apty formulas

Time and Distance

1. Speed, Time and Distance:

Speed =Distance , Time =

Distance , Distance = (Speed x Time).

Time Speed

1. km/hr to m/sec conversion:

x km/hr = x x5

m/sec.18

2. m/sec to km/hr conversion:

x m/sec = x x18 km/hr.5

3. If the ratio of the speeds of A and B is a : b, then the ratio of the

the times taken by then to cover the same distance is 1 : 1 or b : a.a b

4. Suppose a man covers a certain distance at x km/hr and an equal distance at ykm/hr. Then,

the average speed during the whole journey is2xy

km/hr.x + y

Time and Work

1. Work from Days:

If A can do a piece of work in n days, then A's 1 day's work = 1.n

2. Days from Work:

If A's 1 day's work = 1, then A can finish the work in n days.n

3. Ratio:

If A is thrice as good a workman as B, then:

Ratio of work done by A and B = 3 : 1.

Ratio of times taken by A and B to finish a work = 1 : 3.

Problems on Trains

1. km/hr to m/s conversion:

a km/hr = a x5

m/s.18

2. m/s to km/hr conversion:

a m/s = a x18

km/hr.5

3. Formulas for finding Speed, Time and Distance4. Time taken by a train of length l metres to pass a pole or standing man or a signal post is

equal to the time taken by the train to cover l metres.5. Time taken by a train of length l metres to pass a stationery object of length bmetres is the

time taken by the train to cover (l + b) metres.6. Suppose two trains or two objects bodies are moving in the same direction at um/s and v m/s,

where u > v, then their relative speed is = (u - v) m/s.7. Suppose two trains or two objects bodies are moving in opposite directions at um/s and v m/s,

then their relative speed is = (u + v) m/s.8. If two trains of length a metres and b metres are moving in opposite directions atu m/s

and v m/s, then:

The time taken by the trains to cross each other = (a + b) sec.(u + v)

9. If two trains of length a metres and b metres are moving in the same direction atu m/s and v m/s, then:

The time taken by the faster train to cross the slower train = (a + b) sec.(u - v)

10. If two 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) = (b : a)

Permutation and Combination

1. Factorial Notation:

Let n be a positive integer. Then, factorial n, denoted n! is defined as:

n! = n(n - 1)(n - 2) ... 3.2.1.

Examples:

i. We define 0! = 1.ii. 4! = (4 x 3 x 2 x 1) = 24.iii. 5! = (5 x 4 x 3 x 2 x 1) = 120.

2. Permutations:

The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Examples:

i. All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).

ii. All permutations made with the letters a, b, c taking all at a time are:( abc, acb, bac, bca, cab, cba)

3. Number of Permutations:

Number of all permutations of n things, taken r at a time, is given by:

nPr = n(n - 1)(n - 2) ... (n - r + 1) = n!(n - r)!

Examples:

i. 6P2 = (6 x 5) = 30.ii. 7P3 = (7 x 6 x 5) = 210.iii. Cor. number of all permutations of n things, taken all at a time = n!.

4. An Important Result:

If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind, such that (p1 + p2 + ... pr) = n.

Then, number of permutations of these n objects is = n!(p1!).(p2)!.....(pr!)

5. Combinations:

Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.

Examples:

1. Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.

Note: AB and BA represent the same selection.

2. All the combinations formed by a, b, c taking ab, bc, ca.

3. The only combination that can be formed of three letters a, b, c taken all at a time is abc.

4. Various groups of 2 out of four persons A, B, C, D are:

AB, AC, AD, BC, BD, CD.

5. Note that ab ba are two different permutations but they represent the same combination.

6. Number of Combinations:

The number of all combinations of n things, taken r at a time is:

nCr = n! = n(n - 1)(n - 2) ... to r factors .(r!)(n - r)! r!

Note:

i. nCn = 1 and nC0 = 1.ii. nCr = nC(n - r)

Examples:

i. 11C4 = (11 x 10 x 9 x 8) = 330.(4 x 3 x 2 x 1)

ii. 16C13 = 16C(16 - 13) = 16C3 = 16 x 15 x 14 = 16 x 15 x 14 = 560.3! 3 x 2 x 1

Partnership

1. Partnership:

When two or more than two persons run a business jointly, they are calledpartners and the deal is known as partnership.

2. Ratio of Divisions of Gains:

I. When investments of all the partners are for the same time, the gain or loss is distributed among the partners in the ratio of their investments.

Suppose A and B invest Rs. x and Rs. y respectively for a year in a business, then at the end of the year:

(A's share of profit) : (B's share of profit) = x : y.

II. When investments are for different time periods, then equivalent capitals are calculated for a unit of time by taking (capital x number of units of time). Now gain or loss is divided in the ratio of these capitals.

Suppose A invests Rs. x for p months and B invests Rs. y for q months then,

http://www.indiabix.com/aptitude/partnership/

(A's share of profit) : (B's share of profit)= xp : yq.

3. Working and Sleeping Partners:

A partner who manages the the business is known as a working partner and the one who simply invests the money is a sleeping partner.

Percentage

1. Concept of Percentage:

By a certain percent, we mean that many hundredths.

Thus, x percent means x hundredths, written as x%.

To express x% as a fraction: We have, x% = x .100

Thus, 20% = 20 = 1.100 5

To express

aas a percent: We have,

a=

ax 100

%.b b b

Thus,1

=1

x 100%

= 25%.4 4

2. Percentage Increase/Decrease:

If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is:

Rx 100

%(100 + R)

If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is:

Rx 100

%(100 - R)

3. Results on Population:

Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:

1. Population after n years = P 1 +R n100

2. Population n years ago = P

1 +R n100

4. Results on Depreciation:

Let the present value of a machine be P. Suppose it depreciates at the rate of R% per annum. Then:

1. Value of the machine after n years = P 1 -R n100

2. Value of the machine n years ago =P

1 -R n100

3. If A is R% more than B, then B is less than A byR

x 100%.(100 + R)

4. If A is R% less than B, then B is more than A byR

x 100%.(100 - R)

Calendar

1. Odd Days:

We are supposed to find the day of the week on a given date.

For this, we use the concept of 'odd days'.

In a given period, the number of days more than the complete weeks are calledodd days.

2. Leap Year:

(i). Every year divisible by 4 is a leap year, if it is not a century.

(ii). Every 4th century is a leap year and no other century is a leap year.

Note: A leap year has 366 days.

Examples:

i. Each of the years 1948, 2004, 1676 etc. is a leap year.ii. Each of the years 400, 800, 1200, 1600, 2000 etc. is a leap year.iii. None of the years 2001, 2002, 2003, 2005, 1800, 2100 is a leap year.

3. Ordinary Year:

The year which is not a leap year is called an ordinary years. An ordinary year has 365 days.

4. Counting of Odd Days:

1. 1 ordinary year = 365 days = (52 weeks + 1 day.)

1 ordinary year has 1 odd day.

2. 1 leap year = 366 days = (52 weeks + 2 days)

1 leap year has 2 odd days.

3. 100 years = 76 ordinary years + 24 leap years

= (76 x 1 + 24 x 2) odd days = 124 odd days.

= (17 weeks + days) 5 odd days.

Number of odd days in 100 years = 5.

Number of odd days in 200 years = (5 x 2) 3 odd days.

Number of odd days in 300 years = (5 x 3) 1 odd day.

Number of odd days in 400 years = (5 x 4 + 1) 0 odd day.

Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. has 0 odd days.

5. Day of the Week Related to Odd Days:

No. of days: 0 1 2 3 4 5 6

Day: Sun.

Mon.

Tues.

Wed.

Thurs.

Fri.

Sat.

Chain Rule

1. Direct Proportion:

Two quantities are said to be directly proportional, if on the increase (or decrease) of the one, the other increases (or decreases) to the same extent.

Eg. Cost is directly proportional to the number of articles. (More Articles, More Cost)

2. Indirect Proportion:

Two quantities are said to be indirectly proportional, if on the increase of the one, the orther decreases to the same extent and vice-versa.

Eg. The time taken by a car is covering a certain distance is inversely proportional to the speed of the car. (More speed, Less is the time taken to cover a distance.)

Note: In solving problems by chain rule, we compare every item with the term to be found out.

Pipes and Cistern

1. Inlet:

A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.

Outlet:

A pipe connected with a tank or cistern or reservoir, emptying it, is known as an outlet.

2. If a pipe can fill a tank in x hours, then:

part filled in 1 hour = 1.x

3. If a pipe can empty a tank in y hours, then:

part emptied in 1 hour = 1.y

4. If a pipe can fill a tank in x hours and another pipe can empty the full tank in yhours (where y > x), then on opening both the pipes, then

the net part filled in 1 hour =1

-1

.x y

5. If a pipe can fill a tank in x hours and another pipe can empty the full tank in yhours (where x > y), then on opening both the pipes, then

the net part emptied in 1 hour =1

-1

.y x

Probability

1. Experiment:

An operation which can produce some well-defined outcomes is called an experiment.

2. Random Experiment:

An experiment in which all possible outcomes are know and the exact output cannot be predicted in advance, is called a random experiment.

Examples:

i. Rolling an unbiased dice.ii. Tossing a fair coin.iii. Drawing a card from a pack of well-shuffled cards.iv. Picking up a ball of certain colour from a bag containing balls of different colours.

Details:

v. When we throw a coin, then either a Head (H) or a Tail (T) appears.vi. A dice is a solid cube, having 6 faces, marked 1, 2, 3, 4, 5, 6 respectively. When we

throw a die, the outcome is the number that appears on its upper face.vii. A pack of cards has 52 cards.

It has 13 cards of each suit, name Spades, Clubs, Hearts and Diamonds.

Cards of spades and clubs are black cards.

Cards of hearts and diamonds are red cards.

There are 4 honours of each unit.

There are Kings, Queens and Jacks. These are all called face cards.

3. Sample Space:

When we perform an experiment, then the set S of all possible outcomes is called the sample space.

Examples:

1. In tossing a coin, S = {H, T}2. If two coins are tossed, the S = {HH, HT, TH, TT}.3. In rolling a dice, we have, S = {1, 2, 3, 4, 5, 6}.

4. Event:

Any subset of a sample space is called an event.

5. Probability of Occurrence of an Event:

Let S be the sample and let E be an event.

Then, E S.

P(E) = n(E) .n(S)

6. Results on Probability:

i. P(S) = 1ii. 0 P (E) 1iii. P( ) = 0iv. For any events A and B we have : P(A B) = P(A) + P(B) - P(A B)v. If A denotes (not-A), then P(A) = 1 - P(A).

Profit and Loss

IMPORTANT FACTS

Cost Price:

The price, at which an article is purchased, is called its cost price, abbreviated as C.P.

Selling Price:

The price, at which an article is sold, is called its selling prices, abbreviated as S.P.

Profit or Gain:

If S.P. is greater than C.P., the seller is said to have a profit or gain.

Loss:

If S.P. is less than C.P., the seller is said to have incurred a loss.

IMPORTANT FORMULAE

1. Gain = (S.P.) - (C.P.)2. Loss = (C.P.) - (S.P.)3. Loss or gain is always reckoned on C.P.4. Gain Percentage: (Gain %)

Gain % =Gain x 100C.P.

5. Loss Percentage: (Loss %)

Loss % =Loss x 100C.P.

6. Selling Price: (S.P.)

SP =(100 + Gain %)

x C.P100

7. Selling Price: (S.P.)

SP =(100 - Loss %)

x C.P.100

8. Cost Price: (C.P.)

C.P. =100

x S.P.(100 + Gain %)

9. Cost Price: (C.P.)

C.P. =100

x S.P.(100 - Loss %)

10. If an article is sold at a gain of say 35%, then S.P. = 135% of C.P.11. If an article is sold at a loss of say, 35% then S.P. = 65% of C.P.12. When a person sells two similar items, one at a gain of say x%, and the other at a loss of x%,

then the seller always incurs a loss given by:

Loss % =Common Loss and Gain % 2

=x 2

.10 10

13. If a trader professes to sell his goods at cost price, but uses false weights, then

Gain % =Error

x 100%.(True Value) - (Error)

Problems on Ages

Important Formulas on "Problems on Ages" :

1. If the current age is x, then n times the age is nx.

2. If the current age is x, then age n years later/hence = x + n.

3. If the current age is x, then age n years ago = x - n.

4. The ages in a ratio a : b will be ax and bx.

5. If the current age is x, then 1 of the age is x.n n

Area

FUNDAMENTAL CONCEPTS

1. Results on Triangles:i. Sum of the angles of a triangle is 180°.ii. The sum of any two sides of a triangle is greater than the third side.

iii. Pythagoras Theorem:

In a right-angled triangle, (Hypotenuse)2 = (Base)2 + (Height)2.

iv. The line joining the mid-point of a side of a triangle to the positive vertex is called the median.

v. The point where the three medians of a triangle meet, is called centroid.The centroid divided each of the medians in the ratio 2 : 1.

vi. In an isosceles triangle, the altitude from the vertex bisects the base.vii. The median of a triangle divides it into two triangles of the same area.viii. The area of the triangle formed by joining the mid-points of the sides of a given

triangle is one-fourth of the area of the given triangle.2. Results on Quadrilaterals:

i. The diagonals of a parallelogram bisect each other.ii. Each diagonal of a parallelogram divides it into triangles of the same area.iii. The diagonals of a rectangle are equal and bisect each other.iv. The diagonals of a square are equal and bisect each other at right angles.v. The diagonals of a rhombus are unequal and bisect each other at right angles.vi. A parallelogram and a rectangle on the same base and between the same parallels are

equal in area.vii. Of all the parallelogram of given sides, the parallelogram which is a rectangle has the

greatest area.

Important Formulas - Area1. Pythagorean Theorem

(Pythagoras' theorem)

In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides

c2 = a2 + b2

where c is the length of the hypotenuse and a and b are the lengths of the other two sides

2. Pi is a mathematical constant which is the ratio of a circle's circumference to its diameter. It is denoted by π

π≈3.14≈2273. Geometric Shapes and solids and Important Formulas

Geometric Shapes Description Formulas

Rectangle

l = Length

b = Breadth

d= Length of diagonal

Area = lb

Perimeter = 2(l + b)

d = l2+b2−−−−−−√

Square

a = Length of a side

d= Length of diagonal

Area = a2=12d2Perimeter = 4a

d = 2√a

Parallelogram

b and c are sides

b = base

h = height

Area = bh

Perimeter = 2(b + c)

Rhombus

a = length of each side

b = base

h = height

d1, d2 are the diagonal

Area = bh(Formula 1 for area)

Area = 12d1d2 (Formula 2 for area)Perimeter = 4a

Triangle

a , b and c are sides

b = base

h = height

Area = 12bh (Formula 1 for area)

Area = S(S−a)(S−b)(S−c)−−−−−−−−−−−−−−−−−−√ where S is the semiperimeter = a+b+c2(Formula 2 for area - Heron's formula)

Perimeter = a + b + c

Radius of incircle of a triangle of area A = AS where S is the semiperimeter = a+b+c2

Equilateral Triangle

a = side

Area = 3√4a2Perimeter = 3a

Radius of incircle of an equilateral triangle of side a = a23√Radius of circumcircle of an equilateral triangle of side a = a3√

Base a is parallele to base b

Trapezium

(Trapezoid in American English)

h = height

Area = 12(a+b)h

Circle

r = radius

d = diameter

d = 2r

Area = πr2=14πd2

Circumference = 2πr=πdCircumferenced=π

Sector of Circle

r = radius

θ = central angle

Area, A =⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪θ360πr2 (if angle measure is in degrees)12r2θ (if angle measure is in radians)Arc Length, s = ⎧⎩⎨⎪⎪⎪⎪θ180πr (if angle measure is in degrees)rθ (if angle measure is in radians)

Plese note that in the radian system for angular measurement, 2π radians = 360°⇒1 radian = 180°π⇒1° = π180 radiansHence, Angle in Degrees = Angle in Radians ×180°πAngle in Radians = Angle in Degrees ×π180°

Ellipse

Major axis length = 2a

Minor axis length = 2b

Area = πab

Perimeter ≈2πa2+b22−−−−−−−√

Rectangular Solid

l = length

w = width

h = height

Total Surface Area = 2lw + 2wh + 2hl = 2(lw + wh + hl)

Volume = lwh

Cube

s = edge

Total Surface Area = 6s2

Volume = s3

Right Circular Cylinder

h = height

r = radius of base

Lateral Surface Area = (2 π r)h

Total Surface Area = (2 π r)h + 2 (π r2)

Voulme = (π r2)h

Pyramid

h = height

B = area of the base

Total Surface Area = B + Sum of the areas of the trianguar sides

Volume = 13Bh

Right Circular Cone

h = height

r = radius of base

Lateral Surface Area =πrr2+h2−−−−−−√=π rs where s is the slant height = r2+h2−−−−−−√

Total Surface Area = πrr2+h2−−−−−−√+πr2=πrs+πr2

Sphere

r = radius

d = diameter

d = 2r

Surface Area = 4πr2=πd2

Volume = 43πr3=16πd3

4. Important properties of Geometric Shapes

I. Properties of Trianglei. Sum of the angles of a triangle = 180°

ii. Sum of any two sides of a triangle is greater than the third side.

iii. The line joining the midpoint of a side of a triangle to the positive vertex is called the median

iv. The median of a triangle divides the triangle into two triangles with equal areas

v. Centroid is the point where the three medians of a triangle meet.

vi. Centroid divides each median into segments with a 2:1 ratio

vii. Area of a triangle formed by joining the midpoints of the sides of a given triangle is one-fourth of the area of the given triangle.

viii. An equilateral triangle is a triangle in which all three sides are equal

ix. In an equilateral triangle, all three internal angles are congruent to each other

x. In an equilateral triangle, all three internal angles are each 60°

xi. An isosceles triangle is a triangle with (at least) two equal sides

xii. In isosceles triangle, altitude from vertex bisects the base.

II. Properties of Quadrilaterals

A. Rectangle

i. The diagonals of a rectangle are equal and bisect each other

ii. opposite sides of a rectangle are parallel

iii. opposite sides of a rectangle are congruent

iv. opposite angles of a rectangle are congruent

v. All four angles of a rectangle are right angles

vi. The diagonals of a rectangle are congruent

B. Square

vii. All four sides of a square are congruent

viii. Opposite sides of a square are parallel

ix. The diagonals of a square are equal

x. The diagonals of a square bisect each other at right angles

xi. All angles of a square are 90 degrees.

C. Parallelogram

xii. The opposite sides of a parallelogram are equal in length.

xiii. The opposite angles of a parallelogram are congruent (equal measure).

xiv. The diagonals of a parallelogram bisect each other.

xv. Each diagonal of a parallelogram divides it into two triangles of the same area

D. Rhombus

xvi. All the sides of a rhombus are congruent

xvii. Opposite sides of a rhombus are parallel.

xviii. The diagonals of a rhombus are unequal and bisect each other at right angles

xix. Opposite internal angles of a rhombus are congruent (equal in size)

xx. Any two consecutive internal angles of a rhombus are supplementary; i.e. the sum of their angles = 180° (equal in size)

Other properties of quadrilaterals

xxi. The sum of the interior angles of a quadrilateral is 360 degrees

xxii. A square and a rhombus on the same base will have equal areas.

xxiii. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.

xxiv. Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.

xxv. Each diagonal of a parallelogram divides it into two triangles of the same area

III. Sum of Interior Angles of a polygoni. The sum of the interior angles of a polygon = 180(n - 2) degrees where n

= number of sides

Example 1 : Number of sides of a triangle = 3. Hence, sum of the interior angles of a triangle = 180(3 - 2) = 180 × 1 = 180 °

Example 2 : Number of sides of a quadrilateral = 4. Hence, sum of the interior angles of any quadrilateral = 180(4 - 2) = 180 × 2 = 360 °

Average

1. Average:

Average =

Sum of observationsNumber of observations

2. Average Speed:

Suppose a man covers a certain distance at x kmph and an equal distance at ykmph.

Then, the average speed druing the whole journey is2xy

kmph.x + y

Banker's Discount

IMPORTANT CONCEPTS

Banker's Discount:

Suppose a merchant A buys goods worth, say Rs. 10,000 from another merchant B at a credit of say 5 months. Then, B prepares a bill, called the bill of exchange. A signs this bill and allows B to withdraw the amount from his bank account after exactly 5 months.

The date exactly after 5 months is called nominally due date. Three days (known as grace days) are added to it get a date, known as legally due date.

Suppose B wants to have the money before the legally due date. Then he can have the money from the banker or a broker, who deducts S.I. on the face vale (i.e., Rs. 10,000 in this case) for the period from the date on which the bill was discounted (i.e., paid by the banker) and the legally due date. This amount is know as Banker's Discount (B.D.).

Thus, B.D. is the S.I. on the face value for the period from the date on which the bill was discounted and the legally due date.

Banker's Gain (B.G.) = (B.D.) - (T.D.) for the unexpired time.

Note: When the date of the bill is not given, grace days are not to be added.

IMPORTANT FORMULAE

1. B.D. = S.I. on bill for unexpired time.

2. B.G. = (B.D.) - (T.D.) = S.I. on T.D. = (T.D.)2

P.W.

3. T.D. P.W. x B.G.

4. B.D. =Amount x Rate x Time100

5. T.D. =Amount x Rate x Time100 + (Rate x Time)

6. Amount =B.D. x T.D.B.D. - T.D.

7. T.D. =B.G. x 100Rate x Time

Boats and Streams

1. Downstream/Upstream:

In water, the direction along the stream is called downstream. And, the direction against the stream is called upstream.

2. If the speed of a boat in still water is u km/hr and the speed of the stream is vkm/hr, then:

Speed downstream = (u + v) km/hr.

Speed upstream = (u - v) km/hr.

3. If the speed downstream is a km/hr and the speed upstream is b km/hr, then:

Speed in still water = 1(a + b) km/hr.2

Rate of stream = 1(a - b) km/hr.2

Important Formulas - Boats and Streams

1. Let the speed of a boat in still water be u km/hr and the speed of the stream be v km/hr, then

Speed downstream = (u + v) km/hr Speed upstream = (u - v) km/hr.

2. Let the speed downstream be a km/hr and the speed upstream be b km/hr, then

Speed in still water =12(a+b) km/hrRate of stream =12(a−b) km/hr

Some more short-cut methods

3. Assume that a man can row at the speed of x km/hr in still water and he rows the same distance up and down in a stream which flows at a rate of y km/hr. Then his average speed throughout the journey

= Speed downstream × Speed downstreamSpeed in still

water=(x+y)(x−y)x km/hr

4. Let the speed of a man in still water be x km/hr and the speed of a stream be y km/hr. If he takes t hours more in upstream than to go downstream for the same distance, the distance

=(x2−y2)t2y km

5. A man rows a certain distance downstream in t1 hours and returns the same distance upstream in t2hours. If the speed of the stream is y km/hr, then the speed of the man in still water

=y(t2+t1t2−t1) km/hr

6. A man can row a boat in still water at x km/hr. In a stream flowing at y km/hr, if it takes him t hours to row a place and come back, then the distance between the two places

=t(x2−y2)2x km

7. A man takes n times as long to row upstream as to row downstream the river. If the speed of the man is x km/hr and the speed of the stream is y km/hr, then

x=y(n+1n−1)

Problems on H.C.F and L.C.M

1. Factors and Multiples:

If number a divided another number b exactly, we say that a is a factor of b.

In this case, b is called a multiple of a.

2. Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or Greatest Common Divisor (G.C.D.):

The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly.

There are two methods of finding the H.C.F. of a given set of numbers:

I. Factorization Method: Express the each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.

II. Division Method: Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.

Finding the H.C.F. of more than two numbers: Suppose we have to find the H.C.F. of three numbers, then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number.

Similarly, the H.C.F. of more than three numbers may be obtained.

3. Least Common Multiple (L.C.M.):

The least number which is exactly divisible by each one of the given numbers is called their L.C.M.

There are two methods of finding the L.C.M. of a given set of numbers:

I. Factorization Method: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors.

II. Division Method (short-cut): Arrange the given numbers in a rwo in any order. Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.

4. Product of two numbers = Product of their H.C.F. and L.C.M.5. Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.6. H.C.F. and L.C.M. of Fractions:

1. H.C.F. = H.C.F. of NumeratorsL.C.M. of Denominators

2. L.C.M. = L.C.M. of NumeratorsH.C.F. of Denominators

8. H.C.F. and L.C.M. of Decimal Fractions:

In a given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without decimal point, find H.C.F. or L.C.M.

as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.

9. Comparison of Fractions:

Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.

Logarithm

1. Logarithm:

If a is a positive real number, other than 1 and am = x, then we write:m = logax and we say that the value of log x to the base a is m.

Examples:

(i). 103 1000 log10 1000 = 3.

(ii). 34 = 81 log3 81 = 4.

(iii). 2-3 = 1 log21 = -3.

8 8

(iv). (.1)2 = .01 log(.1) .01 = 2.

2. Properties of Logarithms:

1. loga (xy) = loga x + loga y

2. logax

= loga x - loga yy

3. logx x = 1

4. loga 1 = 0

5. loga (xn) = n(loga x)

6. loga x = 1logx a

7. loga x = logb x = log x.logb a log a

3. Common Logarithms:

Logarithms to the base 10 are known as common logarithms.

4. The logarithm of a number contains two parts, namely 'characteristic' and 'mantissa'.

Characteristic: The internal part of the logarithm of a number is called itscharacteristic.

Case I: When the number is greater than 1.

In this case, the characteristic is one less than the number of digits in the left of the decimal point in the given number.

Case II: When the number is less than 1.

In this case, the characteristic is one more than the number of zeros between the decimal point and the first significant digit of the number and it is negative.

Instead of -1, -2 etc. we write 1 (one bar), 2 (two bar), etc.

Examples:-

Number

Characteristic

Number

Characteristic

654.24 2 0.645

3 1

26.649 1 0.061

34 2

8.3547 0 0.001

23 3

Mantissa:

The decimal part of the logarithm of a number is known is its mantissa. For mantissa, we look through log table.

Clock

1. Minute Spaces:

The face or dial of watch is a circle whose circumference is divided into 60 equal parts, called minute spaces.

Hour Hand and Minute Hand:

A clock has two hands, the smaller one is called the hour hand or short hand while the larger one is called minute hand or long hand.

2.i. In 60 minutes, the minute hand gains 55 minutes on the hour on the hour hand.ii. In every hour, both the hands coincide once.iii. The hands are in the same straight line when they are coincident or opposite to each

other.iv. When the two hands are at right angles, they are 15 minute spaces apart.v. When the hands are in opposite directions, they are 30 minute spaces apart.vi. Angle traced by hour hand in 12 hrs = 360°vii. Angle traced by minute hand in 60 min. = 360°.viii. If a watch or a clock indicates 8.15, when the correct time is 8, it is said to be 15

minutes too fast.

On the other hand, if it indicates 7.45, when the correct time is 8, it is said to be 15 minutes too slow.

Simple Interest

1. Principal:

The money borrowed or lent out for a certain period is called the principal or thesum.

2. Interest:

Extra money paid for using other's money is called interest.

3. Simple Interest (S.I.):

If the interest on a sum borrowed for certain period is reckoned uniformly, then it is called simple interest.

Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then

(i). Simple Intereest =P x R x T100

(ii). P =100 x S.I.

; R =100 x S.I.

and T =100 x S.I.

.R x T P x T P x R

Simplification

1. 'BODMAS' Rule:

This rule depicts the correct sequence in which the operations are to be executed, so as to find out the value of given expression.

Here B - Bracket, O - of, D - Division, M - Multiplication,

A - Addition and S - Subtraction

Thus, in simplifying an expression, first of all the brackets must be removed, strictly in the order (), {} and ||.

After removing the brackets, we must use the following operations strictly in the order:

(i) of (ii) Division (iii) Multiplication (iv) Addition (v) Subtraction.

2. Modulus of a Real Number:

Modulus of a real number a is defined as

|a| =a, if a > 0-a, if a < 0

Thus, |5| = 5 and |-5| = -(-5) = 5.

3. Virnaculum (or Bar):

When an expression contains Virnaculum, before applying the 'BODMAS' rule, we simplify the expression under the Virnaculum.

Square Root and Cube Root

1. Square Root:

If x2 = y, we say that the square root of y is x and we write y = x.

Thus, 4 = 2, 9 = 3, 196 = 14.

2. Cube Root:

The cube root of a given number x is the number whose cube is x.

We, denote the cube root of x by x.

Thus, 8 = 2 x 2 x 2 = 2, 343 = 7 x 7 x 7 = 7 etc.

Note:

1. xy = x x y

2. xy

=x

=x

xy

=xy

.y y y y

Height and Distance

1. Trigonometry:

In a right angled OAB, where BOA = ,

i. sin =Perpendicular = AB;Hypotenuse OB

ii. cos =Base

=OA

;Hypotenuse

OB

iii. tan =Perpendicular = AB;Base OA

iv. cosec = 1 = OB;sin AB

v. sec = 1 = OB;cos OA

vi. cot = 1 = OA;tan AB

2. Trigonometrical Identities:i. sin2 + cos2 = 1.ii. 1 + tan2 = sec2 .iii. 1 + cot2 = cosec2 .

3. Values of T-ratios:

0°( /6)

30°

( /4)

45°

( /3)

60°

( /2)

90°

sin 012

32 1

cos 132

12 0

tan 013 1 3 not defined

4. Angle of Elevation:

Suppose a man from a point O looks up at an object P, placed above the level of his eye. Then, the angle which the line of sight makes with the horizontal through O, is called the angle of elevation of P as seen from O.

Angle of elevation of P from O = AOP.

5. Angle of Depression:

Suppose a man from a point O looks down at an object P, placed below the level of his eye, then the angle which the line of sight makes with the horizontal through O, is called the angle of depression of P as seen from O.

Compound Interest

1. Let Principal = P, Rate = R% per annum, Time = n years.2. When interest is compound Annually:

Amount = P 1 +R n100

3. When interest is compounded Half-yearly:

Amount = P 1 +(R/2) 2n100

4. When interest is compounded Quarterly:

Amount = P 1 +(R/4) 4n100

5. When interest is compounded Annually but time is in fraction, say 3 years.

Amount = P 1 + R3

x 1 + R100 100

6. When Rates are different for different years, say R1%, R2%, R3% for 1st, 2ndand 3rd year respectively.

Then, Amount = P 1 +R1

1 +R2

1 +R3

.100 100 100

7. Present worth of Rs. x due n years hence is given by:

Present Worth =x

.1 +

R100

8. Simple Interest and Compound Interest for 1 year at a given rate of interest per annum will be equal.

Partnership

1. Partnership:

When two or more than two persons run a business jointly, they are calledpartners and the deal is known as partnership.

2. Ratio of Divisions of Gains:

I. When investments of all the partners are for the same time, the gain or loss is distributed among the partners in the ratio of their investments.

Suppose A and B invest Rs. x and Rs. y respectively for a year in a business, then at the end of the year:

(A's share of profit) : (B's share of profit) = x : y.

II. When investments are for different time periods, then equivalent capitals are calculated for a unit of time by taking (capital x number of units of time). Now gain or loss is divided in the ratio of these capitals.

Suppose A invests Rs. x for p months and B invests Rs. y for q months then,

(A's share of profit) : (B's share of profit)= xp : yq.

3. Working and Sleeping Partners:

A partner who manages the the business is known as a working partner and the one who simply invests the money is a sleeping partner.

Decimal Fraction

1. Decimal Fractions:

Fractions in which denominators are powers of 10 are known as decimal fractions.

Thus,1

= 1 tenth = .1;1

= 1 hundredth = .01;10

100

99= 99 hundredths = .99;

7= 7 thousandths = .007, etc.;10

01000

2. Conversion of a Decimal into Vulgar Fraction:

Put 1 in the denominator under the decimal point and annex with it as many zeros as is the number of digits after the decimal point. Now, remove the decimal point and reduce the fraction to its lowest terms.

Thus, 0.25 = 25 = 1; 2.008 = 2008 = 251.100 4 1000 125

3. Annexing Zeros and Removing Decimal Signs:

Annexing zeros to the extreme right of a decimal fraction does not change its value. Thus, 0.8 = 0.80 = 0.800, etc.

If numerator and denominator of a fraction contain the same number of decimal places, then we remove the decimal sign.

Thus, 1.84 = 18 = 8 .

42.99 29

913

4. Operations on Decimal Fractions:i. Addition and Subtraction of Decimal Fractions: The given numbers are so placed

under each other that the decimal points lie in one column. The numbers so arranged can now be added or subtracted in the usual way.

ii. Multiplication of a Decimal Fraction By a Power of 10: Shift the decimal point to the right by as many places as is the power of 10.

Thus, 5.9632 x 100 = 596.32; 0.073 x 10000 = 730.

iii. Multiplication of Decimal Fractions: Multiply the given numbers considering them without decimal point. Now, in the product, the decimal point is marked off to obtain as many places of decimal as is the sum of the number of decimal places in the given numbers.

Suppose we have to find the product (.2 x 0.02 x .002).

Now, 2 x 2 x 2 = 8. Sum of decimal places = (1 + 2 + 3) = 6.

.2 x .02 x .002 = .000008

iv. Dividing a Decimal Fraction By a Counting Number: Divide the given number without considering the decimal point, by the given counting number. Now, in the quotient, put the decimal point to give as many places of decimal as there are in the dividend.

Suppose we have to find the quotient (0.0204 Ã· 17). Now, 204 Ã· 17 = 12.

Dividend contains 4 places of decimal. So, 0.0204 Ã· 17 = 0.0012

v. Dividing a Decimal Fraction By a Decimal Fraction: Multiply both the dividend and the divisor by a suitable power of 10 to make divisor a whole number.

Now, proceed as above.

Thus, 0.00066 = 0.00066 x 100 = 0.066 = .0060.11 0.11 x 100 11

5. Comparison of Fractions:

Suppose some fractions are to be arranged in ascending or descending order of magnitude, then convert each one of the given fractions in the decimal form, and arrange them accordingly.

Let us to arrange the fractions 3, 6 and

7 in descending order.5 7 9

Now, 3 = 0.6, 6 = 0.857, 7 = 0.777...5 7 9

Since, 0.857 > 0.777... > 0.6. So, 6 > 7 > 3.7 9 5

6. Recurring Decimal:

If in a decimal fraction, a figure or a set of figures is repeated continuously, then such a number is called a recurring decimal.

n a recurring decimal, if a single figure is repeated, then it is expressed by putting a dot on it. If a set of figures is repeated, it is expressed by putting a bar on the set.

Thus, 1 = 0.333... = 0.3;22 = 3.142857142857.... = 3.142857.

3 7

Pure Recurring Decimal: A decimal fraction, in which all the figures after the decimal point are repeated, is called a pure recurring decimal.

Converting a Pure Recurring Decimal into Vulgar Fraction: Write the repeated figures only once in the numerator and take as many nines in the denominator as is the number of repeating figures.

Thus, 0.5 = 5 ; 0.53 = 53 ; 0.067 = 67 , etc.9 99 999

Mixed Recurring Decimal: A decimal fraction in which some figures do not repeat and some of them are repeated, is called a mixed recurring decimal.

Eg. 0.1733333.. = 0.173.

Converting a Mixed Recurring Decimal Into Vulgar Fraction: In the numerator, take the difference between the number formed by all the digits after decimal point (taking repeated digits only once) and that formed by the digits which are not repeated. In the denominator, take the number formed by as many nines as there are repeating digits followed by as many zeros as is the number of non-repeating digits.

Thus, 0.16 = 16 - 1 =15 = 1 ; 0.2273 = 2273 - 22 = 2251.

90 90

6 9900 9900

7. Some Basic Formulae:i. (a + b)(a - b) = (a2 + b2)ii. (a + b)2 = (a2 + b2 + 2ab)iii. (a - b)2 = (a2 + b2 - 2ab)iv. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)v. (a3 + b3) = (a + b)(a2 - ab + b2)vi. (a3 - b3) = (a - b)(a2 + ab + b2)

vii. (a3 + b3 + c3 - 3abc) = (a + b + c)(a2 + b2 + c2 - ab - bc - ac)viii. When a + b + c = 0, then a3 + b3 + c3 = 3abc.

Volume and Surface Area

1. CUBOID

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

i. Volume = (l x b x h) cubic units.

ii. Surface area = 2(lb + bh + lh) sq. units.

iii. Diagonal = l2 + b2 + h2 units.

2. CUBE

Let each edge of a cube be of length a. Then,

i. Volume = a3 cubic units.

ii. Surface area = 6a2 sq. units.

iii. Diagonal = 3a units.

3. CYLINDER

Let radius of base = r and Height (or length) = h. Then,

i. Volume = ( r2h) cubic units.

ii. Curved surface area = (2 rh) sq. units.

iii. Total surface area = 2 r(h + r) sq. units.

4. CONE

Let radius of base = r and Height = h. Then,

i. Slant height, l = h2 + r2 units.

ii. Volume = r2h cubic units.

iii. Curved surface area = ( rl) sq. units.

iv. Total surface area = ( rl + r2) sq. units.

5. SPHERE

Let the radius of the sphere be r. Then,

i. Volume = r3 cubic units.

ii. Surface area = (4 r2) sq. units.

6. HEMISPHERE

Let the radius of a hemisphere be r. Then,

i. Volume = r3 cubic units.

ii. Curved surface area = (2 r2) sq. units.

iii. Total surface area = (3 r2) sq. units.

Note: 1 litre = 1000 cm3.

Surds and Indices

1. Laws of Indices:i. am x an = am + n

ii.

am

= am - n

an

iii. (am)n = amn

iv. (ab)n = anbn

v.

a n=

an

b bn

vi. a0 = 12. Surds:

Let a be rational number and n be a positive integer such that a(1/n) = a

Then, a is called a surd of order n.

3. Laws of Surds:i. a = a(1/n)

ii. ab = a x biii.

=ab

iv. (a)n = av.vi. (a)m = am

Ratio and Proportion

1. Ratio:

The ratio of two quantities a and b in the same units, is the fraction and we write it as a : b.

In the ratio a : b, we call a as the first term or antecedent and b, the second term or consequent.

Eg. The ratio 5 : 9 represents 5 with antecedent = 5, consequent = 9.9

Rule: The multiplication or division of each term of a ratio by the same non-zero number does not affect the ratio.

Eg. 4 : 5 = 8 : 10 = 12 : 15. Also, 4 : 6 = 2 : 3.

2. Proportion:

The equality of two ratios is called proportion.

If a : b = c : d, we write a : b :: c : d and we say that a, b, c, d are in proportion.

Here a and d are called extremes, while b and c are called mean terms.

Product of means = Product of extremes.

Thus, a : b :: c : d (b x c) = (a x d).

3. Fourth Proportional:

If a : b = c : d, then d is called the fourth proportional to a, b, c.

Third Proportional:

a : b = c : d, then c is called the third proportion to a and b.

Mean Proportional:

Mean proportional between a and b is ab.

4. Comparison of Ratios:

We say that (a : b) > (c : d) a > c.b d

5. Compounded Ratio:6. The compounded ratio of the ratios: (a : b), (c : d), (e : f) is (ace : bdf).7. Duplicate Ratios:

Duplicate ratio of (a : b) is (a2 : b2).

Sub-duplicate ratio of (a : b) is (a : b).

Triplicate ratio of (a : b) is (a3 : b3).

Sub-triplicate ratio of (a : b) is (a1/3 : b1/3).

If a = c , then a + b = c + d. [componendo and dividendo]b d a - b c - d

8. Variations:

We say that x is directly proportional to y, if x = ky for some constant k and we write, x y.

We say that x is inversely proportional to y, if xy = k for some constant k and

we write, x 1.y

Races and Games

1. Races: A contest of speed in running, riding, driving, sailing or rowing is called a race.

2. Race Course: The ground or path on which contests are made is called a race course.

3. Starting Point: The point from which a race begins is known as a starting point.

4. Winning Point or Goal: The point set to bound a race is called a winning point or a goal.

5. Winner: The person who first reaches the winning point is called a winner.

6. Dead Heat Race: If all the persons contesting a race reach the goal exactly at the same time, the race is said to be dead heat race.

7. Start: Suppose A and B are two contestants in a race. If before the start of the race, A is at the starting point and B is ahead of A by 12 metres, then we say that 'A gives B, a start of 12 metres'.

To cover a race of 100 metres in this case, A will have to cover 100 metres while B will have to cover only (100 - 12) = 88 metres.

In a 100 race, 'A can give B 12 m' or 'A can give B a start of 12 m' or 'A beats B by 12 m' means that while A runs 100 m, B runs (100 - 12) = 88 m.

8. Games: 'A game of 100, means that the person among the contestants who scores 100 points first is the winner'.

If A scores 100 points while B scores only 80 points, then we say that 'A can give B 20 points'.

True Discount

IMPORTANT CONCEPTS

Suppose a man has to pay Rs. 156 after 4 years and the rate of interest is 14% per annum. Clearly, Rs. 100 at 14% will amount to R. 156 in 4 years. So, the payment of Rs. now will clear off the debt of Rs. 156 due 4 years hence. We say that:

Sum due = Rs. 156 due 4 years hence;

Present Worth (P.W.) = Rs. 100;

True Discount (T.D.) = Rs. (156 - 100) = Rs. 56 = (Sum due) - (P.W.)

We define: T.D. = Interest on P.W.; Amount = (P.W.) + (T.D.)

Interest is reckoned on P.W. and true discount is reckoned on the amount.

IMPORTANT FORMULAE

Let rate = R% per annum and Time = T years. Then,

1. P.W. = 100 x Amount = 100 x T.D.100 + (R x T) R x T

2. T.D. = (P.W.) x R x T = Amount x R x T100 100 + (R x T)

3. Sum = (S.I.) x (T.D.)(S.I.) - (T.D.)

4. (S.I.) - (T.D.) = S.I. on T.D.

5. When the sum is put at compound interest, then P.W. =Amount

R100

Stocks and Shares

1. Stock Capital:

The total amount of money needed to run the company is called the stock capital.

2. Shares or Stock:

The whole capital is divided into small units, called shares or stock.

For each investment, the company issues a 'share-certificate', showing the value of each share and the number of shares held by a person.

The person who subscribes in shares or stock is called a share holder or stock holder.

3. Dividend:

The annual profit distributed among share holders is called dividend.

Dividend is paid annually as per share or as a percentage.

4. Face Value:

The value of a share or stock printed on the share-certificate is called its Face Value or Nominal Value or Par Value.

5. Market Value:

The stock of different companies are sold and bought in the open market through brokers at stock-exchanges. A share or stock is said to be:

i. At premium or Above par, if its market value is more than its face value.ii. At par, if its market value is the same as its face value.iii. At discount or Below par, if its market value is less than its face value.

Thus, if a Rs. 100 stock is quoted at premium of 16, then market value of the stock = Rs.(100 + 16) = Rs. 116.

Likewise, if a Rs. 100 stock is quoted at a discount of 7, then market value of the stock = Rs. (100 -7) = 93.

6. Brokerage:

The broker's charge is called brokerage.

(i) When stock is purchased, brokerage is added to the cost price.

(ii) When stock is sold, brokerage is subtracted from the selling price.

Remember:

i. The face value of a share always remains the same.ii. The market value of a share changes from time to time.iii. Dividend is always paid on the face value of a share.iv. Number of shares held by a person

= Total Investment = Total Income =Total Face Value .

Investment in 1 share Income from 1 share Face of 1 share

7. Thus, by a Rs. 100, 9% stock at 120, we mean that:i. Face Value of stock = Rs. 100.ii. Market Value (M.V) of stock = Rs. 120.iii. Annual dividend on 1 share = 9% of face value = 9% of Rs. 100 = Rs. 9.iv. An investment of Rs. 120 gives an annual income of Rs. 9.v. Rate of interest p.a = Annual income from an investment of Rs. 100

=9

x 100 % = 71%.120 2

Alligation or Mixture

1. Alligation:

It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of desired price.

2. Mean Price:

The cost of a unit quantity of the mixture is called the mean price.

3. Rule of Alligation:

If two ingredients are mixed, then

Quantity of cheaper = C.P. of dearer - Mean PriceQuantity of dearer Mean price - C.P. of cheaper

We present as under:

C.P. of a unit quantityof cheaperC.P. of a unit quantityof dearer(c) Mean Price

(m)(d)

(d - m) (m - c)

(Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c).

4. Suppose a container contains x of liquid from which y units are taken out and replaced by water.

After n operations, the quantity of pure liquid = x 1 -y n

units.x

Geometry FormulasAngles

Sum of Interior Angles of Polygon = (n-2)(180)n = number of sides of a polygonCentral Angle = 2(Inscribed Angle)

Area

Square: A = a2

Rectangle: A = lwParallelogram: A = bhTrapezoid: A = .5(a+c)h, where a and c are the lengths of the parallel sides

Circles

π = pi = 3.1415Area: A = πr2

Circumference: C = 2πrCentral Angle = 2(Inscribed Angle)Area of Sector = (x/360)πr2

Perimeter

Square: P = 4lRectangle: P = 2w + 2lParallelogram: P = 2b + 2a, where a and b are the lengths of the non-parallel sidesCircle: P = 2πr

Triangles

Area: A = .5bhPythagorean Theorem: A2 + B2 = C2 where A = one leg, B = the other leg, C = hypotenuse

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Volume

Cube: V = a3 where a is the length of a sideRectangular Solid: V = hwlCylinder: V = πr2h

Important Formulas - Probability or Chance1. Probability or Chance

Probability or chance is a common term used in day-to-day life. For example, we generally say, 'it may rain today'. This statement has a certain uncertainty.

Probability is quantitative measure of the chance of occurrence of a particular event.

2. Experiment

An experiment is an operation which can produce well-defined outcomes.

3. Random Experiment

If all the possible outcomes of an experiment are known but the exact output cannot be predicted in advance, that experiment is called a random experiment.

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Examples

i. Tossing of a fair coin

When we toss a coin, the outcome will be either Head (H) or Tail (T)

ii. Throwing an unbiased die

Die is a small cube used in games. It has six faces and each of the six faces shows a different number of dots from 1 to 6. Plural of die is dice.

When a die is thrown or rolled, the outcome is the number that appears on its upper face and it is a random integer from one to six, each value being equally likely.

iii. Drawing a card from a pack of shuffled cards

A pack or deck of playing cards has 52 cards which are divided into four categories as given below

a. Spades (♠)b. Clubs (♣)c. Hearts (♥)d. Diamonds (♦)

Each of the above mentioned categories has 13 cards, 9 cards numbered from 2 to 10, an Ace, a King, a Queen and a jack

Hearts and Diamonds are red faced cards whereas Spades and Clubs are black faced cards.

Kings, Queens and Jacks are called face cards

iv. Taking a ball randomly from a bag containing balls of different colours

4. Sample Space

Sample Space is the set of all possible outcomes of an experiment. It is denoted by S.

Examples

i. When a coin is tossed, S = {H, T} where H = Head and T = Tail

ii. When a dice is thrown, S = {1, 2 , 3, 4, 5, 6}

iii. When two coins are tossed, S = {HH, HT, TH, TT} where H = Head and T = Tail

5. Event

Any subset of a Sample Space is an event. Events are generally denoted by capital letters A, B , C, D etc.

Examples

i. When a coin is tossed, outcome of getting head or tail is an event

ii. When a die is rolled, outcome of getting 1 or 2 or 3 or 4 or 5 or 6 is an event

6. Equally Likely Events

Events are said to be equally likely if there is no preference for a particular event over the other.

Examples

i. When a coin is tossed, Head (H) or Tail is equally likely to occur.

ii. When a dice is thrown, all the six faces (1, 2, 3, 4, 5, 6) are equally likely to occur.

7. Mutually Exclusive Events

Two or more than two events are said to be mutually exclusive if the occurrence of one of the events excludes the occurrence of the other

This can be better illustrated with the following examples

i. When a coin is tossed, we get either Head or Tail. Head and Tail cannot come simultaneously. Hence occurrence of Head and Tail are mutually exclusive events.

ii. When a die is rolled, we get 1 or 2 or 3 or 4 or 5 or 6. All these faces cannot come simultaneously. Hence occurrences of particular faces when rolling a die are mutually exclusive events.

Note : If A and B are mutually exclusive events, A ∩ B = ϕ where ϕ represents empty set.

iii. Consider a die is thrown and A be the event of getting 2 or 4 or 6 and B be the event of getting 4 or 5 or 6. Then

A = {2, 4, 6} and B = {4, 5, 6}

Here A ∩ B ≠ϕ. Hence A and B are not mutually exclusive events.8. Independent Events

Events can be said to be independent if the occurrence or non-occurrence of one event does not influence the occurrence or non-occurrence of the other.

Example : When a coin is tossed twice, the event of getting Tail(T) in the first toss and the event of getting Tail(T) in the second toss are independent events. This is because the occurrence of getting Tail(T) in any toss does not influence the occurrence of getting Tail(T) in the other toss.

9. Simple Events

In the case of simple events, we take the probability of occurrence of single events.

Examples

i. Probability of getting a Head (H) when a coin is tossed

ii. Probability of getting 1 when a die is thrown

10. Compound Events

In the case of compound events, we take the probability of joint occurrence of two or more events.

Examples

i. When two coins are tossed, probability of getting a Head (H) in the first toss and getting a Tail (T) in the second toss.

11. Exhaustive Events

Exhaustive Event is the total number of all possible outcomes of an experiment.

Examples

i. When a coin is tossed, we get either Head or Tail. Hence there are 2 exhaustive events.

ii. When two coins are tossed, the possible outcomes are (H, H), (H, T), (T, H), (T, T). Hence there are 4 (=22) exhaustive events.

iii. When a dice is thrown, we get 1 or 2 or 3 or 4 or 5 or 6. Hence there are 6 exhaustive events.

12. Algebra of Events

Let A and B are two events with sample space S. Then

i. A ∪ B is the event that either A or B or Both occur. (i.e., at least one of A or B occurs)

ii. A ∩ B is the event that both A and B occuriii. A¯ is the event that A does not occuriv. A¯∩B¯ is the event that none of A and B occurs

Example : Consider a die is thrown , A be the event of getting 2 or 4 or 6 and B be the event of getting 4 or 5 or 6. Then

A = {2, 4, 6} and B = {4, 5, 6}

A ∪ B = {2, 4, 5, 6}A ∩ B = {4, 6}A¯ = {1, 3, 5}B¯ = {1, 2, 3}A¯∩B¯ = {1,3}

13. Probability of en Event

Let E be an event and S be the sample space. Then probability of the event E can be defined as

P(E) = n(E)n(S)

where P(E) = Probability of the event E, n(E) = number of ways in which the event can occur and n(S) = Total number of outcomes possibleExamples

i. A coin is tossed once. What is the probability of getting Head?

Total number of outcomes possible when a coin is tossed = n(S) = 2 (∵ Head or Tail)

E = event of getting Head = {H}. Hence n(E) = 1

P(E) = n(E)n(S)=12ii. Two dice are rolled. What is the probability that the sum on the top face of

both the dice will be greater than 9?

Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces)

Hence, total number of outcomes possible two dice are rolled, n(S) = 6 × 6 = 36

E = Getting a sum greater than 9 when the two dice are rolled = {(4, 6), {5, 5}, {5, 6}, {6, 4}, {6, 5}, (6, 6)}

Hence, n(E) = 6

P(E) = n(E)n(S)=636=1614. Important formulas

P(S) = 1

0 ≤ P (E) ≤ 1P(ϕ) = 0 (∵ Probability of occurrence of an impossible event = 0)

15. Addition Theorem

Let A and B be two events associated with a random experiment. Then

P(A U B) = P(A) + P(B) – P(A ∩ B)If A and B are mutually exclusive events, then P(A U B) = P(A) + P(B) because for mutually exclusive events, P(A ∩ B) = 0

16. If A and B are two independents events, then

P(A ∩ B) = P(A).P(B)

Example : Two dice are rolled. What is the probability of getting an odd number in one die and getting an even number in the other die?

Total number of outcomes possible when a die is rolled, n(S) = 6 (∵ any one face out of the 6 faces)

Let A be the event of getting the odd number in one die = {1,3,5}. => n(A)= 3

P(A) = n(A)n(S)=36=12Let B be the event of getting an even number in the other die = {2,4, 6}. => n(B)= 3

P(B) = n(B)n(S)=36=12Required Probability, P(A ∩ B) = P(A).P(B) = 12×12=14Let A be any event and A¯ be its complementary event (i.e., A¯ is the event that A does not occur). ThenP(A¯) = 1 - P(A)

17. Odds on an event

Let E be an event associated with a random experiment. Let x outcomes are favourable to E and y outcomes are not favourable to E, thenOdds in favour of E are x:y, i.e., xy and Odds against E are y:x, i.e., yxP(E) = xx+yP(E¯) =yx+yExample : What are the odds in favour of and against getting a 1 when a die is rolled?

Let E be an event of getting 1 when a die is rolled

Outcomes which are favourable to E, x=1Outcomes which are not favourable to E, y=5Odds in favour of getting 1 = xy=15Odds against getting 1 = xy=yx=51

18. Conditional Probability

Let A and B be two events associated with a random experiment. Then, probability of the occurrence of A given that B has already occurred is called conditional probability and denoted by P(A/B)

Example : A bag contains 5 black and 4 blue balls. Two balls are drawn from the bag one by one without replacement. What is the probability of

drawing a blue ball in the second draw if a black ball is already drawn in the first draw?

Let A be the event of drawing black ball in the first draw and B be the event of drawing a blue ball in the second draw. Then, P(B/A) = Probability of drawing a blue ball in the second draw given that a black ball is already drawn in the first draw.

Total Balls = 5 + 4 = 9

Since a black ball is drawn already, total number of balls left after the first draw = 8total number of blue balls after the first draw = 4

P(B/A) =48=1219. Binomial Probability distribution

A binomial experiment is a probability experiment which satisfies the following requirements.

1. Each trial can have only two outcomes. These outcomes can be considered as either success or failure.2. There must be a fixed number of trials. 3. The outcomes of each trial must be independent of each other. 4. The probability of a success must remain the same for each trial.

In a binomial experiment, The probability of achieving exactly r successes in n trials can be given by

P (r successes in n trials) = (nr)prqn−r

where p = probability of success in one trial

q = 1 - p = probability of failure in one trial

(nr) = nCr=n!(r!)(n−r)!=n(n−1)(n−2)⋯(n−r+1)r!20. More Shortcut Formulas

If n fair coins are tossed,

Total number of outcomes in the sample space = 2nThe probability of getting exactly r-number of heads when n coins are tossed = nCr2n

Basics Concepts and Formulas in Permutations and Combinations1. Fundamental Principles of Counting : Multiplication Theorem

If an operation can be performed in m different ways and following which a second operation can be performed in n different ways, then the two operations in succession can be performed in m × n different ways

2. Fundamental Principles of Counting : Addition Theorem

If an operation can be performed in m different ways and a second independent operation can be performed in n different ways, either of the two operations can be performed in (m+n) ways.

3. Factorial

Let n be a positive integer. Then n factorial (n!) can be defined as

n! = n(n-1)(n-2)...1

Examples

i. 5! = 5 x 4 x 3 x 2 x 1 = 120

ii. 3! = 3 x 2 x 1 = 6

Special Cases

iii. 0! = 1

iv. 1! = 1

4. Permutations

Permutations are the different arrangements of a given number of things by taking some or all at a time

Examples

i. All permutations (or arrangements) formed with the letters a, b, c by taking three at a time are (abc, acb, bac, bca, cab, cba)

ii. All permutations (or arrangements) formed with the letters a, b, c by taking two at a time are (ab, ac, ba, bc, ca, cb)

5. Combinations

Each of the different groups or selections formed by taking some or all of a number of objects is called a combination

Examples

i. Suppose we want to select two out of three girls P, Q, R. Then, possible combinations are PQ, QR and RP. (Note that PQ and QP represent the same selection)

ii. Suppose we want to select three out of three girls P, Q, R. Then, only possible combination is PQR

6. Difference between Permutations and Combinations and How to Address a Problem

Sometimes, it will be clearly stated in the problem itself whether permutation or combination is to be used. However if it is not mentioned in the problem, we have to find out whether the question is related to permutation or combination.

Consider a situation where we need to find out the total number of possible samples of two objects which can be taken from three objects P,Q , R. To understand if the question is related to permutation or combination, we need to find out if the order is important or not.

If order is important, PQ will be different from QP , PR will be different from RP and QR will be different from RQ

If order is not important, PQ will be same as QP, PR will be same as RP and QR will be same as RQ

Hence,If the order is important, problem will be related to permutations. If the order is not important, problem will be related to combinations.

For permutations, the problems can be like "What is the number of permutations the can be made", "What is the number of arrangements that can be made", "What are the different number of ways in which something can be arranged", etc

For combinations, the problems can be like "What is the number of combinations the can be made", "What is the number of selections the can be made", "What are the different number of ways in which something can be selected", etc.

Mostly problems related to word formation, number formation etc will be related to permutations. Similarly most problems related to selection of persons, formation of geometrical figures , distribution of items (there are exceptions for this) etc will be related to combinations.

7. Repetition

The term repetition is very important in permutations and combinations.

Consider the same situation described above where we need to find out the total number of possible samples of two objects which can be taken from three objects P,Q , R.

If repetition is allowed, the same object can be taken more than once to make a sample.

i.e., if repetition is allowed, PP, QQ, RR can also be considered as possible samples.

If repetition is not allowed, then PP, QQ, RR cannot be considered as possible samples

Normally repetition is not allowed unless mentioned specifically.

8. pq and qp are two different permutations ,but they represent the same combination.

9. Number of permutations of n distinct things taking r at a time

Number of permutations of n distinct things taking r at a time can be given by

nPr = n!(n−r)!=n(n−1)(n−2)...(n−r+1)where 0≤r≤n

If r > n, nPr = 0

Special Case: nP0 = 1nPr is also denoted by P(n,r). nPr has importance outside combinatorics as well where it is known as the falling factorial and denoted by (n)r or nr

Examples

i. 8P2 = 8 x 7 = 56

ii. 5P4= 5 x 4 x 3 x 2 = 120

10. Number of permutations of n distinct things taking all at a time

Number of permutations of n distinct things taking them all at a time = nPn = n!

11. Number of Combinations of n distinct things taking r at a time

Number of combinations of n distinct things taking r at a time ( nCr) can be given by nCr = n!(r!)(n−r)!=n(n−1)(n−2)⋯(n−r+1)r!where 0≤r≤n

If r > n, nCr = 0

Special Case: nC0 = 1nCr is also denoted by C(n,r). nCr occurs in many other mathematical contexts as well where it is known as binomial coefficient and denoted by (nr)Examples

i. 8C2 = 8×72×1 = 28

ii. 5C4= 5×4×3×24×3×2×1 = 5Important Concepts and Formulas - Numbers

I. Number Sets and Properties of Numbers

a. Counting Numbers (Natural numbers) : 1, 2, 3 ...

b. Whole Numbers : 0, 1, 2, 3 ...

c. Integers : -3, -2, -1, 0, 1, 2, 3 ...

d. Rational Numbers

Rational numbers can be expressed as ab where a and b are integers and b≠0 Examples: 112, 42, 0, −811 etc.All integers, fractions and terminating or recurring decimals are rational numbers.

e. Irrational Numbers

Any number which is not a rational number is an irrational number. In other words, an irrational number is a number which cannot be expressed as ab where a and b are integers.For instance, numbers whose decimals do not terminate and do not repeat cannot be written as a fraction and hence they are irrational numbers.

Example : π, 2√, (3+5√), 43√ (meaning 4×3√), 6√3 etcPlease note that the value of π = 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59230 78164 06286 20899 86280 34825 34211 70679...We cannot π as a simple fraction (The fraction 22/7 = 3.14.... is just an approximate value of π)

f. Real Numbers

Real numbers include counting numbers, whole numbers, integers, rational numbers and irrational numbers.

g. Surds

Let a be any rational number and n be any positive integer such that a√n is irrational. Then a√n is a surd.Example : 3√, 10−−√6, 43√ etcPlease note that numbers like 9√, 27−−√3 etc are not surds because they are not irrational numbersEvery surd is an irrational number. But every irrational number is not a surd. (eg : π , e etc are not surds though they are irrational numbers.)

II. Addition, Subtraction and Multiplication Rules for Even and Odd Numbers

Addition Rules for Even and Odd Numbers

1. The sum of any number of even numbers is always even

2. The sum of even number of odd numbers is always even

3. The sum of odd number of odd numbers is always odd

Subtraction Rules for Even and Odd Numbers

4. The difference of two even numbers is always even

5. The difference of two odd numbers is always even

Multiplication Rules for Even and Odd Numbers

6. The product of even numbers is always even

7. The product of odd numbers is always odd

8. If there is at least one even number multiplied by any number of odd numbers, the product is always even

III. Divisibility

a. Divisible By

One whole number is divisible by another if the remainder we get after the division is zero.

Examples

i. 36 is divisible by 4 because 36 ÷ 4 = 9 with a remainder of 0.

ii. 36 is divisible by 6 because 36 ÷ 6 = 6 with a remainder of 0.

iii. 36 is not divisible by 5 because 36 ÷ 5 = 7 with a remainder of 1.

b. Divisibility Rules

By using divisibility rules we can easily find out whether a given number is divisible by another number without actually performing the division. This helps to save time especially when working with numbers.

Divisibility Rule Description Examples

Divisibility by 2

A number is divisible by 2 if the last digit is even. i.e., if the last digit is 0 or 2 or 4 or 6 or 8

Example1: Check if 64 is divisible by 2.

The last digit of 64 is 4 (even). Hence 64 is divisible by 2


The last digit of 69 is 9 (not even). Hence 69 is not divisible by 2

Divisibility by 3

A number is divisible by 3 if the sum of the digits is divisible by 3

(Please note that we can apply this rule to the answer again and again if we need)


3 + 8 + 7 = 18. 18 is divisible by 3. Hence 387 is divisible by 3


4 + 2 + 1 = 7. 7 is not divisible by 3. Hence 421 is not

divisible by 3

Divisibility by 4

A number is divisible by 4 if the number formed by the last two digits is divisible by 4.


Number formed by the last two digits = 16. 16 is divisible by 4. Hence 416 is divisible by 4


Number formed by the last two digits = 81. 81 is not divisible by 4. Hence 481 is not divisible by 4

Divisibility by 5

A number is divisible by 5 if the last digit is either 0 or 5.

Example1: Check if 305 is divisible by 5.Last digit is 5. Hence 305 is divisible by 5.

Example2: Check if 420 is divisible by 5.Last digit is 0. Hence 420 is divisible by 5.

Example3: Check if 312 is divisible by 5.Last digit is 2. Hence 312 is not divisible by 5.

Divisibility by 6

A number is divisible by 6 if it is divisible by both 2


and 3.

546 is divisible by 2. 546 is also divisible by 3. (Check the divisibility rule of 2and 3 to find out this)

Hence 546 is divisible by 6


633 is not divisible by 2 though 633 is divisible by 3. (Check the divisibility rule of 2 and 3 to find out this)

Hence 633 is not divisible by 6


635 is not divisible by 2. 635 is also not divisible by 3. (Check the divisibility rule of 2 and 3 to find out this)



428 is divisible by 2 but 428 is not divisible by 3.(Check the divisibility rule of 2 and 3 to find out this)


Divisibility by 7

To find out if a number is divisible by 7, double the last digit and subtact it


Given number = 349

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from the number formed by the remaining digits.

Repeat this process until we get at a smaller number whose divisibility we know.

If this smaller number is 0 or divisible by 7, the original number is also divisible by 7.

34 - (9 × 2) = 34 - 18 = 16 16 is not divisible by 7. Hence 349 is not divisible by 7


Given number = 364 36 - (4 × 2) = 36 - 8 = 28 28 is divisible by 7. Hence 364 is also divisible by 7


Given number = 3374 337 - (4 × 2) = 337 - 8 = 329 32 - (9 × 2) = 32 - 18 = 14 14 is divisible by 7. Hence 329 is also divisible by 7.Hence 3374 is also divisible by 7.

Divisibility by 8

A number is divisible by 8 if the number formed by the last three digits is divisible by 8.


The number formed by the last three digits of 7624 = 624. 624 is divisible by 8. Hence 7624 is also divisible by 8.


The number formed by the last three

digits of 129437464 = 464. 464 is divisible by 8. Hence 129437464 is also divisible by 8.


The number formed by the last three digits of 737460 = 460. 460 is not divisible by 8. Hence 737460 is also not divisible by 8.

Divisibility by 9

A number is divisible by 9 if the sum of its digits is divisible by 9.

(Please note that we can apply this rule to the answer again and again if we need)


3 + 6 + 7 + 8 + 2 + 1 = 2727 is divisible by 9. Hence 367821 is also divisible by 9.


4 + 7 + 1 + 2 + 8 = 2222 is not divisible by 9. Hence 47128 is not divisible by 9.

Example3: Check if 4975291989 is divisible by 9.4 + 9+ 7 + 5 + 2 + 9 + 1 + 9 + 8 + 9= 63Since 63 is big, we can use the same method to see if it is divisible by 9.6 + 3 = 99 is divisible by 9. Hence 63 is also

divisible by 9. Hence 4975291989 is also divisible by 9.

Divisibility by 10

A number is divisible by 10 if the last digit is 0.


Last digit is 0. Hence 2570 is divisible by 10.


Last digit is not 0. Hence 5462 is not divisible by 10

Divisibility by 11

To find out if a number is divisible by 11, find the sum of the odd numbered digits and the sum of the even numbered digits.

Now substract the lower number obtained from the bigger number obtained.

If the number we get is 0 or divisible by 11, the original number is also divisible by 11.


8 + 1 + 6 = 155 + 3 = 815 - 8 = 77 is not divisible by 11. Hence 85136 is not divisible by 11.


2 + 3 + 1 + 2 = 87 + 7 + 5 = 1919 - 8 = 1111 is divisible by 11. Hence 2737152 is

also divisible by 11.


9 + 7 = 165 = 516 - 5 = 1111 is divisible by 11. Hence 957 is also divisible by 11.

Example4: Check if 9548 is divisible by 11.9 + 4 = 135 + 8 = 1313 - 13 = 0We got the difference as 0. Hence 9548 is divisible by 11.

Divisibility by 12

A number is divisible by 12 if the number is divisible by both 3 and 4


720 is divisible by 3 and 720 is also divisible by 4. (Check the divisibility rules of 3 and 4 to find out this)

Hence 720 is also divisible by 12


916 is not divisible by 3 , though 916 is divisible by 4.(Check the divisibility





rules of 3 and 4 to find out this)



921 is divisible by 3. But 921 is not divisible by 4.(Check the divisibility rules of 3 and 4 to find out this)



827 is not divisible by 3. 827 is also not divisible by 4.(Check the divisibility rules of 3 and 4 to find out this)


Divisibility by 13

To find out if a number is divisible by 13, multiply the last digit by 4 and add it to the number formed by the remaining digits.

Repeat this process until we get at a smaller number whose divisibility we know.


Given number = 349 34 + (9 × 4) = 34 + 36 = 70 70 is not divisible by 13. Hence 349 is not divisible by 349










If this smaller number is divisible by 13, the original number is also divisible by 13.

Given number = 572 57 + (2 × 4) = 57 + 8 = 6565 is divisible by 13. Hence 572 is also divisible by 13


Given number = 68172 6817 + (2 × 4) = 6817 + 8 = 6825682 + (5 × 4) = 682 + 20 = 70270 + (2 × 4) = 70 + 8 = 7878 is divisible by 13. Hence 68172 is also divisible by 13.


Given number = 65165 + (1 × 4) = 65 + 4 = 6969 is not divisible by 13. Hence 651 is not divisible by 13

Divisibility by 14

A number is divisible by 14 if it is divisible by both 2 and 7.

Example1: Check if 238 is divisible by 14

238 is divisible by 2 . 238 is also divisible by 7. (Please check the divisibility rule of 2 and 7 to find out this)Hence 238 is also divisible by 14

Example2: Check if 336 is divisible by



14



342 is divisible by 2 , but 342 is not divisible by 7.(Please check the divisibility rule of 2 and 7 to find out this)Hence 342 is not divisible by 12


175 is not divisible by 2 , though it is divisible by 7.(Please check the divisibility rule of 2 and 7 to find out this)Hence 175 is not divisible by 14


337 is not divisible by 2 and also by 7 (Please check the divisibility rule of 2and 7 to find out this)











Divisibility by 15

A number is divisible by 15 If it is divisible by both 3 and 5.






483 is divisible by 3 , but 483 is not divisible by 5. (Please check the divisibility rule of 3 and 5 to find out this)Hence 483 is not divisible by 15


485 is not divisible by 3 , though it is







divisible by 5. (Please check thedivisibility rule of 3 and 5 to find out this)Hence 485 is not divisible by 15


487 is not divisible by 3 . It is also not divisible by 5 (Please check thedivisibility rule of 3 and 5 to find out this)Hence 487 is not divisible by 15

Divisibility by 16

A number is divisible by 16 if the number formed by the last four digits is divisible by 16.


The number formed by the last four digits of 5696512 = 6512 6512 is divisible by 16. Hence 5696512 is also divisible by 16.


The number formed by the last four digits of 3326976 = 6976 6976 is divisible by 16. Hence 3326976 is also divisible by 16.






The number formed by the last three digits of 732374360 = 4360 4360 is not divisible by 16. Hence 732374360 is also not divisible by 16.

Divisibility by 17

To find out if a number is divisible by 17, multiply the last digit by 5 and subtract it from the number formed by the remaining digits.

Repeat this process until you arrive at a smaller number whose divisibility you know.



Given Number = 50032750032 - (7 × 5 )= 50032 - 35 = 499974999 - (7 × 5 ) = 4999 - 35 = 4964496 - (4 × 5 ) = 496 - 20 = 47647 - (6 × 5 ) = 47 - 30 = 1717 is divisible by 17. Hence 500327 is also divisible by 17


Given Number = 52146152146 - (1 × 5 )= 52146 -5 = 521415214 - (1 × 5 ) = 5214 - 5 = 5209520 - (9 × 5 ) = 520 - 45 = 47547 - (5 × 5 ) = 47 - 25 = 2222 is not divisible by 17. Hence 521461 is not divisible by 17

Divisibility by 18

A number is divisible by 18 if it is divisible by both 2 and 9.


31104 is divisible by 2. 31104 is also divisible by 9. (Please check

thedivisibility rule of 2 and 9 to find out this)Hence 31104 is divisible by 18


1170 is divisible by 2. 1170 is also divisible by 9. (Please check the divisibility rule of 2 and 9 to find out this)Hence 1170 is divisible by 18


1182 is divisible by 2 , but 1182 is not divisible by 9. (Please check thedivisibility rule of 2 and 9 to find out this)Hence 1182 is not divisible by 18


1287 is not divisible by 2 though it is divisible by 9. (Please check thedivisibility rule of 2 and 9 to find out this)Hence 1287 is not divisible by 18

Divisibility by 19

To find out if a number is divisible by 19, multiply










the last digit by 2 and add it to the number formed by the remaining digits.

Repeat this process until you arrive at a smaller number whose divisibility you know.


Given Number = 746897468 + (9 × 2 )= 7468 + 18 = 7486748 + (6 × 2 ) = 748 + 12 = 76076 + (0 × 2 ) = 76 + 0 = 7676 is divisible by 19. Hence 74689 is also divisible by 19


Given Number = 712347123 + (4 × 2 )= 7123 + 8 = 7131713 + (1 × 2 )= 713 + 2 = 71571 + (5 × 2 )= 71 + 10 = 8181 is not divisible by 19. Hence 71234 is not divisible by 19

Divisibility by 20

A number is divisible by 20 if it is divisible by 10 and the tens digit is even.

(There is one more rule to see if a number is divisible by 20 which is given below.A number is divisible by 20 if the number is divisible by both 4 and 5)


720 is divisible by 10. (Please check the divisibility rule of 10 to find out this).The tens digit = 2 = even digit.Hence 720 is also divisible by 20


1340 is divisible by 10. (Please check the divisibility rule of 10 to find out this).



The tens digit = 2 = even digit.Hence 1340 is divisible by 20


1350 is divisible by 10. (Please check the divisibility rule of 10 to find out this).But the tens digit = 5 = not an even digit.Hence 1350 is not divisible by 20


1325 is not divisible by 10 (Please check the divisibility rule of 10 to find out this) though the tens digit = 2 = even digit.Hence 1325 is not divisible by 20

IV. What are Factors of a Number and how to find it out?

a. Factors of a number

If one number is divisible by a second number, the second number is a factor of the first number

The lowest factor of any positive number = 1

The highest factor of any positive number = the number itself

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Example

The factors of 36 are 1, 2, 3, 4, 6, 9 12, 18, 36 because each of these numbers divides 36 with a remainder of 0

b. How to find out factors of a number

i. Write down 1 and the number itself (lowest and highest factors).

ii. Check if the given number is divisible by 2 (Reference: Divisibility by 2 rule)

If the number is divisible by 2, write down 2 as the second lowest factor and divide the given number by 2 to get the second highest factor

iii. Check for divisibility by 3, 4,5, and so on. till the beginning of the list reaches the end

Example1: Find out the factors of 72

iv. Write down 1 and the number itself (72) as lowest and highest factors.

1 . . . 72

v. 72 is divisible by 2 (Reference: Divisibility by 2 Rule).72 ÷ 2 = 36. Hence 2nd lowest factor = 2 and 2nd highest factor = 36. So we can write as

1, 2 . . . 36, 72

vi. 72 is divisible by 3 (Reference: Divisibility by 3 Rule).72 ÷ 3 = 24 . Hence 3rd lowest factor = 3 and 3rd highest factor = 24. So we can write as

1, 2, 3, . . . 24, 36, 72





vii. 72 is divisible by 4 (Reference: Divisibility by 4 Rule).72 ÷ 4 = 18. Hence 4th lowest factor = 4 and 4th highest factor = 18. So we can write as

1, 2, 3, 4, . . . 18, 24, 36, 72

viii. 72 is not divisible by 5 (Reference: Divisibility by 5 Rule)

ix. 72 is divisible by 6 (Reference: Divisibility by 6 Rule).72 ÷ 6 = 12. Hence 5th lowest factor = 6 and 5th highest factor = 12. So we can write as

1, 2, 3, 4, 6, . . . 12, 18, 24, 36, 72

x. 72 is not divisible by 7 (Reference: Divisibility by 7 Rule)

xi. 72 is divisible by 8 (Reference: Divisibility by 8 Rule).72 ÷ 8 = 9. Hence 6th lowest factor = 8 and 6th highest factor = 9.

Now our list is complete and the factors of 72 are

1, 2, 3, 4, 6, 8, 9 12, 18, 24, 36, 72

Example2: Find out the factors of 22

xii. Write down 1 and the number itself (22) as lowest and highest factors

1 . . . 22

xiii. 22 is divisible by 2 (Reference: Divisibility by 2 Rule).22 ÷ 2 = 11. Hence 2nd lowest factor = 2 and 2nd highest factor = 11. So we can write as

1, 2 . . . 11, 22

xiv. 22 is not divisible by 3 (Reference: Divisibility by 3 Rule).








xv. 22 is not divisible by 4 (Reference: Divisibility by 4 Rule).

xvi. 22 is not divisible by 5 (Reference: Divisibility by 5 Rule).

xvii. 22 is not divisible by 6 (Reference: Divisibility by 6 Rule).

xviii. 22 is not divisible by 7 (Reference: Divisibility by 7 Rule).

xix. 22 is not divisible by 8 (Reference: Divisibility by 8 Rule).

xx. 22 is not divisible by 9 (Reference: Divisibility by 9 Rule).

xxi. 22 is not divisible by 10 (Reference: Divisibility by 10 Rule).

Now our list is complete and the factors of 22 are

1, 2, 11, 22

c. Important Properties of Factors

If a number is divisible by another number, then it is also divisible by all the factors of that number.

Example : 108 is divisible by 36 because 106 ÷ 38 = 3 with remainder of 0.

The factors of 36 are 1, 2, 3, 4, 6, 9 12, 18, 36 because each of these numbers divides 36 with a remainder of 0.

Hence, 108 is also divisible by each of the numbers 1, 2, 3, 4, 6, 9, 12, 18, 36.

V. What are Prime Numbers and Composite Numbers?

a. Prime Numbers

A prime number is a positve integer that is divisible by itself and 1 only. Prime numbers will have exactly two integer factors.









Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, etc.

Please note the following facts

Zero is not a prime number because zero is divisible by more than two factors. Zero can be divided by 1, 2, 3 etc. (0 ÷ 1 = 0, 0÷ 2 = 0 ...)

One is not a prime number because it does not have two factors. It is divisible by only 1

b. Composite Numbers

Composite numbers are numbers that have more than two factors. A composite number is divisible by at least one number other than 1 and itself.

Examples: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, etc.

Please note that zero and 1 are neither prime numbers nor composite numbers.

Every whole number is either prime or composite, with two exceptions 0 and 1 which are neither prime nor composite

VI. What are Prime Factorization and Prime factors ?

a. Prime factor

The factors which are prime numbers are called prime factors

b. Prime factorization

Prime factorization of a number is the expression of the number as the product of its prime factors

Example 1:Prime factorization of 280 can be written as 280 = 2 × 2 × 2 × 5 × 7 = 23 × 5 × 7 and the prime factors of 280 are 2, 5 and 7


Example 2: Prime factorization of 72 can be written as 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32 and the prime factors of 72 are 2 and 3

c. How to find out prime factorization and prime factors of a number

Repeated Division Method : In order to find out the prime factorization of a number, repeatedly divide the number by the smallest prime number possible(2,3,5,7,11, ...) until the quotient is 1.

Example 1: Find out Prime factorization of 280

2280

2140

2 70

5 35

7 7

1Hence, prime factorization of 280 can be written as 280 = 2 × 2 × 2 × 5 × 7 = 23 × 5 × 7 and the prime factors of 280 are 2, 5 and 7

Example 2: Find out Prime factorization of 72

272

236

2

18

3 9

3 3

1Hence, prime factorization of 72 can be written as 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32 and the prime factors of 72 are 2 and 3

d. Important Properties

Every whole number greater than 1 can be uniquely expressed as the product of its prime factors. For example, 700 = 22 × 52 × 7

VII. Multiples

Multiples of a whole number are the products of that number with 1, 2, 3, 4, and so on

Example : Multiples of 3 are 3, 6, 9, 12, 15, ...

If a number x divides another number y exactly with a remainder of 0, we can say that x is a factor of y and y is a multiple of x

For instance, 4 divides 36 exactly with a remainder of 0. Hence 4 is a factor of 36 and 36 is a multiple of 4

VIII. What is Least Common Multiple (LCM) and how to find LCM a. Least Common Multiple (LCM)

Least Common Multiple (LCM) of two or more numbers is the smallest number that is a multiple of all the numbers

Example: LCM of 3 and 4 = 12 because 12 is the smallest multiple which is common to 3 and 4 (In other words, 12 is the smallest number which is divisible by both 3 and 4)

We can find out LCM using prime factorization method or division method

b. How to find out LCM using prime factorization method

Step1 : Express each number as a product of prime factors.

Step2 : LCM = The product of highest powers of all prime factors

Example 1 : Find out LCM of 8 and 14

Step1 : Express each number as a product of prime factors. (Reference: Prime Factorization andhow to find out Prime Factorization)

8 = 23

14 = 2 × 7


Here the prime factors are 2 and 7

The highest power of 2 here = 23


Hence LCM = 23 × 7 = 56

Example 2 : Find out LCM of 18, 24, 9, 36 and 90

Step1 : Express each number as a product of prime factors (Reference: Prime Factorization andhow to find out Prime Factorization).

18 = 2 × 32

24 = 23 × 3

9 = 32

36 = 23 × 32

90 = 2 × 5 × 32


Here the prime factors are 2, 3 and 5

http://www.careerbless.com/aptitude/qa/numbers_imp.php?src=hcf_lcm#loc3







Hence LCM = 23 × 32 × 5 = 360

c. How to find out LCM using Division Method (shortcut)

Step 1 : Write the given numbers in a horizontal line separated by commas.

Step 2 : Divide the given numbers by the smallest prime number which can exactly divide at least two of the given numbers.

Step 3 : Write the quotients and undivided numbers in a line below the first.

Step 4 : Repeat the process until we reach a stage where no prime factor is common to any two numbers in the row.

Step 5 : LCM = The product of all the divisors and the numbers in the last line.

Example 1 : Find out LCM of 8 and 14

2 8, 14

4, 7Hence Least common multiple (L.C.M) of 8 and 14 = 2 × 4 × 7 = 56

Example 2 : Find out LCM of 18, 24, 9, 36 and 90

2 18, 24, 9, 36, 90

2 9, 12, 9, 18, 45

3 9, 6, 9, 9, 45

3 3, 2, 3, 3, 15

1, 2, 1, 1, 5 Hence Least common multiple (L.C.M) of 18, 24, 9, 36 and 90 = 2 × 2 × 3 × 3 × 2 × 5 = 360

IX. What is Highest Common Factor (HCF) or Greatest Common Measure (GCM) or Greatest Common Divisor (GCD) and How to find it out ?

a. Highest Common Factor(H.C.F) or Greatest Common Measure(G.C.M) or Greatest Common Divisor (G.C.D)

Highest Common Factor(H.C.F) or Greatest Common Measure(G.C.M) or Greatest Common Divisor (G.C.D) of two or more numbers is the greatest number which divides each of them exactly.

Example : HCF or GCM or GCD of 60 and 75 = 15 because 15 is the highest number which divides both 60 and 75 exactly.

We can find out HCF using prime factorization method or division method

b. How to find out HCF using prime factorization method

Step1 : Express each number as a product of prime factors. (Reference: Prime Factorization and how to find out Prime Factorization)

Step2 : HCF is the product of all common prime factors using the least power of each common prime factor.

Example 1 : Find out HCF of 60 and 75 (Reference: Prime Factorization and how to find out Prime Factorization)

Step1 : Express each number as a product of prime factors.

60 = 22 × 3 × 5

75 = 3 × 52





Here, common prime factors are 3 and 5

The least power of 3 here = 3


Hence, HCF = 3 × 5 = 15

Example 2 : Find out HCF of 36, 24 and 12


36 = 22 × 32

24 = 23 × 3

12 = 22 × 3


Here 2 and 3 are common prime factors.



Hence, HCF = 22 × 3 = 12



36 = 22 × 32

27 = 33

80 = 24 × 5





Step2 : HCF = HCF is the product of all common prime factors using the least power of each common prime factor.

Here you can see that there are no common prime factors.

Hence, HCF = 1

c. How to find out HCF using prime factorization method - By dividing the numbers (shortcut)

Step 1 : Write the given numbers in a horizontal line separated by commas.

Step 2 : Divide the given numbers by the smallest prime number which can exactly divide all of the given numbers.

Step 3 : Write the quotients in a line below the first.

Step 4 : Repeat the process until we reach a stage where no common prime factor exists for all of the numbers.

Step 5 :We can see that the factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. Their product is the HCF

Example 1 : Find out HCF of 60 and 75

360, 75

520, 25

4, 5 We can see that the prime factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. no common prime factor is exists for the numbers came at the bottom.

Hence HCF = 3 × 5 =15.


2

36, 24, 12

218, 12, 6

3 9, 6, 3

3, 2, 1 We can see that the prime factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. no common prime factor is exists for the numbers came at the bottom.

Hence HCF = 2 × 2 × 3 = 12.


236, 24, 48

218, 12, 24

3 9, 6, 12

3, 2, 4 We can see that the prime factors mentioned in the left side clearly divides all the numbers exactly and they are common prime factors. no common prime factor is exists for the numbers came at the bottom.

Hence HCF = 2 × 2 × 3 = 12.

d. How to find out HCF using division method (shortcut)

i. To find out HCF of two given numbers using division method

Step 1: Divide the larger number by the smaller number

Step 2: Divisor of step 1 is divided by its remainder

Step 3: Divisor of step 2 is divided by its remainder. Continue this process till we get 0 as remainder.

Step 4: Divisior of the last step is the HCF.

ii. To find out HCF of three given numbers using division method

Step 1: Find out HCF of any two numbers.

Step 2: Find out the HCF of the third number and the HCF obtained in step 1

Step 3: HCF obtained in step 2 will be the HCF of the three numbers

iii. To find out HCF of more than three numbers using division method

In a similar way as explained for three numbers, we can find out HCF of more than three numbers also


60) 75 (1 60

15) 60 (4 60

0Hence HCF of 60 and 75 = 15


12) 48 (4 48



3224) 3556 (1 3224

332) 3224 (9 2988

236) 332 (1 236

96) 236 (2 192

44) 96 (2 88

8) 44 (5 40

4) 8 (2 8


Example 3 : Find out HCF of 9, 27, and 48

Taken any two numbers and find out their HCF first. Say, let's find out HCF of 9 and 27 initially.

9) 27 (3 27


HCF of 9 ,27, 48 = HCF of [(HCF of 9, 27) and 48] = HCF of [9 and 48]

9) 48 (5 45

3) 9 (3 9

0Hence, HCF of 9 ,27, 48 = 3


5) 7 (1 5

2) 5 (2 4

1) 2 (2 2


X. How to calculate LCM and HCF for fractions

Least Common Multiple (L.C.M.) for fractions

LCM for fractions = LCM of NumeratorsHCF of DenominatorsExample 1: Find out LCM of 12, 38, 34LCM = LCM (1, 3, 3)HCF (2, 8, 4)=32

Example 2: Find out LCM of 25, 310LCM = LCM (2, 3)HCF (5, 10)=65

XI. Highest Common Multiple (H.C.F) for fractions

HCF for fractions = HCF of NumeratorsLCM of Denominators

Example 1: Find out HCF of 35, 611, 920HCF = HCF (3, 6, 9)LCM (5, 11, 20)=3220

Example 2: Find out HCF of 45, 23 HCF = HCF (4, 2)LCM (5, 3)=215

XII. How to calculate LCM and HCF for Decimals

Step 1 : Make the same number of decimal places in all the given numbers by suffixing zero(s) in required numbers as needed.

Step 2 : Now find the LCM/HCF of these numbers without decimal.

Step 3 : Put the decimal point in the result obtained in step 2 leaving as many digits on its right as there are in each of the numbers.

Example1 : Find the LCM and HCF of .63, 1.05, 2.1

Step 1 : Make the same number of decimal places in all the given numbers by suffixing zero(s) in required numbers as needed.

i.e., the numbers can be writtten as .63, 1.05, 2.10

Step 2 : Now find the LCM/HCF of these numbers without decimal.

Without decimal, the numbers can be written as 63, 105 and 210 .

LCM (63, 105 and 210) = 630

HCF (63, 105 and 210) = 21

Step 3 : Put the decimal point in the result obtained in step 2 leaving as many digits on its right as there are in each of the numbers. i.e., here, we need to put decimal point in the result obtained in step 2 leaving two digits on its right.

i.e., the LCM (.63, 1.05, 2.1) = 6.30

HCF (.63, 1.05, 2.1) = .21

XIII. How to compare fractions? a. Type 1 : Fractions with same denominators.

Compare 35 and 15These fractions have same denominator. So just compare the numerators. Bigger the numerator, bigger the number.

3 > 1. Hence 35>15Example 2: Compare 27 and 37 and 87These fractions have same denominator. So just compare the numerators. Bigger the numerator, bigger the number.

8 > 3 > 2. Hence 87>37>27

b. Type 2 : Fractions with same numerators.

Example 1: Compare 35 and 38These fractions have same numerator. So just compare the denominators. Bigger the denominator, smaller the number.

8 > 5. Hence 38<35Example 2: Compare 78 and 72 and 75These fractions have same numerator. So just compare the denominators. Bigger the denominator, smaller the number.

8 > 5 > 2. Hence 78<75<72

c. Type 3 : Fractions with different numerators and denominators.

Example 1: Compare 35 and 47To compare such fractions, find out LCM of the denominators. Here, LCM(5, 7) = 35

Now , convert each of the given fractions into an equivalent fraction with 35 (LCM) as the denominator.

The denominator of 35 is 5. 5 needs to be multiplied with 7 to get 35. Hence,35=3×75×7=2135The denominator of 47 is 7. 7 needs to be multiplied with 5 to get 35. Hence,47=4×57×5=20352135>2035Hence, 35>47 Or

Convert the fractions to decimals

35=.647=.5... (Need not find out the complete decimal value; just find out up to what is required for comparison. In this case the first digit itself is sufficient to do the comparison).6 > .5...

Hence, 35>47

XIV. Co-prime Numbers or Relatively Prime Numbers

Two numbers are said to be co-prime (also spelled coprime) or relatively prime if they do not have a common factor other than 1. i.e., if their HCF is 1.

Example1: 3, 5 are co-prime numbers (Because HCF of 3 and 5 = 1)

Example2: 14, 15 are co-prime numbers (Because HCF of 14 and 15 = 1)

A set of numbers is said to be pairwise co-prime (or pairwise relatively prime) if every two distinct numbers in the set are co-prime

Example1 : The numbers 10, 7, 33, 13 are pairwise co-prime, because HCF of any pair of the numbers in this is 1. HCF (10, 7) = HCF (10, 33) = HCF (10, 13) = HCF (7, 33) = HCF (7, 13) = HCF (33, 13) = 1.

Example2 : The numbers 10, 7, 33, 14 are not pairwise co-prime because HCF(10, 14) = 2 ≠ 1 and HCF(7, 14) = 7 ≠ 1.

If a number is divisible by two co-prime numbers, then the number is divisible by their product also.

Example

3, 5 are co-prime numbers (Because HCF of 3 and 5 = 1)

14325 is divisible by 3 and 5.3 × 5 = 15 Hence 14325 is divisible by 15 also

If a number is divisible by more than two pairwise co-prime numbers, then the number is divisible by their product also.

Example1 : The numbers 3, 4, 5 are pairwise co-prime because HCF of any pair of numbers in this is 1

1440 is divisible by 3, 4 and 5.

3 × 4 × 5 = 60. Hence 1440 is also divisible by 60

Example2

The numbers 3, 4, 9 are not pairwise co-prime because HCF (3, 9 ) = 3 ≠ 1

1440 is divisible by 3, 4 and 9.

3 X 4 X 9 = 108. However 1440 is not divisible by 108 as 3, 4, 9 are not pairwise co-prime

XV. Important Points to Note on LCM and HCF

Product of two numbers = Product of their HCF and LCM.

Example

LCM (8, 14) = 56

HCF (8, 14) = 2

LCM (8, 14) × HCF (8, 14) = 56 × 2 = 112

8 × 14 = 112

Hence LCM (8, 14) × HCF (8, 14) = 8 × 14

Important Concepts and Formulas - Algebra

I. Basic Algebraic Formulas 1. a(b+c)=ab+ac(Distributive Law)

2. (a+b)2=a2+2ab+b2

3. (a−b)2=a2−2ab+b2

4. (a+b)2+(a−b)2=2(a2+b2)

5. (a+b)3=a3+3a2b+3ab2+b3=a3+3ab(a+b)+b3

6. (a−b)3=a3−3a2b+3ab2−b3=a3−3ab(a−b)−b3

7. a2−b2=(a−b)(a+b)

8. a3+b3=(a+b)(a2−ab+b2)

9. a3−b3=(a−b)(a2+ab+b2)

10. an−bn=(a−b)(an−1+an−2b+an−3b2+ ... +bn−1)

11. (a+b+c)2=a2+b2+c2+2(ab+bc+ca)

12. (a+b)(a+c)=a2+(b+c)a+bc

13. an=a.a.a ... (n times)

14. am.an...ap=am+n+...+p

15. am.an=am+n

16. aman=am−n

17. (am)n=amn=(an)m

18. (ab)n=anbn

19. (ab)n=anbn

20. a−n=1an

21. an=1a−n

22. a0=1 where a∈ R, a ≠0

23. ap/q=ap−−√q

24. if am=an where a≠0 and a≠±1, then m=n

25. if an=bn where n≠0, then a=±b

26. (x+a)2=x2+2ax+a2

27. (x−a)2=x2−2ax+a2

28. (x+a)(x+b)=x2+(a+b)x+ab

29. x2−a2=(x−a)(x+a)

30. (xn+1) is completely divisible by (x + 1) when n is odd

31. (xn+an) is completely divisible by (x + a) when n is odd

32. (xn−an) is completely divisible by (x−a) for every natural number n

33. (xn−an) is completely divisible by (x+a) when n is even

Binomial Theorem

If a and b are any real numbers and n is a positive integer,

(a+b)n=an+nan−1b+n(n−1)2!an−2b2+n(n−1)(n−2)3!an−3b3+⋯+n(n−1)(n−2)⋯(n−r+1)r!an−rbr+⋯+bn=(n0)an+(n1)an−1b+(n2)an−2b2+(n3)an−3b3+⋯+(nr)an−rbr+⋯+(nn)bn=∑r=0n(nr)an−rbrwhere (nr) is known as Binomial Coefficient and is defined by(nr)=n!(r!)(n−r)!=n(n−1)(n−2)⋯(n−r+1)r!where r = 1,2,...,nand (n0)=1

Binomial Coefficient also occurs in many other mathematical areas than algebra, especially in combinatorics where (nr)represents the number of combinations of n distinct things taking r at a time and is denoted by n C r or C(n,r) .

The properties of binomial coefficients have led to extending its meaning beyond the basic case where n and r are nonnegative integers with r ≤ n; such expressions are also called binomial coefficients.

Binomial Expansions

(a+b)0=1

(a+b)1=a+b

(a+b)2=a2+2ab+b2

(a+b)3=a3+3a2b+3ab2+b3

http://www.careerbless.com/aptitude/qa/permutations_combinations_imp.php#im11

(a+b)4=a4+4a3b+6a2b2+4ab3+b4

... and so on

Binomial Series

One of Newton's achievement was to extend the Binomial Theorem to the case where n can be any real number.

If n is any real number and -1 < x < 1, then

(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+⋯=(n0) x0+(n1) x1+(n2) x2+(n3) x3+⋯=∑r=0∞(nr) xrwhere (nr)=n(n−1)(n−2)⋯(n−r+1)r!where r ≥1and (n0)=1

http://www.careerbless.com/aptitude/qa/algebra_imp.php#bt

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