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Maths Recap Flashcards

1
Q

How to test whether a number is prime or composite

A

Before we start off, what is a prime number and a composite number? (For people who are not sure) 

Quote:
A Prime Number is a positive integer that is divisible by ONLY 2 numbers (1 and itself). Whereas, A composite number is a positive integer which has divisor(s) other than the 2 numbers (1 and itself).
Ok, coming back to the point. I will name the number as n for simplicity. Following are the steps to test whether a number is a prime or composite, 

1. Identify the perfect square (P.S) closest to the n. 
2. Compute the square root of P.S 
3. List all prime numbers upto the computed square root 
4. Check if all listed prime numbers divide n equally. If not, then n is a prime. Even if atleast one of the listed prime numbers divide n, then n is a composite. 


Example: 

Take n as 113. To test whether 113 is a prime, 

1. 100 is the closest perfect square to 113 (Remember that you take a closest perfect square that is smaller than n itself!) 
2. Square root of 100 ==> 10 
3. Prime numbers upto the square root (10) ==> 2,3,5,7. 
4. Check whether 2,3,5,7 divides 113. None of the numbers divide 113. So, 113 is a prime.

There is one interesting cool fact to know. I remember applying this fact in actual GMAT. It’s good to learn if you don’t know. 


Quote:
Product of any 2 numbers = Product of LCM and HCF of those 2 numbers 

Product of any 2 fractions = Product of LCM and HCF of those 2 fractions
I will try to find and post a GMAT problem that uses this concept. Please feel free to post a question if you find it.

Warning: Some people may not find this approach comfortable. Some may find it comfortable. Please follow and practice only if you are comfortable with this approach. Otherwise, please ignore it. 


Sometimes, we get one type of question in GMAT where we need to calculate units digit of integers raised to some power. I found a shortcut where you could save time by remembering some patterns. 

How to find unit digit of powers of numbers: 

Pattern 1: 
Unit’s place that has digits - 2/3/7/8 

Then, unit’s digit repeats every 4th value. Divide the power (or index) by 4. 

After dividing, 
If remainder is 1, unit digit of number raised to the power 1. 
If remainder is 2, unit digit of number raised to the power 2. 
If remainder is 3, unit digit of number raised to the power 3. 
If remainder is 0, unit digit of number raised to the power 4. 

Pattern 2: 
Unit’s place that has digits - 0/1/5/6 

Then, all powers of the number have same digit as unit’s place. 

For e.g., 6^1 = 6, 6^2 = 36, 6^3 = 216, 6^4 = 1296 


Pattern 3: 
Unit’s place that has digit - 4 

Then, 
If power is odd –> unit’s digit will be ‘4’ 
If power is even –> unit’s digit will be ‘6’ 

Similarly, 
Unit’s place that has digit - 9 

Then, 
If power is odd –> unit’s digit will be ‘9 
If power is even –> unit’s digit will be ‘1’ 


Example: 
Let’s take a long number - 122 ^ 94. Find unit’s digit. 

Unit’s place is 2. So, it repeats every 4th term of the power. 
So, divide the power by 4. 94 % 4 ==> 2 (remainder). 

Raise the unit digit of the base number to the power (2 - remainder). 2^2 = 4. 

Thus, 4 is the unit’s digit of 122^94. 


I found this approach very easy and comfortable. So, see how comfortable it is for you and apply. 


Real GMAT Problem: OG-12 PS #190

We are often faced to test the divisibility of some number in the exam. Following points may help you in simplifying the process, 

Divisibility Tests: 

To check whether a number (say n) is divisible 

By 2: unit’s place of n must be 0 (OR) unit’s place of n must be divisible by 2. 

By 3: Sum of the digits of n must be divisible by 3. 

By 4: Last 2 digits (Unit’s place and ten’s place) of n are 0’s (OR) Last 2 digits of n must be divisible by 4. 

By 5: Unit’s digit must be a 5 (OR) a 0. 

By 6: n must be divisible by both 2 and 3 (Follow the method used for 2 and 3). 

By 8: Last 3 digits (units, tens and hundredth place) of n are 0’s (OR) Last 3 digits of n is divisible by 8. 

By 9: Sum of the digits of n must be divisible by 9. 

By 11: (Sum of the digits of n in odd places) - (Sum of the digits of n in even places) ==> Either 0 (OR) divisible by 11. 

By 12: n must be divisible by both 3 and 4 (Follow the method used for 3 and 4). 

By 25: Last 2 digits (units and tens place) of n are 0’s (OR) Last 2 digits of n must be divisible by 25. 

By 75: n must be divisible by both 3 and 25 (Follow the method used for 3 and 25). 

By 125: Last 3 digits of n are 0’s (OR) are divisible by 125. 


Try out examples for each divisibility to grasp better.

How to find number of factors for a POSITIVE INTEGER: 

There are 2 approaches to find number of factors of an integer. 

Approach #1: (Factor Pairs Method) 

i. Let’s take a non-perfect square number such as 32. Keep picking a number (start from 1) that divides 32 until you reach a number that is smaller than the quotient. 

Small Large 
1 32 
2 16 
4 8 

Stop! If you take 8, you get 4 as quotient which is smaller than the number (8). 
Therefore, there are 32 = 6 factor pairs or number of factors of 32. 

ii. Let’s take a perfect square number such as 36. Keep picking a number (start from 1) that divides 36 until you reach a number that is smaller than the quotient. 

Small Large 
1 36 
2 18 
3 12 
4 9 
6 6 

Totally, there are 52 = 10 factor pairs or number of factors of 36. But, (6,6) gets repeated twice. So, deduct 1 from factor pairs i.e. 10-1 = 9 factor pairs or number of factors of 36. 


Approach #2: (RECOMMENDED) 

If N is expresses in terms of its prime factors as a^p * b^q * c^r, where p,q,r are positive integers, then N will have (p+1) * (q+1) * (r+1) positive factors. 

Example: 

i. 32 = 2^5. 
No. of factors = (5+1) = 6. 

ii. 1452 = 2^2 * 3 * 11^2 
No. of factors = (2+1) * (1+1) * (2+1) = 18.

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2
Q

Prime numbers

A

Prime Numbers

A prime number can be divided, without a remainder, only by itself and by 1. For example, 17 can be divided only by 17 and by 1.

Some facts:

The only even prime number is 2. All other even numbers can be divided by 2.
If the sum of a number’s digits is a multiple of 3, that number can be divided by 3.
No prime number greater than 5 ends in a 5. Any number greater than 5 that ends in a 5 can be divided by 5.
Zero and 1 are not considered prime numbers.
Except for 0 and 1, a number is either a prime number or a composite number. A composite number is defined as any number, greater than 1, that is not prime.
To prove whether a number is a prime number, first try dividing it by 2, and see if you get a whole number. If you do, it can’t be a prime number. If you don’t get a whole number, next try dividing it by prime numbers: 3, 5, 7, 11 (9 is divisible by 3) and so on, always dividing by a prime number (see table below).

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3
Q

The Sieve of Eratosthenes

A

The Sieve of Eratosthenes

Eratosthenes (275-194 B.C., Greece) devised a ‘sieve’ to discover prime numbers. A sieve is like a strainer that you use to drain spaghetti when it is done cooking. The water drains out, leaving your spaghetti behind. Eratosthenes’s sieve drains out composite numbers and leaves prime numbers behind.
To use the sieve of Eratosthenes to find the prime numbers up to 100, make a chart of the first one hundred positive integers (1-100):

         1   2   3   4   5   6   7   8   9  10
        11  12  13  14  15  16  17  18  19  20
        21  22  23  24  25  26  27  28  29  30
        31  32  33  34  35  36  37  38  39  40
        41  42  43  44  45  46  47  48  49  50
        51  52  53  54  55  56  57  58  59  60
        61  62  63  64  65  66  67  68  69  70
        71  72  73  74  75  76  77  78  79  80
        81  82  83  84  85  86  87  88  89  90
        91  92  93  94  95  96  97  98  99 100

Cross out 1, because it is not prime.

Circle 2, because it is the smallest positive even prime. Now cross out every multiple of 2; in other words, cross out every second number.

Circle 3, the next prime. Then cross out all of the multiples of 3; in other words, every third number. Some, like 6, may have already been crossed out because they are multiples of 2.

Circle the next open number, 5. Now cross out all of the multiples of 5, or every 5th number.
Continue doing this until all the numbers through 100 have either been circled or crossed out. You have just circled all the prime numbers from 1 to 100!

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4
Q

How to test whether a number is prime or composite

A
  1. Identify the perfect square (P.S) closest to the n. 

  2. Compute the square root of P.S 

  3. List all prime numbers upto the computed square root 

  4. Check if all listed prime numbers divide n equally. If not, then n is a prime. Even if atleast one of the listed prime numbers divide n, then n is a composite.




Example: 

Take n as 113. To test whether 113 is a prime, 



  1. 100 is the closest perfect square to 113 (Remember that you take a closest perfect square that is smaller than n itself!)


2. Square root of 100 ==> 10 


  1. Prime numbers upto the square root (10) ==> 2,3,5,7. 

  2. Check whether 2,3,5,7 divides 113. None of the numbers divide 113. So, 113 is a prime.
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5
Q

How to calculate LCM and HCF of fractions:

A

Quote:
L.C.M of 2 fractions = L.C.M of NUMERATORS / H.C.F of DENOMINATORS

H.C.F of 2 fractions = H.C.F of NUMERATORS / L.C.M of DENOMINATORS

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6
Q

Product of two numbers

Product of two fractions

A

Quote:
Product of any 2 numbers = Product of LCM and HCF of those 2 numbers

Product of any 2 fractions = Product of LCM and HCF of those 2 fractions
I will try to find and post a GMAT problem that uses this concept. Please feel free to post a question if you find it.

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7
Q

How to find unit digit of powers of numbers

A

Pattern 1:
Unit’s place that has digits - 2/3/7/8

Then, unit’s digit repeats every 4th value. Divide the power (or index) by 4.

After dividing,
If remainder is 1, unit digit of number raised to the power 1.
If remainder is 2, unit digit of number raised to the power 2.
If remainder is 3, unit digit of number raised to the power 3.
If remainder is 0, unit digit of number raised to the power 4.

Pattern 2:
Unit’s place that has digits - 0/1/5/6

Then, all powers of the number have same digit as unit’s place.

For e.g., 6^1 = 6, 6^2 = 36, 6^3 = 216, 6^4 = 1296

Pattern 3:
Unit’s place that has digit - 4

Then,
If power is odd –> unit’s digit will be ‘4’
If power is even –> unit’s digit will be ‘6’

Similarly,
Unit’s place that has digit - 9

Then,
If power is odd –> unit’s digit will be ‘9
If power is even –> unit’s digit will be ‘1’

Example:
Let’s take a long number - 122 ^ 94. Find unit’s digit.

Unit’s place is 2. So, it repeats every 4th term of the power.
So, divide the power by 4. 94 % 4 ==> 2 (remainder).

Raise the unit digit of the base number to the power (2 - remainder). 2^2 = 4.

Thus, 4 is the unit’s digit of 122^94.

I found this approach very easy and comfortable. So, see how comfortable it is for you and apply.

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8
Q

Area of triangle
Area of rectangle
Area of trapezoid
Area of elipse

A

Triangle
Area = ½ × b × h
b = base
h = vertical height

Rectangle
Area = w × h
w = width
h = height

Trapezoid (US)
Trapezium (UK)
Area = ½(a+b) × h
h = vertical height

Ellipse
Area = πab

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9
Q

Area of square
Area of parallelogram
Area of circle
Area of sector

A

Square
Area = a2
a = length of side

Parallelogram
Area = b × h
b = base
h = vertical

Circle
Area = π × r2
Circumference = 2 × π × r
r = radius

Sector
Area = ½ × r2 × θ
r = radius
θ = angle in radians

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10
Q 
Recognize multiples of
2
3
4
5
6
9
A
  1. Last digit is even
  2. Sum of digits is a multiple of 3
  3. Last two digits are multiples of 4
  4. Last digit is 5 or 0
  5. Sum of digits is a multiple of 3 and the last digit is even
  6. Sum of digits is a multiple of 9
  7. Last digit is 0
  8. Sum of digits is a multiple of 3 and the last two digits are a multiple of 4
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11
Q 
Isosceles triangle
Equilateral triangle
Pythagorean theorem
30-60-90 triangle
45-45-90 triangle
A

Isosceles triangle - two equal sides and two equal angles
Equilateral triangle- all sides equal and all equal angles
Pythagorean theorem- a^2 + b^2 = c^2
30-60-90 triangle - 1 / root 3 / 2
45-45-90 triangle - 1 / 1 / root 2

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12
Q

Slope
Permutation
Combination

A

Slope = change in y / change in x
Permutation - n! / (n-k)!
Combination- n!/[k!(n-k)!]

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13
Q

Sum of all angles of a regular polygon
Area of sector
Volume of cylinder
Volume of sphere

A

Sum of angles = (n-2)*180
area of Sector - r/360 * Pi * r ^2
Volume of cylinder - Pi * r^2 * h
Volume of sphere- 4/3 * Pi * r^3

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14
Q

Squares of 2 till 10

A 
2-4
3-9
4-32
5-25
6-36
7-49
8-64
9-81
10-100
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15
Q

Squares of 11 to 15

A 
11-121
12-144
13-169
14-212
15-225
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16
Q

Squares of 16 to 20

A 
16-256
17-289
18-324
19-361
20-400
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17
Q

Squares of 21 to 25

A 
21-441
22-484
23-529
24-592
25-625
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18
Q

Cubes of 1 to 5

A 
1-1
2-8
3-27
4-64
5-125
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19
Q

Cubes of 6 to 10

A 
6-216
7-343
8-512
9-729
10-1000
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20
Q

Cubes of 11 to 15

A 
11-1331
12-1728
13-2197
14-2744
15-3375
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21
Q

Divisibility Tests:

A

To check whether a number (say n) is divisible

By 2: unit’s place of n must be 0 (OR) unit’s place of n must be divisible by 2.

By 3: Sum of the digits of n must be divisible by 3.

By 4: Last 2 digits (Unit’s place and ten’s place) of n are 0’s (OR) Last 2 digits of n must be divisible by 4.

By 5: Unit’s digit must be a 5 (OR) a 0.

By 6: n must be divisible by both 2 and 3 (Follow the method used for 2 and 3).

By 8: Last 3 digits (units, tens and hundredth place) of n are 0’s (OR) Last 3 digits of n is divisible by 8.

By 9: Sum of the digits of n must be divisible by 9.

By 11: (Sum of the digits of n in odd places) - (Sum of the digits of n in even places) ==> Either 0 (OR) divisible by 11.

By 12: n must be divisible by both 3 and 4 (Follow the method used for 3 and 4).

By 25: Last 2 digits (units and tens place) of n are 0’s (OR) Last 2 digits of n must be divisible by 25.

By 75: n must be divisible by both 3 and 25 (Follow the method used for 3 and 25).

By 125: Last 3 digits of n are 0’s (OR) are divisible by 125.

Try out examples for each divisibility to grasp better.

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22
Q

How to find number of factors for a POSITIVE INTEGER:

A

There are 2 approaches to find number of factors of an integer.

Approach #1: (Factor Pairs Method)

i. Let’s take a non-perfect square number such as 32. Keep picking a number (start from 1) that divides 32 until you reach a number that is smaller than the quotient.

Small Large
1 32
2 16
4 8

Stop! If you take 8, you get 4 as quotient which is smaller than the number (8).
Therefore, there are 3*2 = 6 factor pairs or number of factors of 32.

ii. Let’s take a perfect square number such as 36. Keep picking a number (start from 1) that divides 36 until you reach a number that is smaller than the quotient.

Small Large 
1 36 
2 18 
3 12 
4 9 
6 6

Totally, there are 5*2 = 10 factor pairs or number of factors of 36. But, (6,6) gets repeated twice. So, deduct 1 from factor pairs i.e. 10-1 = 9 factor pairs or number of factors of 36.

Approach #2: (RECOMMENDED)

If N is expresses in terms of its prime factors as a^p * b^q * c^r, where p,q,r are positive integers, then N will have (p+1) * (q+1) * (r+1) positive factors.

Example:

i. 32 = 2^5.
No. of factors = (5+1) = 6.

ii. 1452 = 2^2 * 3 * 11^2
No. of factors = (2+1) * (1+1) * (2+1) = 18.

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23
Q

Number of factors

A

If N is a perfect square, then the number of factors of N will ALWAYS be an ODD number.

If N is a NON-perfect square, then the number of factors of N will ALWAYS be an EVEN number.

_________________
Download GMAT Math and CR questions with Solutions from Instructors and High-scorers:
http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html

GO GREEN..! GO VEG..!

Daily Quote:
“Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”

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24
Q

How to find Sum of all factors of a POSITIVE integer

A

How to find Sum of all factors of a POSITIVE integer:

If N is expressed in terms of its prime factors as a^p * b^q * c^r, where p,q,r are positive integers, then the sum of all factors of N is

[ (a^(p+1) - 1) / a-1 ] * [ (b^(q+1) - 1) / b-1 ] * [ (c^(r+1) - 1) / c-1 ]

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25
Factorization rule
Any number whose prime factorization contains even powers of primes, then the number must be a perfect square. Any number whose prime factorization contains powers of primes with multiples of 3, then the number must be a perfect cube
26
Remainders
Guys are indeed following the SC thread. I hope people are also following this thread. Let me continue to post flashcards. REMAINDERS: (I) When 2 numbers are divided by same divisor and the remainders obtained are the same, THEN DIFFERENCE b/w 2 numbers is also divisible by that divisor. (II) When 2 positive numbers 'a' and 'b' are divided by the same divisor 'd' and remainders obtained are 'r1' and 'r2' respectively, THEN the remainders obtained when a+b is divided by d will be r1+r2 Quote: NOTE: If r1+r2 >= d, compute (r1+r2) - d as the remainder. (III) When 2 positive numbers 'a' and 'b' are divided by the same divisor 'd' and the remainders obtained are 'r1' and 'r2' respectively, THEN the remainders obtained when a*b is divided by d will be r1*r2 Quote: NOTE: If r1*r2 >= d, compute (r1*r2) / d as the remainder. TAKEAWAY: A remainder can NEVER be greater than or equal to the divisor.
27
How to find REMAINDER for LARGE POWERS of numbers:
There are 2 ways to do so: 1. Pattern Method: Example: What is the remainder when 2^56 / 7 ? ``` Solution: Remainder when 2^1 is divided by 7 is 2 Remainder when 2^2 is divided by 7 is 4 Remainder when 2^3 is divided by 7 is 1 Remainder when 2^4 is divided by 7 is 2 --> Repeats again. ``` The remainder repeats after 3 steps i.e. in the 4th step. Now, Divide the power (or index) by 3 (no of steps after which remainder repeats) and compute a new remainder. 56 % 3 --> 2 (remainder) Now, raise the base (2) to the power 2 (new remainder). 2^2 % 7 --> 4. Thus, 4 is the remainder when 2^56 / 7. 2. Remainder Theorem Method: (NOT RECOMMENDED unless clear) Example: What is the remainder when 2^51 / 7 ? Solution: 2^51 can be changed to (2^3)^17. 7 can be changed to (8-1) OR (2^3 - 1) Substitute 'x' in place of 2^3, x^17 / (x-1) Remainder is f(1). Substitute 1 in 'x', Remainder is 1. Thus, 1 is the remainder when 2^51 / 7.
28
Simple Facts:
Simple Facts: a^n - b^n: 1. ALWAYS divisible by a-b 2. If n is even, it is divisible by a+b 3. If n is odd, it is NOT divisible by a+b a^n + b^n: 1. NEVER divisible by a-b 2. If n is odd, it is divisible by a+b 3. If n is even, it is NOT divisible by a+b
29
Multiples of N
Playing with Multiples of N: (I) If you add/subtract multiples of number 'N', the result is also a multiple of 'N'. Examples: 35+21 = 56 [Multiple of 7] 20-15 = 5 [Multiple of 5] TAKEAWAY: In general, if N is a divisor of both x and y, then N is a divisor of both x+y and x-y. (II) If you add/subtract a multiple of N to/from a non-multiple of N, the result is a non-multiple of N. Example: 9-5 = 4 [(Multiple of 3) - (Non-Multiple of 3) = (Non-multiple of 3)] (III) If you add/subtract 2 non-multiples of N, the result could either be a multiple or a non-multiple of N. Examples: 19+13 = 32 [(Non-Multiple of 3) - (Non-Multiple of 3) = (Non-multiple of 3)] 19+14 = 33 [(Non-Multiple of 3) - (Non-Multiple of 3) = (Multiple of 3)] EXCEPTION: When N = 2, two odds always sum to an even number (Multiple of 2).
30
GCF Facts
GCF Facts: 1. GCF of integers 'm' and 'n' CANNOT be larger than the difference between 'm' and 'n'. Assume that GCF of m and n is 12. m and n are both multiples of 12. Consecutive multiples of 12 are 12 units apart from each other on the number line. Therefore, m and n CANNOT be less than 12 units apart. 2. Consecutive multiples of n have a GCF of n. 4 and 8 are multiples of 4. Thus 4 is a common factor of both the numbers. 4 and 8 are exactly 4 units apart from each other on the number line. Thus, 4 is the greatest common factor (GCF) of 4 and 8. That is why GCF of any 2 consecutive numbers is ALWAYS 1 as both are multiples of 1. _________________ Download GMAT Math and CR questions with Solutions from Instructors and High-scorers: http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html ----------- GO GREEN..! GO VEG..! Daily Quote: “Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”
31
Consecutive integers
Consecutive Integers: 1. In any set of 3 consecutive integers, one of the integers is ALWAYS divisible by 3. 2. (x-1) * x * (x+1) is divisible by 8, if x is odd. (x-1) and (x+1) are even. (x-1) is atleast divisible by 2 (Since it always has a 2). As (x-1) and (x+1) are consecutive multiples of 2, (x+1) will have an additional 2 units apart from another 2 (i.e. 4). So, (x+1) is divisible by 4. Thus, (x-1) * x * (x+1) is divisible by 8
32
Factor foundation rule
Factor Foundation Rule: "If x is a factor of y and y is a factor of z, then x is a factor of z". In other words, any integer is divisible by all of its factors and it is also divisible by all of the factors of its factors. Example: Consider 36. 36 can be broken down into 12 * 3 12 can be broken down into 2^2 * 3 So, 2^2 is also a factor of 36
33
Divisibility rules- odds and even
Divisibility Rules - ODDs and EVENs: In general, Odd integer divided by any other integer CANNOT produce an even integer. Odd integer divided by an even integer CANNOT produce and integer. _________________ Download GMAT Math and CR questions with Solutions from Instructors and High-scorers: http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html ----------- GO GREEN..! GO VEG..! Daily Quote: “Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”
34
Exponent rule
Exponent Rules: 1. x^a * x^b = x^(a+b) 2. (a^x)^y = a^xy = (a^y)^x 3. a^x * b^x = (ab)^x 4. x^(a/b) = b root (x^a) = (b root (x))^a 5. x^a / x^b = x^(a-b) 6. (a/b)^x = a^x / b^x 7. x^-a = 1 / x^a 8. a^x + a^x + a^x = 3 * a^x
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Properties of root
Properties of Roots: 1. n root(x) / n root(y) = n root(x/y) 2. n root(x) * n root(y) = n * root(xy) 3. b root(x^a) = (b root(x)) ^ a = x^(a/b)
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Evenly spaced set
Evenly Spaced Sets: i. Mean and Median are equal ii. Mean and Median of the set = Average (First + Last terms) iii. Sum of the elements = Mean of the set * Number of elements in the set
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Counting integers
Counting Integers: For Consecutive integers: (Last term - First term + 1) For Consecutive multiples: [(Last term - First term) / increment] + 1, where increment is the difference between each consecutive term in the set.
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Average of odd and even numbers
Average of an ODD number of consecutive integers will ALWAYS be an integer. E.g: Average of 5 numbers (1,2,3,4,5) is 3 (an integer). Average of an EVEN number of consecutive integers will NEVER be an integer. E.g: Average of 5 numbers (1,2,3,4,5,6) is 3.5 (NOT an integer).
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Consecutive integers
The product of "k" consecutive integers is ALWAYS divisible by K! Example: Consider numbers 5,6,7,8,9,10 (6 consecutive numbers). Their PRODUCT is divisible by 6! 5*6*7*8*9*10 / 6*5*4*3*2*1 = 210.
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Set of consecutive integers
For any set of consecutive integers with an ODD number of items, the sum of all integers is ALWAYS a multiple of the number of items. Example: Consider 1,2,3,4,5. Number of items is 5. 1+2+3+4+5 = 15. 15 is a multiple of 5 (the number of items). For any set of consecutive integers with an EVEN number of items, the sum of all integers is NEVER a multiple of the number of items. Example: Consider 1,2,3,4,5,6. Number of items is 6. 1+2+3+4+5+6 = 21. 21 is NOT a multiple of 6 (the number of items). --[You can try out any set of numbers to see whether the rule holds]--
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Another GCF
Another GCF: If "n" is NOT divisible by "k" and GCF (n,k) = "z" THEN the remainder (When "n" is divided by "k") will be "z" Example: Assume n=20, k=15. 20 is NOT divisible by 15. GCF (20,15) = 5. Remainder when 20 is divided by 15 is 5 (Which is the GCF(20,15)).
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Transmitting / non transmitting decimals ( very important)
Terminating/Non-Terminating decimals: (VERY IMPORTANT) If the fraction is completely simplified and the denominator only have 2's and 5's (or) only 2's or only 5's, then it is a terminating decimal. If the fraction is completely simplified and the denominator of the fraction has prime factors other than 2's and/or 5's , then it is a repeating/non-terminating decimal.
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Ruthless root
Ruthless ROOTS: If n is a POSITIVE integer, then the nth root of any number (> 1) will ALWAYS be > 1. Example: (Official Problem) What is root (4) + 3root (4) + 4root (4) approximately? root (4) = 2 3root (4) > 1 4root (4) > 1 So, root (4) + 3root (4) + 4root (4) is approximately > 4.
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(n^p - n)
Although this is not very important, it is fun to simply know this point. If p is a prime number, then for any integer n, (n^p - n) is ALWAYS divisible by p. Example: p = 3, n = 2 (2^3 - 2) is divisible by 3 OR 6 is divisible by 3 _________________ Download GMAT Math and CR questions with Solutions from Instructors and High-scorers: http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html ----------- GO GREEN..! GO VEG..! Daily Quote: “Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”
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If x>y.... | If x
If x > y, then x - y > 0 AND x = y + k (Where k > 0) If x < y, then x - y < 0 AND [x = y - k (Where k > 0) OR x = y + k (Where k < 0)]
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If x > y and 'k' is a real number,
If x > y and 'k' is a real number, then x + k > y + k (Irrespective of the value of k)
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X,y and k relationship
I. If x > y and k > 0, then k*x > k*y AND x/k > y/k ``` II. If x > y and k < 0, then k*x < k*y AND x/k < y/k [-- Inequality reverses when k < 0 --] ```
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If x and y are both POSITIVE/NEGATIVE AND x > y THEN .......
(IMPORTANT) If x and y are both POSITIVE/NEGATIVE AND x > y THEN 1/x < 1/y Examples: 1. x = 3, y = 2, x > y So, 1/3 < 1/2 2. x = -2, y=-3, x > y So, -1/2 < -1/3
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If a1 > b1, a2 > b2, ... an > bn,
If a1 > b1, a2 > b2, ... an > bn, Then (a1 + a2 + ... an) > (b1 + b2 + ... bn) If a1 > b1, a2 > b2, ... an > bn, Then (a1 * a2 * ... an) > (b1 * b2 * ... bn) --- [ONLY When all values of a and b are POSITIVE]
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If ax^2+bx+c > 0, where a > 0
If ax^2+bx+c > 0, where a > 0, Then x DO NOT LIE between xL and xU (xL and xU are lower and upper limits of x respectively) If ax^2+bx+c < 0, where a > 0, Then x LIES between xL and xU (xL and xU are lower and upper limits of x respectively)
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i. If | X1 | > | X2 |, Then .......... | ii. If | X1 | < | X2 |, Then .....
i. If | X1 | > | X2 |, Then X1 > X2 OR - X1 > X2 | ii. If | X1 | < | X2 |, Then X1 < X2 OR - X1 < X2
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|x - a| < r, where a is a positive real number and a is a fixed real number, then a-r < x < a+r |x - a| > r, where a is a positive real number and a is a fixed real number, then ..........l
|x - a| < r, where a is a positive real number and a is a fixed real number, then a-r < x < a+r |x - a| > r, where a is a positive real number and a is a fixed real number, then x < a-r (OR) x > a+r
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Golden rule
Golden Rule: When a NEGATIVE value is multiplied both sides in an INEQUALITY, then the INEQUALITY SIGN reverses/flips. Example: -1/x > -3/14 When -ve is multipled both sides, 1/x < 3/14 [-- Not only does the number sign changes, but also the inequality sign --] _________________ Download GMAT Math and CR questions with Solutions from Instructors and High-scorers: http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html ----------- GO GREEN..! GO VEG..! Daily Quote: “Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”
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Always remember
Always Remember to: Plug in the solutions back to the absolute equations to see whether it satisfies, whenever you get multiple solutions while solving absolute equations (A very useful takeaway especially for solving DS questions) Example: Solve, |x| = 3x - 2. i. x = 3x - 2 2x = 2 x = 1. ii. -x = 3x - 2 4x = 2 x = 1/2. You get 2 solutions 1 and 1/2. DO NOT CONCLUDE that there are 2 solutions and move on. Plug in both solutions back to the original equation (|x| = 3x - 2) and see whether the equation holds. Sub. x = 1 in |x| = 3x - 2, Equation SATISFIES! Sub. x = 1/2 in |x| = 3x - 2, Equation DOES NOT SATISFY! Hence, x = 1 is the ONLY solution.
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Another point to remember
Another point to remember: Example: -1/10 < n < 1/10 After taking reciprocal of n, FLIP SIGNS! -10 > n > 10 _________________ Download GMAT Math and CR questions with Solutions from Instructors and High-scorers: http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html ----------- GO GREEN..! GO VEG..! Daily Quote: “Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”
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|x+y| < |x| + |y|
Example: |x+y| < |x| + |y| 1. If x and y have SAME SIGNS, then both sides will be equal. 2. If x and y have DIFFERENT SIGNS, then right side will be greater. _________________ Download GMAT Math and CR questions with Solutions from Instructors and High-scorers: http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html ----------- GO GREEN..! GO VEG..! Daily Quote: “Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life” Last edited by papgust on Wed Jun 16, 2010 9:44 am; edited 1 time in total
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Weighted averages
Weighted Averages: In weighted averages involving 2 groups, if you have any 3 of the following, you can ALWAYS FIND the fourth: - Average of Group #1 - Average of Group #2 - Overall Average - Ratio of Group #1 to Group #2 Solve this GMATPrep question to understand the trick! NOTE: This takeaway is especially useful for Data Sufficiency questions for which you sometimes do not need to solve for the answer. When you see any 3 points in the question and statement and when you are asked to calculate the fourth point, you DON'T need to solve actually. You can straight away conclude that the statements are sufficient. Courtesy: Ian Stewart, GMAT Expert.
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Way to solve weighted averages
A way to solve "Weighted Averages": Avg Group 'A' ........... X ............ Overall Avg ............ Y .................... Avg Group 'B' where, X is the distance b/w Group 'A' and Overall Avg and Y is the distance b/w Group 'B' and Overall Avg. A/B = Y/X ``` Example: A class writes a math test and the overall average is 75%. If the girls average 85% and the boys average 70%, what fraction of the class is boys? ``` Boys (70%) ............ 5 ............... Overall Avg (75%) ............. 10 ............... Girls (85%) ``` B/G = 10 / 5 = 2/1 B:G = 2:1 ``` The class is 2/(2+1) OR 2/3 boys
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Standard deviation
Standard Deviations: In order to calculate Standard Deviations, you need the following data points, - Population of the Data set [-- Total Number of elements in the set --] - Mean of the Data set - Data points/elements themselves EXCEPTION: If the set has CONSECUTIVE INTEGERS, you DO NOT need to know the Data points/elements. But knowing NUMBER OF TERMS (or POPULATION) is must. Courtesy: Ian Stewart, GMAT Expert. _________________ Download GMAT Math and CR questions with Solutions from Instructors and High-scorers: http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html ----------- GO GREEN..! GO VEG..! Daily Quote: “Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life” Last edited by papgust on Mon Jun 14, 2010 7:42 pm; edited 1 time in total
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Important points
--[ IMPORTANT POINTS ]-- If a constant percentage is added/subtracted from each term of the set, then the new SD is also added/subtracted by that constant percentage. If a constant is added/subtracted from each term of the set, then the new SD remains constant.
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Adding data to a group and it's impact on mean
If you add data that is farther from the mean, then SD will increase. If you add data that is closer to the mean, then SD will decrease.
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Multiplication impact on SD
If each term in a set is multiplied by a number greater than 1, then SD will increase. If each term in a set is multiplied by a number less than 1, then SD will decrease. ---- If each term in a set is divided by a number greater than 1, then SD will decrease. If each term in a set is divided by a number that lies between 0 and 1, then SD will increase.
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Does the change of signs in all terms of a set affects SD??
Does the change of signs in all terms of a set affects SD?? Answer is a simple NO. Change of signs has no bearing on the SD.
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Calculating average speed
Calculating Average Speeds: There are 2 ways to calculate average speeds, 1. Typical Distance-Time formula: (Traditional way - RECOMMENDED) Assume S1 and S2 are the speeds of 2 trips. Distance d is the same for both the trips. Assume T as the total time taken for both the trips. Time taken = Distance / Speed. For this scenario, Total Time Taken = (Distance of Trip 1/Speed of Trip 1) + (Distance of Trip 2/Speed of Trip 2) ``` T = d/S1 + d/S2 T = (d*S1 + d*S2) / (S1*S2) T = d*(S1+S2) / (S1*S2) ``` Now calculate Average speed using Total Distance and Total Speed, Avg Speed = Total Distance / Total Time Taken Avg Speed = 2d / T .......... [2d is the distance of Trip 1 and 2] Sub. T = d*(S1+S2) / (S1*S2) in the equation, Avg Speed = 2d*(S1*S2) / d*(S1+S2) Avg Speed for the entire trip = 2*(S1*S2) / (S1+S2) .... [Cancelling out d] NOTE: You don't need to solve this to get average speed. You can apply this formula straight away. Just wanted to show you how the formula is arrived. 2. A cool shortcut: (UNORTHODOX Approach) Assuming Distance is the same for both the trips. S1 and S2 are the speeds of Trips 1 and 2. Step i: Form S1:S2 and reduce the ratio as much as possible to S3:S4. Then, add the parts of ratio -- S3+S4 Step ii: |S1-S2| / (S3+S4) = N Step iii: S1 + (S3*N) = Avg Speed for the entire trip. Example: Calculate Avg Speed of 20 mph and 200 mph Step i: 20:200 = 1:10 = 1+10 = 11 parts. Step ii: |20-200| / 11 = 16.36 Step iii: 20 + (1*16.36) = 36.36 mph -- [Average Speed for the entire trip].
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Ratio rules
RATIOS: 1. a:b = a/b Value of ratio remains unchanged when multiplied/divided, a/b = a*m / b*m = (a/m) / (b/m) 2. If a/b = c/d = e/f, then each of these ratios is equal to a+c+e / b+d+f 3. If a/b = c/d, then b/a = d/c [Invertendo Rule] 4. If a/c = b/d, then a/b = c/d [Alternendo Rule] 5. If a/b = c/d, then a+b / b = c+d / d [Componendo Rule] 6. If a/b = c/d, then a-b / b = c-d / d [Dividendo Rule] 7. The compound of a:b and c:d is a*c:b*d 8. If a:b :: c:d, then a*d = b*c 9. If a:b:c :: d:e:f, then a/d = b/e = c/f = K (a constant)
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Arithmetic progression
Arithmetic Progressions: i. To find a term, tn = a + (n-1)*d, where a is the first term of the series, n is the number of terms in the series and d is the constant difference between any 2 consecutive terms of the series. ii. Average of the AP series = (First Term + Last Term)/2 --[This is applicable to the concept - Evenly Spaced Sets. Please refer this post]-- iii. Sum of the AP series, Sn = n/2 * [2a + (n-1)*d] OR Sn = n/2 * [a + tn]
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If constant number is added or subtracted from AP
If a constant number (C) is added/subtracted from each term of an AP, then i. Resulting sequence is also AP. ii. Common difference (d) remains constant. iii. Sum of the NEW series = Sum of the OLD series + (n * C) [where n -> No. of terms in the series] If a constant number (C) is multiplied/divided by each term of an AP, then i. Resulting sequence is also AP. ii. Common difference (d) = C * Old Common difference iii. Sum of the NEW series = C * Sum of the OLD series
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If a1, a2, ... an and b1, b2, ... bn are two AP's A and B, then
If a1, a2, ... an and b1, b2, ... bn are two AP's A and B, then I. (a1+b1, a2+b2, ... an+bn) is ALSO an AP. II. Common Difference (d) = Sum [d(A) + d(B)] --------- Where d(A) is the common difference of series A and d(B) is the common difference of series B
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Geometric Progressions:
Geometric Progressions: i. To find a term, tn = a * r^(n-1), where a is the first term of the series, n is the number of terms in the series and r is the common ratio of the series. ii. Sum of the GP series, Sn = a * (r^n - 1) / r-1 .....[ If r > 1 ] Sn = a * (1 - r^n) / 1-r .....[ If r < 1 ] iii. Sum of the indefinite GP (A series that doesn't have an end) Sn = a / (1-r)
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If a fixed non-zero constant is multiplied with each term of GP, then
If a fixed non-zero constant is multiplied with each term of GP, then i. Resulting sequence is also GP ii. Same Common ratio. _________________ Download GMAT Math and CR questions with Solutions from Instructors and High-scorers: http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html ----------- GO GREEN..! GO VEG..! Daily Quote: “Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”
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If a1, a2, ... an and b1, b2, ... bn are two GP's A and B, then
If a1, a2, ... an and b1, b2, ... bn are two GP's A and B, then I. (a1*b1, a2*b2, ... an*bn) is ALSO a GP. II. Common Ratio (r) = r(A) * r(B) --------- Where r(A) is the common ratio of series A and r(B) is the common ratio of series B
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Set theory
Well, thanks rockeyb once again! Sorry for a really long break. Got badly held up with work this week. Here we go! -- SET THEORY: For 3 set questions, there are 2 formulae that you can use. Picture a Venn diagram; the first formula is just the sum of all of the various parts: 1. True # of objects = (# only A) + (# only B) + (# only C) + (# only AB) + (# only AC) + (# only BC) + (# only ABC) The second formula is the one we use more often: 2. True # of objects = (total # A) + (total # B) + (total #C) - (# only AB) - (# only AC) - (# only BC) - 2(# ABC) [Note that, technically, we should add a "+ (# with none of ABC)" to the end of each equation, but a 3 set question on the GMAT that had a "none" component is not often seen.] We can simplify the second equation to: True # of objects = (sum of total characteristics) - (sum of doubles) - 2(triples) To understand why we have to subtract the doubles once and the triples twice, again picture a Venn diagram. If an object is in the AB portion of the diagram, it's already been counted in the A circle and the B circle. In other words, it's been counted twice. To get a true count, therefore, we must subtract it once. If an object is in the ABC portion of the diagram, it's already been counted in the A circle, the B circle AND the C circle. In other words, it's been counted three times. To get a true count, therefore, we must subtract it twice. Here's the primary principle we're using: Every object should be counted exactly once. Courtesy: Stuart Kovinsky, GMAT Expert.
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Examples: In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products?
Examples: In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products? Let's say there's 100 people, just to use numbers instead of percents. Since 85% like at least one of 3 products, we'll use 85 as our base number. Since we have "at least one" information, we need to use the second formula: True # of objects = (sum of total characteristics) - (sum of doubles) - 2(triples) ``` plugging in what we know: 85 = 50 + 30 + 20 - doubles - 2(5) 85 = 100 - 10 - doubles 85 = 90 - doubles doubles = 5 ``` The question is what % like more than 1, so we want to solve for: doubles + triples = 5 + 5 = 10 So, 10% is the final answer.
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Examples: (Contd..) There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German.
Examples: (Contd..) There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German. We use the exact same formula: True # of objects = (sum of total characteristics) - (sum of doubles) - 2(triples) ``` 70 = (40 + 30 + 35) - (doubles) - 2(15) 70 = 105 - doubles - 30 70 = 75 - doubles doubles = 5 ```
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Positive and negative slope
POSITIVE/NEGATIVE slope: (A very important concept to know) A line with a POSITIVE slope: i. Passes through 1st and 3rd quadrant definitely. ii. Passes through the 2nd quadrant (IF y-intercept is POSITIVE). iii. Passes through the 4th quadrant (IF y-intercept is NEGATIVE). A line with a NEGATIVE slope: i. Passes through 2nd and 4th quadrant definitely. ii. Passes through the 1st quadrant (IF y-intercept is POSITIVE). iii. Passes through the 3rd quadrant (IF y-intercept is NEGATIVE).
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Distance between two points
Distance b/w 2 points: Distance b/w points (x1,y1) and (x2,y2) = ROOT [(x1-x2)^2 + (y1-y2)^2] _________________ Download GMAT Math and CR questions with Solutions from Instructors and High-scorers: http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html ----------- GO GREEN..! GO VEG..! Daily Quote: “Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”
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Co-ordinates of point 'P' dividing the join of the two points (x1,y1) and (x2,y2) INTERNALLY in the ratio m:n is,
Co-ordinates of point 'P' dividing the join of the two points (x1,y1) and (x2,y2) INTERNALLY in the ratio m:n is, P(x,y) = [(m*x2 + n*x1) / m+n, (m*y2 + n*y1) / m+n] Co-ordinates of point 'P' dividing the join of the two points (x1,y1) and (x2,y2) EXTERNALLY in the ratio m:n is, P(x,y) = [(m*x2 - n*x1) / m-n, (m*y2 - n*y1) / m-n] NOTE: If ratio m:n has both m and n as POSITIVE, then its INTERNAL If ratio m:n has one of m and n as NEGATIVE, then its EXTERNAL
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Mid point of a line
Mid-point of a line: Mid-point of a segment joining (x1,y1) and (x2,y2) is (x1+x2 / 2, y1+y2 / 2)
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Centroid
Centroid (G) = (x1+x2+x3 / 3, y1+y2+y3 / 3) NOTE: Centroid divides each median in the ratio 2:1. _________________ Download GMAT Math and CR questions with Solutions from Instructors and High-scorers: http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html ----------- GO GREEN..! GO VEG..! Daily Quote: “Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”
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Area of a triangle
Area of a triangle: 1/2 * [x1*(y2-y3) + x2*(y3-y1) + x3*(y1-y2)] where (x1,y1), (x2,y2) and (x3,y3) are 3 end-points of a triangle.
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If the origin (0,0) shifts to a point (h,k),
If the origin (0,0) shifts to a point (h,k), then old co-ordinates of point P(x,y) changes to (x-h, y-k)
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If (x1,y1), (x2,y2), (x3,y3) and (x4,y) are the vertices of a PARALLELOGRAM,
If (x1,y1), (x2,y2), (x3,y3) and (x4,y) are the vertices of a PARALLELOGRAM, then ``` x1+x3 = x2+x4 AND y1+y3 = y2+y4 ```
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To prove end-points A,B,C, and D are the vertices of:
To prove end-points A,B,C, and D are the vertices of: i. A PARALLELOGRAM: * Show that diagonals AC and BD bisect each other. ii. A RHOMBUS: * Show that diagonals AC and BD bisect each other. * Show that a pair of adjacent sides are equal. iii. A SQUARE: * Show that diagonals AC and BD bisect each other. * Show that a pair of adjacent sides are equal. * Show that two diagonals AC and BD are equal. iv. A RECTANGLE: * Show that diagonals AC and BD bisect each other. * Show that two diagonals AC and BD are equal.
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SLOPE +ve OR -ve:
SLOPE +ve OR -ve: Slope is tan x where, x is angle of inclination. i. Slope is POSITIVE when x is acute. ii. Slope is NEGATIVE when x is obtuse.
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Finding Slope:
Finding Slope: Slope can be calculated in 2 ways, i. m = y1-y2 / x1-x2 ii. m = - Coeff. of x / coeff. of y (OR) m = -a / b [From ax + by + c = 0]
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Parallel and perpendicular
If two lines l1 and l2 are PARALLEL, then m1 = m2 ....... [Where, m1 and m2 are the slopes of lines l1 and l2 respectively] If two lines l1 and l2 are PERPENDICULAR, then m1 * m2 = -1....... [Where, m1 and m2 are the slopes of lines l1 and l2 respectively]
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Slope of horizontal and vertical lines
Slope of all HORIZONTAL LINES is 0..... [Lines parallel to x-axis] Slope of all VERTICAL LINES is undefined..... [Lines parallel to y-axis. Angle of inclination of ANY line parallel to y-axis is 90 or 270 degrees]
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Finding equation of a line
Finding EQUATION of a LINE: i. PASSING through two points (x1,y1) and (x2,y2): y-y1 / y2-y1 = x-x1 / x2-x1 ii. y = mx+c, when slope (m) and y-intercept (c) are given. iii. y-y1 = m*(x-x1), when slope (m) and one of the points (x1,y1) on the line are given. iv. x/a + y/b = 1, when x-intercept (a) and y-intercept (b) are given.
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Angle of inclination b/w 2 SLOPES:
Angle of inclination b/w 2 SLOPES: Angle (x) between two slopes, tan x = |(m1-m2) / (1+m1*m2)| _________________ Download GMAT Math and CR questions with Solutions from Instructors and High-scorers: http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html ----------- GO GREEN..! GO VEG..! Daily Quote: “Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”
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Length of perpendicular from a point
Length of perpendicular from point (x1,y1) to the line ax+by+c = 0 is |ax1+by1+c / root(a^2+b^2)|
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Distance between two lines [PARALLEL lines] ax+by+c1 = 0 and ax+by+c2 = 0 is
Distance between two lines [PARALLEL lines] ax+by+c1 = 0 and ax+by+c2 = 0 is |c1-c2 / root(a^2+b^2)|
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Types of angles
Types of angles: Figure 1: 1. Adjacent angles: Any 2 angles that share a common side separating the 2 angles and that share a common vertex. E.g.: | 1 and | 2 are adjacent angles. [See Figure 1] 2. Vertical angles: Any 2 angles that are not adjacent angles. Vertical angles are EQUAL in measure. E.g.: | 1 and | 3 are vertical angles. [See Figure 1] 3. Complementary angles: Any 2 angles whose sum is 90 degrees. Complementary angles NEED NOT be adjacent to each other. 4. Supplementary angles: Any 2 angles whose sum is 180 degrees.[img][/img] Figure 2: L and M are parallel lines. T is a traversal. 5. Corresponding Angles: Angles that appear to be in the same relative position in each group of four angles. Corresponding angles are EQUAL when two parallel lines are cut by a traversal. E.g.: | 1 and | 5 are corresponding angles. [See Figure 2] 6. Alternate Interior Angles: Angles within the lines being intersected, on opposite sides of the traversal, and are not adjacent. Alternate interior angles are EQUAL when two parallel lines are cut by a traversal. E.g.: | 4 and | 6 are alternate interior angles. [See Figure 2] 7. Alternate Exterior Angles: Angles outside the lines being intersected, on opposite sides of the traversal, and are not adjacent. Alternate exterior angles are EQUAL when two parallel lines are cut by a traversal. E.g.: | 1 and | 7 are alternate interior angles. [See Figure 2] 8. Consecutive Interior Angles: Angles are same-side interior angles. Consecutive interior angles are SUPPLEMENTARY when two parallel lines are cut by a traversal. E.g.: | 4 and | 5 are consecutive interior angles. [See Figure 2] 9. Consecutive Exterior Angles: Angles are same-side exterior angles. Consecutive exterior angles are SUPPLEMENTARY when two parallel lines are cut by a traversal. E.g.: | 1 and | 8 are alternate interior angles. [See Figure 2]
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Line segments in triangles
Lines segments in Triangles: * ALTITUDES: Altitudes are the perpendicular segments from a vertex to the opposite sides. The three lines containing the altitudes intersect in a single point, which may or may not be inside the triangle. * MEDIANS: A Median is the line segment drawn from a vertex to the mid-point of its opposite side. The three medians meet in one point inside the triangle. * ANGLE BISECTOR: Angle bisector is a segment drawn from a vertex that bisects the vertex angle. The three angle bisectors meet in one point inside the triangle. TAKEAWAY: Altitude drawn from the vertex angle can be proven to be a median as well as an angle bisector in an ISOSCELES triangle. _________________ Download GMAT Math and CR questions with Solutions from Instructors and High-scorers: http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html ----------- GO GREEN..! GO VEG..! Daily Quote: “Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”
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An equiangular quadrilateral vs equilateral. | --]
An equiangular quadrilateral DOES NOT have to be equilateral. An equilateral quadrilateral DOES NOT have to be equiangular. [--- Unlike Equilateral Triangle --]
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Regular polygon
Regular Polygons: * When a polygon is BOTH equilateral and equiangular. * Sum of INTERIOR angles of a convex polygon with 'n' sides = (n-2) * 180 * Sum of EXTERIOR angles of a convex polygon = 360 degrees.
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How to prove a figure as PARALLELOGRAM?
How to prove a figure as PARALLELOGRAM? 5 ways: * If both sides of opposite sides of a quadrilateral are EQUAL. * If both pairs of opposite angles of a quadrilateral are EQUAL. * If all pairs of consecutive angles of a quadrilateral are SUPPLEMENTARY. * If one pair of opposite sides of a quadrilateral is both EQUAL and PARALLEL. * If the diagonals of a quadrilateral BISECT each other. TAKEAWAY: A diagonal of a Parallelogram DIVIDES it into 2 congruent triangles.
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Isosceles Trapezoids:
Isosceles Trapezoids: * If the legs are EQUAL. * Base Angles are EQUAL. * Diagonals are EQUAL. * Median of any trapezoid: 1. Is parallel to both bases. 2. Has length = 1/2 * (Sum of bases)
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Regular polygon
Regular Polygons: * One point in its interior that is equidistant from its vertices is called the center of the regular polygon. * An Apothegm is a line segment that goes from the center and is perpendicular to one of the polygon's sides. Perimeter (regular n-gon) = n * s [n--> no of sides and s--> length of a side] Area (regular n-gon) = 1/2 * a * p [a--> Apothegm length and p--> perimeter of regular n-gon]
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Similar Polygons:
Similar Polygons: * Two polygons with the same shape. * When two polygons are similar, then the following MUST be true. i. Corresponding angles are EQUAL. ii. The ratios of pairs of corresponding sides must all be EQUAL.
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Similar Triangles:
Similar Triangles: If two triangles are similar, then the ratio of any two corresponding segments (such as altitudes, medians, angle bisectors) EQUALS the ratio of any two corresponding sides. Example: If ∆ QRS ~ ∆ TUV, then QR/TU = RS/UV = QS/TV. According to the theorem, Length of altitude RA / Length of altitude UD = QR / TU Length of median QB / Length of median TE = QR / TU Length of bisector CS / Length of bisector FV = QR / TU
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Perimeters and Areas of Similar Triangles:
Perimeters and Areas of Similar Triangles: When two triangles are similar, the reduced ratio of any two corresponding sides is called the scale factor of the similar triangle. i. If two similar triangles have a scale factor of a:b, the the ratio of their perimeters is a:b. Example: 6/3 = 8/4 = 10/5 ==> 2/1 (or) 2:1 (Perimeter) ii. If two similar triangles have a scale factor of a:b, then the ratio of their areas is a^2 : b^2. Example: Area of ∆ ABC / Area of ∆ DEF = 24 / 6 = 4 / 1 (or) 4:1 4:1 is nothing but a^2:b^2 (or) 2^2 : 1^2. _________________ Download GMAT Math and CR questions with Solutions from Instructors and High-scorers: http://www.beatthegmat.com/download-gmat-questions-with-expert-solutions-t59366.html ----------- GO GREEN..! GO VEG..! Daily Quote: “Stop feeling sorry for the Butcher if you had to go Veg. The butcher can find another job but the poor animal cannot get back its life”
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Relationship between square, rectangle and a quadrilateral
A rectangle is a special quadrilateral | A square is a special rectangle.
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What is the short cut method to calculate combinations
C(8,3) = take first three values of 8 factorial / 3 factorial = 8x7x6/ 3!

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