The distances are in a ratio of
 to 5 to 2, or 1 to 15 to 6. The swim distance
represents
 of the total distance.
The correct answer is (B).
 60. (D) 10 ≤ q ≤ 55: The problem does not specify the extent to which each person paints
over the area that was already painted by the previous person. For instance, when
Jamaica painted 70 square feet (sq ) of the wall, she might have painted over the
entire 60 sq already painted by Nelson, or she might have painted over only some
of it.
However, by the time Bryan finished painting, the entire wall had been painted with
at least one coat of paint, and some of it might have been painted with as many as
three coats. The question asks how much of the wall had three coats, and the
answer choices represent ranges from minimum to maximum area.
First, make q as large as possible. In order to do so, the amount of paint used on the
remainder of the wall must be minimized. The area q will be as large as possible
when the remaining 100 – q square foot area is covered in only a single coat.


The entire amount of paint applied by all three painters is 60 + 70 + 80 = 210 sq . If
q sq of the wall has three coats of paint, and the remaining 100 – q has only one
coat, the following is true:
3(area with 3 coats) + 1(area with 1 coat) = total paint applied
Turn this into an equation and solve for q.
The greatest possible value of q is 55 sq . Therefore, answer (D) must be correct.
It is also possible to determine the greatest possible value of q by Working
Backwards from the choices, testing their extreme values using pictures. If the
painting is done as shown below, then 55 sq of the wall will have three coats of
paint and the rest will have only one coat.


To find the smallest possible number of square feet with three coats, do the
opposite: Assume that the rest of the wall has as much paint as possible. In that
case, the remaining 100 – q sq of wall will have two coats of paint, so the following
will be true:
The smallest possible value of q is 10. Here is another way to visualize the
minimization of the three-coat area.
The correct answer is (D).


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GMAT QUANTITATIVE REVIEW


GMAT STRATEGY/TIPS
DO NOT assume that a number is an integer unless explicitly stated in the problem or if the object of a problem has a physical restriction of being divided (such as votes, cards, pencils etc.)
Example: if p < q and p < r, is pqr < p?
Statement (1): pq < 0
Statement (2): pr < 0
Statement (1) INSUFFICIENT: We learn from this statement that either p or q is negative, but since we know from the question that p < q, p must be negative. To determine whether pqr < p, let's test values for p, q, and r. Our test values must meet only 2 conditions: p must be negative and q must be positive. However, r could be either positive or negative, and hence, the product could be either negative (may or may not be less than p) or positive (greater than p)
Statement (2) INSUFFICIENT: We learn from this statement that either p or r is negative, but since we know from the question that p < r, p must be negative and r must be positive. However, q could be either positive or negative.
If we look at both statements together, we know that p is negative and that both q and r are positive. To determine whether pqr < p, let's test values for p, q, and r. Our test values must meet 3 conditions: p must be negative, q must be positive, and r must be positive. For instance, if p = -2 , q = 10, r = 5; pqr = -100, which is less than p. Taking another example, if p = -2 , q = 7, r = 4; pqr = -56, which is also less than p. Therefore, at first glance, it may appear that we will always get a "YES" answer. But don't forget to test out fractional (decimal) values as well. The problem never specifies that p, q, and r must be integers. If p = -2, q = 0.3 and r = 0.4, pqr = -0.24, which is NOT less than p. Hence, the correct answer is E – Both statements together are insufficient


ers. If p = -2, q = 0.3 and r = 0.4, pqr = -0.24, which is NOT less than p. Hence, the correct answer is E – Both statements together are insufficient

GMAT can also hide POSITIVE CONSTRAINTS. Sides of a square, number of votes, etc. will always be positive. When you have a positive constraint, you can:
Eliminate negative solutions from a quadratic function
Multiply or divide an inequality by a variable
Cross-multiply inequalities: x/y < y/x  x² < y²
Change an inequality sign for reciprocals: x<y; 1/x > 1/y

DO NOT ASSUME THE SIGN OF A VARIABLE
Example: Is T/S > F/G?
(1) T < S
(2) F > G
If one assumes the signs for all four variables to be positive, then T/S (an improper fraction) is NOT greater than F/G (an improper fraction). However, if negative signs are taken into account, we cannot conclusively ascertain whether T/S > F/G


GMAT STRATEGY/TIPS


GMAT STRATEGY/TIPS

64 is the only number (besides 1) under 100 that is the square of an integer and the cube of an integer
16 and 81 are the only numbers (again, besides 1) that are both the square of an integer and the square of a square
30 and 42 are the smallest integers with at least three prime factors
The only integers with exactly three factors are the squares of prime numbers
When odd number n is doubled, 2n has twice as many factors as n
When even number is doubled, 2n has 1.5 more factors as n
All prime numbers above 3 are of the form 6n – 1 or 6n + 1 , because all other numbers are divisible by 2 or 3
Xn – yn is always divisible by x – y and is divisible by x + y if n is even
Generally the last digit of 1! + 2! … + N!   can take ONLY 3 values:A. N=1 --> last digit 1;B. N=3 --> last digit 9;C. N=any other value --> last digit 3 (N=2 --> 1!+2!=3 and for N=4 --> 1!+2!+3!+4!=33,   the terms after N=4 will end by 0 thus not affect last digit and it'll remain 3).
Finding the number of powers of a prime number P in n!
n/p + n/p2 + n/p3 …. Till px < n
For example, what is the power of 2 in 25!?
25/2 + 25/4 + 25/8 + 25/16  12+ 6 + 3 + 1  22
Finding the power of non-prime in n!
For example, how many powers of 900 are in 50!?
Step 1: Determine the prime factorization the number: PF of 900 is 22*32*52
Step 2: Find the powers of these prime numbers in n! (refer above)
Power of 2 = 47, power of 3 = 22 and power of 5 = 12
Step 3: 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!


OVERLAPPING SETS


2 OVERLAPPING SETS – VENN DIAGRAMS
X
Formula: Total – Neither = A + B – X
Formula for elements who are AT LEAST one of the two groups: A + B – X
Formula for elements who are in ONLY one of the two groups: A + B – 2X
Only in group A = a
Only in group B = b
Both in group A & B = X
Neither = N


DOUBLE SET MATRIX – 2 MUTUALLY EXCLUSIVE OPTIONS FOR A DECISION
For GMAT problems involving only two categorizations or decisions, the most efficient tool is the Double Set Matrix: a table whose rows correspond to the options for one decision and whose columns correspond to the options for the other decision.
The rows should correspond to the mutually exclusive options for one decision. Likewise, the columns should correspond to the mutually exclusive options for the other. For instance, if a problem deals with students getting either right or wrong answers on problems 1 and 2, the columns should not be "problem 1" and "problem 2," and the rows should not be "right" and "wrong." Instead, the columns should list options for one decision-problem 1 correct, problem 1 incorrect, total-and the rows should list options for the other decision-problem 2 correct, problem 2 incorrect, total


3 OVERLAPPING SETS
Formula: Total – Neither = A + B + C – (a+b+c+2x)

Formula for elements who are AT LEAST one of the three groups: A + B + C – (a + b + c + 2x)
It is important to note that “a” represents elements that belong ONLY to A and C (not B)

Total = A + B + C – (a+ x + b + x + c + x) + x

Formula for elements in any two of the groups: a+ b + c

Note: Work from the inside out

Only A = A – a – b – x
Only B = B – b – c – x
Only C = C – a – c – x

x
a
c
b


3 OVERLAPPING SETS


STATISTICS


MEAN – THE AVERAGE
MEAN = MEDIAN in a set of consecutive integers
If each one of the given numbers is increased (or decreased) by K, their average is increased (or decreased) by k
If each one of some given numbers is multiplied by K, their average is multiplied by K
The sum of first “n” natural numbers is given by n (n + 1)/2
If the average of a few consecutive integers is 0, then either all the numbers are zero or there will be an odd number of integers
DETERMINING THE MEAN
Mean = Sum of terms/ number of terms
For an evenly spaced set (i.e., series where the difference between any two consecutive numbers is the same – 3, 6, 9), or a set of consecutive integers
Mean = (first term + last term) / 2
Mean = middle term (if the number of terms in the set is even, then the mean is the average of the middle two terms)
IMPORTANT: The mean of a set of consecutive integers will be an INTEGER only if the number of elements in the set is ODD or if the difference between/sum of the “consecutive” terms is divisible by 2. For example, in the case of 2,4,6 and 8, the average is 5

CHANGE IN MEAN
Change in mean = New term – Old mean/ New number of terms


is divisible by 2. For example, in the case of 2,4,6 and 8, the average is 5

CHANGE IN MEAN
Change in mean = New term – Old mean/ New number of terms

If a set currently has 9 terms and a mean value of 56, and you add a tenth term of 71, then the mean will increase by (71 - 56)/10 = 1.5
You can interpret this formula conceptually as taking the "excess" or "deficit" in the new term relative to the mean and redistributing that excess (or deficit) evenly to all of the terms in the set, including that new term. For instance, in the example above, the new term (71) exceeds the average by 71 - 56 = 15. When you add the new term to the set, those 15 "extra points" are distributed evenly among all ten data points, increasing each by 1.5


y 71 - 56 = 15. When you add the new term to the set, those 15 "extra points" are distributed evenly among all ten data points, increasing each by 1.5

RESIDUALS
Residuals = Data point – Mean
For any set, the residuals sum to zero. Alternatively, the positive residuals ("overs") and negative residuals ("unders") for any set will cancel out
If the mean of the set {97, 100, 85, 90, 94, 80, 92, x} is 91, what is the value of x?
You could certainly solve this problem with the traditional formula for averages, but you would waste a lot of time on arithmetic. Instead, just use the given mean of 91 to compute the residuals for all the terms except x: +6, +9, -6, -1, +3, -11, + 1. These residuals sum to +1.Therefore, x must leave a residual of -1, since all the residuals sum to zero. As a result, x is one less than the mean, or 90


MEDIAN – THE MIDDLE TERM

MEAN = MEDIAN in a set of consecutive integers. However, just because the mean and the median are equal does not mean that the set comprises of evenly spaced/consecutive integers
Ultimately, if your mean and median are the same, what it tells you is that your data is arranged symmetrically around the median. For every bit over the mean on one side, there's a corresponding bit under the mean on the other side, balancing it out. When they're different, that tells you that this symmetry has been broken

DETERMINING THE MEDIAN
To find the median, list the numbers in ascending or descending order, and find the middle term


this symmetry has been broken

DETERMINING THE MEDIAN
To find the median, list the numbers in ascending or descending order, and find the middle term

When number of terms N is large, then the following two formulae are useful for finding the middle term:
When N is odd, the middle term is (N+1)/2th term
When N is even, the median will be the average of the middle two terms, i.e., average of the (N/2)th term and the (N/2)th + 1 term
VALUE OF THE MEDIAN
When the number of items in a set of integers is odd, the median is an integer
When the number of terms in a set of integers is even, the median can only end with 0.5 as it is the average of the middle two terms. However, it is important to note that the median does not ALWAYS end in 0.5 when the number of terms in the set is even. For example, the median for 2,4,6,8,10,12 is 7
ADVANCED PROPERTIES OF MEDIANS
Median of combined set will always lie between the two medians or could be equal to one or both the medians of the two sets


CED PROPERTIES OF MEDIANS
Median of combined set will always lie between the two medians or could be equal to one or both the medians of the two sets

For example: a,b,c are integers and a<b<c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4)*b. The median of set Q is (7/8)*c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?

Since we the sets contain consecutive integers, the median = (largest + smallest term)/2
Therefore, equate (a+b)/2 to (3/4)*b for Set S and (b+c)/2 to (7/8)*c for Set Q. Similarly, equate (a+c)/2


WEIGHTED AVERAGE
(weight1*data point 1) + (weight2*data point 2) (weight 1 + weight 2)
SUM OF THE WEIGHTS
The sum of the weights must equal to 1. However, if the sum of the weights does not equal 1, we need to divide by the sum of the weights. If the "weights" are fractions or percentages, then the denominator of the weighted-average expression (sum of weights) will already be 1

Example:
If we put a weight of 2 on 20 and a weight of 3 on 30, we have to divide by 5 to get weights that sum to 1, as shown below:

[2(20) + 3(30)]/ 5 = (2/5)*20 + (3/5)* 30
Average = 26

ADVANCED PROPERTIES
A weighted average of only two values will fall closer to whichever value is weighted more heavily
For instance, if a drink is made by mixing 2 shots of a liquor containing 15% alcohol with 3 shots of a liquor containing 20% alcohol, then the alcohol content of the mixed drink will be closer to 20% than to 15%


containing 15% alcohol with 3 shots of a liquor containing 20% alcohol, then the alcohol content of the mixed drink will be closer to 20% than to 15%

The ratio that determines how to weight the averages of two or more subgroups in a weighted average ALSO REFLECTS the ratio of the distances from the weighted average to each subgroup’s average
If the average weight of the entire group was twice as close to the average weight of the men as it was to the average weight of the women, there must be twice as many men as women

In case two groups have an equal number of elements, then the weighted average of the two groups combined will be the average of the individual averages of the two groups


ave an equal number of elements, then the weighted average of the two groups combined will be the average of the individual averages of the two groups

DATA SUFFICIENCY QUESTIONS
Importantly for Data Sufficiency problems-you do not necessarily need concrete values for the weights in a weighted-average problem. Having just the ratios of the weights will allow you to find a weighted average. Simply write the ratio as a fraction, and use the numerator and the denominator as weights
A mixture of "lean" ground beef (10% fat) and "super-lean" ground beef (4% fat) contains twice as much lean beef as super-lean beef. What is the percentage of fat in the mixture?
The ratio of lean beef to super-lean beef is 2 : 1, and so we can use 2 as the weight for the lean beef and 1 as the weight for the super-lean beef
The percentage of fat in the mixture is [ (10%)*2} + (4%)*1 ]/ (2+ 1) = 24%/3 = 8%


MODE, RANGE AND STANDARD DEVIATION
Mode is the term that occurs most frequently. There may be no mode, or more than one mode in a given series

RANGE
Maximum value – minimum value
Range does not change when a constant value is either added to or subtracted from every term in a set
Range is multiplied or divided by that constant value which is used to multiply/ divide every term in a set
Range = 0 when all values are the same
Range is always >= 0
Range of a set has to be equal or greater than difference between any two numbers in the set


ge = 0 when all values are the same
Range is always >= 0
Range of a set has to be equal or greater than difference between any two numbers in the set

STANDARD DEVIATION
Formula: √ (term1 – average)2 + (term2 – average)2 + (term3 – average)2 / √number of terms
Rules
The set in which numbers are farther away from each other than the numbers in some other set will have greater value of standard deviation
A = {2,4,6,8,10}
B = {3,6,9,12,15}
In set B, the numbers are wider away from each other and hence will have greater value of standard deviation
Standard deviation does not change when a constant value is either added to or subtracted from every term in a set
Standard deviation is multiplied or divided by that constant value which is used to multiply or divide every term in a set
Taking the absolute value of the elements of a set will not change the standard deviation if all elements are positive or negative
If y=ax + b, and if the standard deviation of x series is ‘S’, then the standard deviation of y series will be a*s
The SD of any list is not dependent on the average, but on the deviation of the numbers from the average. So just by knowing that two lists having different averages doesn't say anything about their standard deviation - different averages can have the same SD
If mean = maximum value it means that all values are equal and SD is 0


ything about their standard deviation - different averages can have the same SD
If mean = maximum value it means that all values are equal and SD is 0

VARIANCE = STANDARD DEVIATION2


FRACTIONS,DIGITS AND DECIMALS AND PERCENTAGES


FRACTIONS
PROPER FRACTIONS
Proper fractions fall between 0 and 1. In proper fractions, the numerator is always smaller than the denominator

IMPROPER FRACTIONS
In improper fractions, the numerator is always greater than the denominator. Improper fractions are therefore always greater than 1. For example, 3/2 = 1.5, 5/4 = 1.25
Improper fractions can be rewritten as mixed numbers, i.e. as an integer and a proper fraction

COMPARING FRACTIONS
The traditional method of comparing fractions entails finding a common denominator and comparing the numerators
The shortcut method entails multiplying the numerator of one fraction with the denominator of the other fraction


ominator and comparing the numerators
The shortcut method entails multiplying the numerator of one fraction with the denominator of the other fraction

FRACTION RULES FOR POSITIVE, PROPER FRACTIONS
As the numerator increases, the value of the fraction increases as it approaches 1, assuming that the denominator is held constant
As the denominator increases, the value of the fraction decreases as it approaches 1, assuming that the numerator is held constant
Increasing both the numerator and denominator by the same value brings the fraction closer to 1, regardless of the original value of the fraction. Hence, a proper fraction will increase in value, while an improper fraction will decrease in value
Example: 3/2 > 4/3 > 13/12 > 1,013/1,012


fraction. Hence, a proper fraction will increase in value, while an improper fraction will decrease in value
Example: 3/2 > 4/3 > 13/12 > 1,013/1,012

IMPACT OF ARITHMETIC OPERATIONS ON PROPER FRACTIONS
Adding fractions increase their value
Subtracting fractions decrease their value
Multiplying fractions decrease their value
Squaring a positive fraction reduces its value. For example, (1/2)2 = (1/4), where 1/ 2 > 1/ 4
Squaring a negative fraction increases its value. For example, (-1/2)2 = (1/4), where 1/ 4 > - 1/ 2
Cubing a positive fraction reduces its value
Cubing a negative fraction increases its value (i.e., it becomes a smaller NEGATIVE number)
Multiplying the numerator of a positive, proper fraction increases the value of the fraction as it approaches 1
There will be no change in the value of the fraction if both the numerator and the denominator are multiplied by the same constant
Dividing fractions increase their value
Taking the square root of a fraction increases its value

If there is an addition or subtraction operation in the denominator, the fraction CANNOT be simplified further


DIGITS AND DECIMALS


DECIMALS – ADVANCED CONCEPTS


DIGITS – LAST DIGIT
Last digit of (xyz)n is the same as that of Zn

Determine the cyclicity number of Z
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
2 – 2, 4, 8, 6
3 – 3, 9, 7, 1
7 – 7, 9, 3, 1
8 – 8, 4, 2, 6
 - Integers ending with 4 have a cyclicity of 2. When n is odd, last digit will be 4 and when n is even, last digit will be 6
- Integers ending with 9 have a cyclicity of 2. When n is odd, last digit will be 9 and when n is even, last digit will be 1
Find the remainder when n is divided by the cyclicity
When the remainder > 0, the last digit will be the same as Zr
When remainder = 0, the last digit will be the ZC, where c is the cyclicity number

Example # 1:
What is the last digit of 12739?

Last digit of 12739 is the same as that of 739. First we need to determine the cyclicity of 7, which is 4.

Now divide 39 (power) by 4 (cyclicity), remainder is 3. Therefore, the last digit of 12739 is the same as the last digit of 73 which is 3

Example # 2:
Given that x is a positive integer, what is the units digit of 24(2x+1)* 33 (x+1)* 17(x+2)* 9(2x)


ame as the last digit of 73 which is 3

Example # 2:
Given that x is a positive integer, what is the units digit of 24(2x+1)* 33 (x+1)* 17(x+2)* 9(2x)

x is a positive integer, so 2x is always even, while 2x + 1 is always odd

Thus,
4(2x + 1) = (4)(odd), which always has a units digit of 4
9(2x) = 9(even), which always has a units digit of 1

That leaves us to find the units digit of (3)(x + 1)*(7)(x + 2). Rewriting, and dropping all but the units digit at each intermediate step,
= (3)(x + 1)(7)(x + 2)
= (3)(x + 1)(7)(x + 1)(7)
= (3 × 7)(x + 1)(7)
= (21)(x + 1)(7)
= (1)(x + 1)(7) = 7, for any value of x

So, the units digit of is (4)(7)(1) = 28, then once again drop all but the units digit to get
Solve the problem step by step. However, pay attention only to the last digit of the intermediate product

Note 1: you just need a 5 and a 2 (or a multiple of 2) to decide the units digit, as it will be always be 0 if you find these two numbers in a product

Example:
What is the units digit of 5*5*9*9*4*4*4?
5*5 = 25; 9*9 = 81; 4*4*4= 64; Multiply the last digit of each of the intermediary products 5*1*4 = 20; Units digit will therefore be 0


*5*9*9*4*4*4?
5*5 = 25; 9*9 = 81; 4*4*4= 64; Multiply the last digit of each of the intermediary products 5*1*4 = 20; Units digit will therefore be 0

Last digit of a product
Last digit of a power
What is the units digit of 1727 ?
1727 will end in the same units digit as 727
27 / 4 results in a remainder of 3. Therefore, the units digit of 1727 will be the same as the units digit of 73 , which is 3

If x is a positive integer, what is the units digit of 24(2x+1) 33(x+1) 17(x+2) 9 (2x)
Note that x is a positive integer, so 2x is always even, while 2x + 1 is always odd.  Thus,(4)(2x + 1) = (4)(odd), which always has a units digit of 4(9)(2x) = (9)(even), which always has a units digit of 1That leaves us to find the units digit of (3)(x + 1)(7)(x + 2).  Rewriting, and dropping all but the units digit at each intermediate step,(3)(x + 1)(7)(x + 2)= (3)(x + 1)(7)(x + 1)(7)= (3 × 7)(x + 1)(7)= (21)(x + 1)(7)= (1)(x + 1)(7) = 7, for any value of x.
So, the units digit of (4)(2x + 1)(3)(x + 1)(7)(x + 2)(9)(2x) is (4)(7)(1) = 28, then once again drop all but the units digit to get 8


PERCENTAGES – 1


PERCENTAGES – 2
MIXTURE PROBLEMS
The first type of problem will have a mixture of two components and will ask you to alter the make-up of the mixture by adding or subtracting one of the components

For example, a 50-ounce mixture of sugar and water is made up of 40% sugar and 60% water. How much sugar should you add so that the mixture is 60% sugar and 40% water
Always remember: the key to these problems is the component that does not change.  In this case, we are adding sugar, while water remains constant.  Therefore, we will focus on the water for most of the problem
First, determine how much water is in the mixture; 60% of 50 is 30 ounces
Because we are not adding or subtracting any water, we will still have 30 ounces of it after we add more sugar. However, we now want that 30 ounces to represent 40% of the total, rather than 60%. Therefore, 40% * x = 30; x = 75
Since the increase in total volume is only made up of additional sugar, and we went from 50 total ounces to 75 total ounces, we must add 25 ounces of sugar


the increase in total volume is only made up of additional sugar, and we went from 50 total ounces to 75 total ounces, we must add 25 ounces of sugar

In the second type of mixture problem, substances with different characteristics are combined, and it is necessary to determine the characteristics of the resulting mixture
This is similar to a weighted average problem

For example, mixture X is 10% acid and mixture Y is 30% acid. If mixture X and mixture Y are combined, the new mixture is 15% acid. What percentage of the new mixture is mixture X?
(0.1X + 0.30Y)/ (X+Y) = 0.15
Solving for X and Y, we get 0.05 X = 0.15 Y
X/Y = 3 / 1, which means that for every 3 parts of X, there is 1 part of Y
The ratio of mixture X to the total ratio is 3/4. Therefore, mixture X constitutes 75% of the new mixture


or every 3 parts of X, there is 1 part of Y
The ratio of mixture X to the total ratio is 3/4. Therefore, mixture X constitutes 75% of the new mixture

How many liters of a solution that is 15 percent salt must be added to 5 liters of a solution that is 8 percent salt so that the resulting solution is 10 percent salt?
Let n represent the number of liters of the 15% solution. The amount of salt in the 15% solution [0.15n] plus the amount of salt in the 8% solution [(0.08)(5)] must be equal to the amount of salt in the 10% mixture [0.10 (n + 5)]. Therefore,
0.15n + 0.08(5) = 0.10(n + 5)
15n + 40 = 10n + 50
5n = 10
n = 2 liters


PERCENTAGES – 3
DISCOUNT PROBLEMS
If a price is discounted by n percent, then the price becomes (100 – n) percent of the original price
Discount is calculated on Marked price and NOT on Cost price

Example # 1
A certain customer paid $24 for a dress. If that price represented a 25 percent discount on the original price of the dress, what was the original price of the dress?
Solution # 1
If p is the original price of the dress, then 0.75p is the discounted price and 0.75p = $24, or p = $32. The original price of the dress was $32.

Example #2
The price of an item is discounted by 20 percent and then this reduced price is discounted by an additional 30 percent. These two discounts are equal to an overall discount of what percent?
Solution # 2
If p is the original price of the item, then 0.8p is the price after the first discount. The price after the second discount is (0.7)(0.8)p = 0.56p. This represents an overall discount of 44 percent (100% – 56%)


e after the first discount. The price after the second discount is (0.7)(0.8)p = 0.56p. This represents an overall discount of 44 percent (100% – 56%)

SIMPLE INTEREST
SI = Simple Interest, P = Principal, T = Time period in years, R = Rate of interest per annum
S. I. = (PTR)/100
If simple interest for 2 years is 200, then S.I for 4 years is 400, for 1 year it is 100, for 10 years it is 1000. If S.I. for any length of time is known, then it can be calculated for any other length of time

A man invests $1000 in a bank which pays him 10% p.a. rate of simple interest for 2 years. How much does the man get after 2 years?

S.I. = (1000 ×2 ×10)/100 = 200
Therefore, amount after 2 years = P + S.I. = 1000 + 200 = 1200

COMPOUND INTEREST
A = P × [1 + (R/100)] ^n

If interest is compounded monthly, divide the annual rate by 12, and multiply n by 12
If interest is compounded quarterly, divide the annual rate by 4, and multiply n by 4
If interest is compounded semi-annually, divide the annual rate by 2, and multiply n by 2


PERCENTAGES – 4


PERCENTAGES – 5
All of the furniture for sale at Al’s Discount Furniture is offered for less than the manufacturer’s suggested retail price. Once a year, Al’s holds a clearance sale. If jamie purchased a certain desk during the sale, did she get a discount of more than 50% of Al’s regular price for the desk?
Al’s regular price for the desk is 60%, rounded to the nearest percent, of the MSRP of $2000
The sale price was $601 less than Al’s regular price for the desk


ular price for the desk is 60%, rounded to the nearest percent, of the MSRP of $2000
The sale price was $601 less than Al’s regular price for the desk

In order to determine the percent discount received by Jamie, we need to know two things: the regular price of the desk and the sale price of the desk. Alternatively, we could calculate the percent discount from the price reduction and either the regular price or the sale price
(1) INSUFFICIENT: This statement tells us the regular price of the desk at Al’s, but provides no information about how much Jamie actually paid for the desk during the annual sale.(2) INSUFFICIENT: This statement tells us how much the price of the desk was reduced during the sale, but provides no information about the regular price. For example, if the regular price was $6010, then the discount was only 10%. On the other hand, if the regular price was $602, then the discount was nearly 100%. (1) AND (2) INSUFFICIENT: At first glance, it seems that the statements together provide enough information


$602, then the discount was nearly 100%. (1) AND (2) INSUFFICIENT: At first glance, it seems that the statements together provide enough information

Statement (1) seems to provide the regular price of the desk, while statement (2) provides the discount in dollars. However, pay attention to the words “rounded to the nearest percent” in statement (1). This indicates that the regular price of the desk at Al’s is 60% of the MSRP, plus or minus 0.5% of the MSRP. Rather than clearly stating that the regular price is (0.60)($2000) = $1200, this statement gives a range of values for the regular price: $1200 plus or minus $10 (0.5% of 2000), or between $1190 and $1210. If the regular price was $1190, then the discount was ($601/$1190) × 100% = 50.5% (you can actually see that this is greater than 50% without calculating). If the regular price was $1210, then the discount was ($601/$1210) × 100% = 49.7% (you can actually see that this is less than 50% without calculating).The uncertainty about the regular price means that we cannot answer with certainty whether the discount was more than 50% of the regular price. The correct answer is E


COMMON FRACTIONS, DECIMALS AND PERCENTAGE EQUIVALENTS


RATIO AND PROPORTION


RATIO AND PROPORTION
A ratio expresses a particular relationship between two or more quantities. Here are some examples of ratios:
The two partners spend time working in the ratio of 1 to 3. For every 1 hour the first partner works, the second partner works 3 hours.
Three sisters invest in a certain stock in the ratio of 2 to 3 to 8. For every $2 the first sister invests, the second sister invests $3, and the third sister invests $8
The ratio of men to women in the room is 3 to 4. For every 3 men, there are 4 women
Ratios can express a part-part relationship or a part-whole relationship
A part-part relationship: The ratio of men to women in the office is 3:4
A part-whole relationship: There are 3 men for every 7 employees
Notice that if there are only two parts in the whole, you can derive a part-whole ratio from a part-part ratio, and vice versa
Remember that ratios only express a relationship between two or more items; they do not provide enough information, on their own, to determine the exact quantity for each item. For example, knowing that the ratio of men to women in an office is 3 to 4 does NOT tell us exactly how many men and how many women are in the office. All we know is that the number of men is less than the number of women. However, ratios are surprisingly powerful on Data Sufficiency - they often provide enough information to answer the question
Multiple ratios: Make a common Term
If C:A = 3:2, and C:L = 5:4, what is A:L?
C : A : L C : A : L
3 : 2 :  Multiply by 5  15 : 10 :
5 : : 4  Multiply by 3  15 : : 12
Once you have converted C to make both ratios equal (15), you can combine the ratios
Combined Ratio C:A:L = 15:10:12, and ratio A:L = 10:12 or 5:6
If a/b = c/d, then:
(a+b)/b = (c+d)/d
(a – b)/b = (c – d)/d
(a + b)/(a – b) = (c + d)/(c – d)


EXPONENTS & ROOTS


EXPONENTS


EXPONENTS
Simplifying exponential expressions
Always try to simplify exponential expressions when they have the same base or the same exponent. You can only simplify exponential expressions that are linked by multiplication or division. When expressions with the same base are linked by a sum, you cannot simplify but you can factor the expression:
7² + 7³ = 7².(7 + 1) = 49.8
Example: what is the largest prime factor of 4²¹ + 4²2 + 4²³?
Answer: 4²¹ + 42² + 4²³ = 4²¹(1 + 4 + 16) = 4²¹.(21) = 4²¹(3.7). The largest prime factor of 4²¹ + 422 + 4²³ is 7

*** When both sides of the equation are broken down to the product of prime bases, the respective exponents of like bases must be equal


ROOTS
A ROOT is the inverse operation of raising a number to an exponent, and answers the question: Which number do I multiply by itself n times in order to get a product of b ?

EVEN ROOTS ONLY HAVE ONE SOLUTION – A POSITIVE VALUE
Unlike even exponents, which yield both a positive and a negative solution, even roots have only one solution. √4 can only be 2, and not -2
A root can only have a negative value if it is an odd root, and the base is negative

SIMPLIFYING A ROOT
GMAT very often tries to trick you by giving a root linked by addition where it is tempting to simplify the terms, for example: √(25 + 16). It is tempting to think that this will result into 5 + 4, but you can only simplify roots when the terms inside/outside are linked by multiplication or division
Simplifying radicals in the denominator with conjugate radical expressions is very useful on challenging GMAT radical questions
Examples of roots simplification:
√25.16 = √25 . √16 = 5.4
√50 . √18 = √50.18 = √900 = 30
√144:16 = √144 : √16 = 12/4 = 3
√25+16 = √41  you cannot simplify this one
2√7 + 3√7 = 5√7


n:
√25.16 = √25 . √16 = 5.4
√50 . √18 = √50.18 = √900 = 30
√144:16 = √144 : √16 = 12/4 = 3
√25+16 = √41  you cannot simplify this one
2√7 + 3√7 = 5√7

KNOWING COMMON ROOTS
GMAT requires you to know all the perfect square roots from 1 to 30 and also the following imperfect rules:
√2 ≈ 1.4
√3 ≈ 1.7
√5 ≈ 2.25

SUM OF THE ROOTS OF AN EQUATION ax2+bx+c=0 IS (-B/A)


COMMON EXPONENTS AND ROOTS


RATE AND WORK
Basic motion problems (speed, distance, time)
Average rate problems
Simultaneous motion problems
Work problems
Population problems


RATE, DISTANCE TIME
RATE
Rate is expressed as a ratio of distance and time, with two corresponding units. Some examples of rates include: 30 miles per hour, 10 meters/second etc.

DISTANCE
Distance is expressed using a unit of distance. Some examples of distances include: 18 miles, 20 meters, 100 kilometres

TIME
Time is expressed using a unit of time. Some examples of times include: 6 hours, 23 seconds, 5 months, etc.

RATE = DISTANCE / TIME
RATE * TIME = DISTANCE

Speed and time are inversely proportional to each other. Therefore, if a man increases his speed by 3/2, then the time taken will become 2/3*original time
If three men cover the same distance with speeds in the ratio a : b : c, the times taken by these three will be in the ratio 1/a : 1/b : 1/c respectively


MULTIPLE RATE, DISTANCE TIME PROBLEMS – SPEED RELATIONS


MULTIPLE RATE, DISTANCE TIME PROBLEMS


MULTIPLE RATE, DISTANCE TIME PROBLEMS – THE KISS/CRASH


MULTIPLE RATE, DISTANCE TIME PROBLEMS – CATCHING UP/OVERTAKING


MULTIPLE RATE, DISTANCE TIME PROBLEMS – ROUND TRIP


MULTIPLE RATE, DISTANCE TIME PROBLEMS – SAMPLE SITUATIONS


MISCELLANEOUS PROBLEMS
BOAT PROBLEMS

If the speed of a boat (or man) in still water is X km/hour, and the speed of the stream (or current) is Y km/hour, then:
Speed of boat with the stream (or Downstream or D/S) = (X + Y) km/hour
Speed of boat against the stream (or upstream or U/S) = (X – Y) km/hour

X = [(X + Y) + (X – Y)] / 2 and Y = [(X + Y) – (X – Y)] / 2
Boat’s speed in still water = [Speed downstream + Speed upstream] / 2
Speed of current = [Speed downstream – Speed upstream] / 2


AVERAGE RATE / RELATIVE SPEED
If an object moves the same distance twice, but at different rates, then the average rate will NEVER be the average of the two rates given for the two legs of the journey. In fact, because the object spends more time traveling at the slower rate, the average rate will be closer to the slower of the two rates than to the faster
It is not necessary to know the total distance travelled or the total time taken while computing the average speed
When two objects travel in the same direction, the time to catch up/overtake will be = lead distance/ difference of speeds (note: in order for A to overtake B, A needs to be traveling at a higher speed than B otherwise A will never catch up)
When two objects travel in different directions, time to meet will be = lead distance/ sum of speeds
AVERAGE RATE = TOTAL DISTANCE/ TOTAL TIME
OR
AVERAGE SPEED = (2*SPEED OF A*SPEED OF B) / (SPEED A + B)  only to be used when the distance travelled is the same


SPEED , DISTANCE, TIME PROBLEMS


WORK

In distance problems, if the rate (speed) is known, it will normally be given to you as a ready-to-use number. In work problems, though, you will usually have to figure out the rate from some given information about how many jobs the agent can complete in a given amount of time
For example, if Oscar can perform one surgery in 1.5 hours, the work rate = # of given jobs/ given amount of time i.e., 1/1.5 hours, which is equal to 2/3 surgeries per hour


an perform one surgery in 1.5 hours, the work rate = # of given jobs/ given amount of time i.e., 1/1.5 hours, which is equal to 2/3 surgeries per hour

WORKING TOGETHER – ADD WORK RATES
If A can finish a job alone in x days and B can finish the same job in y days alone, then
One-day work of both A and B is (x + y)/x*y
Number of days A and B take to finish the same job working together is: (x*y) / (x + y)
Exception: one agent undoes the other’s work, like a pump putting water into a tank and another drawing water out. Rate will be x – y
Example: Working alone, A finish a job in 2 days and B can finish a job in 3 days. If A and B work together, in how many days will they finish the work?
A’s rate = 1/ 2
B’s rate = 1/ 3
In one day, A and B finish 5 / 6 of the job; therefore, they take 6/ 5 days to complete the job
Example: Four men working together all day, can finish a piece of work in 11 days; but two of them having other engagements can work only one half–time and quarter time respectively. How long will it take them to complete the work ?
Each man will take 11 x 4 = 44 days to complete the work. If one man works half day/day he will take 44 x 2 = 88 days to finish the work. Similarly, a man working quarter day/day will take 44 x 4 = 176 days to finish the work. When these work together they will require 1 / [ (1/44) + (1/44) + (1/88) + (1/176) ] = 16 days
Men * hours * days
Example: If 2 men can finish a job in 3 days working 4 hours per day, then how many days will 3 men working 2 hours per day take to finish the same job?
Solution: In the first set the product is: 2 men ×3 days× 4 hours. This must remain constant in the second set in which we need to find number of days: 2 men × 3 days × 4 hours = 3 men × N days × 2 hours
2 × 3 × 4 = 3 × N ×2
Therefore, N = 4 days


the second set in which we need to find number of days: 2 men × 3 days × 4 hours = 3 men × N days × 2 hours
2 × 3 × 4 = 3 × N ×2
Therefore, N = 4 days

If a man can do a piece of work in N days (or hours or any other unit of time), then the work done by him in one day will be 1/N of the total work
If A is twice as good a workman as B, then A will take half the time B takes to finish a piece of work


WORK – 2


NUMBER PROPERTIES OF INTEGERS
IMPORTANT CONCEPTS TO ALWAYS KEEP IN MIND
An integer is any number in the following set – {..-3,-2,-1,0,1,2,3..}
An integer can be either positive or negative, even or odd, a prime or a consonant (0 and 1 are neither prime nor consonant)
Never assume a number to be an integer unless EXPLICITLY stated in the problem
If n2 is an integer, it is not necessary that n is also an integer. For example, the square of √2 is 2.

Impact of arithmetic operations on integers
When integers are added to, subtracted from, or multiplied with each other, the result is ALWAYS an integer. However, during division, it is not necessary that the result will be an integer – when an even number is divided by an odd number, it may result in a non – integer (Example: 10/4). Similarly, an odd integer divided by an odd integer may result in a non – integer (Example: 11/3). An odd integer divided by an even integer ALWAYS results in a non – integer


r divided by an odd integer may result in a non – integer (Example: 11/3). An odd integer divided by an even integer ALWAYS results in a non – integer

Divisibility
The product of 2 even numbers will always be divisible by 4
The product of x consecutive integers will always be divisible by x and x!. For example, x*(x +1)* (x+2) will be divisible by 3
The sum of x consecutive integers
If x is odd, the sum of the integers is always divisible by x. Example, 2+3+4 = 9, which is divisible by 3
If x is even, the sum of the integers is never divisible by x. Example, 5+6+7+8 = 26, which is not divisible by 4

Prime factorization – Lowest Common Multiple (LCM), Greatest Common Factor (GCF)
LCM: The smallest number that has all given numbers as a factor
GCF: The largest number which is a factor of all given numbers
Product of two numbers = LCM * GCF


PROPERTIES OF ZERO


PRIME NUMBERS
POSITIVE INTEGER greater than 1 that has EXACTLY TWO different positive factors – ONE AND ITSELF

Note: If a positive integer greater than or equal to 2 has only two factors, then the integer must be a prime number
TWO is the smallest and only EVEN prime number

If the product of two primes is even or if the sum of two primes is odd, one of the primes has to be 2
Conversely, if you know that 2 CANNOT be one of the primes in the sum, then the sum of the two primes must be even
Method to determine a prime number
Find approximate square root of the number. Then check if all the prime numbers below the square root are factors of the given number. If none are then the number is prime else not
For example: the approximate square root of 91 is 10, and the prime numbers below 10 are 2,3,5,7. 91 is not divisible by 2,3 or 5, but it is divisible by 7. Hence, it is not a prime number
First 100 prime numbers: 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
ONE is NOT a prime number
EXAMPLES

What is the value of integer x?

(1) x has exactly 2 factors.
(2) When x is divided by 2, the remainder is 0


89,97
ONE is NOT a prime number
EXAMPLES

What is the value of integer x?

(1) x has exactly 2 factors.
(2) When x is divided by 2, the remainder is 0

Statement (1) indicates that x is prime, because it has only 2 factors. This statement is insufficient by itself, since there are infinitely many prime numbers
Statement (2) indicates that 2 divides evenly into x, meaning that x is even; that is also insufficient by itself
Taken together, however, the two statements reveal that x must be an even prime-and the only even prime number is 2. The answer is (C): BOTH statements TOGETHER are sufficient


DIVISIBILITY RULES
For any integer n, n3 – n is divisible by 3, n5 – n is divisible by 5, n11 – n is divisible by 11, n13 – n is divisible by 13. In general, if p is a prime number then for any whole number a, ap – a is divisible by p. For example, 53 – 5 is divisible by 3: (53 – 5) = 120, which is divisible by 3

FACTORIALS: The factorial of N, symbolized by N!, is the product of all integers from 1 up to and including N. For instance, 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720. Because it is the product of all the integers from 1 to N, any factorial N! must be divisible by all integers from 1 to N. Another way of saying this is that N! is a multiple of all the integers from 1 to N


PRIME FACTORIZATION
COMMON FACTOR
Break down both numbers to their prime factors to see what factors they have in common. Multiply shared prime factors to find all common factors

What factors greater than 1 do 135 and 225 have in common?
135 = 3 x 3 x 3 x 5
225 = 3 x 3 x 5 x 5
Both share 3 x 3 x 5 in common—find all combinations of these numbers: 3 x 3 = 9; 3 x 5 = 15; 3 x 3 x 5 = 45

GREATEST COMMON FACTOR (GCF)
The number that contains all the factors common to both numbers (i.e., biggest number that will divide into both the numbers)

Note: If the numbers do not share any common factors, then the GCF is 1

The GCF for two consecutive numbers is always 1. However, just because the GCF is 1 doesn’t guarantee that the numbers are consecutive

Example:
The prime factorizations for 2940 & 3150 are:

2940 = 2 x 2 x 3 x 5 x 7 x 7
3150 = 2 x 3 x 3 x 5 x 5 x 7

Therefore, GCF is 2 x 3 x 5 x 7 = 210

LOWEST COMMON MULTIPLE (LCM)
The LCM is the smallest number that contains both numbers as factors i.e. the smallest number that is a multiple of both these values

L.C.M of two numbers x and y is maximum power of the prime factors in x and y


actors i.e. the smallest number that is a multiple of both these values

L.C.M of two numbers x and y is maximum power of the prime factors in x and y

Example:
The prime factorizations for 2940 & 3150 are:

2940 = 2 x 2 x 3 x 5 x 7 x 7
3150 = 2 x 3 x 3 x 5 x 5 x 7

Therefore, LCM is 2 x 2 x 3 x 3 x 5 x 5 x 7 x 7 = 44,100

LCM * GCF = PRODUCT OF TWO NUMBERS (DOES NOT WORK FOR MORE THAN TWO NUMBERS)

GCF of fractions = GCF of numerators ÷ LCM of denominators

LCM of fractions = LCM of numerators ÷ GCF of denominators


PRIME FACTORIZATION AND DIVISIBILITY – EXAMPLES


FACTORS AND MULTIPLES
FACTORS/DIVISORS
X IS A FACTOR OF Y

Essentially, a factor is a POSITIVE integer that divides evenly into an integer. Ex: 1,2,4 and 8 are all the factors of 8

One is a factor of all numbers

The largest factor of a number is the number itself

MULTIPLES
Y IS A MULTIPLE OF X

Essentially, a multiple of an integer is formed by multiplying that integer by any integer, so 8, 16,24, and 32 are some of the multiples of 8

For the purpose of the GMAT, multiples are always POSITIVE

** Divisibility can be stated in multiple ways. For example, 12 items can be shared among 3 people so that each person has the same number of items. This just means that 12 (y) is divisible by 3 (x), and that each of the individuals has 4 items each (n)

The smallest multiple of a number is the number itself

Y = X*INTEGER
Where x and y are both integers, and x is not equal to zero


FACTORS AND MULTIPLES – IMPORTANT CONCEPTS
PERFECT SQUARES
Always have an odd number of factors. Therefore, if an integer has an odd number of factors, it has to be a perfect square
The prime factorization of a perfect square contains only even powers of primes. This is because perfect squares are formed from the product of two copies of the same prime factors. For example, 902 = (2*32*5) * (2*32*5). By contrast, if a number's prime factorization contains any odd powers of primes, then the number is not a perfect square. For instance, 132,300 = 22 x 33 X 52 X 72 is not a perfect square, because the 3 is raised to an odd power. If this number is multiplied by 3, then the result, 396,900, is a perfect square: 396,900 = 22 x 34 X 52x 72
The sum of all distinct factors of a perfect square is always odd

PERFECT CUBES
- If a number is a perfect cube, then it is formed from three identical sets of primes, so all the powers of primes are multiples of 3 in the factorization of a perfect cube. For instance, 903 = (2 X 32 x 5) * (2 X 32 x 5) * (2 X 32 x 5) = 23 x 36 x 53


ers of primes are multiples of 3 in the factorization of a perfect cube. For instance, 903 = (2 X 32 x 5) * (2 X 32 x 5) * (2 X 32 x 5) = 23 x 36 x 53

THE SUM OR DIFFERENCE OF TWO MULTIPLES OF A NUMBER IS ALSO A MULTIPLE OF THAT NUMBER
Algebraically, if N is a divisor of X and of Y, then N is a divisor of (X+Y)
Example: 55 and 155 are both multiples of 5. Thus, 55 + 155 = 210 and 155 – 55 = 100 are also multiples of 5
If you add two non-multiples of N, the result could be either a multiple of N or a non-multiple of N
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)
If you add a multiple of N to a non-multiple of N, the result is a non-multiple of N

IF A NUMBER IS A FACTOR OF TWO OTHER NUMBERS, IT IS A FACTOR OF THEIR SUM OR DIFFERENCE
Example: 4 is a factor of 64 and 44. Thus, 4 is a factor of 64 + 44 = 108 and 64 – 44 = 20


WO OTHER NUMBERS, IT IS A FACTOR OF THEIR SUM OR DIFFERENCE
Example: 4 is a factor of 64 and 44. Thus, 4 is a factor of 64 + 44 = 108 and 64 – 44 = 20

FACTOR FOUNDATION RULE
If a is a factor of b, and b is a factor of c, then a is actor of c
All the factors of both x and y must be factors of the product, x*y
Example:
Given that the integer n is divisible by 3, 7, and 11, what other numbers must be divisors (factors) of n?
Since we know that 3, 7, and 11 are prime factors of n, we know that n must also be divisible by all the possible products of the primes (3, 7 and 11): 21, 33, 77, and 231 (given that n > 231)

IN ANY LIST OF N CONSECUTIVE INTEGERS, EXACTLY ONE OF THE INTEGERS IS A MULTIPLE OF N. AS A RESULT, THE PRODUCT OF N CONSECUTIVE INTEGERS IS ALWAYS A MULTIPLE OF N (OR, IT IS DIVISIBLE BY N)
In any list of three consecutive integers, one number (33) is a multiple of 3  33, 34, 35
Products of 113 × 114 × 115 and (y + 5)(y + 6)(y + 7) are both divisible by 3


ee consecutive integers, one number (33) is a multiple of 3  33, 34, 35
Products of 113 × 114 × 115 and (y + 5)(y + 6)(y + 7) are both divisible by 3

THE NUMBER OF FACTORS A NUMBER HAS (INCLUDING 1 AND THE NUMBER ITSELF) CAN BE COMPUTED BY FOLLOWING THE STEPS DETAILED BELOW:
First, determine all the prime factors of the number through LCM
Second, determine the number of copies of each of the prime factors and express them in the following format - p^a * q^b * .... *r^k
Third, the number of factors would be (a+1)*(b+1)*...(k+1)
 Example
If your number is 360, the factors are 8, 9 and 5
8*9*5 = 2^3 * 3^2 * 5^1
The number of factors is (3+1)*(2+1)*(1+1) = 24


QUOTIENTS AND REMAINDERS
Y = (QUOTIENT * DIVISOR) + REMAINDER


QUOTIENTS AND REMAINDERS


EVEN AND ODD INTEGERS
ONLY integers can be even/odd
An odd integer can be expressed as 2n + 1 and an even integer can be expressed as 2n
Sum of two different prime numbers will ALWAYS be EVEN, unless one of the numbers is 2
If two even integers are multiplied, the result will be divisible by 4, and if three even integers are multiplied, the result will be divisible by 8 etc.
Example: If x is even, and N = (x)(x+1)(x+2), is N divisible by 24?
Yes. The terms x and x+2 must each have a factor of 2, and the product (x)(x+1)(x+2) must have factors 1, 2, 3 and 4 Therefore, the prime factorization of N must be a multiple of 1*2*3*2*2 = 24


POSITIVE AND NEGATIVE INTEGERS
ZERO is neither positive nor negative – be very careful !


CONSECUTIVE INTEGERS
SUM OF INTEGERS IN A SET = AVERAGE * # OF ITEMS

Example

What is the sum of all integers from 20 to 100, inclusive?
Average the first and last term to find the precise "middle" of the set  (100+20)/2 = 60
Count the number of terms: 100 - 20 = 80, plus 1 yields 81
Multiply the "middle" number by the number of terms to find the sum: 60 x 81 = 4,860


EQUATIONS


BASIC EQUATIONS (THOSE WITHOUT EXPONENTS)
SIMULTANEOUS EQUATIONS – THREE VARIABLES
Look for ways to simplify the work. Look especially for shortcuts or symmetries in the form of the equations to reduce the number of steps needed to solve the system
Example: What is the sum of x, y and z?
x + y=8; x + z=1; y + z=7
In this case, DO NOT try to solve for x, y, and z individually. Instead, notice the symmetry of the equations-each one adds exactly two of the variables-and add them all together. Therefore, 2x + 2y + 2z = 26; and x + y+ z = 13


ce the symmetry of the equations-each one adds exactly two of the variables-and add them all together. Therefore, 2x + 2y + 2z = 26; and x + y+ z = 13

MISMATCH PROBLEMS
MISMATCH problems, which are particularly common on the Data Sufficiency portion of the test, are those in which the number of unknown variables does NOT correspond to the number of given equations
A MASTER RULE for determining whether 2 equations involving 2 variables (say, x and y) will be sufficient to solve for the variables is this: (1) If both of the equations are linear-that is, if there are no squared terms (such as x2 or y2) and no xy terms-the equations will be sufficient UNLESS the two equations are mathematically identical (e.g., x +y = 10 is identical to 2x + 2y = 20) and (2) If there are ANY non linear terms in either of the equations (such as x2, y2, x/y, or xy), there will USUALLY be two (or more) different solutions for each of the variables and the equations will not be sufficient.
Example:
What is X?
3x/ (3y + 5z) = 8 ; 6y + 10 z = 18
It is tempting to say that these two equations are not sufficient to solve for x, since there are 3 variables and only 2 equations. However, the question does NOT ask you to solve for all three variables. It only asks you to solve for x, which IS possible. First, get the x term on one side of the equation: 3x = 8(3y + 5z). Then, notice that the second equation gives us a value for 3y + 5z, which we can substitute into the first equation in order to solve for x. Thus, BOTH statements TOGETHER are sufficient


n gives us a value for 3y + 5z, which we can substitute into the first equation in order to solve for x. Thus, BOTH statements TOGETHER are sufficient

COMBO PROBLEMS – MANIPULATION
The GMAT often asks you to solve for a combination of variables, called COMBO problems. For example, a question might ask, what is the value of x +y?
In these cases, since you are not asked to solve for one specific variable, you should generally NOT try to solve for the individual variables right away. Instead, you should try to manipulate the given equation(s) so that the COMBO is isolated on one side of the equation. There are four easy manipulations that are the key to solving most COMBO problems:
M: Multiply or divide the whole equation by a certain number
A: Add or subtract a number on both sides of the equation
D: Distribute or factor an expression on ONE side of the equation.
S: Square or un square both sides of the equation
Combo problems occur most frequently in Data Sufficiency. Whenever you detect this is a Data Sufficiency question that may involve a combo, you should try to manipulate the given equation(s) in either the question or the statement, so that the combo is isolated on one side of the equation. Then, if the other side of an equation from a statement contains a VALUE, that equation is SUFFICIENT. If the other side of the equation contains a VARIABLE EXPRESSION, that equation is NOT SUFFICIENT


ABSOLUTE VALUE EQUATIONS
Absolute value refers to the POSITIVE value of the expression within the absolute value brackets. Equations that involve absolute value generally have TWO SOLUTIONS. In other words, there are TWO numbers that the variable could equal in order to make the equation true. The reason is that the value of the expression inside the absolute value brackets could be POSITIVE OR NEGATIVE. For instance, if we know |x| = 5, then x could be either 5 or -5, and the equation would still be true
If |x| > x then x must be negative
The following three-step method should be used when solving for a variable expression inside absolute value brackets. Consider the following example: Solve for w, given that 12 + |w – 4| = 30.
Isolate the expression within the absolute value brackets: |w – 4| = 18
Remove the absolute value brackets and solve the equation for 2 different cases (positive and negative)
w – 4 = 18; w = 22
- (w – 4 ) = - 18; - w + 4 = 18; - w = 14; w = -14
3. Check to see whether each solution is valid by putting each one back into the original equation and verifying that the two sides of the equation are in fact equal. The possibility of a failed solution is a peculiarity of absolute value equations. For most other types of equations, it is good to check your solutions, but doing so is less critical


EXPONENTIAL AND QUADRATIC EQUATIONS


INEQUALITIES


INEQUALITIES
COMBINING INEQUALITIES
Many GMAT inequality problems involve more than one inequality. To solve such problems, you may need to convert several inequalities to a compound inequality, which is a series of inequalities strung together
For example, If x > 8, x < 17, and x + 5 < 19, what is the range of possible values for x?
First, solve any inequalities that need to be solved. In this example, only the last inequality needs to be solved: x < 14
Second, simplify the inequalities so that all the inequality symbols point in the same direction, preferably to the left (less than): 8 < x, x <17, x < 14
Third, combine the inequalities by taking the more limiting upper and lower extremes: 8 < x < 14


ferably to the left (less than): 8 < x, x <17, x < 14
Third, combine the inequalities by taking the more limiting upper and lower extremes: 8 < x < 14

Another helpful approach is to combine inequalities by adding the inequalities together. However, note that we should never subtract or divide two inequalities, and can only multiply inequalities together under certain circumstances
For example, is mn < 10 given that (1) m < 2 (2) n < 5
It is tempting to multiply these two statements together and conclude that mn < 10. That would be a mistake, however, because both m and n could be negative numbers that yield a number larger than 10 when multiplied together. For example, if m = -2 and n = -6, then mn = 12, which is greater than 10. Since you can find cases with mn < 10 and cases with mn > 10, the correct answer is (E): The two statements together are INSUFFICIENT
Consider the following variation: if both m and n are positive, is mn < 10 given that (1) m < 2 (2) n < 5 . Since the variables are positive, we can multiply these inequalities together and conclude that mn < 10. The correct answer is (C)


) m < 2 (2) n < 5 . Since the variables are positive, we can multiply these inequalities together and conclude that mn < 10. The correct answer is (C)

MANIPULATING COMPOUND INEQUALITIES
Sometimes a problem with compound inequalities will require you to manipulate the inequalities in order to solve the problem. You can perform operations on a compound inequality as long as you remember to perform those operations on every term in the inequality, not just the outside terms.
For example, consider the equation x + 3 <Y < x + 5
Wrong: x < y < x + 2 ( 3 needs to be subtracted from each term)
Correct: x < y - 3 < x + 2


terms.
For example, consider the equation x + 3 <Y < x + 5
Wrong: x < y < x + 2 ( 3 needs to be subtracted from each term)
Correct: x < y - 3 < x + 2

ADDING INEQUALITIES – very powerful technique!
In order to add inequalities, we must make sure the inequality signs are facing the same direction. If x < y and w < z, then x + w < y + z
For example, is a < c, given that (1) b > d and (2) ab2 – b > b2c – d
This is a multiple variable inequality problem, so you must solve it by doing algebraic manipulations on the inequalities
 (1) INSUFFICIENT: Statement (1) relates b to d, while giving us no knowledge about a and c
(2) INSUFFICIENT: Statement (2) does give a relationship between a and c, but it still depends on the values of b and d. One way to see this clearly is by realizing that only the right side of the equation contains the variable d. Perhaps ab2 – b is greater than b2c – d simply because of the magnitude of d. Therefore there is no way to draw any conclusions about the relationship between a and c.
(1) AND (2) SUFFICIENT: By adding the two inequalities from statements (1) and (2) together, we can come to the conclusion that a > c. Two inequalities can always be added together as long as the direction of the inequality signs is the same:ab2 – b > b2c – d(+)     b >  d           -----------------------      ab2 > b2cNow divide both sides by b2. Since b2 is always positive, you don't have to worry about reversing the direction of the inequality. The final result: a > c


INEQUALITIES – USING EXTREME VALUES
One effective technique for solving GMAT inequality problems is to focus on the EXTREME VALUES of a given inequality. This is particularly helpful when solving the following types of inequality problems:
Problems with multiple inequalities where the question involves the potential range of values for variables in the problem
Problems involving both equations and inequalities


INEQUALITIES AND ABSOLUTE VALUES
Solving inequalities
√x2 = |x|
It is often helpful to try to visualize the problem with a number line. For example, take the inequality |x| = 5. One way to understand this inequality is to say "x must be less than 5 units from zero on the number line." Indeed, one interpretation of absolute value is simply distance on the number line. For a simple absolute value expression such as lx|, we are evaluating distance from zero. The range therefore, would be -5 < x < 5
Equations involving absolute value require you to consider two scenarios: one where the expression inside the absolute value brackets is positive, and one where the expression is negative. The same is true for inequalities. Note that you should never change I x – 5 |to x + 5. Remember, when you drop the absolute value signs, you either leave the expression alone or enclose the ENTIRE expression in parentheses and put a negative sign in front.
Example: Given that | x – 2 | < 5, what is the range of possible values of x?
To work out the FIRST scenario, we simply remove the absolute value brackets and solve: x – 2 < 5 ; x < 7
To work out the SECOND scenario, we reverse the signs of the terms inside the absolute value brackets, remove the brackets, and solve again: - (x - 2) < 5 ; - x + 2 < 5 ; - x < 3 ; x > - 3
Therefore, range is -3 < x < 7
iii. Inequalities with two absolute value expressions
Because there are two absolute value expressions, each of which yields two algebraic cases, it seems that we need to test four cases overall: positive/positive, positive/negative, negative/positive, and negative/negative
Example: Given that | x – 2 | = |2x – 3 |, what is the range of possible values of x?
The positive/positive case: (x- 2) = (2x- 3)
The positive/negative case: (x - 2) = -(2x - 3)
The negative/positive case: - (x - 2) = (2x- 3)
The negative/negative case: - (x- 2) = - (2x- 3)


INEQUALITIES AND ABSOLUTE VALUES – EXAMPLES


SEQUENCES AND FUNCTIONS
Sequences may be of two major types – arithmetic or geometric
Arithmetic progression is a series of numbers in which every term after the first can be derived from the immediately preceding term by adding to it a fixed quantity called common difference (d)
Geometric progression is a series of numbers in which each term is formed from the preceding by multiplying it by a constant factor ( r ). The constant factor is called the common ratio and is formed by dividing any term by the term which precedes it
To answer a question about a sequence, we not only need to know the rule the sequence follows, but also at least one term in the sequence


PROGRESSIONS AND SEQUENCES – ARITHMETIC PROGRESSION

Arithmetic progression is a series of numbers in which every term after the first can be derived from the immediately preceding term by adding to it a fixed quantity called common difference (d)
a, a + d, a + 2d, a + 3d are in Arithmetic Progression
Three numbers in an A.P should be taken as a – d, a, a+ d
Four numbers in an A.P. should be taken as a – 3d, a – d, a + d, a + 3d
A series of quantities is said to be in harmonic progression when their reciprocals are in A.P


PROGRESSIONS AND SEQUENCES – GEOMETRIC PROGRESSION
Geometric progression is a series of numbers in which each term is formed from the preceding by multiplying it by a constant factor ( r ). The constant factor is called the common ratio and is formed by dividing any term by the term which precedes it. For example, in the series 3,9,27,81 and 243, the factor is 3
Three numbers in a G.P. should be taken as a/r, a, ar
Four numbers in a G.P. should be taken as a/r3, a/r, ar, ar3


FUNCTIONS
The "domain" of a function indicates the possible inputs and the "range" of a function indicates the possible outputs. For instance, the function f(x) = x2 can take any input but never produces a negative number. So the domain is all numbers, but the range is f(x) >= O

COMPOUND FUNCTIONS – work from the INSIDE out
If f(x) = x³ + √x and g(x) = 4x – 3, what is f(g(3))?
First, solve g(3): 4.3 – 3 = 9. Now, plug-in 9 on f(x): 9³ + √9 = 729 + 3


k from the INSIDE out
If f(x) = x³ + √x and g(x) = 4x – 3, what is f(g(3))?
First, solve g(3): 4.3 – 3 = 9. Now, plug-in 9 on f(x): 9³ + √9 = 729 + 3

DETERMINING WHETHER A FUNCTION IS EVEN/ODD - To do this, you take the function and plug –x in for x, and then simplify. If you end up with the exact same function that you started with (that is, if f(–x) = f(x), so all of the signs are the same), then the function is even. If you end up with the exact opposite of what you started with (that is, if f(–x) = –f(x), so all of the "plus" signs become "minus" signs, and vice versa), then the function is odd. So I'll plug –x in for x, and simplify: In all other cases, the function is "neither even nor odd“
Example # 1 : Determine algebraically whether f(x) = –3x2 + 4 is even, odd, or neither
f(–x) = –3(–x)2 + 4 = –3(x2) + 4 = –3x2 + 4 Since the new expression is the same as the initial expression, the function is even


n, odd, or neither
f(–x) = –3(–x)2 + 4 = –3(x2) + 4 = –3x2 + 4 Since the new expression is the same as the initial expression, the function is even

COMMON FUNCTION TYPES
i) Direct proportionality means that the two quantities always change by the same factor and in the same direction. For instance, tripling the input will cause the output to triple as well. Direct proportionality relationships are of the form Y = kx, where x is the input value, y is the output value and k is the proportionality constant. This equation can also be written as y/x = k
Example: The maximum height reached by an object thrown directly upward is directly proportional to the square of the velocity with which the object is thrown. If an object thrown upward at 16 feet per second reaches a maximum height of 4 feet, with what speed must the object be thrown upward to reach a maximum height of 9 feet?
Solution: Typically with direct proportion problems, you will be given "before" and "after" values. Simply set up ratios to solve the problem for example, y1 and x1 can be used for the "before” values and y2 and x2 can be used for the "after" values. We then write y1/x1 = y2/x2 , since both ratios are equal to the same constant k. In the problem given above, note that the direct proportion is between the height and the square of the velocity, not the velocity itself. Therefore, write the proportion as (4/162) = 9/v2 ; v2 = 576; v = 24
ii) Inverse proportionality means that the two quantities change by RECIPROCAL factors. Cutting the input in half will actually double the output. Inverse proportionality relationships are of the form yx = k. In these cases, set up the ratio as a product, and then solve
iii) Linear growth - Many GMAT problems, especially word problems, feature quantities with linear growth (or decay), i.e., they grow at a constant rate. Such quantities are determined by the linear function: Y = mx + b. In this equation, the slope m is the constant rate at which the quantity grows. The y-intercept b is the value of the quantity at time zero, and the variable (in this case, x) stands for time. For instance, if a baby weighs 9 pounds at birth and gains 1.2 pounds per month, the function of his weight is: w = 1.2t + 9 (t = nº of months)


COMBINATORICS AND PROBABILITY


COMBINATORICS
PERMUTATION
ORDER MATTERS - a permutation is nothing but an ordered combination. Essentially, permutation questions are about taking a group of objects and totaling the number of ways in which they can be arranged in specific/pre-defined ways
The number of permutations of r items, chosen from a pool of n items, is n! / (n-r)!
The number of permutations when things are not all different : If there be n things, p of them of one kind, q of another kind, r of still another kind and so on, then the total number of permutations is given by n! / (p! q! r!...)
As a special case, the number of permutations of all n items in a pool of n items is just n! , and the number of ways to arrange n distinct objects along a fixed circle is: (n – 1)!
COMBINATION
ORDER DOES NOT MATTER (i.e., whether the events are taking place simultaneously or sequentially). Since the order does not matter, we need to remove all the “duplicate” cases
The number of combinations of r items, chosen from a pool of n items, is n! / (n-r)! r!


does not matter, we need to remove all the “duplicate” cases
The number of combinations of r items, chosen from a pool of n items, is n! / (n-r)! r!

“Does it matter if the two items (or however many items you have) positions are changed?” If the answer is yes, it is a permutations problem, if not, it is a combinations problem
Replacement / Repetition

Potential outcomes that are replaced or constant
Examples
How many possibilities for a combination lock with 40 numbers that requires 3 selections

Non – replacement
Potential outcomes that decrease with each selection
Examples
How many possibilities for a combination lock with 40 numbers that requires 3 selections and cannot have the same number twice
Combinations from multiple groups
In this situation, combinations are being drawn from several groups to form a complete set. Figure out the combinations from each group and then multiply them together


, combinations are being drawn from several groups to form a complete set. Figure out the combinations from each group and then multiply them together

Combinations from pairings
These are pairing questions – so they must take place in two
Example: 7 basketball teams with five players each are at a tournament. If each player shakes hands with every other player NOT on his own team, how many handshakes take place?
How many people take part in a handshake? Two. There are 35 people who could be in the first spot, but that person cannot shake hands with anyone on his own team. So that person has only 30 people who’s hands he can shake. Therefore, the number of combinations is (35*30)/2


PERMUTATIONS VS. COMBINATIONS
One concept that you need to know for the exam is that when dealing with combinations and permutations, each result corresponds to a unique set of circumstances. For example, if you have z people and know that choosing two of them would result in 15 different possible groups of two, it must be true that z = 6. No other value of z would yield exactly 15 different groups of two. So if you know how many subgroups of a certain size you can choose from an unknown original larger group, you can deduce the size of the larger group
Seating 5 people in 3 chairs, is the same as seating 3 people in 5 chairs


SOLVING COMBINATORICS PROBLEMS – BASIC STRATEGIES
FUNDAMENTAL COUNTING PRINCIPLE
If you must make a number of separate decisions, then multiply the number of ways to make each individual decision to find the number of ways to make all decisions together
Example: If you have 4 types of bread, 3 types of cheese and 2 types of ham and wish to make a sandwich, you can make it in 4*3*2 = 24 different ways


Example: If you have 4 types of bread, 3 types of cheese and 2 types of ham and wish to make a sandwich, you can make it in 4*3*2 = 24 different ways

SLOT METHOD
For problems where some choices are restricted and/or affect other choices, choose most restricted options first (slot method)
For example, you must insert a 5-digit lock code, but the first and last numbers need to be odd, and no repetition is allowed. How many codes are possible?
In this problem, you do not want to choose the five digits in order. Instead, you should start by picking the first and last digits (which must be odd), because these digits are the most restricted. Because there are 5 different odd digits (1, 3, 5, 7, and 9), there are 5 ways of picking the first digit. Since no repetition is allowed, there are only 4 odd digits left for the last digit. You can then pick the other three digits in any order, but make sure you account for the lack of repetition. For those three choices, you have only 8, 7, and 6 digits available. Therefore, the total number of lock codes is 4 x 5 x 8 x 7 x 6 = 6,720


epetition. For those three choices, you have only 8, 7, and 6 digits available. Therefore, the total number of lock codes is 4 x 5 x 8 x 7 x 6 = 6,720

SIMPLE FACTORIALS
The number of ways in which n different things can be arranged in a straight line, such that no one thing is allowed to appear more than once in any of the arrangements (i.e. when repetitions are not allowed) is n!. n! counts the rearrangements of n distinct objects as a special (but very common) application of the Slot Method
For example, In staging a house, a real-estate agent must place six different books on a bookshelf. In how many different orders can she arrange the books? Using the Fundamental Counting Principle, we see that we have 6 choices for the book that goes first, 5 choices for the book that goes next, and so forth. Ultimately, we have this total: 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720 different orders


hat goes first, 5 choices for the book that goes next, and so forth. Ultimately, we have this total: 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720 different orders

ANAGRAMS
An anagram is a rearrangement of the letters in a word or phrase. For instance, the word DEDUCTIONS is an anagram of DISCOUNTED, and so is the gibberish "word“ CDDEINOSTU
Anagrams with repeated words
The number of anagrams of a word is the factorial of the total number of letters, divided by the factorial(s) corresponding to each set of repeated letters


umber of anagrams of a word is the factorial of the total number of letters, divided by the factorial(s) corresponding to each set of repeated letters

Example # 1
How many different anagrams (meaningful or nonsense) are possible for the word PIZZAZZ?
We might expect 7! possible combinations from the 7 letters in the word PIZZAZZ. However, there are only 210. Why are there, relatively speaking, so few anagrams? The answer lies in repetition: the four Z's are indistinguishable from each other. If the four Z's were all different letters, then we would have 7! = 5,040 different anagrams. To picture that scenario, imagine labeling the Z's with subscripts: Z1, Z2, Z3 and Z4. We could then list the 5,040 anagram. Now erase the subscripts from those 5,040 anagrams. You will notice that many "different“ arrangements-like AIPZ1Z2Z3Z4, AIPZ4Z2Z3Z1' and so on-are now the same. In fact, for any genuinely unique anagram-like AIPZZZZ-there are now 4! = 24 identical copies in the list of 5,040 anagrams, because there are 4! = 24 ways to rearrange the four Z's in the word PIZZAZZ without changing anything. Because this 24-fold repetition occurs for every unique anagram of PIZZAZZ, we take 7! (which counts the arrangements as if the letters were all distinct) and divide by 4! (= 24) to account for the 4 repeated Z's


SOLVING COMBINATORICS PROBLEMS – ADVANCED STRATEGIES


PROBABILITY
PROBABILITY = NUMBER OF FAVORABLE OUTCOMES TOTAL NUMBER OF OUTCOMES
PROBABILITY OF MULTIPLE EVENTS
AND
If two events have to occur together, generally an "and" is used. For example: "I will only be happy today if I get email and win the lottery." The "and" means that both events are expected to happen together

In the case of “and,” we multiply probabilities together to get a lower overall probability

Example:
If a coin is tossed twice, what is the probability that on the first toss the coin lands heads and on the second toss the coin lands tails?
The probability that the coin will land on heads is 1/2. The probability that the coin will land on Tails is also 1/2. Since we want them to happen together, we multiply individual probabilities 1/2 × 1/2 = 1/4

OR
If both events do not necessarily have to occur together, an “or” may be used. For example, I will be happy today if I win the lottery OR have email

“OR” means that we add probabilities together to get a higher overall probability


r example, I will be happy today if I win the lottery OR have email

“OR” means that we add probabilities together to get a higher overall probability

Example:
John will win $100 if, from a deck of 52 standard playing cards, he chooses either a 7 or a 9 when pulling a single card from the deck. What is the probability that John will win $100?
How can John win? He can win by pulling out either a 7 or a 9. His chances of doing that are higher than if he could win only by pulling out a 7. In that case, he’d only have 4 cards that would make him win $100 (because there are 4 7's in a standard deck), now he has 8 cards. To find the total probability, we need to figure out the probability of each event and then add them together. Therefore probability = (4/52) + (4/52) = 2/13

INDEPENDENT EVENTS

Events where the outcome of first event does not affect the probability of the second event
Example: coin toss
DEPENDENT EVENTS

Events where the probability of the second event is affected by the outcome of the first event
Example: Drawing from a pack of cards without replacing the card drawn
MUTUALLY EXCLUSIVE EVENTS


event is affected by the outcome of the first event
Example: Drawing from a pack of cards without replacing the card drawn
MUTUALLY EXCLUSIVE EVENTS

Two events cannot occur together  Add the probabilities of individual events

EVENTS CAN OCCUR TOGETHER

P (A or B) = P(A) + P(B) – P (A&B)

If you simply add P(A) and P(B) in situations when A and B can occur together, then you are overestimating the probability of either A or B (or both) occurring


PROBABILITY – 2
The greatest probability- the certainty that an event will occur-is 1. Thus, a probability of 1 means that the event must occur
The lowest probability-the impossibility that an event will occur is 0. Thus, a probability of 0 means that an event will NOT occur. For example, the probability that you roll a fair die once and it lands on the number 9 is impossible, or 0
If on a GMAT problem, "success“ contains multiple possibilities –especially if the wording contains phrases such as “at least" and “at most“ then consider finding the probability that success does not happen. If you can find this “failure" probability more easily (call it x) the probability you really want to find will be 1 – x
Example: What is the probability that, on three rolls of a single fair die, at least of the rolls will be a six?
Failure: What is the probability that NONE of the rolls will yield a 6? On each roll, there is a 5/6 probability that the die will not yield a 6. Thus, the probability that on all 3 rolls the die will not yield a 6 is (5/6)*(5/6)*(5*6) = 125/216
Now, we originally defined success as rolling at least one six. Since we have found the probability of failure, we answer the original question by subtracting this probability from 1: 1 – (125/216) = 91/256


GEOMETRY


LINES AND ANGLES
Two straight lines which meet at a point form an angle between them
Acute angle is between 0 and 90 degrees
Right angle is 90 degrees
Obtuse angle is between 90 degrees and 180 degrees
Reflex angle is between 180 degrees and 360 degrees
A straight line is the shortest distance between two points and measures 180 degrees
Parallel lines are lines that never intersect, no matter how far they are stretched, and are equidistant
Perpendicular lines intersect at 90 degrees (may or may not bisect the line segment)
The sum of angles at a point is 360 degrees

INTERSECTING LINES

EXTERIOR ANGLES OF A TRIANGLE


LINES AND ANGLES
PARALLEL LINES CUT BY A TRANSVERSAL

GMAT can disguise the lines, by not drawing them completely. Extending them can help you to find out whether they are transversal or parallel. GMAT uses the symbol II to indicate 2 lines are parallel. MN II OP means those segments are parallel

Properties
Alternate interior angles are equal
Angles 3 and 6 are equal; angles 4 and 5 are equal

Alternate exterior angles are equal
Angles 1 and 8 are equal; angles 2 and 7 are equal

Corresponding angles are equal (corresponding angles are one interior and one exterior angle that are on the same side of the transversal)
Angles 1 and 5
Angles 2 and 6
Angles 3 and 7
Angles 4 and 8

It is important to note that if two lines cut by a transversal have any of the above properties, then the two lines must be parallel. For example, if alternate interior angles are equal, then the two lines cut by a transversal must be parallel


TWO – DIMENSIAL POLYGONS

TWO DIMENSIONAL POLYGONS

TRIANGLES – 3 SIDED

QUADRILATERALS – 4 SIDED

OTHERS – N SIDED

Equilateral
Isosceles
Scalene

Pentagon – 5 sides

Hexagon – 6 sides
Comprised of 6 equilateral triangles. Therefore, the area of a hexagon will be 6 * area of an equilateral triangle

Octagon – 8 sides

Irregular quadrilaterals

Parallelogram
Rectangle
Rhombus
Square

All quadrilaterals are not necessarily parallelograms. For a quadrilateral to be a parallelogram opposite sides and opposite angles have to be equal and parallel

Trapezium

SUM OF THE INTERIOR ANGLES OF A POLYGON – This depends on the number of sides (n) a polygon has: (n-2)*180
SUM OF THE EXTERIOR ANGLES OF A POLYGON = 360


TRIANGLES
EQUILATERAL
ISOSCELES
SCALENE
TYPES OF TRIANGLES – SIDES
TWO SIDES ARE EQUAL AND THE THIRD SIDE IS LESS THAN THE SUM OF THE OTHER TWO SIDES
All THREE SIDES ARE EQUAL
ALL THREE SIDES ARE DIFFERENT
ACUTE
OBTUSE
RIGHT – ANGLED
TYPES OF TRIANGLES – ANGLES
ONE ANGLE IS GREATER THAN 90 DEGREES

In a triangle ABC with the vertex at B, AB2 + BC2 < AC2
ALL ANGLES ARE LESS THAN 90 DEGREES

In a triangle ABC with the vertex at B, AB2 + BC2 > AC2
ONE ANGLE IS 90 DEGREES

The 45-45-90 Triangle is known as a right-angled isosceles triangle

In a triangle ABC with the vertex at B, AB2 + BC2 = AC2


BASIC PROPERTIES OF TRIANGLES – A


BASIC PROPERTIES OF TRIANGLES – C


BASIC PROPERTIES OF TRIANGLES – B
SUM OF THE ANGLES OF A TRIANGLE = 180 DEGREES

When one side is extended in any direction, an angle is formed with another side. This is called the exterior angle. There are six exterior angles in a triangle

ANGLES CORRESPOND TO THEIR OPPOSITE SIDE

This means that the largest angle is opposite the longest side, while the smallest angle is opposite the shortest side. Additionally, if two angles are equal, their sides are also equal

SUM OF ANY TWO SIDES OF A TRIANGLE MUST BE GREATER THAN THE THIRD SIDE

DIFFERENCE BETWEEN ANY TWO SIDES IS LESS THAN THE THIRD SIDE

THE LENGTH OF THE THIRD SIDE MUST LIE BETWEEN THE SUM AND THE DIFFERENCE OF THE TWO GIVEN SIDES


SIMILAR AND CONGRUENT TRIANGLES
Similarity – if any of the conditions are met –
Angle, angle, angle – if the corresponding angles of each triangle have the same measurement. In other words, the above triangles are similar if:Angle L = Angle O; Angle N = Angle Q; Angle M = Angle P
Side, angle, side - An angle in one triangle is the same measurement as an angle in the other triangle and the two sides containing these angles have the same ratio. In other words, the above triangles are similar if:Angle L = Angle O; Side LM/Side OP = Side LN/Side OQ
Side, side, side - Each pair of corresponding sides have the same ratio. In other words, the above triangles are similar if:Side LM/Side OP = Side LN/Side OQ = Side MN/Side PQ


RIGHT- ANGLED TRIANGLE AND THE PYTHAGOREAN THEOREM
Pythagorean Theorem: c2 = a2 + b2
a and b are the legs, while c is the hypotenuse

THE 45-45-90 ISOSCELES TRIANGLE

The 45-45-90 triangle is a special triangle with 2 equal sides and a relation between each side. If you are given one dimension on a 45-45-90 triangle, you can find the others.

Relationship between sides
45° - 45° - 90°
Leg - leg - hypotenuse
: 1 : √2
x : x : x√2

Therefore, in a 45-45-90 degree triangle, you only need to know the value of one side to determine the others

A 45-45-90 is exactly half of a square. Two 45-45-90 triangles put together make up a square. So, if you are given the diagonal of a square, you can find the side by using the relation above. For example, if the diagonal of a square is 6√2, find the length of each of the two sides.
(6√2)2 = x2 + x2
36*2 = 2x2
36 = x2
 X = 6


HYPOTENUSE
HYPOTENUSE
EQUILATERAL TRIANGLES AND THE 30-60-90 TRIANGLE
An equilateral triangle is one in which all three sides and all three angles are equal
Equilateral Triangles and the 30-60-90 Triangle:

30
30
60
60
SHORT
L
ONG
L
ONG
SHORT


INSCRIBED AND CIRCUMSCRIBED TRIANGLES AND CIRCLES
A right triangle inscribed in a semi- circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle


EXAMPLES


QUADRILATERALS
How to prove a quadrilateral is a parallelogram
If both pairs 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

The area of a parallelogram is twice the area of a triangle created by one of its diagonals
Of all quadrilaterals with a given perimeter, the square has the largest area. Conversely, of all the quadrilaterals with a given area, the square is the one with the smaller perimeter
The diagonal formed by joining vertices of two smaller angles will be greater than the diagonal formed by joining vertices of two greater angles in a rhombus
The diagonals of a kite are perpendicular bisectors of each other. The kite also has two pairs of adjacent, congruent sides (i.e., all sides are not equal)– THE KITE IS NOT A PARALLELOGRAM


QUADRILATERALS