MyMaths MyLife

Pure mathematics is more like art. Pure mathematicians work on building a foundation for a theory. One nice feature about pure mathematics is that it is free from argument. When a mathematician makes a discovery there is no opposition, as in science. And his theory stands the test of time, unlike science where one law is shown to be wrong in special cases. But once a foundation is build (like complex analysis) applied mathematicians take its result and use it to solve important problems.

Pure math is much more difficult. Classes in applied math consist of memorizing the steps to solve problems. However, classes in pure math involve proofs, which implies a good understanding of the subject matter is required. In pure math you need to justify everything you do. Which can sometimes make a simple argument long and complicated. It is easier for someone in pure math to learn applied math rather than someone in applied math to learn pure math.

Determinants and matrices, they look alike. Their similarities caught many unaware and results in "excitements" and much interests.

Both contain numbers within. But ......

- determinants are bounded by two straights lines whereas matrices are by square braces

- determinant resulted in a single numerical value, whereas matrices are sets of numbers grouped within the braces

- determinant can be extracted from matrix, but not the other way round

- there are inverse matrix but not inverse determinant

- a scalar multiplier affects only a single row or single column of a determinant, but affects all the numbers within a matrix

The problems relating to geometry cover mostly triangles and circles. Even though polygons also are covered, the emphasis on polygons is not as much as on triangles and circles.

An angle of 90⁰ is a right angle; an angle less than 90⁰ is an acute angle; an angle between 90⁰ and 180⁰ is an obtuse angle; and angle between 180⁰ and 360⁰ is a reflex angle.
The sum of all angles on one side of a straight line AB at a point O by any number of lines joining the line AB at O is 180⁰. When any number of straight lines join at a point, the sum of all the angles around that point is 360⁰.
Two angles whose sum is 90⁰ are said to be complementary to each other and two angles whose sum is 180⁰ are said to be supplementary angles.



When two straight lines intersect, vertically opposite angles are equal. In the figure given alongside,

Facts And Formulae:

Highest Common Factor:(H.C.F) or Greatest Common Meaure(G.C.M) :
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 :

i.Factorization method: Express each one of the given numbers as
the product of prime factors. The product of least powers of common
prime factors gives HCF.

Example : Find HCF of 26 * 32*5*74 , 22 *35*52 * 76 ,
2*52 *72
Solution: The prime numbers given common numbers are 2,5,7
Therefore HCF is 22 * 5 *72 .

ii.Division Method : Divide the larger number by smaller one. Now
divide the divisor by remainder. Repeat the process of dividing
preceding number last obtained till zero is obtained as number. The
last divisor is HCF.


Least common multiple[LCM] : The least number which is
divisible by each one of given numbers is LCM.

There are two methods for this:

i.Factorization method : Resolve each one into product of prime
factors. Then LCM is product of highest powers of all factors.

ii.Common division method.

When a straight line XY cuts two parallel line PQ and RS [as shown in figure], the following are the relationships between various angles that are formed. [M and N are the points of intersection of XY with PQ and RS respectively].

a) alternate angles are equal
i.e, angle PMN = angle MNS
angle QMN = angle MNR
b) corresponding angles are equal
i.e, angle XMQ = angle MNS
angle QMN = angle SNY
angle XMP = angle MNR
angle PMN = angle RNY
c) Sum of interior angles on the same side of cutting line is equal to 180⁰
i.e, angle QMN + angle MNS = 180⁰
angle PMN + angle MNR = 180⁰
d) Sum of exterior angles on the same side of transversal line is equal to 180⁰
i.e, angle XMQ + angle SNY = 180⁰
angle XMP + angle RNY = 180⁰
If three or more parallel lines make equal intercepts on one transversal, they make equal intercepts on any other transversal as well.

Important Facts:

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.

Ex:(i) Cost is directly proportional to the number of articles.
(More articles, More cost).

(ii) Work done is directly proportional to the number of men
working on it. (More men, more work).

Indirect Proportion: Two Quantities are said to be
indirectly proportional,if on the increase of the one , the other
decreases to the same extent and vice-versa.

Ex:(i) The time taken by a car covering a certain distance is
inversely proportional to the speed of the car.(More speed,
less is the time taken to cover the distance).

(ii) Time taken to finish a work is inversely proportional to
the number of persons working at it.
(More persons, less is the time taken to finish a job).

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

We all know about cm, mm, Km ..................., but we dont know much about inches, yard .............
you might have got a question that " what is the relationship between inch and cm, inch and yard"...........
This is going to list all the conversions of such units

LINEAR

Inch=Basic Unit

1 hand = 4 inches
1 link = 7.92 inches
1 span = 9 inches
1 foot = 12 inches
1 yard = 3 feet
1 fathom = 2 yards
1 rod = 5.5 yards
1 chain = 100 links=22 yards
1 furlong = 20 yards
1 mile = 1760 yards
1 knot mile= 6076.1155 feet
1 league = 3 miles
1 inch = 2.54 cm

Cartesian Product:

Let A and B be any two sets. Then the Cartesian product of A and B is the set of all ordered pairs of the form (a, b), where aЄA and bЄB
The product is denoted by A×B
A×B = {(a, b)/ aЄA, bЄB }

Example
A = {a, b, c} and B={1,2}, then
A×B = {(a, 1), (b, 1), (c, 1), (a, 2), (b, 2), (c, 2)}
B×A = {(1, a), (2, a), (1, b), (2, b), (1, c), (2, c)}

Mathematical concept means just about anything with a mathematical name. For example, some of the mathematical concepts we learn in high school are: constant, variable, polynomial, factor, factoring, equation, solving an equation, logarithm, sine, cosine, tangent, etc., point, line, triangle, square, and other geometric figures, area, perimeter of a geometric figure, etc., and many others. Among the mathematical concepts we learn in our first years of college mathematics are: set, operation, limit, function, and, specifically, continuous function, derivative, integral, theorem, proof, countable infinity, uncountable infinity, algebra, linear algebra, vector space, group, ring, field, and many others.

Now one thing that makes the understanding of these concepts difficult is that they are defined in terms of other concepts.

Thus, e.g., a vector space is defined in terms of the concepts of vector, set, function, abelian group, field, and others. How does the typical mathematics textbook, and mathematics course, deal with this fact? It attempts to teach the concepts in logical order, i.e., it assumes that, e.g., when you begin your study of vector spaces, you will already know — through having remembered what you learned in previous courses — the meaning of each of the concepts in terms of which a vector space is defined. And, indeed, one of the things that makes mathematics such a frightening subject to many students, is the grandiose manner with which these assumptions are set forth in the list of prerequisites for the course.

Combinations:
Each of different groups or selections which can be formed by
taking some or all of a number of objects,is called a combination.
eg:- Suppose we want to select two out of three boys A,B,C .
then ,possible selection are AB,BC & CA.
Note that AB and BA represent the same selection.

Number of Combination:
The number of all combination of n things taken r at a time is:
nCr = n! / (r!)(n-r)!
= n(n-1)(n-2). . . . . . . tor factors / r!
Note: nCn = 1 and nC0 =1

An Important Result:
nCr = nC(n-r)

For problems click Here.

An equation which has the unknown quantity raised only to powers which are whole numbers and the highest power being the square of the unknown quantity, is called a quadratic equation.
The most general form of a quadratic equation is ax^2 + bx + c = 0.
There are two values of x that satisfy such a quadratic equation. These values are called the roots of the quadratic equation.

The roots of the above quadratic equation are given by (-b±√(b^2-4ac))/2a

For ax^2 + bx + c = 0, sum of the roots = -b/a; Product of the roots = c/a

cube and cuboid:

Consider the fallowing objects: a brick, a box of matches, a die, a text book, a room in the house. They have a common shape, though their sizes are different. The geometrical name that we give to each of these objects is the cuboid.
It has six rectangular faces. There are in all 12 edges of the cuboid. A cuboid has 8 corners called vertices.
The total area of all the six faces of a cuboid is called the total surface area of the cuboid.
Let l,b, and h, be the length, the breadth and the height of a cuboid,
then the lateral surface area= 2h(l+b)

The total surface area
=(the lateral surface area)+(area of ABCD)+(area of EFGH)
=2h(l+b)+lb+lb
=2lh+2bh+2lb
=2(lb+bh+hl)

The Relation Between G.C.D and L.C.M:

For GCD concept click here:http://business-maths.blogspot.com/2009/02/greatest-common-divisor.html
For LCM concept click here:http://business-maths.blogspot.com/2009/02/least-common-multiple-lcm.html
Find the G.C.D and L.C.M of 30 and 48 and it shows that the product of GCD and LCM is equal to the product of the two given numbers.
GCD of 30,48 is 6.
And LCM of 30,40 is 240.
LCM*GCD=240*6=1440
Product of 30 and 48= 30*48=1440.
Hence the product of the two numbers is equal to the product of their G.C.D and L.C.M.
If a and b are any two natural numbers and L and G are respectively their L.C.M and G.C.D., then a*b=L*G

->(a+b)²=a²+b²+2ab
->(a-b)²=a²+b²-2ab
->(a+b)²-(a-b)²=4ab
->(a+b)²+(a-b)²=2(a²+b²)
->a²-b²=(a+b)(a-b)
->(a-+b+c)²=a²+b²+c²+2(ab+b c+ca)
->a³+b³=(a+b)(a²+b²-ab)
->a³-b³=(a-b)(a²+b²+ab)
->a³+b³+c³-3a b c=(a+b+c)(a²+b²+c²-ab-b c-ca)
->If a+b+c=0 then a³+b³+c³=3a b c

Addition and Multiplication are said to be commutative, because
A+B = B+A
A*B = B*A

Addition and Multiplication are said to be Associative, because
(A+B)+C = A+(B+C)
(A*B)*C = A*(B*C)

Multiplication is Distributive over Addition, because
A*(B+C) = (A*B)+(A*C)

here A, B, C represent any Real Number

Example problems:

1.Find the average of all these numbers.142,147,153,165,157.

Solution:
142 147 153 165 157
Here consider the least number i.e, 142
comparing with others,
142 147 153 165 157
+5 +11 +23 +15
Now add 5+11+23+15 = 52/5 = 10.8
Now add 10.8 to 142 we get 152.8
(Average of all these numbers).
Answer is 152.8


2.Find the average of all these numbers.4,10,16,22,28
Solution:
4,10,16,22,28
As the difference of number is 6
Then the average of these numbers is central one i.e, 16.
Answer is 16.

3.Find the average of all these numbers.4,10,16,22,28,34.

Solution:
Here also difference is 6.
Then middle numbers 16,22 take average of these
two numbers 16+22/2=19
Therefore the average of these numbers is 19.
Answer is 19.

4.The average marks of a marks of a student in 4 Examination
is 40.If he got 80 marks in 5th Exam then what is
his new average.

Solution:
4*40+80=240
Then average means 240/5=48.
Answer is 48.

5.In a group the average income of 6 men is 500 and that
of 5 women is 280, then what is average income of the group.

Solution:
6*500+5*280=4400
then average is 4400/11=400.
Another Method: here consider for 6 men
6 men – each 500.
so 5th women is 280.
then 500-280=220.
then 220*6/11=120.
therefore 120+280=400.
Answer is 400.

6.The average weight of a class of 30 students is 40 kgs if the
teacher weight is included then average increases by 2 kgs then
find the weight of the teacher?

Solution:
30 students average weight is 40 kgs.
So,when teacher weight is added it increases by 2 kgs
so total 31 persons ,therefore 31*2=62.
Now add the average weight of all student to it
we get teachers weight i.e, 62+40=102 kgs.
Answer is 102 kgs.

7.The average age of Mr and Mrs Sharma 4 years ago is 28 years .
If the present average age of Mr and Mrs Sharma and their son
is 22 years. What is the age of their son.

Solution:
4 years ago their average age is 28 years.
So their present average age is 32 years.
32 years for Mr and Mrs Sharma then 32*2=64 years.
Then present age including their son is 22 years.
So 22*3 =66 years.
Therefore son age will be 66-64 = 2 years.
Answer is 2 years.

8.The average price of 10 books is increased by 17 Rupees when
one of them whose value is Rs.400 is replaced by a new book.
What is the price of new book?

Solution:
10 books Average increases by 17 Rupees
so 10*17= 170.
so the new book cost is more and by adding its cost average
increase,therefore the cost of new book is 400+170=570Rs.
Answer is 570 Rs.

9.The average marks of girls in a class is 62.5. The average marks
of 4 girls among them is 60.The average marks of remaining girls
is 63,then what is the number of girls in the class?

Solution:
Total number of girls be x+4.
Average marks of 4 girls is 60.
therefore 62.5-60=2.5
then 4*2.5 =10.
the average of remaining girls is 63
here 0.5 difference therefore 0.5*x=10(since we got from 4 girls)
(this is taken becoz both should be equal)
x=10/0.5
x=20.
This clear says that remaining are 20 girls
therefore total is x+4=20+4=24 girls
Answer is 24 girls.

10.Find the average of first 50 natural numbers.

Solution:
Sum of the Natural Numbers is n(n+1)/2
therefore for 50 Natural numbers 50*51/2=775.
the average is 775/50=15.5
Answer is 15.5 .

11.The average of the first nine prime number is?

Solution:
Prime numbers are 2,3,5,7,11,13,17,19,23
therefore 2+3+5+7+11+13+17+19+23=100
then the average 100/9= 11 1/9.
Answer is 11 1/9.

12.The average of 2,7,6 and x is 5 and the average of and the
average of 18,1,6,x and y is 10 .what is the value of y?

Solution:
2+7+6+x/4=5
=>15+x=20
=>x=5.
18+1+6+x+y/5=10
=>25+5+y=50
=>y=20.

13.The average of a non-zero number and its square is 5 times the
number.The number is

Solution:
The number be x
then x+x2/2=5x
=>x2-9x=0
=>x(x-9)=0
therefore x=0 or x=9.
The number is 9.

14.Nine persons went to a hotel for taking their meals . Eight of
them spent Rs.12 each on their meals and the ninth spent Rs.8 then
the average expenditure of all the nine. What was the total money
spent by them?

Solution:
The average expenditure be x.
then 8*12+(x+8)=9x
=>96+x+8=9x.
=>8x=104
=>x=13
Total money spent =9x=>9*13=117
Answer is Rs.117


15.The average weight of A.B.C is 45 Kgs.If the average weight of
A and B be 40 Kgs and that of Band C be 43 Kgs. Find the weight of B?

Solution:
The weight of A,B,Care 45*3=135 Kgs.
The weight of A,B are 40*2=80 Kgs.
The weight of B,C are 43*2=86 Kgs.
To get the Weight of B.
(A+B)+(B+C)-(A+B+C)=80+86-135
B=31 kgs.
Answer is 31 Kgs.

16.The sum of three consecutive odd number is 48 more than the average
of these number .What is the first of these numbers?

Solution:
let the three consecutive odd numbers are x, x+2, x+4.
By adding them we get x+x+2+x+4=3x+6.
Then 3x+6-(3x+6)/3=38(given)
=>2(3x+6)=38*3.
=>6x+12=114
=>6x=102
=>x=17.
Answer is 17.

17.A family consists of grandparents,parents and three grandchildren.
The average age of the grandparents is 67 years,that of parents is 35
years and that of the grand children is 6 years . What is the average
age of the family?

Solution:
grandparents age is 67*2=134
parents age is 35*2=70
grandchildren age is 6*3=18
therefore age of family is 134+70+18=222
average is 222/7=31 5/7 years.
Answer is 31 5/7 years.

18.A library has an average of 510 visitors on Sundays and 240 on
other days .The average number of visitors per day in a month 30
days beginning with a Sunday is?

Solution:
Here specified that month starts with Sunday
so, in a month there are 5 Sundays.
Therefore remaining days will be 25 days.
510*5+240*25=2550+6000
=8550 visitors.
The average visitors are 8550/30=285.
Answer is 285.

19.The average age of a class of 39 students is 15 years .
If the age of the teacher be included ,then average
increases by 3 months. Find the age of the teacher.

Solution: Total age for 39 persons is 39*15=585 years.
Now 40 persons is 40* 61/4=610 years
(since 15 years 3 months=15 3/12=61/4)
Age of the teacher =610-585 years
=>25 years.
Answer is 25 years.

20.The average weight of a 10 oarsmen in a boat is increases
by 1.8 Kgs .When one of the crew ,who weighs 53 Kgs is
replaced by new man. Find the weight of the new man.

Solution: Weight of 10 oars men is increases by 1.8 Kgs
so, 10*1.8=18 Kgs
therefore 53+18=71 Kgs will be the weight of the man.
Answer is 71 Kgs.

21.A bats man makes a score of 87 runs in the 17th inning
and thus increases his average by 3. Find the average
after 17th inning.

Solution: Average after 17 th inning =x
then for 16th inning is x-3.
Therefore 16(x-3)+87 =17x
=>x=87-48
=>x=39.
Answer is 39.

22.The average age of a class is 15.8 years .The average age
of boys in the class is 16.4 years while that of the girls
is 15.4 years .What is the ratio of boys to girls in the class.

Solution: Ratio be k:1 then
k*16.4 + 1*15.4 = (k+1)*15.8
=>(16.4-15.8)k=15.8-15.4
=>k=0.4/0.6
=>k=2/3
therefore 2/3:1=>2:3
Answer is 2:3

23.In a cricket eleven ,the average of eleven players is
28 years .Out of these ,the average ages of three groups
of players each are 25 years,28 years, and 30 years
respectively. If in these groups ,the captain and the
youngest player are not included and the captain is
eleven years older than the youngest players ,
what is the age of the captain?

Solution: let the age of youngest player be x
then ,age of the captain =(x+11)
therefore 3*25 + 3*28 + 3*30 + x + x+11=11*28
=>75+84+90+2x+11=308
=>2x=48
=>x=24.
Therefore age of the captain =(x+11)= 24+11= 35 years.
Answer is 35 years.

24.The average age of the boys in the class is twice
the number of girls in the class .If the ratio of
boys and girls in the class of 36 be 5:1, what is
the total of the age (in years) of the boys in the class?

Solution: Number of boys=36*5/6=30
Number of girls =6
Average age of boys =2*6=12 years
Total age of the boys=30*12=360 years
Answer is 360 years.

25.Five years ago, the average age of P and Q was
15 years ,average age of P,Q, and R today is
20 years,how old will R be after 10 years?

Solution: Age of P and Q are 15*2=30 years
Present age of P and Q is 30+5*2=40 years.
Age of P Q and R is 20*3= 60 years.
R ,present age is 60-40=20 years
After 10 years =20+10=30 years.
Answer is 30 years.

26.The average weight of 3 men A,B and C is 84 Kgs.
Another man D joins the group and the average now
becomes 80 Kgs.If another man E whose weight is
3 Kgs more than that of D ,replaces A then the
average weight B,C,D and E becomes 79 Kgs.
The weight of A is.

Solution:Total weight of A, B and C is 84 * 3 =252 Kgs.
Total weight of A,B,C and Dis 80*4=320 Kgs
Therefore D=320-252=68 Kgs.
E weight (68+3)=71 kgs
Total weight of B,C,D and E = 79*4=316 Kgs
(A+B+C+D)-(B+C+D+E)=320-316 =4Kgs
A-E=4Kgs
A-71=4 kgs
A=75 Kgs
Answer is 75 kgs

27.A team of 8 persons joins in a shooting competition.
The best marksman scored 85 points.If he had scored
92 points ,the average score for the team would
have been 84.The team scored was.

Solution: Here consider the total score be x.
therefore x+92-85/8=84
=>x+7=672
=>x=665.
Answer is 665

28.A man whose bowling average is 12.4,takes 5 wickets
for 26 runs and there by decrease his average by 0.4.
The number of wickets,taken by him before his last match is:

Solution: Number of wickets taken before last match be x.
therefore 12.4x26/x+5=12(since average decrease by 0.4
therefore 12.4-0.4=12)
=>12.4x+2612x+60
=>0.4x=34
=>x=340/4
=>x=85.
Answer is 85.

29.The mean temperature of Monday to Wednesday was 37 degrees
and of Tuesday to Thursday was 34 degrees .If the
temperature on Thursday was 4/5th that of Monday.
The temperature on Thursday was:


Solution:
The total temperature recorded on Monday,Wednesday was 37*3=111.
The total temperature recorded on Tuesday,
Wednesday,Thursday was 34*3=102.
and also given that Th=4/5M
=>M=5/4Th
(M+T+W)-(T+W+Th)=111-102=9
M-Th=9
5/4Th-Th=9
Th(1/4)=9
=>Th=36 degrees.

30. 16 children are to be divided into two groups A and B
of 10 and 6 children. The average percent marks obtained
by the children of group A is 75 and the average percent
marks of all the 16 children is 76. What is the average
percent marks of children of groups B?

Solution: Here given average of group A and whole groups .
So,(76*16)-(75*10)/6
=>1216-750/6
=>466/6=233/3=77 2/3
Answer is 77 2/3.

31.Of the three numbers the first is twice the second and
the second is twice the third .The average of the reciprocal
of the numbers is 7/72,the number are.

Solution:Let the third number be x
Let the second number be 2x.
Let the first number be 4x.
Therefore average of the reciprocal means
1/x+1/2x+1/4x=(7/72*3)
7/4x=7/24
=>4x=24
x=6.
Therefore
First number is 4*6=24.
Second number is 2*6=12
Third number is 1*6=6
Answer is 24,12,6.

32.The average of 5 numbers is 7.When 3 new numbers
are added the average of the eight numbers is 8.5.
The average of the three new number is:

Solution: Sum of three new numbers=(8*8.5-5*7)=33
Their average =33/3=11.
Answer is 11.

33.The average temperature of the town in the first
four days of a month was 58 degrees. The average
for the second ,third,fourth and fifth days was
60 degree .If the temperature of the first and
fifth days were in the ratio 7:8 then what is
the temperature on the fifth day?

Solution :
Sum of temperature on 1st 2nd 3rd
and 4th days =58*4=232 degrees.
Sum of temperature on 2nd 3rd 4th
and 5th days =60*4=240 degrees
Therefore 5th day temperature is 240-232=8 degrees.
The ratio given for 1st and 5th days be 7x and 8x degrees
then 8x-7x=8
=>x=8.
therefore temperature on the 5th day =8x=8*8=64 degrees.

Least Common Multiple (L.C.M):


Rules:

The smallest of the common multiples of two natural numbers (a and b) is called the least common multiple(LCM) of the numbers a and b.

The smallest of the common multiples of two or more natural numbers is called the least common multiple(LCM).

If two numbers are c0-prime, then their LCM is equal to their product.

Given two numbers, if the first number is a multiple of the second number, then their LCM is equal to the first number.

Relationship Between GCD and LCM:
If a and b are any two natural numbers and L and G are respectively their LCM and GCD, then a*b=L*G

Example:

LCM of 30,48

2 |30,48
_______
3 |15,24
_____
5,8

LCM of 30,48 = 2*3*5*8= 240

Factorial is defined for any positive integer. It is denoted by !. Thus “Factorial n” is written as n!. n! is defined as the product of all the integers from 1 to n.

Thus n! = 1.2.3.. ... (n-1),n.

Example 5! = 1*2*3*4*5 = 120

0! is defined to be equal to 1.
Therefore 0! = 1 and 1! = 1

n!=n*(n-1)!
eg: 10!=10*(10-1)!=10*(9)!

Area of the four walls of a room:

If we look around and observe the walls of a room, we find that generally the walls are in the shape of a rectangle the floor and the ceiling of the room are also of rectangular shape.
Let l, b, h be the lengths of AB,AD and AE as shown in the figure. Here l and b are the length and breadth of the floor and h the height of the room.
For the rectangle ABFE, the lengths of two adjacent sides are l and h. its area = lh
For the rectangle BCGF, the lengths of two adjacent sides are b and h. its area = bh
For the rectangle CDHG, the lengths of two adjacent sides are l and h. its area = lh
For the rectangle DAEH, the lengths of two adjacent sides are b and h. its area = bh
Observe the opposite walls being of the same size and shape have the same area too.

Hence the total area of the four walls = lh+bh+lh+bh
= 2lh+2bh
= 2h(l+b)