GMAT Club
Math Book
8th Edition
May 2024


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
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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
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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 =
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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
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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
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power of the common factors).
• Every common divisor of a and b is a divisor of gcd(a, b).
• a*b=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.
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• 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
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=
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 3*4*5*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
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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
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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...
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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
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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
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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 845*9512*408*613 would be the last 2 digits of
45*12*8*13=540*104=40*4=160=60
Example: The last digit of 85945*89*58307=5*9*7=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
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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


, 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: $15,000*0.1*9/12 = $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


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
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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


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 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
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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
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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
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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 2*4=8 (basically if then will be
divisible by 8*3=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)
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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
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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


(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
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


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
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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
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27


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


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− −√
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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√
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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
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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
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
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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)
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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(50*2) = 10 ;
the HM = (2*50*2)/(52) = 3.85
AM > GM > HM
AM*HM = 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
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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
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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
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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
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
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"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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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|
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.
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
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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
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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.
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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’.
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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:
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Length = 2*width
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 = 2*width
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.
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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 = 2*W
‘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.
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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
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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