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NguyenHaiHa
17-08-2006, 04:25 PM
During my preparation for GRE, I complied this stuff. These experiences were contributed by korean and chinese GRE aspirants who succeeded in GRE. However, learning Math formulae by heart is not the rule of thumb for GRE. For those who are new bees of GRE, the best advice is that you should read GRE Math Review of ETS in an effort to understand the range of math knowledge tested in GRE exams, familarize yourself with Math vocabulary and then brush up on the basic Math concepts

NguyenHaiHa

Mixtures first....

1. when you mix different quantities (say n1 and n2) of A and B, with different strengths or values v1 and v2 then their mean value vm after mixing will be:
Vm = (v1.n1 + v2.n2) / (n1 + n2)

you can use this to find the final price of say two types of rice being mixed or final strength of acids of different concentration being mixed etc....

the ratio in which they have to be mixed in order to get a mean value of vm can be given as:
n1/n2 = (v2 - vm)/(vm - v1)

When three different ingredients are mixed then the ratio in which they have to be mixed in order to get a final strength of vm is:
n1 : n2 : n3 = (v2 - vm)(v3 - vm) : (vm - v1)(v3 - vm) : (vm - v2)(vm - v1)

2. If from a vessel containing M units of mixtures of A & B, x units of the mixture is taken out & replaced by an equal amount of B only .And If this process of taking out & replacement by B is repeated n times , then after n operations,

Amount of A left/ Amount of A originally present = (1-x/M)^n

3. If the vessel contains M units of A only and from this x units of A is taken out and replaced by x units of B. if this process is repeated n times, then:

Amount of A left = M [(1 - x/M)^n]

This formula can be applied to problem involving dilution of milk with water, etc...

EXPLAINATION TO THE ABOVE FORMULA

when you mix different quantities (say n1 and n2) of A and B, with different strengths or values v1 and v2 then their mean value vm after mixing will be:
Vm = (v1.n1 + v2.n2) / (n1 + n2) (I assume that you understood this... )
vm (n1 + n2) = v1 n1 + v2 n2
n1 (vm - v1) = n2 (v2 - vm)
so, n1/n2 = (v2 - vm)/(vm - v1) ----> (1)

similarly if you mix n2 and n3, then their ratio would be given by
n2/n3 = (v3 - vm)/(vm - v2) ----> (2)

now assume we mix n1, n2 and n3 of different ingredients of value v1, v2 and v3. the individual ratios (1) and (2) will still be the same.

now combine these ratios to get n1:n2:n3 by making the denominators common
n1/n2 = (v2 - vm)(v3 - vm)/(vm - v1)(v3 - vm) and
n2/n3 = (v3 - vm)(vm - v1)/(vm - v2)(vm - v1)

rearrange this and you will get the formula:
n1 : n2 : n3 = (v2 - vm)(v3 - vm) : (vm - v1)(v3 - vm) : (vm - v2)(vm - v1)
Hope this is clear...
PROGRESSION:

Sum of first n natural numbers: 1 +2 +3 + .... + n = [n(n+1)]/2
Sum of first n odd numbers: 1 + 3 + 5 + .... upto n terms = n^2
Sum of first n even numbers: 2 + 4 + 6 + ... upto n terms = n(n+1)

ARITHMETIC PROGRESSION

nth term of an Arithmetic progression = a + (n-1)d
Sum of n terms in an AP = s = n/2 [2a + (n-1)d]
where, a is the first term and d is the common differnce.

If a, b and c are any three consecutive terms in an AP, then 2b = a + c

GEOMETRIC PROGRESSION

nth term of a GP is = a[r^(n-1)]
sum of n terms of a GP:
s = a [(r^n - 1)/(r-1)] if r > 1
s = a [(1 - r^n)/(r-1)] if r < 1]

sum of an infinite number of terms of a GP is
s(approx.) = a/ (1-r) if r <1

If a, b and c are any three consequtive terms in a GP, then b^2 = ac

HARMONIC PROGRESSION

A series of non-zero numbers is said to be harmonic progression (abbreviated H.P.) if the series obtained by taking reciprocals of the corresponding terms of the given series is an arithmetic progression.
For example, the series 1 +1/4 +1/7 +1/10 +..... is an H.P. since the series obtained by taking reciprocals of its corresponding terms i.e. 1 +4 +7 +10 +... is an A.P.

A general H.P. is 1/a + 1/(a + d) + 1(a + 2d) + ...

nth term of an H.P. = 1/[a +(n -1)d]

Three numbers a, b, c are in H.P. iff 1/a, 1/b, 1/c are in A.P.
i.e. iff 1/a + 1/c = 2/b
i.e. iff b= 2ac/(a + c)
Thus the H.M. between a and b is H = 2ac/(a + c)
----------------------------------------------------------------------------------------
If A, G, H are arithmetic, geometric and harmonic means between two distinct, positive real numbers a and b, THEN
1. G² = AH i.e. A, G, H are in G.P.
2. A, G, H are in descending order of magnitude i.e. A > G > H.

NguyenHaiHa
17-08-2006, 04:30 PM
Problems on trains

a km/hr = (a* (5/18)) m/s

a m/s = (a* (18/5)) km/hr

Time taken by a train of length l metres to pass a pole or a standing man or a signal post is equal to the time taken by the train to cover l metres.

Time taken by a train of length l metres to pass a stationary object of length b metres is the time taken by the train to cover (l+b) metres.

Suppose 2 trains or 2 bodies are moving in the same direction at u m/s and v m/s, where u>v, then their relative speed = (u-v) m/s

Suppose 2 trains or 2 bodies are moving in the opposite direction at u m/s and v m/s, where u>v, then their relative speed = (u+v) m/s

If 2 trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then the time taken by the trains to cross each other = (a+b)/(u+v) sec

If 2 trains of length a metres and b metres are moving in same directions at u m/s and v m/s, then the time taken by the faster train to cross the slower train = (a+b)/(u-v) sec

If 2 trains(or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then

(A's speed): (B's speed) = (root(B):root(A))

EXPLAINATION TO THE ABOVE FORMULA

EXP 1: It is Speed of train A: Speed of train B = sqrt(b) : sqrt(a)

EXP2: If 2 trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then

(A's speed) : (B's speed) = (root(b) : root(a))
where a and b number of seconds.

FORMULA FOR FINDING THE UNIQUE DIVISORS AND THEIR SUMS

This one is a nice formula for finding the number of unique divisors for any number and also the sum of those divisors.... such questions are there in powerprep and so you might also get it in your real GRE.

If N is a number such that
N = (a^m) (b^n) (c^p)....
where, a, b, c, ... are prime numbers, then the number of divisors of N, including N itself is equal to:
(m+1) (n+1) (p+1)....
and the sum of the divisors of N is given by:
S = [(a^m+1) - 1]/[a - 1] * [(b^n+1) - 1]/[b - 1] * [(c^p+1) - 1]/[c- 1].....

Example:
for say N = 90, on factorizing you get 90 = 3*3*5*2= (3^2)*(5^1)*(2^1)
then the number of divisors of 90 are (2+1)(1+1)(1+1) = 12
the 12 divisors are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
And the sum of the divisors would be
[(3^3) - 1]/[3 - 1] * [(5^2) - 1]/[5 - 1] * [(2^2) - 1]/[2 - 1]
= (26/2) (24/4) (3/1)
= 234

Though this method looks more complicated than listing the factors and adding them, once you get used to this formula, it saves lot of time..

LINE FORMULAE

Let the coordinates of P1 be (x1,y1) and of P2 be (x2,y2)

- The distance from P1 to P2 is:
d = sqrt[(x1-x2)2+ (y1-y2)2]
- The coordinates of the point dividing the line segment P1P2 in the ratio r/s are:
([r x2+s x1]/[r+s], [r y2+s y1]/[r+s])
- As a special case, when r = s, the midpoint of the line segment has coordinates:
([x2+x1]/2,[y2+y1]/2)
- The slope m of a non-vertical line passing through the points P1 and P2:
slope = m = (y2 -y1)/(x2 -x1)
 Two (non-vertical) lines are parallel if their slopes are equal.
 Two (non-vertical) lines are perpendicular if the product of their slopes = -1.
 Slope of a perpendicular line is the negative reciprocal of the slope of the given line.
FORMULAE FOR POPULATION RELATED QUESTIONS
The population of a town decreases by 'x%' during the first year, decreases by 'y%' during the second year and again decreases by 'z%' during the third year. If the present population of the town is 'P', then the population of the town three years ago was::

P*100*100*100
-----------------------
(100-x)(100-y)(100-z).

The population of a town is 'P'.It decreases by 'x%' during the first year, decreases by 'y%' during the second year and again deceases by 'z%' during the third year. The population after 3 years will be:
P*(100-x)(100-y)(100-z)
--------------------------
100*100*100.

If 'X' litres of oil was poured into a tank and it was still 'x%' empty, then the quantity of oil that must be poured into the tank in order to fill it to the brim is:
X*x
------- litres.
100 - x

If 'X' liters of oil was poured into a tank and it was still 'x%' empty, then the capacity of the tank is:
X*100
---------- litres.
100 - x

A candidate scoring x% in an examination fails by "a" marks, while another candidate who scores y% marks gets "b" marks more than the minimum required pass marks. Then the maximum marks for that exam =
100(a+b)
----------
y-x .

The pass marks in an examination is x%. If a candidate who secures y marks fails by z marks, then the maximum marks is given by

100(y+z)
-----------
x.

NguyenHaiHa
17-08-2006, 04:31 PM
Permutations And Combinations

1. If one operation can be performed in m ways and another operation in n ways, then the two operations in succesion can be done in m*n ways

2. The linear permutation of n distinct objects (that is, the number of ways in which these n objects can be arranged is n! and the circular permutation of n distinct objects is (n-1)! But if the clockwise and anticlockwise directions are indistinguishable then the circular permutations of n different things taken at a time is (n-1)!/2

3. But out of these n objects, if there are n1 objects of a certain type, n2 of another type and n3 of another, and so on, Then the number of arrangements (linear permutations) possible is n!/n1!n2!...nz!

4. The total number of ways of arranging r things from n things is given by nPr = n!/(n-r)!

5. The number of ways to select r things out of n things is given by nCr = n!/(r!*(n-r)!)

6. nPr = r! * nCr

Suppose you have a name with n letters, and there are k1 of one letter, k2 of another letter, and so
on, up to kz. For example, in ELLEN,

n = 5, k1 = 2 [two E's], k2 = 2 [two L's], k3 = 1 [one N]).

Then the number of rearrangements is n!/k1!k2!...kz!

Numbers And Percentages

COUNTING

SUM OF FIRST n NATURAL NUMBERS = n(n+1)/2
Sum of first n ODD integers = n*n
Sum of first n EVEN integers = n(n+1)
Sum of the squares of the first n integers = n(n+1)(2n+1)/6
Sum of the cubes of first n integers =(n(n+1)/2)^2
IF n is even, then
No. of odd no.s from 1 to n is n/2
No. of even no.s from 1 to n is n/2

If n is odd then,
No. of odd no.s from 1 to n is (n+1)/2
No. of even no.s from 1 to n is (n-1)/2

POWERS AND INDICES

To find the unit digit of p^n

If there is an odd no. in the unit place of p eg 741,843 etc
Multiply the unit digit by itself until u get 1.

Example:
If u need to find the unit digit of (743)^38:

Multiply 3 four times to get 81.

(743)^38=(743)^36 X (743)^2

36 is a multiple of 4, and 3 when multiplied 4 times gives 1 in the unit digit.Therefore, when multiplied 9 x 4 times, it will still give 1 in the unit digit.
the unit digit of (743)^38, hence will be 1 x 9 =9

In short
(..1)^n =(..1)
(.3)^4n=(..1)
(.7)^4n=(..1)
(.9)^2n=(.1)

If the unit digit of p is even and u need to find the unit digit of (p)^n

Multiply the unit digit of p by itself until a 6 is in the unit place
(2)^4n=(.6)
(.4)^2n=(.6)
(.6)^n=(.6)
(.8)^4n=(.6)

For numbers ending with 1,5,6, after any times of multiplication, you get only 1, 5, 6 respectively.

Number of numbers divisible by a certain integer:
How many numbers upto 100 are divisible by 6?

Soln:
Divide 100 by 6, the resulting quotient is the required answer
Here,
100/6 = 16x6+4 16 is the quotient and 6 is the remainder.
Therefore, there are 16 numbers within 100 which are divisible by 6.

PERCENTAGES
 If the value of a number is first increased by x% and later decreased by x%, the net change is always A DECREASE= (x^2)/100
 if the value of a number is first increased by x% and then decreased by y%, then there is (x-y-(xy/100))% increase if positive , and decrease if negative
 If the order of increase or decrease is changed, THE RESULT IS UNAFFECTED
 If the value is increased successively by x% and y%,then final increase is given by x+y+(xy/100) %

NguyenHaiHa
17-08-2006, 04:33 PM
GEOMETRY - TRIANGLES
1. Area of a triangle with base b and height h = (1/2)*b*h
2. The area of an equilateral triangle with side a is [sqrt(3)/4]*a^2
3. The area of any triangle given the length of its 3 sides a, b and c:is sqrt[s(s-a)(s-b)(s-c)] where s= (a+b+c)/2

SOME USEFUL FACTS

Number Theory

The product of any three consecutive integers is divisible by 6.

Similarly, the product of any four consecutive integers is divisible by 24.

Permutation and Combination

When n dice (n > 1) are rolled simultaneously, the number of outcomes in which all n dice show the same number is 6, irrespective of the value of n.

Similarly, when n fair coins (n > 1) are tossed simultaneously, the number of outcomes in which all n coins turn up as heads or as tails is 1, irrespective of the value of n.

Speed Time

When an object travels the first x hours at p km/hr and the next x hours at q km/hr, the average speed of travel is the arithmetic mean of p and q.

However, when the object travels the first x kms at p km/hr and the next x kms at q km/hr, the average speed is the harmonic mean of p and q.
Number Theory

Any perfect square has an odd number of factors including 1 and the number itself and a composite number has an even number of factors including 1 and itself.

Any perfect square can be expressed in the form 4n or 4n+1.

Profit Loss

If the selling price of 2 articles are equal and 1 of them is sold at a profit of p% and the other at a loss of p%, then the 2 trades will result in a cumulative loss of ((p^2)/100)%.

If the cost of price of 2 articles are equal and 1 of them is sold at a profit of p% and the other at a loss of p%, then the 2 trades will result in no profit or loss.

Progressions

Arithmetic mean of 'n' numbers will always be greater than the geometric mean of those 'n' numbers which will be greater than the Harmonic mean of those 'n' numbers.

Arithmetic mean of 2 numbers = geometric mean of '2' numbers = harmonic mean of '2' numbers if both the numbers are equal.
TIME AND WORK
If A can do a piece of work in x days, then As one days work=1/x
 If the ratio of time taken by A and B in doing a work is x:y, then, ratio of work done is 1/x :1/y=y:x. And the ratio in which the wages is to be distributed is y:x
 If A can do a work in x days and B can do the same work in y days, then A and B can together do the work in (xy)/(x+y) days
 If a men or b women can do a piece of work in x days, then m men and n women can together finish the work in (abx)/(an+bm) days
 If A is x times efficient than B, and working together, they finish the work in y days, then Time taken by A=y(x+1)/(x), Time taken by B=y(x+1)
 If A and B can finish a work in x and ax days respectively, that is if A is a times efficient than B, then working together, they can finish the work in (ax)/(a+1) days
 If A and B working together can complete a work in x days, whereas B working alone can do the same work in y days, ten, A alone will complete the work in (xy)/(y-x) days.
 Pipe A can fill a tank in x hrs and B can empty a tank in y hrs.If both pipes are opened together, the tank will be filled in (xy)/(y-x) hrs
 A pipe can fill a cistern in x hrs but due to leakage in the bottom, it is filled in y hrs, then the time taken by the leak to empty the cistern is (xy)/(y-x) hrs
GEOMETRY STUFF

Prisms

Volume = Base area X Height
Surface = 2b + Ph (b is the area of the base P is the perimeter of the base)

Cylinder

Volume = r2 h
Surface = 2rh

Pyramid

V = 1/3 bh
b is the area of the base
Surface Area: Add the area of the base to the sum of the areas of all of the triangular faces. The areas of the triangular faces will have different formulas for different shaped bases.

Cones

Volume = 1/3 r2 x height = 1/3 r2h
Surface = r2 + rs = r2 + r

Sphere

Volume = 4/3 r^3
Surface area = 4r^2

Distance of a Point from a Line

The perpendicular distance d of a point P (x 1, y 1) from the line ax +by +c = 0 is given by:
d =| ax1 +by1 +c|/[ (a² +b²)]

Simple And Compound Interest

1. Simple Interest = PNR/100

where, P --> Principal amount
N --> time in years
R --> rate of interest for one year

2. Compound interest (abbreviated C.I.) = A -P =

where A is the final amount, P is the principal, r is the rate of interest compounded yearly and n is the number of years

3. When the interest rates for the successive fixed periods are r1 %, r2 %, r3 %, ..., then the final amount A is given by A =

4. S.I. (simple interest) and C.I. are equal for the first year (or the first term of the interest period) on the same sum and at the same rate.

5. C.I. of 2nd year (or the second term of the interest period) is more than the C.I. of Ist year (or the first term of the interest period), and C.I. of 2nd year -C.I. of Ist year = S.I. on the interest of the first year.

NguyenHaiHa
17-08-2006, 04:36 PM
STATISTICS
1. Mean.
(i) Mean (for ungrouped data) = where x1, x2, x3, ..., xn are the observations and n is the total no. of observations.
(ii) Mean (for grouped data) = , where x1, x2, x3, ..., xn are different variates with frequencies f1, f2, f3, ..., fn respectively.
(iii) Mean for continuous distribution.
Let there be n continuous classes, yi be the class mark and fi be the frequency of the ith class, then
mean = (Direct method)
Let A be the assumed mean, then
mean = A + , where di = yi -A (Short cut method)
If the classes are of equal size, say c, then
mean = A +c x , where ui = (Step deviation method)
ANALYTICAL GEOMETRY

LINES - BASICS:

1. The equation of X axis: y =0
2. The equation of Y axis: x = 0
3. Equation of straight line parallel to X axis: y =a, where a is any constant
4. Equation of straight line parallel to Y axis: x =a, where a is any constant
5. Equation of a straight line through a given point (x1, y1) and having a given slope m: y -y1 = m (x - x1)
6. Equation of a straight line through a given point (0, 0) and having a given slope m: y = m x
7. Equation of a straight line with a slope m and y-intercept c is: y = mx + c
8. Equation of a straight line passing through two points (x1, y1) and (x2, y2) is:
(y -y1)/(y2 - y1) = (x -x1)/(x2 -x1)
9. Equation of a straight line whose x and y intercepts are a and b is:
x/a + y/b = 1
10. The length of the perpendicular drawn from origin (0,0) to the line Ax + By + C =0 is :
C/ sqrt(A^2 + B^2)
11. Length of the perpendicular from (x1, y1) to the line Ax + By + C =0 is:
Ax1 + By1 +C / sqrt(A^2 + B^2)
12. The point of intersection of two lines a1x + b1y +c1 = 0 and a2x + b2y +c2 = 0 is :
([b1*c2 - b2*c1]/[a1*b2 - a2*b1], [c1*a2 - c2*a1]/[a1*b2 - a2*b1])
13. The condition for concurrency of three lines a1x + b1y +c1 = 0, a2x + b2y +c2 = 0 and a3x + b3y +c3 = 0 is (in determinant form)
| a1 b1 c1 |
| a2 b2 c2 | = 0
| a3 b3 c3 |
14. The angle between two lines y = m1x + c1 and y = m2x + c2 is tan inverse of the modulus of :
[(m1 - m2)/(1 + m1*m2)]
15. Condition for parallelism of two lines with slopes m1 and m2 is m1 = m2
16. Condition for perpendicularity of two lines with slopes m1 and m2 is m1*m2=-1

CIRCLES:

17. General equation of a circle with centre (x1, y1) and radius r is:
(x - x1)^2 + (y - y1)^2 = r^2
18. The equation of a circle whose diameter is the line joining the points (x1, y1) and (x2, y2) is :
(x - x1)(x - x2) + (y - y1)(y - y2) = 0
19. The equation of the tangent to the circle x^2 + y^2 = a^2 (where a is the radius of the circle) at the point (x1, y1) on it is :
x*x1 + y*y1 + a^2
20. The condition for y = mx + c to be a tangent to the circle x^2 + y^2 = a^2 is :
c^2 = a^2 (1 + m^2)

Multiplication of 2digit by 2 digit number

ab
x cd
------
pqrs

1. first multiply bd - write down the unit fig at s carry over the tens fig.
2. Multiply axd & bxc add them together and also add the carry over from step 1 write down the units fig at r and carry over the tens fig.
3. Multiply axc and add the carry over from step 2. write down at pq.

TRIGONOMETRY

For angle A = 0, 30° (π/6), 45° (π/4), 60° (π/3), 90° (π/2):

sin A = (sqrt0)/2, (sqrt1)/2, (sqrt2)/2, (sqrt3)/2, (sqrt4)/2
cos A = (sqrt4)/2, (sqrt3)/2, (sqrt2)/2, (sqrt1)/2, (sqrt0)/2
tan A = 0, (sqrt3)/3, 1, sqrt3, undefined

In any triangle:
sine = (opposite side) / hypotenuse
cosine = (adjacent side) / hypotenuse
tan = (opposite side)/(adjacent side) = (sine/cosine)

Probability - 'The Rules'

1. If two events are mutually exclusive (i.e. they cannot occur at the same time), then the probability of them both occurring at the same time is 0. then: P(A and B) = 0 and P(A or B) = P(A) + P(B)

2. if two events are not-mutually exclusive (i.e. there is some overlap) then: P(A or B) = P(A) + P(B) - P(A and B)

3. If events are independent (i.e. the occurrence of one does not change the probability of the other occurring), then the probability of them both occurring is the product of the probabilities of each occurring. Then: P(A and B) = P(A) * P(B)

4. If A, B and C are not mutually exclusive events, then P(A or B or C) = P(A) + P(B) + P(C) - P(A and B) - P(B and C) - P(C and A) + P(A and B and C)

and = intersection
or = union

NguyenHaiHa
17-08-2006, 04:37 PM
STATISTICS  HARMONIC MEAN

The harmonic mean of x1,...,xn is
n / (1/x1 + ... + 1/xn)
As the name implies, it's a mean (between the smallest and largest values). An example of the use of the harmonic mean: Suppose we're driving a car from Amherst (A) to Boston (B) at a constant speed of 60 miles per hour. On the way back from B to A, we drive a constant speed of 30 miles per hour (damn Turnpike). What is the average speed for the round trip?

We would be inclined to use the arithmetic mean; (60+30)/2 = 45 miles per hour. However, this is incorrect, since we have driven for a longer time on the return leg. Let's assume the distance between A and B is n miles. The first leg will take us n/60 hours, and the return leg will take us n/30 hours. Thus, the total round trip will take us (n/60) + (n/30) hours to cover a distance of 2n miles. The average speed (distance per time) is thus:
2n / {(n/60) + (n/30)} = 2 / (1/20) = 40 miles per hour.
The reason that the harmonic mean is the correct average here is that the numerators of the original ratios to be averaged were equal (i.e. n miles at 60 miles/hour versus n miles at 30 miles/hour). In cases where the denominators of two ratios are averaged, we can use the arithmetic mean.

SOME USEFUL SHORTCUTS

* Product of 2 numbers is the produst of their LCM & HCF.

* LCM of a fraction = LCM of numerator/HCF 0f denominator.

*HCF of a fraction = HCF of numer./LCM of denom.
Ratio & Proportion:

* if a/b = c/d = e/f = .....

then, a/b = c/d = e/f =(a+c+e+...)/(b+d+f+...)

* If a/b = c/d,

Then,

i) b/a = d/c

ii) a/c = b/d

iii) (a+b)/ b = (c+d)/d

iv) (a-b)/b = (c-d)/d

v) (a+b)/(a-b) = (c+d)/(c-d)

FORMULAE ON INTEREST

Money in Compound Interest gets doubled in 70/r years (approximately)

ie. P(1+r/100)^N = 2P when N=70/r

DIVISIBILITY RULES

Divisibility by:

2 If the last digit is even, the number is divisible by 2.
3 If the sum of the digits is divisible by 3, the number is also.
4 If the last two digits form a number divisible by 4, the number is also.
5 If the last digit is a 5 or a 0, the number is divisible by 5.
6 If the number is divisible by both 3 and 2, it is also divisible by 6.
7 Take the last digit, double it, and subtract it from the rest of the number; if the answer is divisible by 7 (including 0), then the number is also.
8 If the last three digits form a number divisible by 8, then so is the whole number.
9 If the sum of the digits is divisible by 9, the number is also.
10 If the number ends in 0, it is divisible by 10.
11 Alternately add and subtract the digits from left to right. If the result (including 0) is divisible by 11, the number is also.
Example: to see whether 365167484 is divisible by 11, start by subtracting:
3-6+5-1+6-7+4-8+4 = 0; therefore 365167484 is divisible by 11.
12 If the number is divisible by both 3 and 4, it is also divisible by 12.
13 Delete the last digit from the number, then subtract 9 times the deleted digit from the remaining number. If what is left is divisible by 13,then so is the original number.

NguyenHaiHa
17-08-2006, 04:40 PM
STATISTICS AGAIN

1. Mean of a distribution x1, x2, x3, ......, xn is given by the formula:

where n is the number of terms in the given set.

2. Median value of an ordered distribution y1, y2, y3, ......., yn-1, yn can be given as:

3. Mode is the most common value obtained (or value that occurs at highest frequency) in a set of observations.

4. The sample variance may be computed as

where is the sample mean.

5. The square root of the sample variance of a set of N values is the sample standard deviation

EXPLAINATION OF THE ABOVE TOPIC (FOR NON-MATHEMATICAL BACKGROUND STUDENTS

Say that x1, x2, x3, x4, x5, ...., xn are n draws from a (random) sample. Then:
Step 1: Compute the mean, i.e. m =[ Sum xi (i=1,..., n) ] / n
Step 2: Compute the squared deviation of each observation from its mean, i.e.
For x1 --------> (x1-m)^2
For x2---------> (x2-m)^2
.....
For xn---------> (xn-m)^2
Step 3: The variance is V= [(x1-m)^2 + (x2-m)^2 + .... + (xn-m)^2 ] / n
Step 4: The s.d. is s.d. = V^(1/2)

Example: Let x1=10, x2= 20 and x3=30
Then:
(1) m=20
(3) V = [ (10-20)^2 + 0 + (30-20)^2] / 3 = 200/3
(4) s.d. = (20/3) ^ (1/2)

- If perimeters of a square and parallelogram are equal, then area of a square is always greater than area of a parallelogram.
- Similarly, if perimeters of a square and circle are same, then area of a circle is greater than area of a square.

Standard deviation- normal distribution

Z values needed to be memorized (see illustration):

- Area within .5 Standard Deviation above and below the mean is 38%
- Area within 1 Standard Deviation above and below the mean is 68%
- Area within 2 Standard Deviation above and below the mean is 95%
- Area within 3 Standard Deviation above and below the mean is 99.7

- Area below 1 standard deviation is 84%
- Area below 2 standard deviation is 97.7%
- Area above 1 standard deviation is 15.8%
- Area above 2 standard deviation is 2.27%

NUMBER FACTS

If a < b, then a + c < b + c
If a < b, then a - c < b  c

Multiplication/Division Properties for Inequalities
when multiplying/dividing by a positive value
If a < b AND c is positive, then ac < bc
If a < b AND c is positive, then a/c < b/c
when multiplying/dividing by a negative value
If a < b AND c is negative, then ac > bc
If a < b AND c is negative, then a/c > b/c

Natural (or Counting) Numbers : N = {1, 2, 3, 4, 5, ...}

Whole Numbers : {0, 1, 2, 3, 4, 5, ...}

Integers : Z = {..., -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ...}

Real Numbers : R = {x | x corresponds to point on the number line}

If a cube of sides n*n*n is painted and then divided into 1*1*1 size cubes then number cubes with NO face painted is given by (n-2)^2

If SD of x1, x2, x3, ... xn is sigma then SD of x1+k , x2+k, x3+k ... Xn+k is also sigma

If SD of x1, x2, x3, ... xn is sigma then SD of x1*k , x2*k, x3*k ... Xn*k is k*sigma

Variance (kx) = k^2 Variance(x)

Binomial probability mass function: P(x) = nCx * p^x * q ^ (n-x)
where x is happening event, n is total number of event, p is probability of happening of event and q is probability of not happening.

A remainder rule to remember:

If a product of 2 integers, x and y is divided by an integer n, then the remainder that you get will be the product of the remainders when x is divided by n and y is divided by n.

R[] ---> remainder function

R[(1046*1047*1048)/11] = R[1046/11]*R[1047/11]*[1048/11] = 1*2*3 = 6

Note: Sometimes the product of the remainders will be greater than the original divisor. In this case you'll have to repeat the process.

Pick's theorem
Pick's theorem provides an elegant formula for the area of a simple lattice polygon: a lattice polygon whose boundary consists of a sequence of connected nonintersecting straight-line segments.

The formula is Area = I +B/2  1where I = number of interior lattice points ( ) and
B = number of boundary lattice points ( )
For example, the area of the simple lattice polygon in the figure is
31 + 15 /2  1 = 37.5.

The interior and boundary lattice points of the fourteen pieces of the Stomachion are indicated on the second figure. Using Pick's theorem the areas of the fourteen pieces can be determined as in the above example; e.g., the blue piece in the upper right-hand corner has area
18 + 14 /2  1 = 24

NguyenHaiHa
17-08-2006, 04:41 PM
MORE ON PERCENTAGES

ONE VARIABLE INCREASED/ DECREASED PROBLEMS:

PRICE INCREASED AND REDUCTION OF THE CONSUMPTION:

1. Price of sugar is increased 25%. How much percent must a house hold must reduce his consumption of sugar so as not to increase his expenditure?

how much time u require to this problem? just try this short cut less then 5 sec u will get the answer

% REDUCTION= (INCREASE/100+INCREASE)* 100
lets try this with short cut

increase = 25% so reduction = (25/ 100+25 ) * 100
= (25/125) * 100
= 1/5* 100
= 20 %
so house hold have to decrease 20% of their consuption to keep constant .

PRICE DECREASED INCREASE IN CONSUPTION:

PROBLEM : 2
Certain familyhave fixed budget for ice cream purchase for year . but,Ice cream price decreased by 20% due to winter season. find by how much % a consumer must increase his consumpion of ice creame so as not to decrease his expenditure.

Here the short cut

%INCREASE IN CONSUMPTION = (REDUCTION/ 100- REDUCTION) *100

just as mentioned above

decrease icecreame price= 20

= 20/(100-20) *100
=20/80 * 100
=1/4 *100
= 25%

BOTH VARIABLES INCREASED/ DECREASED PROBLEMS

TYPE3:

Petrol tax is increased by 20% and the costumer comsumption also increased by 20%.Find the % increase or decrease in the expenditure
OR
Water tax is increased by 20% and consumption also increased by 20% find what is the net effect in change?

the short cut: [(A+B) + AB]/100

increase A: 20
increase B: 20
= (20+20) + (20*20)/100
= 40+ 400/100
= 40 +4
= 44 % net increase

ONE INCREASED ONTHER DECREASED PROBLEMS:

Shop keeper decreased the price of a article by 20% and then increased the artical by 30% what is the net effect of the artical is it increased or decreased?

SHORT CUT IS SAME AS ABOVE

so first decreased the price so we have to take as negative value for A
decrease A : -20%
increase A : 30%
= (-20 + 20)+ (-20)*20/100
= (0) +(-400)/100
= -4%
so net effect is 4% loss to the shop keeper.

Nothingness
03-10-2006, 01:08 PM
Thanks in advance for your share! I'll try. Hope that works to me!

yamafusi
13-08-2007, 05:23 PM
cam on nhung thong tin ve GRE cua ban. minh cung muon thi GRE nhung chua biet gi ve GRE ca.