All Math Formulas(1)

8/13/2019 All Math Formulas(1)

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Volumes

a) Triangle :Area = ½ base * height - universal

Area of equilateral triangle = sqrt(3)/4 side2

b) Rectangle : length * breadthc) Square : side 2

Diagonal = side * sqrt(2)

Area = ½ product of diagonal

d)

Parallelogram : base * heighte) Rhombus: ½ product of diagonalsf) Circle : Area PiR2 Circumference 2PiRg) Cuboid : is the rectangular solid having 6 faces with all the faces as rectangles

Volumes : l*b*h

Area for 4 walls : 2 (i+b)* h

Total surface area of cuboid : 2 (lb + bh + lh)

Body diagonals of cuboid : sqrt(l2+b2+h2)

h) Cube

Volume = a3

Total surface area of Cube 6 * a2

i) Cylinder

Volume Pi R2H

Curved surface= 2PiRH

Total surface = 2PiR(R+H)

j) Cone

Volume = 1/3 PiR2

h

Curved surface area = PiRL where L=Sqrt (R2+H2)

Total surface area : PiR(R+L)

k) SphereVolume : 4/3 PiR3

Surface : 4 Pi R2

Triangle

GEOMETRY FORMULA



a) Sum of angles is 180b) Exterior angle is equal to sum of interior angle non adjacent to it i.e. angles other

than the complementary angle of the exterior anglec) Sum of any two sides is more than the third sided) Equilateral triangle is the triangle with all the sides as same

Area = √ 3/4side2 Height √ 3/2side

Perimeter = 3Side

e) Right angle triangle

45 -90-45 triangle Hypotenuse = √ 2 * Side

30-60-90 triangle 30 side = ½ hypotenuse

60 side = ( √ 3/2) hypotenuse

f) If the angles of two triangles are same then they are similar then all the attributesthat they have will have same proportion – heights, sides etc.

Rectangle

a) Diagonals are equal and bisect each otherb) Diagonal = √ (a2+b2)

c) Of all the given rectangles of same area or perimeter square will have the maximumarea

Parallelogram

a) Diagonals bisect each other

b) Opposite angles are same

c) Each diagonal divides the parallelogram in triangles of same areaTrapezium

a) Only one pair of opposite side are parallel to each otherb) Area = ½ * (sum of parallel sides) * height

c) Isosceles trapezium is the one that is inscribed in a circle. The oblique sides areequal. The opposite angles made by oblique sides with the parallel side are equal.

Circle

a) Tangents drawn from an external side are equal

CUBOID: Let length=l, breadth=b and height=h units. Then,

Volume=(l*b*h) cubic unitsSurface area=2(lb+bh+hl) squints

Length of longest diagonal =√ (l² + b² +h²).

CUBE: let each edge of a cube be of length ‘a’. Then,



V olume =a*a*a cubic units

Surface area =6(a*a) squints

Length of longest diagonal =√3 a.

CYLINDER: let radius of base= r &height =h Then,

Volume = πr2 h

Surface = 2πrh

CONE: Let radius of base=r & height = h. Then,

Slant height, l = √ [(h 2)+(r 2)]

Volume = 1/3 πr2h

Curved Surface area = 2πrl sq. units

SRHERE: Let the radius of the sphere be ‘r’. Then,

Volume = 4/3 πr3

Surface area = 4πr

2

Prisms

Volume = Base area X Height

Surface = 2b + Ph (b is the area of the base P is the perimeter of the base)

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.

HEXAGON

Area of the Regular Hexagon is = 3 √ 3 (side)²

Trapezeum

Length of Arc ABC = 2Βr & 360º Where B is the angle made at the center.

Area of the Sector AOC = Βr² & 360º

Where B is the angle made at the center.

a + c = b + d = 180º

Area of the Cyclic Quadrilateral = √ ( s - a) ( s – b) ( s – c) ( s – d)

Where a, b, c and d are the lengths of the Cyclic Quadrilateral

Length of Direct common tangent



Length of Direct common tangent = √ d ² - (r1 – r2) ²Where r1 and r2 is radius of the circles.. d is the distance between the centers.

Length of Transverse common tangent

Length of Transverse common tangent= √ d ² - (r1 – r2) ²Where r1 and r2 is radius of the circles.. d is the distance between the centers.

(AB) ² = DB x CB

Length of common chord:

Where o and p are center of the circles.

.r 1 and r 2 are radius of the circles with center o

and p respectively.

Area of the triangle Δ opa is given by

=> √ S ( s - a) ( s – b) ( s – c) = ½ x h x d

In the above equation we know S, a, b, c, and d , so we can find out h.

Length of common chord is given by 2h



Length of body diagonal or longest diagonal of a cuboid = √ ( l ² + h ² + b ² )

Length of body diagonal or longest diagonal of a cube = √3 (side)

SOME USEFUL INFORMATION ABOUT GEOMETRY

- 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.

For any regular polygon, the sum of the exterior angles is equal to 360 degrees,

hence measure of any external angle is equal to 360/n (where n is the number of

sides)

For any regular polygon, the sum of interior angles =(n-2)*180 degrees

So measure of one angle is (n-2)/n *180

If any parallelogram can be inscribed in a circle, it must be a rectangle.

If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i.e.

oblique sides equal).

52. For an isosceles trapezium, sum of a pair of opposite sides is equal in length to

the sum of the other pair of opposite sides (i.e. AB+CD = AD+BC, taken in order)

53. For any quadrilateral whose diagonals intersect at right angles, the area of the

quadrilateral is :

0.5*d1*d2, where d1, d2 are the length of the diagonals.

For a cyclic quadrilateral, area = √((s-a) * (s-b) * (s-c) * (s-d)),



where s=(a + b + c + d)/2

Further, for a cyclic quadrilateral, the measure of an external angle is equal to the

measure of the interior opposite angle.

Area of a Rhombus = Product of Diagonals/2

54. Given the coordinates (a, b); (c, d); (e, f); (g, h) of a parallelogram , the

coordinates of the meeting point of the diagonals can be found out by solving for

[(a + e)/2, (b + f)/2] = [(c + g)/2, (d + h)/2]

55. Area of a triangle

1/2*base*altitude

1/2*a*b*sin C (or) 1/2*b*c*sin A (or) 1/2*c*a*sin B

root(s*(s-a)*(s-b)*(s-c)) where s=(a+b+c)/2

a*b*c/(4*R) where R is the circumradius of the triangle

r*s ,where r is the inradius of the triangle

Equilateral Tr iangle

Height = √3/2 ( side) Area = √3/4 ( side ) ² R = 2/3 Height

. r = 1/3 Height

56. In any triangle

a=b*cos C + c*cos B

b=c*cos A + a*cos C

c=a*cos B + b*cos A

a/sin A=b/sin B=c/sin C=2R, where R is the circumradius

cos C = (a2 + b

2 - c

2)/2ab

sin 2A = 2 sin A * cos A

cos 2A = cos2 A - sin2A

57. The ratio of the radii of the circumcircle and incircle of an equilateral triangle

is 2:1

58. Appollonius Theorem

In a triangle ABC, if AD is the median to side BC, thenAB2 + AC

2 = 2(AD

2 + BD

2) or 2(AD

2 + DC

2)



59. I sosceles Tr iangleArea = b/4 √ (4 a² - b²)

In an isosceles triangle, the perpendicular from the vertex to the base or the

angular bisector from vertex to base bisects the base.

In any triangle the angular bisector of an angle bisects the base in the ratio of the

other two sides.

60. The quadrilateral formed by joining the angular bisectors of another

quadrilateral is always a rectangle.

61. Let W be any point inside a rectangle ABCD, then,

WD2 + WB

2 = WC

2 + WA

2

62. Let a be the side of an equilateral triangle, then, if three circles are drawn

inside this triangle such that they touch each other, then each circle’s radius is

given by a/(2( √3 +1)).

63. Distance between a point (x1, y1) and a line represented by the equation

ax + by + c=0 is given by |ax1+by1+c|/Sq(a2+b2)Distance between 2 points (x1, y1) and (x2, y2) is given by

Sq((x1-x2)2+ (y1-y2)2)

64. Where a rectangle is inscribed in an isosceles right angled triangle, then, the

length of the rectangle is twice its breadth and the ratio of area of rectangle to

area of triangle is 1:2.

Co-Ordinate Geometry

A ( Xı, Yı) and B ( X2, Y2 ) are 2 points

Distance between AB = √ (Y2 – Yı ) ² + (X2 - Xı ) ²Slope of Line AB, m = (Y2 – Yı)

(X2 - Xı) Equation of the straight line passing through A ( Xı, Yı) and B ( X2, Y2 ) is

given by



(Y – Yı ) = m (X - Xı ) where

m = Y2 – Yı

X2 -

Xı In Equation of the straight line y = m x + c. m is the slope of the line

. c is the Y intercept .

a is x intercept.b is y intercept

If two lines are parallel then the slopes are equal.

M1 = M 2

Where M1 and M 2 are slopes of the lines.

If two lines are perpendicular then the product of the slopes is -1.

M 1 x M 2 = -1.

Where M1 and M 2 are slopes of the lines

Equations of straight line parallel to a x + b y + c 1 = 0 is a x + b y + c 2 = 0

Equations of straight line perpendicular to a x + b y + c 1 = 0 is

b x - a y + c 2 = 0

Perpendicular distance between two parallel lines

a x + b y + c 1 = 0 is a x + b y + c 2 = 0 is

= │c 1 - c 2 │

√ a² + b²

Arithmetic Progression. .

a , a +d, a+2d, a+3d, a+4d,………………………….. a is the first term of the series.. d is the common difference.

Nth term of the series is given by tn = a + ( n-1) d.

Sum of N terms is given by Sn = n/2 [ 2a + ( n – 1) d]

Sn = n/2 [ L + D ]

L is the first termD is the last term



Geometric Progression. a, ar, ar², ar³, ………………….

Sum of N terms is given by Sn = a (r ⁿ - 1)/ ( r – 1)

Sum of infinite terms is given by S∞ = a/ ( 1 – r)

Where r is less than 1

Sum of infinite terms is given by S∞ = a/( r - 1)

Where r is greater than 1

Speed and Distance:

If a person is traveling from ‘A’ to ‘B’ with a km/ hr, and in return from

B to A with b km/hr his average speed is given by = 2 x a x b / (a+b) km/hr

If the distance traveled is same with two different speeds then average speed

is given by = 2 x a x b / (a+b)km/ hr

If the time taken is same with different speeds then average speed is given by = (a + b)/ 2 km/ hr. a km/ hr and b km/hr are different speeds.

Trains

If speed of the train is ‘A’ km/hr, length of the train is ‘B’ km. Then time taken to cross a pole in hrs = B/ A km/hr

If speed of the train is ‘A’ km/hr, length of the train is ‘B’ km.

Then time taken to cross a platform of length ‘C’ km in hrs =(B +C) km / A km/ hr .

Two trains ‘A’ and ‘B’ are traveling with a km/hr and b km/hrrespectively. Lengths of the train A and B are X km and Y km respectively.

Time taken to cross the slower train by faster train = (X +Y) km / (a+b) km/hr.

(If they are traveling in opposite direction)

Time taken to cross the slower train by faster train (If they are traveling in same

direction, Starting point of the faster train is at the end of the slower train)= (X +Y) km / │a - b│ km/hr.

Time taken to cross the slower train by faster train (If they are traveling in samedirection, Starting point of the faster train and slower train are on the same line)



= (Length of the faster train) km / │a - b│ km/hr.

A person is roving the boat at speed of ‘a’ km/hr. length of the river he has tocross the river is x km. Water flows in the river at ‘b’ km/hr.

Then time taken to cross the river (With the current) = x km / (a + b) km /hr

Then time taken to cross the river (Against the current) = x km/ (a - b) km /hr

Simple InterestSimple Interest = P x N x R / 100

P = Principal N = Number of YearsR = Rate of Interest

Compound Interest = P ( 1 + R / 100 )ⁿ -- PP = Principaln = Number of Years

R = Rate of Interest

Interest Compounded Half yearly

P ( 1 + (R/2) / 100 )²ⁿ - P

Rate of Interest for Installment Interest = 24 x I x 100 / N ( F + L)

I = Installment Charge N = Number of InstallmentsF = Principal left after first Installment

L = Principal left after last Installment

Mixtures :

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*(n1 + n2) = (v1.n1 + v2.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 = n2

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 difference.

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[rn-1

]sum of n terms of a GP: s = a [(r

n - 1)/(r-1)] if r > 1

s = a [(1 - rn)/(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 consecutive terms in a GP, then b2 = 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. if 1/a, 1/b, 1/c are in A.P.

i.e. if 1/a + 1/c = 2/b

i.e. if 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.

----------------------------------------------------------------------------------------

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 asignal 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 = (am

) (bn) (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.

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 bynPr = 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 ofanother 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/2No. 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) %

Conversions1 mile = 1609 meters

1 mile = 5280 feet.

1mile = 8 furlong1 furlong = 220 feet

1 km/hr = (5 / 18 ) m/s

Percentages

1. If the commodity price increases by x% then the consumption has to be reduced by

x/(100+x) to maintain the amount spent2. If the commodity price decreases by x% then the consumption has to be increased

by x/(100-x) to maintain the amount spent3. if a’s income is x% more than b on b’s income then b’s income is x/(100+x)% less

than a on a’s income4. if a’s income is x% less than b on b’s income then b’s income is x/(100-x)% more

than a on a’s income5. When it is stated that a is x% more than b then it means on the base of b. thus

a = b+x%b

Simple & compound interest

1. Simple interest = PNR/100 where p = principal; N=period (years) and R = rateAmount = P+I

2. compound interest = A = P (1+R/100)^NThis in compound interest the formula will give the final amount. Principal will have

to be deducted from it to get the interest portion. In case the interest is half yearly

then rate should be halfed and period should be doubled. Similarly for quarterly.Then the interest rate for different years is different then P (1+R1/100) (1+R2/100)

(1+R3/100)

3. population. The formula for compound interest can also be used for populations. Butwhere the populations is decreasing then the sign will change to ‘- ‘ instead of ‘+’

Averages

1.

Mode is the number that occurs most no of times in given sample



2. Median is the middle number of the given sample. Where the number of items ingiven sample is odd then (n+1)/2th number and if the number is even it is simpleaverage between n/2 and n+2/2th numbers

3. Arithmetic mean : it is sum of all numbers / no of numbers4. weighted average mean: it is sum of product of numbers and their respective

weights / sum of weights

5. Geometric mean: between two numbers it is x = Sqrt (ab). If geometric mean of onegroup of numbers (a) is X and that of group (b) is Y then geometric mean of boththe groups will be (X+Y)/ (a+b)

6. Harmonic mean : between two numbers = number of numbers / (sum of reciprocal of

numbers) ie. 2ab/(a+b). This also gives the average speed when same lengthdistances are covered in different speeds

7. GM^2 = AM*HM

Ratio proportion and variation

1. comparing two quantities as ratios:a. both the quantities should be of same kindb. both should have the same measurement per unitc. ratio is a pure number i.e. it does not have any measurement. It just denotes

how many times one quantity is of one of other2. compounding : if two different ratios (say a:b and c:d) are expressed in different

units then if we require to combine these two ratios then it will be AC:BD3. if a/b=c/d=e/f then the ratio is equal to a+c+e / b+d+f

Mixtures and allegation

1. Alligation ruleQuantity of cheap = Price of dear – average price

Quantity of dear Average price – price of cheap

2. if a vessel contains ‘a’ litres of liquid A and if ‘b’ litres are withdrawn and replaced byliquid B then if ‘b’ litres of the mixture is again withdrawn and replaced by liquid B.the operation is repeated ‘n’ number of times thenLiquid a left in vessel = ((a-b)/a)^n

Initial liquid in vessel

Profit or loss

If the a and b are two successive discounts that have been given then effective discount rate

will (a+b-(ab/100))

Time speed and distance



Km/hr * (5/18) = m/s

Km = 5/8 mile

1. while traveling if a person changes his speed in m:n ratio then the time taken willalso change in n:m ratio

2. if the A to B is traveled in T1 time and a speed and B to A if T2 time and b speedthen the average speed is give by(2ab) / (a+b) ………. Harmonic mean

Distance is given by (T1+T2) (2ab/(a+b))

Or (T1-T2) (2ab/(a-b))

Or (a-b) (T1T2 / (T1+T2))

3. if two persons start towards each other from different points and arrive at twopoints in a hrs and b hrs respectively after having met then ratio of their speed isgiven by SQRT (b) / SQRT (a) = a’s speed / b’s speed

Work

1. if A can do a work in x days then 1/xth work is done in one day2. if A is X times better workman than B then A will take 1/xth time of that taken by

B3. if A and B do work in X and Y days then they will complete the same work in XY /

(X+Y) days and in one day (X+Y)/ XY days work will be done4. if A and B can do a piece of work in X days and if A alone will be able to complete

the work in a days more than X and b can in b days more than X then X2 = ab5. if a pipe can fill a vessel in x hrs then 1/xth part of the vessel is filled in one hour6. if A pipe is X times bigger than B then A will take 1/X times lesser time than B7. if A and B fill the pipe in m and n hours respectively then both will fill the pipe in

MN / (m+n) hours and (m+n) / mn th part of vessel will be filled in one hour8. if one inlet pipe fills the vessels in M hrs and other pipe empties the vessel in N

hrs then the vessel will be filled in MN / (N-M) hrs. and (N-M)/MN the part will befilled

9. if an inlet pipes taken X minutes to fill the cistern and has taken a minutes longerthen the leak will empty the cistern in a*(1+a/x) minutes

10. A and B can fill the cistern in X hrs and A alone will fill the same in a minutesmore than X and b can fill it in b minuted more than X then X= sqrt (ab)

Clocks and Calendars

a) A dial of the clock is divided into 60 parts each called minute spaces

b) The hour hand goes 5 minute spaces in one hour and minutes hand goes 60 minutespaces in one hour. Thus the minute hand gains 55 minute spaces over the hourhand in one hour

c) When two hands are in 90 degree they are 15 minute spaces apart. This occurs twicein an hour.



d) When the two hands are in opposite directions they are 30 minute hands apart thisoccurs once in an hour

e) Two hands are in straight line when they coincide or are in opposite directionsf) The angle between the two hands = 6(x-11/12m)

X= hour hand convert into minute spaces i.e.* 5 of the earlier clock

M = the later part of the time i.e. minutes

g) The years that are divisible by 400 are the only ones that are leap year.

When mentioned RST then S will be the top vertices

Numbers

a) ODD +/- ODD = EVEN

b) ODD +/- EVEN = ODDc) EVEN +/- EVEN = EVENd) ODD * ODD = ODDe) ODD * EVEN = EVEN

f) EVEN * EVEN = EVENg) HCF of two numbers is the number that divides both the numbers exactlyh) LCM of two numbers is the number that is divided by both the numbers exactlyi) HCF*LCM= product of both the numbers

j) HCF of fraction is HCF of the numerators / LCM of denominatorsk) LCF of fractions is LCM of numerators and HCF of denominatorsl) if three numbers a,b,c are divided by N in such manner that r is the remainder each

time then smallest value of N is LCM of (a,b,c)+r

m) if three numbers a,b,c divide N is such manner that remainders are p,q,r then if (a-p)= (b-q) = (c-r) then the smallest value of N is LCM of (a,b,c) – (a-p)

Indices

a) Am * An=A(m+n) b) Am / An=A(m-n) c) (Am)n=A(m*n)

d) Nth root of A = A1/n e) 1/A = A-n

f) AnBn=(AB)n g) (A+B)2=A2+B2+2abh) (A-B)2=A2+B2—2abi) (A+B)2-(A-B)2=4AB

j) (A+B)2+ (A-B)2=2(A+B)2

Inequalities

a) If a>b then a+m>b+m

b) If a>b then am>bm for m>0 and am<bm for m<0. whenever inequality is multipliedby a negative quantity then the sign reverses



c) If a>b then 1/a<1/b

Logarithms

a) Log n A = logm A/logmn

Progressions

a) Arithematic progressionsSum = (n/2)*[2A+(n-1)d]

= (n/2) * (a+l)

Nth Term = A+(N-1)D

N = number of terms

D is the common differences

A is the first term

L is the last term

b) Geometric progression

Arn-1=Nth Term

Sum = A(1-rn)/(1-r)

Geometric mean = (ab)1/2

c) Harmonic mean = it is the arithmetic mean of reciprocals of numbersSum and nth number of harmonic mean is reciprocal of arithmetic mean

Harmonic mean of two numbers is 2ab/(a+b)

Permutation and combination

a) Fundamental principal of addition: if one thing can be done m number of ways andother thing can be done in n number of ways independent of other. Then either ofthem can be done in (m+n) ways

b) Fundamental principal of multiplication: if one thing can be done m number of ways

and other thing can be done in n number of ways independent of other. Then eitherof them can be done in (m*n) ways

c) Permutation : permutation of n objects taken r at a time is the arrangement in a

straight line of r objects taken at a time denoted by N!/(N-R)!d) The number of permutation of n objects taken all at a time = n!e) The number of permutations of n objects taken all at a time when p of them are like,

q are like = n!/p!q!

f) Combination is the selection of r objects in n objects. Denoted as N!/(n-r)!r!

g) Number of permutations of n objects taken all at a time in circle (n-1)!h) When the repetition of allowed then permutation nr



Some points that can be helpful

1. 2N-1 or 2m+1 is always odd

2. If √ N is an integer then N is always an integer

3. N^3-n = N(n-1)(n+1)4. Think negative also as the maximum traps are account of positive to negative

changes – best input nos 2-,2,3,-2,0.5,-0.5

5. Distance between two points on coordinate is given by the formula √ ((A-x)^2+(B-

Y)^2)) where x.y and a,b are the pair of coordinates6.

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