . .ai1 ai2 ai3 i1, i2, i3 = 1, d = 2a1 , , , . . .a5 a10 a15 = 9a1 d = 10 , , , . . .bi1 bi2 bi3 i1, i2, i3 GMAT Club Math Book 35
GMAT Club Math Book 2024 v8_p36_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	36	36	sequences_patterns	For Example : Consider the GP with {1,2,4,8,16,32,...}, so b_n=2^(n-1) Pick out the subsequence of terms New sequence is {4,16,64,...} which is a GP with and The special sequence in which each term is the sum of previous two terms is known as the fibonacci sequence. It is neither an AP nor a GP. The first two terms are 1. {1,1,2,3,5,8,13,...} In a finite AP, the mean of all the terms is equal to the mean of the middle two terms if n is even and the middle term if n is odd. In either case this is also equal to the mean of the first and last terms Some examples Example 1 A coin is tossed repeatedly till the result is a tails, what is the probability that the total number of tosses is less than or equal to 5 ? Solution P(<=5 tosses) = P(1 toss)+...+P(5 tosses) = P(T)+P(HT)+P(HHT)+P(HHHT)+P(HHHHT) We know that P(H)=P(T)=0.5 So Probability = 0.5 + 0.5^2 + ... + 0.5^5 This is just a finite GP, with first term = 0.5, n=5 and ratio = 0.5. Hence : Probability = Example 2 In an arithmetic progression a1,a2,...,a22,a23, the common difference is non-zero, how many terms are greater than 24 ? (1) a1 = 8 (2) a12 = 24 Solution (1) a1=8, does not tell us anything about the common difference, so impossible to say how many terms are greater than 24 (2) a12=24, and we know common difference is non-zero. So either all the terms below a12 are greater than 24 and the terms above it less than 24 or the other way around. In either case, there are exactly 11 terms either side of a12. Sufficient Answer is B Example 3 For positive integers a,b (a<b) arrange in ascending order the quantities a, b, sqrt(ab), avg(a,b), 2ab/(a+b) Solution Using the inequality AM>=GM>=HM, the solution is : a <= 2ab/(a+b) <= Sqrt(ab) <= Avg(a,b) <= b = 1, r = 2b1 , , , . . .b2 b4 b6 = 4b1 r = 4 0.5 ∗ = ∗ =1−0.55 1−0.5 1 2 31 32 1 2 31 32 GMAT Club Math Book 36
GMAT Club Math Book 2024 v8_p37_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	37	37	sequences_patterns	Example 4 For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) *(1/2^k). If T is the sum of the first 10 terms in the sequence then T is a)greater than 2 b)between 1 and 2 c)between 1/2 and 1 d)between 1/4 and 1/2 e)less than 1/4. Solution The sequence given has first term 1/2 and each subsequent term can be obtained by multiplying with -1/2. So it is a GP. We can use the GP summation formula 1023/1024 is very close to 1, so this sum is very close to 1/3 Answer is d Example 5 The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression? A. 300 B. 120 C. 150 D. 170 E. 270 Solution Now we need the sum of first 15 terms, which is given by : Answer is (c) S = b = ∗ = ∗1−rn 1−r 1 2 1−(−1/2)10 1−(−1/2) 1 3 1023 1024 + 2 = 20a4 a1 = + 3 d, 2 = + 11da4 a1 a1 a1 2 + 14d = 20a1 (2 + (15 − 1) d) = ∗ (2 + 14d ) = 15015 2 a1 15 2 a1 GMAT Club Math Book 37
GMAT Club Math Book 2024 v8_p38_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	38	38	sequences_patterns	Practice Questions Easy: 1. https://gmatclub.com/forum/the-infinite ... 00090.html 2. https://gmatclub.com/forum/in-an-increa ... 43749.html 3. https://gmatclub.com/forum/the-sequence ... 46630.html 4. https://gmatclub.com/forum/the-sequence ... 34498.html 5. https://gmatclub.com/forum/in-a-certain ... 44376.html 6. https://gmatclub.com/forum/what-is-the- ... 44459.html 7. https://gmatclub.com/forum/in-the-infin ... 93065.html 8. https://gmatclub.com/forum/the-sequence ... 66830.html 9. https://gmatclub.com/forum/in-the-seque ... 98247.html 10. https://gmatclub.com/forum/for-any-sequ ... 28100.html Medium: 1. https://gmatclub.com/forum/in-a-certain ... 19235.html 2. https://gmatclub.com/forum/in-the-seque ... 68359.html 3. https://gmatclub.com/forum/what-is-the- ... 30041.html 4. https://gmatclub.com/forum/in-the-seque ... 98182.html 5. https://gmatclub.com/forum/a-sequence-i ... 71548.html 6. https://gmatclub.com/forum/the-sequence ... 60525.html 7. https://gmatclub.com/forum/if-the-sum-o ... 98380.html 8. https://gmatclub.com/forum/if-s1-s2-s3- ... 02927.html 9. https://gmatclub.com/forum/around-the-w ... 16033.html 10. https://gmatclub.com/forum/gmat-club-wo ... 95168.html Hard: 1. https://gmatclub.com/forum/a-sequence-o ... 20319.html 2. https://gmatclub.com/forum/a-certain-mu ... -3940.html 3. https://gmatclub.com/forum/the-sequence ... 27175.html 4. https://gmatclub.com/forum/if-the-sum-o ... 98379.html 5. https://gmatclub.com/forum/the-sum-of-n ... 00640.html 6. https://gmatclub.com/forum/x-y-z-is-an- ... 98430.html 7. https://gmatclub.com/forum/gmat-club-ol ... 67365.html 8. https://gmatclub.com/forum/if-the-sum-o ... 98381.html 9. https://gmatclub.com/forum/around-the-w ... 16261.html 10. https://gmatclub.com/forum/around-the-w ... 16134.html GMAT Club Math Book 38
GMAT Club Math Book 2024 v8_p39_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	39	39	coordinate_geometry	GMAT Club Math Book Algebra Functions & Coordinate Geometry https://gmatclub.com/forum/algebra-functions-87652.html FUNCTIONS & COORDINATE GEOMETRY Definition Coordinate geometry, or Cartesian geometry, is the study of geometry using a coordinate system and the principles of algebra and analysis. The Coordinate Plane In coordinate geometry, points are placed on the "coordinate plane" as shown below. The coordinate plane is a two-dimensional surface on which we can plot points, lines and curves. It has two scales, called the x-axis and y-axis, at right angles to each other. The plural of axis is 'axes' (pronounced "AXE-ease"). A point's location on the plane is given by two numbers, one that tells where it is on the x-axis and another which tells where it is on the y-axis. Together, they define a single, unique position on the plane. So in the diagram above, the point A has an x value of 20 and a y value of 15. These are the coordinates of the point A, sometimes referred to as its GMAT Club Math Book 39
GMAT Club Math Book 2024 v8_p40_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	40	40	coordinate_geometry	"rectangular coordinates". X axis The horizontal scale is called the x-axis and is usually drawn with the zero point in the middle. Values to the right are positive and those to the left are negative. Y axis The vertical scale is called the y-axis and is also usually drawn with the zero point in the middle. Values above the origin are positive and those below are negative. Origin The point where the two axes cross (at zero on both scales) is called the origin. Quadrants When the origin is in the center of the plane, they divide it into four areas called quadrants. The first quadrant, by convention, is the top right, and then they go around counter- clockwise. In the diagram above they are labeled Quadrant 1, 2 etc. It is conventional to label them with numerals but we talk about them as "first, second, third, and fourth quadrant". Point (x,y) The coordinates are written as an "ordered pair". The letter P is simply the name of the point and is used to distinguish it from others. The two numbers in parentheses are the x and y coordinate of the point. The first number (x) specifies how far along the x (horizontal) axis the point is. The second is the y coordinate and specifies how far up or down the y axis to go. It is called an ordered pair because the order of the two numbers matters - the first is always the x (horizontal) coordinate. The sign of the coordinate is important. A positive number means to go to the right (x) or GMAT Club Math Book 40
GMAT Club Math Book 2024 v8_p41_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	41	41	coordinate_geometry	up (y). Negative numbers mean to go left (x) or down (y). Distance between two points Given coordinates of two points, distance D between two points is given by: (where is the difference between the x-coordinates and is the difference between the y-coordinates of the points) As you can see, the distance formula on the plane is derived from the Pythagorean theorem. Above formula can be written in the following way for given two points and : Vertical and horizontal lines If the line segment is exactly vertical or horizontal, the formula above will still work fine, but there is an easier way. For a horizontal line, its length is the difference between the x- coordinates. For a vertical line its length is the difference between the y-coordinates. Distance between the point A (x,y) and the origin As the one point is origin with coordinate O (0,0) the formula can be simplified to: Example #1 Q: Find the distance between the point A (3,-1) and B (-1,2) Solution: Substituting values in the equation we'll get D = d + dx2 y2− − − − − − − −√ dx dy ( , )x1 y1 ( , )x2 y2 D = ( − + ( −x2 x1)2 y2 y1)2− −− − − − − − − − − − − − − − − − − √ D = +x2 y2− − − − − −√ D = ( − + ( −x2 x1)2 y2 y1)2− −− − − − − − − − − − − − − − − − −√ GMAT Club Math Book 41
GMAT Club Math Book 2024 v8_p42_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	42	42	coordinate_geometry	Midpoint of a Line Segment A line segment on the coordinate plane is defined by two endpoints whose coordinates are known. The midpoint of this line is exactly halfway between these endpoints and it's location can be found using the Midpoint Theorem, which states: • The x-coordinate of the midpoint is the average of the x-coordinates of the two endpoints. • Likewise, the y-coordinate is the average of the y-coordinates of the endpoints. Coordinates of the midpoint of the line segment AB, ( and ) are and Lines in Coordinate Geometry In Euclidean geometry, a line is a straight curve. In coordinate geometry, lines in a Cartesian plane can be described algebraically by linear equations and linear functions. Every straight line in the plane can represented by a first degree equation with two variables. D = = = 5(−1 − 3 + (2 − (−1))2 )2− −− − − − − − − − − − − − − − − − − √ 16 + 9− − − − −√ M( , )xm ym A( , )x1 y1 B( , )x2 y2 =xm +x1 x2 2 =ym +y1 y2 2 GMAT Club Math Book 42
GMAT Club Math Book 2024 v8_p43_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	43	43	coordinate_geometry	There are several approaches commonly used in coordinate geometry. It does not matter whether we are talking about a line, ray or line segment. In all cases any of the below methods will provide enough information to define the line exactly. 1. General form. The general form of the equation of a straight line is Where , and are arbitrary constants. This form includes all other forms as special cases. For an equation in this form the slope is and the y intercept is . 2. Point-intercept form. Where: is the slope of the line; is the y-intercept of the line; is the independent variable of the function . 3. Using two points In figure below, a line is defined by the two points A and B. By providing the coordinates of the two points, we can draw a line. No other line could pass through both these points and so the line they define is unique. ax+ by+ c = 0 a b c − a b − c b y = mx+ b m b x y GMAT Club Math Book 43
GMAT Club Math Book 2024 v8_p44_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	44	44	algebra	The equation of a straight line passing through points and is: Example #1 Q: Find the equation of a line passing through the points A (17,4) and B (9,9). Solution: Substituting the values in equation we'll get: --> --> OR if we want to write the equation in the slope-intercept form: 4. Using one point and the slope Sometimes on the GMAT you will be given a point on the line and its slope and from this information you will need to find the equation or check if this line goes through another point. You can think of the slope as the direction of the line. So once you know that a line goes through a certain point, and which direction it is pointing, you have defined one unique line. In figure below, we see a line passing through the point A at (14,23). We also see that it's slope is +2 (which means it goes 2 up for every one across). With these two facts we can establish a unique line. ( , )P1 x1 y1 ( , )P2 x2 y2 = y−y1 x−x1 −y1 y2 −x1 x2 = y−y1 x−x1 −y1 y2 −x1 x2 =y−4 x−17 4−9 17−9 =y−4 x−17 −5 8 8y− 32 = −5 x+ 85 8y+ 5x− 117 = 0 y = − x+5 8 117 8 GMAT Club Math Book 44
GMAT Club Math Book 2024 v8_p45_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	45	45	algebra	The equation of a straight line that passes through a point with a slope m is: Example #2 Q: Find the equation of a line passing through the point A (14,23) and the slope 2. Solution: Substituting the values in equation we'll get --> 4. Intercept form. The equation of a straight line whose x and y intercepts are a and b, respectively, is: Example #3 Q: Find the equation of a line whose x intercept is 5 and y intercept is 2. Solution: Substituting the values in equation we'll get --> OR if we want to write the equation in the slope-intercept form: Slope of a Line The slope or gradient of a line describes its steepness, incline, or grade. A higher slope value indicates a steeper incline. The slope is defined as the ratio of the "rise" divided by the "run" between two points on a line, or in other words, the ratio of the altitude change to the horizontal distance between any two points on the line. ( , )P1 x1 y1 y− = m(x− )y1 x1 y− = m(x− )y1 x1 y− 23 = 2( x− 14) y = 2 x− 5 + = 1x a y b + = 1x a y b + = 1x 5 y 2 5y+ 2x− 10 = 0 y = − x+ 22 5 GMAT Club Math Book 45
GMAT Club Math Book 2024 v8_p46_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	46	46	algebra	Given two points and on a line, the slope of the line is: If the equation of the line is given in the Point-intercept form: , then is the slope. This form of a line's equation is called the slope-intercept form, because can be interpreted as the y-intercept of the line, the y-coordinate where the line intersects the y-axis. If the equation of the line is given in the General form: , then the slope is and the y intercept is . SLOPE DIRECTION The slope of a line can be positive, negative, zero or undefined. ( , )x1 y1 ( , )x2 y2 m m = −y2 y1 −x2 x1 y = mx+ b m b ax+ by+ c = 0 − a b − c b GMAT Club Math Book 46
GMAT Club Math Book 2024 v8_p47_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	47	47	algebra	Positive slope Here, y increases as x increases, so the line slopes upwards to the right. The slope will be a positive number. The line below has a slope of about +0.3, it goes up about 0.3 for every step of 1 along the x-axis. Negative slope Here, y decreases as x increases, so the line slopes downwards to the right. The slope will be a negative number. The line below has a slope of about -0.3, it goes down about 0.3 for every step of 1 along the x-axis. Zero slope Here, y does not change as x increases, so the line in exactly horizontal. The slope of any horizontal line is always zero. The line below goes neither up nor down as x increases, so its slope is zero. Undefined slope When the line is exactly vertical, it does not have a defined slope. The two x coordinates are the same, so the difference is zero. The slope calculation is then something like When you divide anything by zero the result has no meaning. The line above is exactly vertical, so it has no defined slope. SLOPE AND QUADRANTS: 1. If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and Y intercepts of the line with negative slope have the same sign. Therefore if X and Y intersects are positive, the line intersects quadrant I; if negative, quadrant III. 2. If the slope of line is positive, line WILL intersect quadrants I and III. Y and X intercepts of the line with positive slope have opposite signs. Therefore if X intersect is negative, line intersects the quadrant II too, if positive quadrant IV. 3. Every line (but the one crosses origin OR parallel to X or Y axis OR X and Y axis themselves) crosses three quadrants. Only the line which crosses origin OR is parallel to either of axis crosses only two quadrants. 4. If a line is horizontal it has a slope of , is parallel to X-axis and crosses quadrant I and II if the Y intersect is positive OR quadrants III and IV, if the Y intersect is negative. Equation of such line is y=b, where b is y intersect. 5. If a line is vertical, the slope is not defined, line is parallel to Y-axis and crosses quadrant I and IV, if the X intersect is positive and quadrant II and III, if the X intersect is negative. Equation of such line is , where a is x-intercept. 6. For a line that crosses two points and , slope slope = 15 0 (0, 0) 0 x = a ( , )x1 y1 ( , )x2 y2 m = −y2 y1 −x2 x1 GMAT Club Math Book 47
GMAT Club Math Book 2024 v8_p48_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	48	48	algebra	7. If the slope is 1 the angle formed by the line is degrees. 8. Given a point and slope, equation of a line can be found. The equation of a straight line that passes through a point with a slope is: Vertical and horizontal lines A vertical line is parallel to the y-axis of the coordinate plane. All points on the line will have the same x-coordinate. A vertical line has no slope. Or put another way, for a vertical line the slope is undefined. The equation of a vertical line is: Where: x is the coordinate of any point on the line; a is where the line crosses the x-axis (x intercept). Notice that the equation is independent of y. Any point on the vertical line satisfies the equation. A horizontal line is parallel to the x-axis of the coordinate plane. All points on the line will have the same y-coordinate. 45 ( , )x1 y1 m y− = m(x− )y1 x1 x = a GMAT Club Math Book 48
GMAT Club Math Book 2024 v8_p49_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	49	49	algebra	A horizontal line has a slope of zero. The equation of a horizontal line is: Where: y is the coordinate of any point on the line; b is where the line crosses the y-axis (y intercept). Notice that the equation is independent of x. Any point on the horizontal line satisfies the equation. Parallel lines Parallel lines have the same slope. The slope can be found using any method that is convenient to you: y = b GMAT Club Math Book 49
GMAT Club Math Book 2024 v8_p50_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	50	50	algebra	From two given points on the line. From the equation of the line in slope-intercept form From the equation of the line in point-slope form The equation of a line through the point and parallel to line is: Distance between two parallel lines and can be found by the formula: Example #1 Q:There are two lines. One line is defined by two points at (5,5) and (25,15). The other is defined by an equation in slope-intercept form form y = 0.52x - 2.5. Are two lines parallel? Solution: For the top line, the slope is found using the coordinates of the two points that define the line. For the lower line, the slope is taken directly from the formula. Recall that the slope intercept formula is y = mx + b, where m is the slope. So looking at the formula we see that the slope is 0.52. So, the top one has a slope of 0.5, the lower slope is 0.52, which are not equal. Therefore, the lines are not parallel. Example #2 Q: Define a line through a point C parallel to a line passes through the points A and B. ( , )P1 x1 y1 ax+ by+ c = 0 a(x− ) + b(y− ) = 0x1 y1 y = mx+ b y = mx+ c D = \|b−c\| +1m2− − − − − √ Slope = = 0.515−5 25−5 GMAT Club Math Book 50
GMAT Club Math Book 2024 v8_p51_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	51	51	algebra	Solution: We first find the slope of the line AB using the same method as in the example above. For the line to be parallel to AB it will have the same slope, and will pass through a given point, C(12,10). We therefore have enough information to define the line by it's equation in point-slope form form: --> Perpendicular lines For one line to be perpendicular to another, the relationship between their slopes has to be negative reciprocal . In other words, the two lines are perpendicular if and only if the product of their slopes is . SlopeAB = = −0.5220−7 5−30 y = −0.52( x− 12) + 10 y = −0.52x + 16.24 − 1 m −1 GMAT Club Math Book 51
GMAT Club Math Book 2024 v8_p52_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	52	52	algebra	The two lines and are perpendicular if . The equation of a line passing through the point ) and perpendicular to line is: Example #1: Q: Are the two lines below perpendicular? Solution: To answer, we must find the slope of each line and then check to see if one slope is the negative reciprocal of the other or if their product equals to -1. x+ y+ = 0a1 b1 c1 x+ y+ = 0a2 b2 c2 + = 0a1a2 b1b2 ( ,P1 x1 y1 ax+ by+ c = 0 b(x− ) − a(y− ) = 0x1 y1 SlopeAB = = = 0.3585−19 9−48 −14 −39 GMAT Club Math Book 52
GMAT Club Math Book 2024 v8_p53_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	53	53	algebra	If the lines are perpendicular, each will be the negative reciprocal of the other. It doesn't matter which line we start with, so we will pick AB: Negative reciprocal of 0.358 is So, the slope of CD is -2.22, and the negative reciprocal of the slope of AB is -2.79. These are not the same, so the lines are not perpendicular, even though they may look as though they are. However, if you looked carefully at the diagram, you might have noticed that point C is a little too far to the left for the lines to be perpendicular. Example # 2. Q: Define a line passing through the point E and perpendicular to a line passing through the points C and D on the graph above. Solution: The point E is on the y-axis and so is the y-intercept of the desired line. Once we know the slope of the line, we can express it using its equation in slope-intercept form y=mx+b, where m is the slope and b is the y-intercept. First find the slope of line CD: The line we seek will have a slope which is the negative reciprocal of: Since E is on the Y-axis, we know that the intercept is 10. Plugging these values into the line equation, the line we need is described by the equation This is one of the ways a line can be defined and so we have solved the problem. If we wanted to plot the line, we would find another point on the line using the equation and then draw the line through that point and the intercept. Intersection of two straight lines The point of intersection of two non-parallel lines can be found from the equations of the two lines. SlopeCD = = = −2.2224−4 22−31 20 −9 − = −2.791 0.358 SlopeCD = = = −2.2224−4 22−31 20 −9 − = 0.451 −2.22 y = 0.45x + 10 GMAT Club Math Book 53
GMAT Club Math Book 2024 v8_p54_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	54	54	algebra	To find the intersection of two straight lines: 1. First we need their equations 2. Then, since at the point of intersection, the two equations will share a point and thus have the same values of x and y, we set the two equations equal to each other. This gives an equation that we can solve for x 3. We substitute the x value in one of the line equations (it doesn't matter which) and solve it for y. This gives us the x and y coordinates of the intersection. Example #1 Q: Find the point of intersection of two lines that have the following equations (in slope- intercept form): Solution: At the point of intersection they will both have the same y-coordinate value, so we set the equations equal to each other: This gives us an equation in one unknown (x) which we can solve: To find y, simply set x equal to 10 in the equation of either line and solve for y: Equation for a line (Either line will do) Set x equal to 10: We now have both x and y, so the intersection point is (10, 27) Example #2 Q: Find the point of intersection of two lines that have the following equations (in slope- y = 3 x− 3 y = 2.3 x+ 4 3x− 3 = 2.3 x+ 4 x = 10 y = 3 x− 3 y = 30 − 3 y = 27 GMAT Club Math Book 54
GMAT Club Math Book 2024 v8_p55_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	55	55	coordinate_geometry	intercept form): and (A vertical line) Solution: When one of the lines is vertical, it has no defined slope. We find the intersection slightly differently. On the vertical line, all points on it have an x-coordinate of 12 (the definition of a vertical line), so we simply set x equal to 12 in the first equation and solve it for y. Equation for a line Set x equal to 12 So the intersection point is at (12,33). Note: If both lines are vertical or horizontal, they are parallel and have no intersection Distance from a point to a line The distance from a point to a line is the shortest distance between them - the length of a perpendicular line segment from the line to the point. The distance from a point to a line is given by the formula: When the line is horizontal the formula transforms to: Where: is the y-coordinate of the given point P; is the y-coordinate of any point on the given vertical line L. \| \| the vertical bars mean "absolute value" - make it positive even if it calculates to a negative. When the line is vertical the formula transforms to: Where: is the x-coordinate of the given point P; is the x-coordinate of any point on the given vertical line L. \| \| the vertical bars mean "absolute value" - make it positive even if it calculates to a negative. When the given point is origin, then the distance between origin and line ax+by+c=0 is given by the formula: Circle on a plane In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that: y = 3 x− 3 x = 12 y = 3 x− 3 y = 36 − 3 y = 33 ( , )x0 y0 ax+ by+ c = 0 D = \|a +b +c\|x0 y0 +a2 b2− − − − − −√ D = \| − \|Py Ly Py Ly D = \| − \|Px Lx Px Lx D = \|c\| +a2 b2− − − − − − √ (x− a + (y− b =)2 )2 r2 GMAT Club Math Book 55
GMAT Club Math Book 2024 v8_p56_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	56	56	geometry	This equation of the circle follows from the Pythagorean theorem applied to any point on the circle: as shown in the diagram above, the radius is the hypotenuse of a right-angled triangle whose other sides are of length x-a and y-b. If the circle is centered at the origin (0, 0), then the equation simplifies to: Number line A number line is a picture of a straight line on which every point corresponds to a real number and every real number to a point. On the GMAT we can often see such statement: is halfway between and on the number line. Remember this statement can ALWAYS be expressed as: . Also on the GMAT we can often see another statement: The distance between and on the number line is the same as the distance between and . Remember this statement can ALWAYS be expressed as: . Parabola A parabola is the graph associated with a quadratic function, i.e. a function of the form + =x2 y2 r2 k m n = km+n 2 p m p n \|p− m\| = \|p − n\| GMAT Club Math Book 56
GMAT Club Math Book 2024 v8_p57_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	57	57	coordinate_geometry	. The general or standard form of a quadratic function is , or in function form, , where is the independent variable, is the dependent variable, and , , and are constants. The larger the absolute value of , the steeper (or thinner) the parabola is, since the value of y is increased more quickly. If is positive, the parabola opens upward, if negative, the parabola opens downward. x-intercepts: The x-intercepts, if any, are also called the roots of the function. The x- intercepts are the solutions to the equation and can be calculated by the formula: and Expression is called discriminant: If discriminant is positive parabola has two intercepts with x-axis; If discriminant is negative parabola has no intercepts with x-axis; If discriminant is zero parabola has one intercept with x-axis (tangent point). y-intercept: Given , the y-intercept is , as y intercept means the value of y when x=0. Vertex: The vertex represents the maximum (or minimum) value of the function, and is very important in calculus. y = a + bx+ cx2 y = a + bx+ cx2 f(x) = a + bx+ cx2 x y a b c a a 0 = a + bx+ cx2 =x1 −b− −4acb2− − − − − −√ 2a =x2 −b+ −4acb2− − − − − −√ 2a − 4acb2 y = a + bx+ cx2 c GMAT Club Math Book 57
GMAT Club Math Book 2024 v8_p58_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	58	58	coordinate_geometry	The vertex of the parabola is located at point . Note: typically just is calculated and plugged in for x to find y. Practice Questions Easy: 1. https://gmatclub.com/forum/which-of-the ... 29938.html 2. https://gmatclub.com/forum/in-the-recta ... 65534.html 3. https://gmatclub.com/forum/in-a-rectang ... 94392.html 4. https://gmatclub.com/forum/in-the-xy-pl ... 05023.html 5. https://gmatclub.com/forum/in-the-figur ... 68654.html 6. https://gmatclub.com/forum/in-the-xy-pl ... 30899.html 7. https://gmatclub.com/forum/in-the-xy-pl ... 21712.html 8. https://gmatclub.com/forum/in-the-xy-pl ... 40727.html 9. https://gmatclub.com/forum/if-each-of-t ... 20583.html 10. https://gmatclub.com/forum/in-the-xy-pl ... 93838.html 11. https://gmatclub.com/forum/in-the-line- ... 31295.html Medium: 1. https://gmatclub.com/forum/which-of-the ... 55961.html 2. https://gmatclub.com/forum/line-m-lies- ... 98105.html 3. https://gmatclub.com/forum/in-the-coord ... 67670.html 4. https://gmatclub.com/forum/in-the-coord ... 44795.html 5. https://gmatclub.com/forum/in-the-xy-co ... 07226.html 6. https://gmatclub.com/forum/in-the-recta ... 20818.html 7. https://gmatclub.com/forum/on-the-graph ... 36560.html 8. https://gmatclub.com/forum/in-the-xy-pl ... 42863.html 9. https://gmatclub.com/forum/the-graph-of ... 07199.html 10. https://gmatclub.com/forum/in-the-xy-co ... 35012.html Hard: 1. https://gmatclub.com/forum/right-triang ... 71597.html 2. https://gmatclub.com/forum/for-every-po ... 91527.html 3. https://gmatclub.com/forum/point-1-0-is ... 82265.html 4. https://gmatclub.com/forum/if-equation- ... 01963.html 5. https://gmatclub.com/forum/in-the-recta ... 44774.html 6. https://gmatclub.com/forum/in-the-xy-pl ... 26463.html 7. https://gmatclub.com/forum/in-the-figur ... 39117.html 8. https://gmatclub.com/forum/the-graph-of ... 07473.html 9. https://gmatclub.com/forum/the-center-o ... 89027.html 10. https://gmatclub.com/forum/in-the-xy-pl ... 00739.html (− ,b 2a c− )b2 4a − ,b 2a GMAT Club Math Book 58
GMAT Club Math Book 2024 v8_p59_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	59	59	word_problems	GMAT Club Math Book Word Problems Made Easy https://gmatclub.com/forum/word-problems-made-easy-87346.html Word Problems Made Easy created by: sriharimurthy edited by: bb, walker, Bunuel This is an introductory post to word problems. It deals primarily with the translation of word problems into equations. Discussions relating to specific types of word problems will be dealt with separately (see end of post). The Following Points Outline a General Approach to Word Problems: 1) Read the entire question carefully and get a feel for what is happening. Identify what kind of word problem you're up against. 2) Make a note of exactly what is being asked. 3) Simplify the problem - this is what is usually meant by 'translating the English to Math'. Draw a figure or table. Sometimes a simple illustration makes the problem much easier to approach. 4) It is not always necessary to start from the first line. Invariably, you will find it easier to define what you have been asked for and then work backwards to get the information that is needed to obtain the answer. 5) Use variables (a, b, x, y, etc.) or numbers (100 in case of percentages, any common multiple in case of fractions, etc.) depending on the situation. 6) Use SMART values. Think for a moment and choose the best possible value that would help you reach the solution in the quickest possible time. DO NOT choose values that would serve only to confuse you. Also, remember to make note of what the value you selected stands for. GMAT Club Math Book 59
GMAT Club Math Book 2024 v8_p60_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	60	60	word_problems	7) Once you have the equations written down it's time to do the math! This is usually quite simple. Be very careful so as not to make any silly mistakes in calculations. 8) Lastly, after solving, cross check to see that the answer you have obtained corresponds to what was asked. The makers of these GMAT questions love to trick students who don’t pay careful attention to what is being asked. For example, if the question asks you to find ‘what fraction of the remaining...’ you can be pretty sure one of the answer choices will have a value corresponding to ‘what fraction of the total…’ Translating Word Problems These are a few common English to Math translations that will help you break down word problems. My recommendation is to refer to them only in the initial phases of study. With practice, decoding a word problem should come naturally. If, on test day, you still have to try and remember what the math translations to some English term is, you haven’t practiced enough! ADDITION: increased by ; more than ; combined ; together ; total of ; sum ; added to ; and ; plus SUBTRACTION: decreased by ; minus ; less ; difference between/of ; less than ; fewer than ; minus ; subtracted from MULTIPLICATION: of ; times ; multiplied by ; product of ; increased/decreased by a factor of (this type can involve both addition or subtraction and multiplication!) DIVISION: per ; out of ; ratio of ; quotient of ; percent (divide by 100) ; divided by ; each EQUALS: is ; are ; was ; were ; will be ; gives ; yields ; sold for ; has ; costs ; adds up to ; the same as ; as much as VARIABLE or VALUE: a number ; how much ; how many ; what Some Tricky Forms: 'per' means 'divided by' Jack drove at a speed of 40 miles per hour OR 40 miles/hour. 'a' sometimes means 'divided by' Jack bought twenty-four eggs for $3 a dozen. 'less than' In English, the ‘less than’ construction is reverse of what it is in math. For example, ‘3 less than x’ means ‘x – 3’ NOT ‘3 – x’ Similarly, if the question says ‘Jack’s age is 3 less than that of Jill’, it means that Jacks age is ‘Jill’s age – 3’. GMAT Club Math Book 60
GMAT Club Math Book 2024 v8_p61_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	61	61	word_problems	The ‘how much is left’ construction Sometimes, the question will give you a total amount that is made up of a number of smaller amounts of unspecified sizes. In this case, just assign a variable to the unknown amounts and the remaining amount will be what is left after deducting this named amount from the total. Consider the following: A hundred-pound order of animal feed was filled by mixing products from Bins A, B and C, and that twice as much was added from Bin C as from Bin A. Let "a" stand for the amount from Bin A. Then the amount from Bin C was "2a", and the amount taken from Bin B was the remaining portion of the hundred pounds: 100 – a – 2a. In the following cases, order is important: ‘quotient/ratio of’ construction If a problems says ‘the ratio of x and y’, it means ‘x divided by y’ NOT 'y divided by x' ‘difference between/of’ construction If the problem says ‘the difference of x and y’ it means \|x – y\| Now that we have seen how it is possible, in theory, to break down word problems, lets go through a few simple examples to see how we can apply this knowledge. Example 1. The length of a rectangular garden is 2 meters more than its width. Express its length in terms of its width. Solution: Key words: more than (implies addition); is (implies equal to) Thus, the phrase ‘length is 2 more than width’ becomes: Length = 2 + width Example 2. The length of a rectangular garden is 2 meters less than its width. Express its length in terms of its width. Solution: Key words: less than (implies subtraction but in reverse order); is (implies equal to) Thus, the phrase ‘length is 2 less than width’ becomes: Length = width - 2 Example 3. The length of a rectangular garden is 2 times its width. Express its length in terms of its width. Solution: Key words: times (implies multiplication); is (implies equal to) Thus, the phrase ‘length is 2 times width’ becomes: GMAT Club Math Book 61
GMAT Club Math Book 2024 v8_p62_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	62	62	fractions_decimals_percents	Length = 2width Example 4. The ratio of the length of a rectangular garden to its width is 2. Express its length in terms of its width. Solution: Key words: ratio of (implies division); is (implies equal to) Thus, the phrase ‘ratio of length to width is 2’ becomes: Length/width = 2 → Length = 2width Example 5. The length of a rectangular garden surrounded by a walkway is twice its width. If difference between the length and width of just the rectangular garden is 10 meters, what will be the width of the walkway if just the garden has width 6 meters? Solution: Ok this one has more words than the previous examples, but don’t worry, lets break it down and see how simple it becomes. Key words: and (implies addition); twice (implies multiplication); difference between (implies subtraction where order is important); what (implies variable); is, will be (imply equal to) Since this is a slightly more complicated problem, let us first define what we want. 'What will be the width of the walkway' implies that we should assign a variable for width of the walkway and find its value. Thus, let width of the walkway be ‘x’. Now, in order to find the width of walkway, we need to have some relation between the total length/width of the rectangular garden + walkway and the length/width of just the garden. Notice here that if we assign a variables to the width and length of either garden+walkway or just garden, we can express every thing in terms of just these variables. So, let length of the garden+walkway = L And width of garden+walkway = W Thus length of just garden = L – 2x Width of just garden = W - 2x Note: Remember that the walkway completely surrounds the garden. Thus its width will have to be accounted for twice in both the total length and total width. GMAT Club Math Book 62
GMAT Club Math Book 2024 v8_p63_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	63	63	algebra	Now let’s see what the question gives us. ‘Garden with width 6 meters’ translates to: Width of garden = 6 W – 2x = 6 Thus, if we know W we can find x. ‘Length of a rectangular garden surrounded by walkway is twice its width’ translates to: Length of garden + length of walkway = 2(width of garden + width of walkway) L = 2W ‘Difference between the length and width of just the rectangular garden is 10 meters’ translates to: Length of garden – width of garden = 10 (L – 2x) – (W – 2x) = 10 L – W = 10 Now, since we have two equations and two variables (L and W), we can find their values. Solving them we get: L = 20 and W = 10. Thus, since we know the value of W, we can calculate ‘x’ 10 – 2x = 6 2x = 4 x = 2 Thus, the width of the walkway is 2 meters. Easy wasn't it? With practice, writing out word problems in the form of equations will become second nature. How much you need to practice depends on your own individual ability. It could be 10 questions or it could be 100. But once you’re able to effortlessly translate word problems into equations, more than half your battle will already be won. GMAT Club Math Book 63
GMAT Club Math Book 2024 v8_p64_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	64	64	word_problems	GMAT Club Math Book Work Problems https://gmatclub.com/forum/work-problems-87357.html ‘Work’ Word Problems Made Easy created by: sriharimurthy edited by: bb, walker, Bunuel NOTE: In case you are not familiar with translating word problems into equations please go through THIS POST FIRST. What is a ‘Work’ Word Problem? It involves a number of people or machines working together to complete a task. We are usually given individual rates of completion. We are asked to find out how long it would take if they work together. Sounds simple enough doesn’t it? Well it is! There is just one simple concept you need to understand in order to solve any ‘work’ related word problem. The ‘Work’ Problem Concept STEP 1: Calculate how much work each person/machine does in one unit of time (could be days, hours, minutes, etc). How do we do this? Simple. If we are given that A completes a certain amount of work in X hours, simply reciprocate the number of hours to get the per hour work. Thus in one hour, A would complete of the work. But what is the logic behind this? Let me explain with the help of an example. Assume we are given that Jack paints a wall in 5 hours. This means that in every hour, he completes a fraction of the work so that at the end of 5 hours, the fraction of work he has completed will become 1 (that means he has completed the task). Thus, if in 5 hours the fraction of work completed is 1, then in 1 hour, the fraction of work 1 X GMAT Club Math Book 64
GMAT Club Math Book 2024 v8_p65_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	65	65	word_problems	completed will be (1*1)/5 STEP 2: Add up the amount of work done by each person/machine in that one unit of time. This would give us the total amount of work completed by both of them in one hour. For example, if A completes of the work in one hour and B completes of the work in one hour, then TOGETHER, they can complete of the work in one hour. STEP 3: Calculate total amount of time taken for work to be completed when all persons/machines are working together. The logic is similar to one we used in STEP 1, the only difference being that we use it in reverse order. Suppose . This means that in one hour, A and B working together will complete of the work. Therefore, working together, they will complete the work in Z hours. Advice here would be: DON'T go about these problems trying to remember some formula. Once you understand the logic underlying the above steps, you will have all the information you need to solve any ‘work’ related word problem. (You will see that the formula you might have come across can be very easily and logically deduced from this concept). Now, lets go through a few problems so that the above-mentioned concept becomes crystal clear. Lets start off with a simple one : Example 1. Jack can paint a wall in 3 hours. John can do the same job in 5 hours. How long will it take if they work together? Solution: This is a simple straightforward question wherein we must just follow steps 1 to 3 in order to obtain the answer. STEP 1: Calculate how much work each person does in one hour. Jack → (1/3) of the work John → (1/5) of the work STEP 2: Add up the amount of work done by each person in one hour. Work done in one hour when both are working together = STEP 3: Calculate total amount of time taken when both work together. If they complete of the work in 1 hour, then they would complete 1 job in hours. 1 X 1 Y +1 X 1 Y + =1 X 1 Y 1 Z 1 Z + =1 3 1 5 8 15 8 15 15 8 GMAT Club Math Book 65
GMAT Club Math Book 2024 v8_p66_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	66	66	word_problems	Example 2. Working, independently X takes 12 hours to finish a certain work. He finishes 2/3 of the work. The rest of the work is finished by Y whose rate is 1/10 of X. In how much time does Y finish his work? Solution: Now the only reason this is trickier than the first problem is because the sequence of events are slightly more complicated. The concept however is the same. So if our understanding of the concept is clear, we should have no trouble at all dealing with this. ‘Working, independently X takes 12 hours to finish a certain work’ This statement tells us that in one hour, X will finish of the work. ‘He finishes 2/3 of the work’ This tells us that of the work still remains. ‘The rest of the work is finished by Y whose rate is (1/10) of X’ Y has to complete of the work. ‘Y's rate is (1/10) that of X‘. We have already calculated rate at which X works to be . Therefore, rate at which Y works is . ‘In how much time does Y finish his work?’ If Y completes of the work in 1 hour, then he will complete of the work in 40 hours. So as you can see, even though the question might have been a little difficult to follow at first reading, the solution was in fact quite simple. We didn’t use any new concepts. All we did was apply our knowledge of the concept we learnt earlier to the information in the question in order to answer what was being asked. Example 3. Working together, printer A and printer B would finish a task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A? Solution: This problem is interesting because it tests not only our knowledge of the concept of word problems, but also our ability to ‘translate English to Math’ ‘Working together, printer A and printer B would finish a task in 24 minutes’ This tells us that A and B combined would work at the rate of per minute. ‘Printer A alone would finish the task in 60 minutes’ This tells us that A works at a rate of per minute. At this point, it should strike you that with just this much information, it is possible to 1 12 1 3 1 3 1 12 ∗ =1 10 1 12 1 120 1 120 1 3 1 24 1 60 GMAT Club Math Book 66
GMAT Club Math Book 2024 v8_p67_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	67	67	word_problems	calculate the rate at which B works: Rate at which B works = . ‘B prints 5 pages a minute more than printer A’ This means that the difference between the amount of work B and A complete in one minute corresponds to 5 pages. So, let us calculate that difference. It will be ‘How many pages does the task contain?’ If of the job consists of 5 pages, then the 1 job will consist of pages. Example 4. Machine A and Machine B are used to manufacture 660 sprockets. It takes machine A ten hours longer to produce 660 sprockets than machine B. Machine B produces 10% more sprockets per hour than machine A. How many sprockets per hour does machine A produce? Solution: The rate of A is sprockets per hour; The rate of B is sprockets per hour. We are told that B produces 10% more sprockets per hour than A, thus --> --> the rate of A is sprockets per hour. As you can see, the main reason the 'tough' problems are 'tough' is because they test a number of other concepts apart from just the ‘work’ concept. However, once you manage to form the equations, they are really not all that tough. And as far as the concept of ‘work’ word problems is concerned – it is always the same!
GMAT Club Math Book 2024 v8_p67_c2	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	67	67	algebra	age to form the equations, they are really not all that tough. And as far as the concept of ‘work’ word problems is concerned – it is always the same! − =1 24 1 60 1 40 − =1 40 1 60 1 120 1 120 = 600(5∗1) 1 120 660 t+10 660 t ∗ 1.1 =660 t+10 660 t t = 100 = 6660 t+10 GMAT Club Math Book 67
GMAT Club Math Book 2024 v8_p68_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	68	68	word_problems	GMAT Club Math Book Distance, Speed Time https://gmatclub.com/forum/distance-speed-time-87481.html 'Distance/Speed/Time' Word Problems Made Easy created by: sriharimurthy edited by: bb, walker, Bunuel NOTE: In case you are not familiar with translating word problems into equations please go through this post first : https://gmatclub.com/forum/word-problem ... 87346.html What is a ‘D/S/T’ Word Problem? Usually involve something/someone moving at a constant or average speed. Out of the three quantities (speed/distance/time), we are required to find one. Information regarding the other two will be provided in the question stem. The ‘D/S/T’ Formula: Distance = Speed x Time I’m sure most of you are already familiar with the above formula (or some variant of it). But how many of you truly understand what it signifies? When you see a ‘D/S/T’ question, do you blindly start plugging values into the formula without really understanding the logic behind it? If then answer to that question is yes, then you would probably have noticed that your accuracy isn’t quite where you’d want it to be. My advice here, as usual, is to make sure you understand the concept behind the formula rather than just using it blindly. So what’s the concept? Lets find out! The Distance = Speed x Time formula is just a way of saying that the distance you travel depends on the speed you go for any length of time. If you travel at 50 mph for one hour, then you would have traveled 50 miles. If you GMAT Club Math Book 68
GMAT Club Math Book 2024 v8_p69_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	69	69	word_problems	travel for 2 hours at that speed, you would have traveled 100 miles. 3 hours would be 150 miles, etc. If you were to double the speed, then you would have traveled 100 miles in the first hour and 200 miles at the end of the second hour. We can figure out any one of the components by knowing the other two. For example, if you have to travel a distance of 100 miles, but can only go at a speed of 50 mph, then you know that it will take you 2 hours to get there. Similarly, if a friend visits you from 100 miles away and tells you that it took him 4 hours to reach, you will know that he AVERAGED 25 mph. Right? All calculations depend on AVERAGE SPEED. Supposing your friend told you that he was stuck in traffic along the way and that he traveled at 50 mph whenever he could move. Therefore, although practically he never really traveled at 25 mph, you can see how the standstills due to traffic caused his average to reduce. Now, if you think about it, from the information given, you can actually tell how long he was driving and how long he was stuck due to traffic (assuming; what is false but what they never worry about in these problems; that he was either traveling at 50 mph or 0 mph). If he was traveling constantly at 50 mph, he should have reached in 2 hours. However, since he took 4 hours, he must have spent the other 2 hours stuck in traffic! Now lets see how we can represent this using the formula. We know that the total distance is 100 miles and that the total time is 4 hours. BUT, his rates were different AND they were different at different times. However, can you see that no matter how many different rates he drove for various different time periods, his TOTAL distance depended simply on the SUM of each of the different distances he drove during each time period? E.g., if you drive a half hour at 60 mph, you will cover 30 miles. Then if you speed up to 80 mph for another half hour, you will cover 40 miles, and then if you slow down to 30 mph, you will only cover 15 miles in the next half hour. But if you drove like this, you would have covered a total of 85 miles (30 + 40 + 15). It is fairly easy to see this looking at it this way, but it is more difficult to see it if we scramble it up and leave out one of the amounts and you have to figure it out going "backwards". That is what word problems do. Further, what makes them difficult is that the components they give you, or ask you to find can involve variable distances, variable times, variable speeds, or any two or three of these. How you "reassemble" all this in order to use the d = s*t formula takes some reflection that is "outside" of the formula itself. You have to think GMAT Club Math Book 69
GMAT Club Math Book 2024 v8_p70_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	70	70	word_problems	about how to use the formula. So the trick is to be able to understand EXACTLY what they are giving you and EXACTLY what it is that is missing, but you do that from thinking, not from the formula, because the formula only works for the COMPONENTS of any trip where you are going an average speed for a certain amount of time. ONCE the conditions deal with different speeds or different times, you have to look at each of those components and how they go together. And that can be very difficult if you are not methodical in how you think about the components and how they go together. The formula doesn't tell you which components you need to look at and how they go together. For that, you need to think, and the thinking is not always as easy or straightforward as it seems like it ought to be. In the case of your friend above, if we call the time he spent driving 50 mph, T1; then the time he spent standing still is (4 - T1) hours, since the whole trip took 4 hours. So we have 100 miles = (50 mph x T1) + (0 mph x [4 - T1]) which is equivalent then to: 100 miles = 50 mph x T1 So, T1 will equal 2 hours. And, since the time he spent going zero is (4 - 2), it also turns out to be 2 hours. Sometimes the right answers will seem counter-intuitive, so it is really important to think about the components methodically and systematically. There is a famous trick problem: To qualify for a race, you need to average 60 mph driving two laps around a 1 mile long track. You have some sort of engine difficulty the first lap so that you only average 30 mph during that lap; how fast do you have to drive the second lap to average 60 for both of them? I will go through THIS problem with you because, since it is SO tricky, it will illustrate a way of looking at almost all the kinds of things you have to think about when working any of these kinds of problems FOR THE FIRST TIME (i.e., before you can do them mechanically because you recognize the TYPE of problem it is). Intuitively it would seem you need to drive 90, but this turns out to be wrong for reasons I will give in a minute. The answer is that NO MATTER HOW FAST you do the second lap, you can't make it. And this SEEMS really odd and that it can't possibly be right, but it is. The reason is that in order to average at least 60 mph over two one-mile laps, since 60 mph is one mile per minute, you will need to do the whole two miles in two minutes or less. But if you drove the first mile at only 30, you used up the whole two minutes just doing IT. So you have run out of time to qualify. To see this with the d = s*t formula, you need to look at the overall trip and break it into components, and that is the hardest part of doing this (these) problem(s), because (often) the components are difficult to figure out, and because it is hard to see which ones you need to put together in which way. GMAT Club Math Book 70
GMAT Club Math Book 2024 v8_p71_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	71	71	word_problems	In the next section we will learn how to do just that. Resolving the Components When you first start out with these problems, the best way to approach them is by organizing the data in a tabular form. Use a separate column each for distance, speed and time and a separate row for the different components involved (2 parts of a journey, different moving objects, etc.). The last row should represent total distance, total time and average speed for these values (although there might be no need to calculate these values if the question does not require them). Assign a variable for any unknown quantity. If there is more than one unknown quantity, do not blindly assign another variable to it. Look for ways in which you can express that quantity in terms of the quantities already present. Assign another variable to it only if this is not possible. In each row, the quantities of distance, speed and time will always satisfy d = s*t. The distance and time column can be added to give you the values of total distance and total time but you CANNOT add the speeds. Think about it: If you drive 20 mph on one street, and 40 mph on another street, does that mean you averaged 60 mph? Once the table is ready, form the equations and solve for what has been asked! GMAT Club Math Book 71
GMAT Club Math Book 2024 v8_p72_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	72	72	word_problems	Warning: Make sure that the units for time and distance agree with the units for the rate. For instance, if they give you a rate of feet per second, then your time must be in seconds and your distance must be in feet. Sometimes they try to trick you by using the wrong units, and you have to catch this and convert to the correct units. A Few More Points to Note Motion in Same Direction (Overtaking): The first thing that should strike you here is that at the time of overtaking, the distances traveled by both will be the same. Motion in Opposite Direction (Meeting): The first thing that should strike you here is that if they start at the same time (which they usually do), then at the point at which they meet, the time will be the same. In addition, the total distance traveled by the two objects under consideration will be equal to the sum of their individual distances traveled. Round Trip: The key thing here is that the distance going and coming back is the same. Now that we know the concept in theory, let us see how it works practically, with the help of a few examples. Note for tables : All values in black have been given in the question stem. All values in blue have been calculated. Example 1. To qualify for a race, you need to average 60 mph driving two laps around a 1-mile long track. You have some sort of engine difficulty the first lap so that you only average 30 mph during that lap; how fast do you have to drive the second lap to average 60 for both of them? Solution: Let us first start with a problem that has already been introduced. You will see that by clearly listing out the given data in tabular form, we eliminate any scope for confusion. GMAT Club Math Book 72
GMAT Club Math Book 2024 v8_p73_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	73	73	word_problems	In the first row, we are given the distance and the speed. Thus it is possible to calculate the time. Time(1) = Distance(1)/Speed(1) = 1/30 In the second row, we are given just the distance. Since we have to calculate speed, let us give it a variable 'x'. Now, by using the 'D/S/T' relationship, time can also be expressed in terms of 'x'. Time(2) = Distance(2)/Speed(2) = 1/x In the third row, we know that the total distance is 2 miles (by taking the sum of the distances in row 1 and 2) and that the average speed should be 60 mph. Thus we can calculate the total time that the two laps should take. Time(3) = Distance(3)/Speed(3) = 2/60 = 1/30 Now, we know that the total time should be the sum of the times in row 1 and 2. Thus we can form the following equation : Time(3) = Time(1) + Time(2) ---> 1/30 = 1/30 + 1/x From this, it becomes clear that '1/x' must be 0. Since 'x' is the reciprocal of 0, which does not exist, there can be no speed for which the average can be made up in the second lap. Example 2. An executive drove from home at an average speed of 30 mph to an airport where a helicopter was waiting. The executive boarded the helicopter and flew to the corporate offices at an average speed of 60 mph. The entire distance was 150 miles; the entire trip took three hours. Find the distance from the airport to the corporate offices. Solution: Since we have been asked to find the distance from the airport to the corporate office (that is the distance he spent flying), let us assign that specific value as 'x'. GMAT Club Math Book 73
GMAT Club Math Book 2024 v8_p74_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	74	74	word_problems	Thus, the distance he spent driving will be '150 - x' Now, in the first row, we have the distance in terms of 'x' and we have been given the speed. Thus we can calculate the time he spent driving in terms of 'x'. Time(1) = Distance(1)/Speed(1) = (150 - x)/30 Similarly, in the second row, we again have the distance in terms of 'x' and we have been given the speed. Thus we can calculate the time he spent flying in terms of 'x'. Time(2) = Distance(2)/Speed(2) = x/60 Now, notice that we have both the times in terms of 'x'. Also, we know the total time for the trip. Thus, summing the individual times spent driving and flying and equating it to the total time, we can solve for 'x'. Time(1) + Time(2) = Time(3) --> (150 - x)/30 + x/60 = 3 --> x = 120 miles Answer : 120 miles Note: In this problem, we did not calculate average speed for row 3 since we did not need it. Remember not to waste time in useless calculations! Example 3. A passenger train leaves the train depot 2 hours after a freight train left the same depot. The freight train is traveling 20 mph slower than the passenger train. Find the speed of the passenger train, if it overtakes the freight train in three hours. Solution: Since this is an 'overtaking' problem, the first thing that should strike us is that the distance traveled by both trains is the same at the time of overtaking. Next we see that we have been asked to find the speed of the passenger train at the time of overtaking. So let us represent it by 'x'. Also, we are given that the freight train is 20 mph slower than the passenger train. Hence its speed in terms of 'x' can be written as 'x - 20'. Moving on to the time, we are told that it has taken the passenger train 3 hours to reach the freight train. This means that the passenger train has been traveling for 3 hours. We are also given that the passenger train left 2 hours after the freight train. This means that the freight train has been traveling for 3 + 2 = 5 hours. Now that we have all the data in place, we need to form an equation that will help us solve for 'x'. Since we know that the distances are equal, let us see how we can use this to our advantage. From the first row, we can form the following equation : Distance(1) = Speed(1) * Time(1) = x*3 GMAT Club Math Book 74
GMAT Club Math Book 2024 v8_p75_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	75	75	word_problems	From the second row, we can form the following equation : Distance(2) = Speed(2) * Time(2) = (x - 20)5 Now, equating the distances because they are equal we get the following equation : 3x = 5(x - 20) --> x = 50 mph. Answer : 50 mph. Example 4. Two cyclists start at the same time from opposite ends of a course that is 45 miles long. One cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long after they begin will they meet? Solution: Since this is a 'meeting' problem, there are two things that should strike you. First, since they are starting at the same time, when they meet, the time for which both will have been cycling will be the same. Second, the total distance traveled by the will be equal to the sum of their individual distances. Since we are asked to find the time, let us assign it as a variable 't'. (which is same for both cyclists) In the first row, we know the speed and we have the time in terms of 't'. Thus we can get the following equation : Distance(1) = Speed(1) Time(1) = 14t In the second row, we know the speed and again we have the time in terms of 't'. Thus we can get the following equation : Distance(2) = Speed(2) Time(2) = 16t Now we know that the total distance traveled is 45 miles and it is equal to the sum of the two distances. Thus we get the following equation to solve for 't' : Distance(3) = Distance(1) + Distance(2) --> 45 = 14t + 16*t --> t = 1.5 hours Answer : 1.5 hours. Example 5. A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in calm water? Solution: Since this is a question on round trip, the first thing that should strike us is that the distance going and coming back will be the same. Now, we are required to find out the boats speed in calm water. So let us assume it to be 'b'. Now if speed of the current is 3 mph, then the speed of the boat while going downstream and upstream will be 'b + 3' and 'b - 3' respectively. GMAT Club Math Book 75
GMAT Club Math Book 2024 v8_p76_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	76	76	word_problems	In the first row, we have the speed of the boat in terms of 'b' and we are given the time. Thus we can get the following equation : Distance(1) = Speed(1) * Time(1) = (b + 3)3 In the second row, we again have the speed in terms of 'b' and we are given the time. Thus we can get the following equation : Distance(2) = Speed(2) Time(2) = (b - 3)4 Since the two distances are equal, we can equate them and solve for 'b'. Distance(1) = Distance(2) --> (b + 3)3 = (b - 3)*4 --> b = 21 mph. Answer : 21 mph. GMAT Club Math Book 76
GMAT Club Math Book 2024 v8_p77_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	77	77	sets_probability_counting	GMAT Club Math Book Overlapping Sets https://gmatclub.com/forum/overlapping-sets-144260.html ADVANCED OVERLAPPING SETS Some hard GMAT quantitative questions will require you to know and understand the formulas for set theory, presenting three sets and asking various questions about them. There are two main formulas to solve questions involving three overlapping sets. Consider the diagram below: FIRST FORMULA . Let's see how this formula is derived. When we add three groups A, B, and C some sections are counted more than once. For instance: sections d, e, and f are counted twice and section g thrice. Hence we need to subtract sections d, e, and f ONCE (to count section g only once) and subtract section g TWICE (again to count section g only once). In the formula above, , where AnB means intersection of A and B (sections d, and g), AnC means intersection of A and C (sections e, and g), and BnC means intersection of B and C (sections f, and g). Now, when we subtract (d, and g), (e, and g), and (f, and g) from , we are subtract sections d, e, and f ONCE BUT section g THREE TIMES (and we need to subtract section g only twice), therefor we should add only section g, which is intersection of A, B and C (AnBnC) again to get . SECOND FORMULA . Notice that EXACTLY (only) 2-group overlaps is not the same as 2-group overlaps: Elements which are common only for A and B are in section d (so elements which are common ONLY for A and B refer to the elements which are in A and B but not in C); Elements which are common only for A and C are in section e; Elements which are common only for B and C are in section f. Let's see how this formula is derived. Total = A + B + C − (sum of 2 −group overlaps) + (all three) +Neither sum of 2 −group overlaps = AnB + AnC + BnC AnB AnC BnC A + B + C Total = A + B + C − (sum of 2 −group overlaps) + (all three) +Neither Total = A + B + C − (sum of EXACTLY 2 −group overlaps) − 2 ∗ (all three) +Neither GMAT Club Math Book 77
GMAT Club Math Book 2024 v8_p78_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	78	78	geometry	Again: when we add three groups A, B, and C some sections are counted more than once. For instance: sections d, e, and f are counted twice and section g thrice. Hence we need to subtract sections d, e, and f ONCE (to count section g only once) and subtract section g TWICE (again to count section g only once). When we subtract from A+B+C we subtract sections d, e, and f once (fine) and next we need to subtract ONLY section g ( ) twice. That's it. Now, how this concept can be represented in GMAT problem? Example 1: Workers are grouped by their areas of expertise, and are placed on at least one team. 20 are on the marketing team, 30 are on the Sales team, and 40 are on the Vision team. 5 workers are on both the Marketing and Sales teams, 6 workers are on both the Sales and Vision teams, 9 workers are on both the Marketing and Vision teams, and 4 workers are on all three teams. How many workers are there in total? Translating: "are placed on at least one team": members of none =0; "20 are on the marketing team": M=20; "30 are on the Sales team": S=30; "40 are on the Vision team": V=40; "5 workers are on both the Marketing and Sales teams": MnS=5, note here that some from these 5 can be the members of Vision team as well, MnS is sections d an g on the diagram (assuming Marketing = A, Sales = B and Vision = C); "6 workers are on both the Sales and Vision teams": SnV=6 (the same as above sections f an g); "9 workers are on both the Marketing and Vision teams": MnV=9. "4 workers are on all three teams": MnSnV=4, section 4. Question: Total=? Applying first formula as we have intersections of two groups and not the number of only (exactly) 2 group members: . Answer: 74. Discuss this question HERE. Example 2: Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 22 students sign up for the poetry club, 27 students for the history club, and 28 students for the writing club. If 6 students sign up for exactly two clubs, how many students sign up for all three clubs? Translating: "Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs": Total=59, Neither=0 (as members are required to sign up for a minimum of one); "22 students sign up for the poetry club": P=22; "27 students for the history club": H=27; "28 students for the writing club": W=28; "6 students sign up for exactly two clubs": (sum of EXACTLY 2-group overlaps)=6, so the sum of sections d, e, and f is given to be 6, (among these 6 students there are no one who is a member of ALL 3 clubs) Question:: "How many students sign up for all three clubs?" --> Apply second formula: --> --> . Answer: 6. Discuss this question HERE. sum of EXACTLY 2 −group overlaps AnBnC Total = M + S + V − (MnS + MnV + SnV ) +MnSnV + Neither = 20 + 30 + 40 − (5 + 6 + 9) + 4 + 0 = 74 PnHnW = g =? Total = P + H + W − (sum of EXACTLY 2 −group overlaps) − 2 ∗PnHnW + Neither 59 = 22 + 27 + 28 − 6 − 2 ∗x + 0 x = 6 GMAT Club Math Book 78
GMAT Club Math Book 2024 v8_p79_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	79	79	statistics	Example 3: Of 20 Adults, 5 belong to A, 7 belong to B, and 9 belong to C. If 2 belong to all three organizations and 3 belong to exactly 2 organizations, how many belong to none of these organizations? Translating: "20 Adults": Total=20; "5 belong to A, 7 belong to B, and 9 belong to C": A=5, B=7, and C=9; "2 belong to all three organizations": AnBnC=g=2; "3 belong to exactly 2 organizations": (sum of EXACTLY 2-group overlaps)=3, so the sum of sections d, e, and f is given to be 3, (among these 3 adults there are no one who is a member of ALL 3 clubs) Question:: Neither=? Apply second formula: --> --> . Answer: 6. Discuss this question HERE. Example 4: This semester, each of the 90 students in a certain class took at least one course from A, B, and C. If 60 students took A, 40 students took B, 20 students took C, and 5 students took all the three, how many students took exactly two courses? Translating: "90 students": Total=90; "of the 90 students in a certain class took at least one course from A, B, and C": Neither=0; "60 students took A, 40 students took B, 20 students took C": A=60, B=40, and C=20; "5 students took all the three courses": AnBnC=g=5; Question:: (sum of EXACTLY 2-group overlaps)=? Apply second formula: --> --> . Answer: 20. Discuss this question HERE. Example 5: In the city of San Durango, 60 people own cats, dogs, or rabbits. If 30 people owned cats, 40 owned dogs, 10 owned rabbits, and 12 owned exactly two of the three types of pet, how many people owned all three? Translating: "60 people own cats, dogs, or rabbits": Total=60; and Neither=0; "30 people owned cats, 40 owned dogs, 10 owned rabbits": A=30, B=40, and C=10; "12 owned exactly two of the three types of pet": (sum of EXACTLY 2-group overlaps)=12; Question:: AnBnC=g=? Apply second formula: --> --> . Answer: 4. Discuss this question HERE. Example 6: When Professor Wang looked at the rosters for this term's classes, she saw that the roster for her economics class (E) had 26 names, the roster for her marketing class (M) had 28, and the roster for her statistics class (S) had 18. When she compared the rosters, she saw that E and M had 9 names in common, E and S had 7, and M and S had 10. She also saw that 4 names were on all 3 rosters. If the rosters for Professor Wang's 3 classes are combined with no student's name listed more than once, how many names will be on the combined roster? Translating: "E had 26 names, M had 28, and S had 18": E=26, M=28, and S=18; "E and M had 9 names in common, E and S had 7, and M and S had 10": EnM=9, EnS=7, and MnS=10; Total = A + B + C − (sum of EXACTLY 2 −group overlaps) − 2 ∗AnBnC + Neither 20 = 5 + 7 + 9 − 3 − 2 ∗ 2 +Neither Neither = 6 Total = A + B + C − (sum of EXACTLY 2 −group overlaps) − 2 ∗AnBnC + Neither 90 = 60 + 40 + 20 −x − 2 ∗ 5 + 0 x = 20 Total = A + B + C − (sum of EXACTLY 2 −group overlaps) − 2 ∗AnBnC + Neither 60 = 30 + 40 + 10 − 12 − 2 ∗x + 0 x = 4 GMAT Club Math Book 79
GMAT Club Math Book 2024 v8_p80_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	80	80	fractions_decimals_percents	"4 names were on all 3 rosters": EnMnS=g=4; Question:: Total=? Apply first formula: --> --> . Answer: 50. Discuss this question HERE. Example 7: There are 50 employees in the office of ABC Company. Of these, 22 have taken an accounting course, 15 have taken a course in finance and 14 have taken a marketing course. Nine of the employees have taken exactly two of the courses and 1 employee has taken all three of the courses. How many of the 50 employees have taken none of the courses? Translating: "There are 50 employees in the office of ABC Company": Total=50; "22 have taken an accounting course, 15 have taken a course in finance and 14 have taken a marketing course"; A=22, B=15, and C=14; "Nine of the employees have taken exactly two of the courses": (sum of EXACTLY 2-group overlaps)=9; "1 employee has taken all three of the courses": AnBnC=g=1; Question:: None=? Apply second formula: --> --> . Answer: 10. Discuss this question HERE. Example 8 (hard): In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products? Translating: "85% of those surveyed liked at least one of three products: 1, 2, and 3": Total=100%. Also, since 85% of those surveyed liked at least one of three products then 15% liked none of three products, thus None=15%; "5% of the people in the survey liked all three of the products": AnBnC=g=5%; Question:: what percentage of the survey participants liked more than one of the three products? Apply second formula: Total = {liked product 1} + {liked product 2} + {liked product 3} - {liked exactly two products} - 2*{liked exactly three product} + {liked none of three products} --> , so 5% liked exactly two products. More than one product liked those who liked exactly two products, (5%) plus those who liked exactly three products (5%), so 5+5=10% liked more than one product. Answer: 10%. Discuss this question HERE. Example 9 (hard): In a class of 50 students, 20 play Hockey, 15 play Cricket and 11 play Football. 7 play both Hockey and Cricket, 4 play Cricket and Football and 5 play Hockey and football. If 18 students do not play any of these given sports, how many students play exactly two of these sports? Translating: "In a class of 50 students...": Total=50; "20 play Hockey, 15 play Cricket and 11 play Football": H=20, C=15, and F=11; "7 play both Hockey and Cricket, 4 play Cricket and Football and 5 play Hockey and football": HnC=7, CnF=4, and HnF=5. Notice that "7 play both Hockey and Cricket" does not mean that out of those 7, some does not play Football too. The same for Cricket/Football and Hockey/Football; "18 students do not play any of these given sports": Neither=18. Total = A + B + C − (sum of 2 −group overlaps) + (all three) +Neither Total = 26 + 28 + 18 − (9 + 7 + 10) + 4 + 0Total = 50 Total = A + B + C − (sum of EXACTLY 2 −group overlaps) − 2 ∗AnBnC + None 50 = 22 + 15 + 14 − 9 − 2 ∗ 1 +None None = 10 100 = 50 + 30 + 20 −x − 2 ∗ 5 + 15 x = 5 GMAT Club Math Book 80
GMAT Club Math Book 2024 v8_p81_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	81	81	fractions_decimals_percents	Question:: how many students play exactly two of these sports? Apply first formula: {Total}={Hockey}+{Cricket}+{Football}-{HC+CH+HF}+{All three}+{Neither} 50=20+15+11-(7+4+5)+{All three}+18 --> {All three}=2; Those who play ONLY Hockey and Cricket are 7-2=5; Those who play ONLY Cricket and Football are 4-2=2; Those who play ONLY Hockey and Football are 5-2=3; Hence, 5+2+3=10 students play exactly two of these sports. Answer: 10. Discuss this question HERE. Example 10 (hard DS question on three overlapping sets): A student has decided to take GMAT and TOEFL examinations, for which he has allocated a certain number of days for preparation. On any given day, he does not prepare for both GMAT and TOEFL. How many days did he allocate for the preparation? (1) He did not prepare for GMAT on 10 days and for TOEFL on 12 days. (2) He prepared for either GMAT or TOEFL on 14 days We have: {Total} = {GMAT } + {TOEFL} - {Both} + {Neither}. Since we are told that "on any given day, he does not prepare for both GMAT and TOEFL", then {Both} = 0, so {Total} = {GMAT } + {TOEFL} + {Neither}. We need to find {Total} (1) He did not prepare for GMAT on 10 days and for TOEFL on 12 days --> {Total} - {GMAT } = 10 and {Total} - {TOEFL} =12. Not sufficient. (2) He prepared for either GMAT or TOEFL on 14 days --> {GMAT } + {TOEFL} = 14. Not sufficient. (1)+(2) We have three linear equations ({Total} - {GMAT } = 10, {Total} - {TOEFL} =12 and {GMAT } + {TOEFL} = 14) with three unknowns ({Total}, {GMAT }, and {TOEFL}), so we can solve for all of them. Sufficient. Just to illustrate. Solving gives: {Total} = 18 - he allocate total of 18 days for the preparation; {GMAT } = 8 - he prepared for the GMAT on 8 days; {TOEFL} = 6 - he prepared for the TOEFL on 6 days; {Neither} = 4 - he prepared for neither of them on 4 days. Answer: C. Discuss this question HERE. Example 11 (disguised three overlapping sets problem): Three people each took 5 tests. If the ranges of their scores in the 5 practice tests were 17, 28 and 35, what is the minimum possible range in scores of the three test-takers? A. 17 B. 28 C. 35 D. 45 E. 80 Consider this problem to be an overlapping sets problem: # of people in group A is 17; # of people in group B is 28; # of people in group C is 35; What is the minimum # of total people possible in all 3 groups? Clearly if two smaller groups A and B are subsets of bigger group C (so if all people who are in A are also in C and all people who are in B are also in C), then total # of people in all 3 groups will be 35. Minimum # of total people cannot possibly be less than 35 since there are already 35 people in group C. GMAT Club Math Book 81
GMAT Club Math Book 2024 v8_p82_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	82	82	sets_probability_counting	Answer: C. P.S. Notice that max range for the original question is not limited when the max # of people in all 3 groups for revised question is 17+28+35 (in case there is 0 overlap between the 3 groups). Answer: C. Discuss this question HERE. ____________________________________________________________________________________________________________ For more questions on overlapping sets check our Question Banks Problem Solving Questions on Overlapping Sets GMAT Club Math Book 82
GMAT Club Math Book 2024 v8_p83_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	83	83	sets_probability_counting	GMAT Club Math Book Probability https://gmatclub.com/forum/probability-87244.html PROBABILITY Definition A number expressing the probability (p) that a specific event will occur, expressed as the ratio of the number of actual occurrences (n) to the number of possible occurrences (N). A number expressing the probability (q) that a specific event will not occur: Examples Coin There are two equally possible outcomes when we toss a coin: a head (H) or tail (T). Therefore, the probability of getting head is 50% or and the probability of getting tail is 50% or . All possibilities: {H,T} Dice p = n N q = = 1 − p(N−n) N 1 2 1 2 GMAT Club Math Book 83
GMAT Club Math Book 2024 v8_p84_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	84	84	sets_probability_counting	There are 6 equally possible outcomes when we roll a die. The probability of getting any number out of 1-6 is . All possibilities: {1,2,3,4,5,6} Marbles, Balls, Cards... Let's assume we have a jar with 10 green and 90 white marbles. If we randomly choose a marble, what is the probability of getting a green marble? The number of all marbles: N = 10 + 90 =100 The number of green marbles: n = 10 Probability of getting a green marble: There is one important concept in problems with marbles/cards/balls. When the first marble is removed from a jar and not replaced, the probability for the second marble differs ( vs. ). Whereas in case of a coin or dice the probabilities are always the same ( and ). Usually, a problem explicitly states: it is a problem with replacement or without replacement. Independent events Two events are independent if occurrence of one event does not influence occurrence of other events. For n independent events the probability is the product of all probabilities of independent events: p = p1 * p2 * ... * pn-1 * pn or P(A and B) = P(A) * P(B) - A and B denote independent events 1 6 p = = =n N 10 100 1 10 9 99 10 100 1 6 1 2 GMAT Club Math Book 84
GMAT Club Math Book 2024 v8_p85_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	85	85	sets_probability_counting	Example #1 Q:There is a coin and a die. After one flip and one toss, what is the probability of getting heads and a "4"? Solution: Tossing a coin and rolling a die are independent events. The probability of getting heads is and probability of getting a "4" is . Therefore, the probability of getting heads and a "4" is: Example #2 Q: If there is a 20% chance of rain, what is the probability that it will rain on the first day but not on the second? Solution: The probability of rain is 0.2; therefore probability of sunshine is q = 1 - 0.2 = 0.8. This yields that the probability of rain on the first day and sunshine on the second day is: P = 0.2 * 0.8 = 0.16 Example #3 Q:There are two sets of integers: {1,3,6,7,8} and {3,5,2}. If Robert chooses randomly one integer from the first set and one integer from the second set, what is the probability of getting two odd integers? Solution: There is a total of 5 integers in the first set and 3 of them are odd: {1, 3, 7}. Therefore, the probability of getting odd integer out of first set is . There are 3 integers in the second set and 2 of them are odd: {3, 5}. Therefore, the probability of getting an odd integer out of second set is . Finally, the probability of of getting two odd integers is: Mutually exclusive events Shakespeare's phrase "To be, or not to be: that is the question" is an example of two mutually exclusive events. Two events are mutually exclusive if they cannot occur at the same time. For n mutually exclusive events the probability is the sum of all probabilities of events: p = p1 + p2 + ... + pn-1 + pn or P(A or B) = P(A) + P(B) - A and B denotes mutually exclusive events Example #1 Q: If Jessica rolls a die, what is the probability of getting at least a "3"? Solution: There are 4 outcomes that satisfy our condition (at least 3): {3, 4, 5, 6}. The probability of each outcome is 1/6. The probability of getting at least a "3" is: 1 2 1 6 P = ∗ =1 2 1 6 1 12 3 5 2 3 P = ∗ =3 5 2 3 2 5 GMAT Club Math Book 85
GMAT Club Math Book 2024 v8_p86_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	86	86	sets_probability_counting	Combination of independent and mutually exclusive events Many probability problems contain combination of both independent and mutually exclusive events. To solve those problems it is important to identify all events and their types. One of the typical problems can be presented in a following general form: Q: If the probability of a certain event is p, what is the probability of it occurring k times in n-time sequence? (Or in English, what is the probability of getting 3 heads while tossing a coin 8 times?) Solution: All events are independent. So, we can say that: (1) But it isn't the right answer. It would be right if we specified exactly each position for events in the sequence. So, we need to take into account that there are more than one outcomes. Let's consider our example with a coin where "H" stands for Heads and "T" stands for Tails: HHHTTTTT and HHTTTTTH are different mutually exclusive outcomes but they both have 3 heads and 5 tails. Therefore, we need to include all combinations of heads and tails. In our general question, probability of occurring event k times in n-time sequence could be expressed as: (2) In the example with a coin, right answer is Example #1 Q.:If the probability of raining on any given day in Atlanta is 40 percent, what is the probability of raining on exactly 2 days in a 7-day period? Solution: We are not interested in the exact sequence of event and thus apply formula #2: A few ways to approach a probability problem There are a few typical ways that you can use for solving probability questions. Let's consider example, how it is possible to apply different approaches: Example #1 Q: There are 8 employees including Bob and Rachel. If 2 employees are to be randomly chosen to form a committee, what is the probability that the committee includes both Bob and Rachel? Solution: P = + + + =1 6 1 6 1 6 1 6 2 3 = ∗ (1 − pP ′ pk )n−k P = ∗ ∗ (1 − pCn k pk )n−k P = ∗ ∗ = ∗C8 3 0.53 0.55 C8 3 0.58 P = ∗ ∗C7 2 0.42 0.65 GMAT Club Math Book 86
GMAT Club Math Book 2024 v8_p87_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	87	87	sets_probability_counting	1) combinatorial approach: The total number of possible committees is . The number of possible committee that includes both Bob and Rachel is . 2) reversal combinatorial approach: Instead of counting probability of occurrence of certain event, sometimes it is better to calculate the probability of the opposite and then use formula p = 1 - q. The total number of possible committees is . The number of possible committee that does not includes both Bob and Rachel is: where, - the number of committees formed from 6 other people. - the number of committees formed from Rob or Rachel and one out of 6 other people. 3) probability approach: The probability of choosing Bob or Rachel as a first person in committee is 2/8. The probability of choosing Rachel or Bob as a second person when first person is already chosen is 1/7. The probability that the committee includes both Bob and Rachel is. 4) reversal probability approach: We can choose any first person. Then, if we have Rachel or Bob as first choice, we can choose any other person out of 6 people. If we have neither Rachel nor Bob as first choice, we can choose any person out of remaining 7 people. The probability that the committee includes both Bob and Rachel is. Example #2 Q: Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other? Solution: 1) combinatorial approach: - we choose 3 couples out of 5 couples. - we chose one person out of a couple. - we have 3 couple and we choose one person out of each couple. - the total number of combinations to choose 3 people out of 10 people. 2) reversal combinatorial approach: In this example reversal approach is a bit shorter N = C8 2 n = 1 P = = =n N 1 C8 2 1 28 N = C8 2 m = + 2 ∗C6 2 C6 1 C6 2 2 ∗ C6 1 P = 1 − = 1 −m N +2∗C6 2 C6 1 C8 2 P = 1 − = 1 − =15+2∗6 28 27 28 1 28 P = ∗ = =2 8 1 7 2 56 1 28 P = 1 − ( ∗ + ∗ 1) = =2 8 6 7 6 8 2 56 1 28 C5 3 C2 1 (C2 1 )3 C10 3 p = = = ∗(C5 3 C2 1 )3 C10 3 10∗8 10∗3∗4 2 3 GMAT Club Math Book 87
GMAT Club Math Book 2024 v8_p88_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	88	88	sets_probability_counting	and faster. - we choose 1 couple out of 5 couples. - we chose one person out of remaining 8 people. - the total number of combinations to choose 3 people out of 10 people. 3) probability approach: 1st person: - we choose any person out of 10. 2nd person: - we choose any person out of 8=10-2(one couple from previous choice) 3rd person: - we choose any person out of 6=10-4(two couples from previous choices). Probability tree Sometimes, at 700+ level you may see complex probability problems that include conditions or restrictions. For such problems it could be helpful to draw a probability tree that include all possible outcomes and their probabilities. Example #1 Q: Julia and Brian play a game in which Julia takes a ball and if it is green, she wins. If the first ball is not green, she takes the second ball (without replacing first) and she wins if the two balls are white or if the first ball is gray and the second ball is white. What is the probability of Julia winning if the jar contains 1 gray, 2 white and 4 green balls? Solution: Let's draw all possible outcomes and calculate all probabilities. Now, It is pretty obvious that the probability of Julia's win is: C5 1 C8 1 C10 3 p = 1 − = 1 − = ∗C5 1 C8 1 C10 3 5∗8 10∗3∗4 2 3 = 110 10 8 9 6 8 p = 1 ∗ ∗ =8 9 6 8 2 3 P = + ∗ + ∗ =4 7 2 7 1 6 1 7 2 6 2 3 GMAT Club Math Book 88
GMAT Club Math Book 2024 v8_p89_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	89	89	word_problems	Tips and Tricks: Symmetry Symmetry sometimes lets you solve seemingly complex probability problem in a few seconds. Let's consider an example: Example #1 Q: There are 5 chairs. Bob and Rachel want to sit such that Bob is always left to Rachel. How many ways it can be done ? Solution: Because of symmetry, the number of ways that Bob is left to Rachel is exactly 1/2 of all possible ways: N = ∗ = 101 2 P 5 2 GMAT Club Math Book 89
GMAT Club Math Book 2024 v8_p90_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	90	90	number_theory	Practice Questions Easy: 1. https://gmatclub.com/forum/in-a-box-of- ... 07395.html 2. https://gmatclub.com/forum/among-a-grou ... 34499.html 3. https://gmatclub.com/forum/the-probabil ... 44730.html 4. https://gmatclub.com/forum/xavier-yvonn ... 65822.html 5. https://gmatclub.com/forum/carol-purcha ... 21728.html 6. https://gmatclub.com/forum/two-integers ... 43451.html 7. https://gmatclub.com/forum/raffle-ticke ... 36563.html 8. https://gmatclub.com/forum/sixty-percen ... 43900.html 9. https://gmatclub.com/forum/in-a-set-of- ... 68915.html 10. https://gmatclub.com/forum/a-committee- ... 44443.html Medium: 1. https://gmatclub.com/forum/if-x-is-to-b ... 67089.html 2. https://gmatclub.com/forum/a-gardener-i ... 99822.html 3. https://gmatclub.com/forum/a-contest-wi ... 38710.html 4. https://gmatclub.com/forum/in-a-certain ... 69115.html 5. https://gmatclub.com/forum/a-committee- ... 81051.html 6. https://gmatclub.com/forum/two-thirds-o ... 10059.html 7. https://gmatclub.com/forum/a-deck-of-ca ... 21976.html 8. https://gmatclub.com/forum/set-s-consis ... 24713.html 9. https://gmatclub.com/forum/john-throws- ... 43960.html 10. https://gmatclub.com/forum/a-fair-die-w ... 84968.html Hard: 1. https://gmatclub.com/forum/a-couple-dec ... 68730.html 2. https://gmatclub.com/forum/mary-and-joe ... 86407.html 3. https://gmatclub.com/forum/a-box-contai ... 27699.html 4. https://gmatclub.com/forum/two-dice-are ... 26477.html 5. https://gmatclub.com/forum/the-table-ab ... 05918.html 6. https://gmatclub.com/forum/minimum-of-h ... 73869.html 7. https://gmatclub.com/forum/kate-and-dav ... 97177.html 8. https://gmatclub.com/forum/when-an-unfa ... 98795.html 9. https://gmatclub.com/forum/artificial-i ... 21978.html 10. https://gmatclub.com/forum/a-deck-of-ca ... 21977.html GMAT Club Math Book 90
GMAT Club Math Book 2024 v8_p91_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	91	91	sets_probability_counting	GMAT Club Math Book Combinations https://gmatclub.com/forum/combinations-87345.html COMBINATORICS created by: walker edited by: bb, Bunuel Definition Combinatorics is the branch of mathematics studying the enumeration, combination, and permutation of sets of elements and the mathematical relations that characterize their properties. Enumeration Enumeration is a method of counting all possible ways to arrange elements. Although it is the simplest method, it is often the fastest method to solve hard GMAT problems and is a pivotal principle for any other combinatorial method. In fact, combination and permutation is shortcuts for enumeration. The main idea of enumeration is writing down all possible ways and then count them. Let's consider a few examples: Example #1 Q:. There are three marbles: 1 blue, 1 gray and 1 green. In how many ways is it possible to arrange marbles in a row? Solution: Let's write out all possible ways: Answer is 6. In general, the number of ways to arrange n different objects in a row Example #2 Q:. There are three marbles: 1 blue, 1 gray and 1 green. In how many ways is it possible GMAT Club Math Book 91
GMAT Club Math Book 2024 v8_p92_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	92	92	statistics	to arrange marbles in a row if blue and green marbles have to be next to each other? Solution: Let's write out all possible ways to arrange marbles in a row and then find only arrangements that satisfy question's condition: Answer is 4. Example #3 Q:. There are three marbles: 1 blue, 1 gray and 1 green. In how many ways is it possible to arrange marbles in a row if gray marble have to be left to blue marble? Solution: Let's write out all possible ways to arrange marbles in a row and then find only arrangements that satisfy question's condition: Answer is 3. Arrangements of n different objects Enumeration is a great way to count a small number of arrangements. But when the total number of arrangements is large, enumeration can't be very useful, especially taking into account GMAT time restriction. Fortunately, there are some methods that can speed up counting of all arrangements. The number of arrangements of n different objects in a row is a typical problem that can be solve this way: 1. How many objects we can put at 1st place? n. 2. How many objects we can put at 2nd place? n - 1. We can't put the object that already placed at 1st place. ..... n. How nany objects we can put at n-th place? 1. Only one object remains. Therefore, the total number of arrangements of n different objects in a row is N = n ∗ ( n − 1) ∗ ( n − 2). . . .2 ∗ 1 = n! GMAT Club Math Book 92
GMAT Club Math Book 2024 v8_p93_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	93	93	sets_probability_counting	Combination A combination is an unordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as: knowing how to find the number of arrangements of n distinct objects we can easily find formula for combination: 1. The total number of arrangements of n distinct objects is n! 2. Now we have to exclude all arrangements of k objects (k!) and remaining (n-k) objects ((n-k)!) as the order of chosen k objects and remained (n-k) objects doesn't matter. Permutation A permutation is an ordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as: knowing how to find the number of arrangements of n distinct objects we can easily find formula for combination: 1. The total number of arrangements of n distinct objects is n! 2. Now we have to exclude all arrangements of remaining (n-k) objects ((n-k)!) as the order of remained (n-k) objects doesn't matter. If we exclude order of chosen objects from permutation formula, we will get combination formula: Cn k =Cn k n! k!(n−k)! P n k =P n k n! (n−k)! = P n k k! Cn k GMAT Club Math Book 93
GMAT Club Math Book 2024 v8_p94_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	94	94	statistics	Circular arrangements Let's say we have 6 distinct objects, how many relatively different arrangements do we have if those objects should be placed in a circle. The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have: Tips and Tricks Any problem in Combinatorics is a counting problem. Therefore, a key to solution is a way how to count the number of arrangements. It sounds obvious but a lot of people begin approaching to a problem with thoughts like "Should I apply C- or P- formula here?". Don't fall in this trap: define how you are going to count arrangements first, realize that your way is right and you don't miss something important, and only then use C- or P- formula if you need them. Resources Walker's post with Combinatorics/probability problems: Combinatorics/probability Problems R = = ( n − 1)!n! n GMAT Club Math Book 94
GMAT Club Math Book 2024 v8_p95_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	95	95	general	Practice Questions Easy: 1. https://gmatclub.com/forum/there-are-8- ... 34582.html 2. https://gmatclub.com/forum/there-are-5- ... 88100.html 3. https://gmatclub.com/forum/a-certain-un ... 68638.html 4. https://gmatclub.com/forum/the-diagram- ... 44271.html 5. https://gmatclub.com/forum/in-a-tournam ... 55939.html 6. https://gmatclub.com/forum/how-many-way ... 47677.html 7. https://gmatclub.com/forum/in-how-many- ... 05506.html 8. https://gmatclub.com/forum/a-certain-ba ... 98055.html 9. https://gmatclub.com/forum/how-many-dif ... 54633.html 10. https://gmatclub.com/forum/if-a-person- ... 24843.html Medium: 1. https://gmatclub.com/forum/the-letters- ... 20320.html 2. https://gmatclub.com/forum/team-a-and-t ... 41461.html 3. https://gmatclub.com/forum/the-subsets- ... 08256.html 4. https://gmatclub.com/forum/each-in-the- ... 95162.html 5. https://gmatclub.com/forum/a-company-th ... 68427.html 6. https://gmatclub.com/forum/clarissa-wil ... 42452.html 7. https://gmatclub.com/forum/in-the-table ... 66572.html 8. https://gmatclub.com/forum/if-a-code-wo ... 26652.html 9. https://gmatclub.com/forum/how-many-two ... 28130.html 10. https://gmatclub.com/forum/john-has-12- ... 07307.html Hard: 1. https://gmatclub.com/forum/pat-will-wal ... 68374.html 2. https://gmatclub.com/forum/five-integer ... 05923.html 3. https://gmatclub.com/forum/in-how-many- ... 85707.html 4. https://gmatclub.com/forum/how-many-fiv ... 91597.html 5. https://gmatclub.com/forum/nine-family- ... 75071.html 6. https://gmatclub.com/forum/how-many-pos ... 55804.html 7. https://gmatclub.com/forum/how-many-odd ... 94655.html 8. https://gmatclub.com/forum/how-many-dif ... 71774.html 9. https://gmatclub.com/forum/a-committee- ... 04966.html 10. https://gmatclub.com/forum/in-how-many- ... 56400.html
GMAT Club Math Book 2024 v8_p95_c2	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	95	95	word_problems	forum/how-many-dif ... 71774.html 9. https://gmatclub.com/forum/a-committee- ... 04966.html 10. https://gmatclub.com/forum/in-how-many- ... 56400.html Page 1 of 1 All times are UTC - 8 hours [ DST ] GMAT Club Math Book 95
GMAT Club Math Book 2024 v8_p96_c1	GMAT Club Math Book 2024 v8	GMAT Club Math Book 2024 v8.pdf	pdf	96	96	statistics	GMAT Club Math Book Standard Deviation https://gmatclub.com/forum/standard-deviation-87905.html STANDARD DEVIATION Definition Standard Deviation (SD, or STD or ) - a measure of the dispersion or variation in a distribution, equal to the square root of variance or the arithmetic mean (average) of squares of deviations from the arithmetic mean. In simple terms, it shows how much variation there is from the "average" (mean). It may be thought of as the average difference from the mean of distribution, how far data points are away from the mean. A low standard deviation indicates that data points tend to be very close to the mean, whereas high standard deviation indicates that the data are spread out over a large range of values. σ variance = ∑( −xi xav)2 N σ = =variance− − − − − − −√ ∑( −xi xav)2 N − − − − − − − − − √ GMAT Club Math Book 96

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