papgusts math guide

8/10/2019 Papgusts Math Guide

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I will post my flashcards based on the core topics that are covered in GMAT,

1. Number Properties2. Inequalities

3. Averages4. Ratios5. Sequences & Progressions6. Set Theory7. Co-ordinate Geometry8. Geometry

How to test whether a number is prime or composite:

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 testwhether 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.S3. List all prime numbers upto the computed square root4. Check if all listed prime numbers divide n equally. If not, then n is a prime. Even if atleast one of thelisted 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 issmaller than n itself!)2. Square root of 100 ==> 103. 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 tolearn 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


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I will try to find and post a GMAT problem that uses this concept. Please feel free to post a question ifyou find it.

Warning: Some people may not find this approach comfortable. Some may find it comfortable. Pleasefollow 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 integersraised 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).


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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 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 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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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 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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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.

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, thenthe 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 ]

Any number whose prime factorization contains even powers of primes, then the number must be aperfect square.

Any number whose prime factorization contains powers of primes with multiples of 3, then the numbermust be a perfect cube.

REMAINDERS:

(I)

When 2 numbers are divided by same divisor and the remainders obtained are the same,THENDIFFERENCE 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,THENthe remainders obtained when a+b is divided by d will be r1+r2

Quote:

NOTE: If r1+r2 >= d, compute (r1+r2) - das 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,THENthe remainders obtained when a*b is divided by d will be r1*r2

Quote:

NOTE: If r1*r2 >= d, compute (r1*r2) / das the remainder.


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TAKEAWAY:

A remainder can NEVER be greater than or equal to the divisor.

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 2^4 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,


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x^17 / (x-1)

Remainder is f(1). Substitute 1 in 'x',

Remainder is 1.

Thus, 1 is the remainder when 2^51 / 7.

Simple Facts:

a^n - b^n:

1. ALWAYS divisible by a-b2. 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-b2. If n is odd, it is divisible by a+b3. If n is even, it is NOT divisible by a+b

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-multipleof N, the result is a non-multipleof N.

Example:9-5 = 4 [(Multiple of 3) - (Non-Multiple of 3) = (Non-multiple of 3)]

(III)


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If you add/subtract 2 non-multiplesof 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).

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 12units apart from each other on the number line. Therefore, m and n CANNOT be less than 12 unitsapart.

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 unitsapart from each other on the number line. Thus, 4 is the greatest common factor (GCF) of 4 and 8. Thatis why GCF of any 2 consecutive numbers is ALWAYS 1 as both are multiples of 1.

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) areconsecutive multiples of 2, (x+1) will have an additional 2 units apart from another 2 (i.e. 4). So, (x+1) isdivisible by 4.Thus, (x-1) * x * (x+1) is divisible by 8

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 itsfactors.

Example:Consider 36.

36 can be broken down into 12 * 3


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12 can be broken down into 2^2 * 3

So, 2^2 is also a factor of 36

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.

Exponent Rules:

1. xâ * 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 (xâ) = (b root (x))â

5. xâ / x^b = x^(a-b)

6. (a/b)^x = a^x / b^x

7. x^-a = 1 / xâ

8. a^x + a^x + a^x = 3 * a^x

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(xâ) = (b root(x)) ^ a = x^(a/b)
http://www.postimage.org/


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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

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.

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).

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.

For any set of consecutive integerswith an ODD number of items, the sum of all integersis

ALWAYS a multiple of the number of items.


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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 integerswith an EVEN number of items, the sum of all integersis

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]--

Another GCF:

If "n" is NOT divisible by "k" and GCF (n,k) = "z"THENthe 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)).

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 than2's and/or 5's , then it is a repeating/non-terminating decimal.

Ruthless ROOTS:

If n is a POSITIVE integer, then the nth root of any number (> 1) will ALWAYS be > 1.


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Example: (Official Problem)

What is root (4) + 3root (4) + 4root (4) approximately?

root (4) = 23root (4) > 1

4root (4) > 1

So, root (4) + 3root (4) + 4root (4) is approximately > 4.

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 3OR6 is divisible by 3

"Number Properties" flashcards are over by now.

I will start posting flashcards in "Inequalities".

If x >y,

then x - y > 0 AND x = y + k (Where k > 0)

If x 0) OR x = y + k (Where k < 0)]

If x > y and 'k' is a real number,thenx + k > y + k (Irrespective of the value of k)


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I.

Ifx > yand k > 0,

then

k*x > k*y AND x/k > y/k

II.

Ifx > yand k < 0,

then

k*x < k*y AND x/k < y/k

[-- Inequality reverses when k < 0 --]

(IMPORTANT)

If x and y are both POSITIVE/NEGATIVEAND x > y

THEN

1/x < 1/y

Examples:1. x = 3, y = 2, x > ySo, 1/3 < 1/2

2. x = -2, y=-3, x > ySo, -1/2 < -1/3

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]

If ax^2+bx+c > 0, where a > 0,Thenx DO NOT LIE between xL and xU


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(xL and xU are lower and upper limits of x respectively)

If ax^2+bx+c < 0, where a > 0,Thenx LIES between xL and xU

(xL and xU are lower and upper limits of x respectively)

i. If | X1 | > | X2 |, Then X1 > X2 OR- X1 > X2

ii. If | X1 | < | X2 |, Then X1 < X2 OR- X1 < X2

|x - a| < r, where a is a positive real number and a is a fixed real number,thena-r < x < a+r

|x - a| > r, where a is a positive real number and a is a fixed real number,thenx < a-r (OR) x > a+r

Golden Rule:When a NEGATIVE value is multiplied both sides in an INEQUALITY, then the INEQUALITY SIGNreverses/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--]

Always Remember to:

Plug in the solutions back to the absolute equations to see whether it satisfies, whenever youget multiple solutions while solving absolute equations (A very useful takeaway especially for solving DSquestions)

Example:Solve, |x| = 3x - 2.

i. x = 3x - 22x = 2x = 1.


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ii. -x = 3x - 24x = 2x = 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 = 1in |x| = 3x - 2,Equation SATISFIES!

Sub.x = 1/2in |x| = 3x - 2,Equation DOES NOT SATISFY!

Hence, x = 1 is the ONLY solution.

Another point to remember:

Example:

-1/10 < n < 1/10

After taking reciprocal of n, FLIP SIGNS!

-10 > n > 10

|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.

That is the end of "INEQUALITIES". I will start posting on the chapter "AVERAGES".

Forthcoming points on "AVERAGES" are all very important takeaways. I'm sure that you will definitelybenefit from these points.

Weighted Averages:

In weighted averages involving 2 groups, if you have any 3of the following, you can ALWAYS FIND thefourth:


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- Average of Group #1- Average of Group #2- Overall Average- Ratio of Group #1 to Group #2

Solve thisGMATPrep questionto understand the trick!

NOTE:This takeaway is especially useful for Data Sufficiency questions for which you sometimes do notneed to solve for the answer. When you see any 3 points in the question and statement and when youare asked to calculate the fourth point, you DON'T need to solve actually. You can straight away concludethat the statements are sufficient.

Courtesy:Ian Stewart, GMAT Expert.

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 OverallAvg.

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 = 2/1B:G = 2:1

The class is 2/(2+1) OR 2/3 boys

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:
http://www.beatthegmat.com/mba/2009/12/13/breaking-down-gmatprep-weighted-average-problems-part-1-of-2http://www.beatthegmat.com/mba/2009/12/13/breaking-down-gmatprep-weighted-average-problems-part-1-of-2http://www.beatthegmat.com/mba/2009/12/13/breaking-down-gmatprep-weighted-average-problems-part-1-of-2http://www.beatthegmat.com/mba/2009/12/13/breaking-down-gmatprep-weighted-average-problems-part-1-of-2


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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.

--[ IMPORTANT POINTS ]--

If a constant percentageis added/subtracted from each term of the set, then the new SD is alsoadded/subtractedby that constant percentage.

If a constantis added/subtracted from each term of the set, then the new SD remains constant.

If you add data that is fartherfrom the mean,then SD will increase.

If you add data that is closerto the mean,then SD will decrease.

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.

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 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


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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].

Folks, That's the end of "Averages" chapter. I will start posting in "RATIOS". There are a few importantpoints to be noted which may be useful to you in the exam.

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


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9. If a:b:c :: d:e:f, then a/d = b/e = c/f = K (a constant)

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 constantdifference 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 referthis post]--

iii. Sum of the AP series, Sn = n/2 * [2a + (n-1)*d]ORSn = n/2 * [a + tn]

If a constant number (C) is added/subtractedfrom 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/dividedby each term of an AP, then

i. Resulting sequence is also AP.ii. Common difference (d) = C * Old Common differenceiii. Sum of the NEW series = C * Sum of the OLD series

If a1, a2, ... anand b1, b2, ... bnare 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 Aand d(B) is the common difference of series B

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.
http://www.beatthegmat.com/papgust-s-gmat-math-flashcards-directory-t58902-15.html#260738http://www.beatthegmat.com/papgust-s-gmat-math-flashcards-directory-t58902-15.html#260738http://www.beatthegmat.com/papgust-s-gmat-math-flashcards-directory-t58902-15.html#260738http://www.beatthegmat.com/papgust-s-gmat-math-flashcards-directory-t58902-15.html#260738


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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)

If a fixed non-zero constant is multipliedwith each term of GP, then

i. Resulting sequence is also GP

ii. Same Common ratio.

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 thecommon ratio of series B

That's the end of "Series & Progressions". I will start posting flashcards on the next chapter "SETTHEORY".

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 3set question on the GMAT that had a "none" component is not often seen.]

We can simplify the second equation to:


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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 Venndiagram.

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 circleAND the C circle. In other words, it's been counted three times. To get a true count, therefore, we mustsubtract it twice.

Here's the primary principle we're using:

Every object should be counted exactly once.

Examples:

In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% ofthose asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in thesurvey liked all three of the products, what percentage of the survey participants liked more than one ofthe three products?

Let's say there's 100 people, just to use numbers instead of percents. Since 85% like at least one of 3products, we'll use 85 as our base number.

Since we have "at least one" information, we need to use the second formula:


plugging in what we know:85 = 50 + 30 + 20 - doubles - 2(5)85 = 100 - 10 - doubles85 = 90 - doublesdoubles = 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.

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.


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We use the exact same formula:


70 = (40 + 30 + 35) - (doubles) - 2(15)

70 = 105 - doubles - 30

70 = 75 - doublesdoubles = 5

That's the end of "Set Theory". Really simple isn't it.. Knowing this stuff alone is more than sufficient forGMAT (trust me!)

POSITIVE/NEGATIVE slope:(A very important concept to know)

A line with a POSITIVE slope:

i. Passes through 1st and 3rd quadrantdefinitely.

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 NEGATIVEslope:

i. Passes through 2nd and 4th quadrantdefinitely.

ii. Passes through the 1st quadrant(IF y-intercept is POSITIVE).

iii. Passes through the 3rd quadrant(IF y-intercept is NEGATIVE).

Distance b/w 2 points:

Distance b/w points (x1,y1) and (x2,y2) = ROOT [(x1-x2)^2 + (y1-y2)^2]

Co-ordinates of point 'P' dividing the join of the two points (x1,y1) and (x2,y2) INTERNALLY in the ratiom:n is,

P(x,y) = [(m*x2 + n*x1) / m+n, (m*y2 + n*y1) / m+n]


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Co-ordinates of point 'P' dividing the join of the two points (x1,y1) and (x2,y2) EXTERNALLYin the ratiom: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 INTERNALIf ratio m:n has one of m and n as NEGATIVE, then its EXTERNAL

Mid-point of a line:

Mid-point of a segment joining (x1,y1) and (x2,y2) is

(x1+x2 / 2, y1+y2 / 2)

Centroid (G) = (x1+x2+x3 / 3, y1+y2+y3 / 3)

NOTE:Centroid divides each median in the ratio 2:1.

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.

If the origin (0,0) shiftsto a point (h,k),

then old co-ordinates of point P(x,y) changes to(x-h, y-k)

If (x1,y1), (x2,y2), (x3,y3) and (x4,y) are the vertices of a PARALLELOGRAM,

then

x1+x3 = x2+x4 ANDy1+y3 = y2+y4

To prove end-points A,B,C, and D are the vertices of:


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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.

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.

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]

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 all HORIZONTAL LINESis 0..... [Lines parallel to x-axis]

Slope of allVERTICAL LINESis undefined..... [Lines parallel to y-axis. Angle of inclination of ANY line

parallel to y-axis is 90 or 270 degrees]

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.

Angle of inclination b/w 2 SLOPES:

Angle (x) between two slopes, tan x = |(m1-m2) / (1+m1*m2)|

Length of perpendicular from point (x1,y1) to the line ax+by+c = 0 is

|ax1+by1+c / root(a^2+b^2)|

Distance between two lines [PARALLEL lines] ax+by+c1 = 0 and ax+by+c2 = 0 is

|c1-c2 / root(a^2+b^2)|

Really sorry for a long break. I will be posting flashcards on "Geometry", the last topic of theseflashcards.

Types of angles:

Figure 1:


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1.Adjacent angles:Any 2 angles that share a common side separating the 2 angles and that share acommon 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 beadjacent 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 fourangles. 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]
http://www.postimage.org/image.php?v=aV1yxxJhttp://www.postimage.org/image.php?v=aV1yizS


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8. Consecutive Interior Angles:Angles are same-side interior angles. Consecutive interior angles areSUPPLEMENTARY 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]

Lines segments in Triangles:

* ALTITUDES:

Altitudes are the perpendicular segments from a vertex to the opposite sides. The three lines containingthe 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 threemedians 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 bisectorsmeet 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 anISOSCELES triangle.

An equiangular quadrilateral DOES NOT have to be equilateral.

An equilateral quadrilateral DOES NOT have to be equiangular. [--- Unlike Equilateral Triangle --]

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?

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.

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)

Regular Polygons:

* One point in its interior that is equidistant from its vertices is called the center of the regular polygon.

* AnApothegm 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]

Similar Polygons:

* Two polygons with the same shape.


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* 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.

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

Perimeters and Areas of Similar Triangles:

When two triangles are similar, the reduced ratio of any two corresponding sides is called thescalefactor of the similar triangle.
http://www.postimage.org/image.php?v=aVnHblShttp://www.postimage.org/image.php?v=PqCpLl9


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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.

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