Simplifying Mathematics in a simple way

Simplifying Mathematics in a simple way
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Thursday, 9 August 2018

WHAT IS SET, TYPES OF SETS ,UNION ,INTERSECTION AND VENN DIAGRAMS

In this post we shall discuss What is Set , types of sets, representation of sets, subset , power set , universal set ,cardinal number , union, intersection , disjoint sets , complement and difference of sets .And we shall be able to solve Puzzles based on set like given below


WHAT IS SET, TYPES OF SETS ,UNION ,INTERSECTION AND VENN  DIAGRAMS

WHAT IS SET

Def : Any collection of well defined and distinct objects is called a set. 

        By well defined object we mean , for a given set and object, it must be possible to decide whether the object belonging to the set or not . i .e it is accepted by everyone as an element for that Set  or  not .The object in a set are called  Elements or Members.


Set are usually denoted by Capital letter A , B , C , D Etc, and their element are represented by small letters.


For any set A let 'a' , 'b'  and 'c' are members of that Set, then a ∈ A  , b ∈ A and  c ∈ A . These special character is read as "belongs to " .So a ∈ A  will be read as  " a belongs to A" , b ∈ A means "b belongs to A" and  c ∈ A read as " c belongs to A".

Here are some examples of  the Set

(1)   The collection of vowels in English Alphabet.
This is Set because every one will accept the letters from a , e , i , o , u  as the Vowels in Enlgish Alphabet
(2)   The collection of  Mountains in India.

(3)   The collection of all districts in India.

(4)   The collection of  even numbers .

(5)   The collection of all the M L As and M Ps in  India .

(6)   The collection of  all the football players of India.

(7)   The collection of hundred  novels of Indian writers.

(8)   The collection of Natural numbers from 100 to 55555.

(9)   The collection of prime numbers less than 100.

(10) The collection of  Prime Ministers of India .
All those person who have accepted the post of Prime Minister are the elements of the set of Prime Ministers of India .On this point there will be no debate,The debate can be on the topic that particular Prime Minister is/was a good , popular or is/was a caring Prime minister.

These were examples of set as all the examples given above were well defined and easily acceptable to everyone.

Here are some examples which are not  Set

(1)   The collection of   beautiful  Mountains in India.
Because the word beautiful is not acceptable to all , Some peoples can says that these Mountains are beautiful But at the same time some other peoples can says that particular Mountains are not beautiful in their opinions .
  
(2)   The collection of   Popular  M L A and M P in  India .
Similar is the case with Popular Leaders, as some people likes  MLAs Or MPs But at the same time many other peoples hate them. So the word Popular is not acceptable to all .
(3)   The collection of   best football players of India.

(4)   The collection of  an Interesting novels of Indian writers.

(5)   The collection of   handsome students in any class .

(6)   The collection of  dangerous animals in this world.

(7)   The collection of  rich person in India.

(8)   The collection of  an easy subjects.

(9)   The collection of an  intelligent  students in any class.

(10) The collection of  beautiful animals .


These were not  examples of set as all the examples given above were not  well defined and were not easily acceptable to everyone. Because the some mountains are  beautiful for some person but at the same time ,same mountains are not beautiful for some persons ,  Similarly some peoples find any book interesting but at the same time same book may not be  interesting for others.


Representation of Sets


The set can be represented in two types

1  Roster Form  or Tabulation Form
2  Set builder form or Rule form

Roster Form 

To represent elements in Roster form ,we separate every element by comma and all the elements are listed in this form.
 For example

Let A is the collection of  even Natural Numbers . Then
A = {2, 4, 6, 8, 10, ..............}

Let P is the collection of prime numbers less than 100
P = {2, 3, 5 , 7, 11, 13, 17, 23, 29, ........, 97}

Let V is the collection vowels in English Alphabet.Then
V = {a, e, i, o, u }

Let S is set of square of the natural number between 1 to 10 Then 
S= {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}

Set Builder form

In this form of set , we write the elements of set by special rule, and that rule must be satisfied bu every elements of the set.in this form all the elements of the set are not written.

Let A is the collection of  even Natural numbers . then
A = {x : x = 2n , n ∈ N } 

Let V is the collection vowels in English Alphabet.Then
V = { x : x is a vowel in English Alphabet }

Let S is set of square of the natural number between 1 to 10 Then 
S= { x : x = n2  :  n  ∈ N and   1 ≤ n    10 }


Empty (Null Set or Void Set)

When a set having No element in it then that Set is Called an 

Empty Set and it is denoted by ф  or { }.

Some examples of an Empty set are given below

(1)  ф = {x : x ∈ N ,  5 < x < 6}  

(2)   ф = {x : x  R ,  x2 +3 = 0 }

(3)   ф = { x ; 99 >  x and  x > 101}

Non  Empty Set

Any set which contain at least one element is called Non empty set .
Some examples of Non Empty set are given below

(1)  A = { 0 } 

(2)   B = { x : x ∈ N ,  5 < x < 8}   

(3)  C = { x ; 99 <  x and  x < 105}

(4)  D = { x : x  R ,  x2 - 3 = 0 }

Singleton Set

Any set which contain only one element is called Singleton Set .
e. g


(1)  A = { 0 } 



(2)   B = { x : x ∈ N ,  6< x < 8}   



(3)   C = { x ; ∈ R , 99 <  x and  x < 101}


(4)   D = { x : x ∈ R , x - 3 = 0 }

Finite Set 

The set in which total numbers of elements can be counted, is called Finite set.
e. g.
(1) The set of  all the countries in the world.

(2) The set containing all the prime number less than 1000.

(3) C  = { x ; ∈ N , 99 <  x and  x < 109}.

(4) D  = { x : x  R ,  x2 - 3 = 0 }.

(5) E  = The set of all the even Natural Numbers less than 100.

Infinite Set

The set in which total numbers of elements can 
Not be counted, i. e. their counting of elements 
do not come to an end ,is called Finite set.

e. g.


(1) The set of all points in a plane.
(2) The set containing all the prime number .

(3) C = { x ; x ∈ R , 99 < x and x < 109}.

(4) D = Set of lines parallel to given line


(5) E  = The set of all the even numbers greater than 100.

(6)  S = Set of stars in the sky,
as the stars can nit be counted , they are numberless.

Cardinal Number of a Set

The numbers of distinct elements present in any finite set is called cardinal number of finite set. And it is denoted by n(A) .

If A =  { 1 , 3 , 5 , 8 , 9 } then n(A) = 5 and if  

B= {a , b , c , d , e , f , g , h , i , j } then  n(B) = 10 and if

C = {1 , 2 , 3 , 4 , 5 , 6 , ...... , 100 }then  n(C) = 100

Because The set A , B and C have 5 ,10 and 100 elements respectively .

Subset

The set B is said to be the subset of A if every element of set B is also  an element of set A . It is denoted by ⊆ , if we write A  B that means A is subset of B and if we write B ⊆ A means B is subset of A .

If A =  { 1 , 3 , 5 , 8 , 9 } and  B = { 1 , 3 , 5 } then B is called subset of A as all the elements of set B are in Set A.


Proper Subset

Any set B is called proper subset of set A if every element of set B is an element of set A whereas every element of set A is not an element of Set B . It is denoted by ⊂ .

Let A = { 1 , 2 , 3}  and B = { 1 , 2 } and C = { 1 , 3 } then B ⊂ A  and also  C  ⊂ A . It is to be noted that A  A i.e. every set is a subset of itself but it is not a proper subset.

Power Set

If we form a set which consist of all the subset of a given set A , then that set is called Power Set and is denoted by P(A) .
e.g  if A = { a , b , c } then

P(A) = { ф , { a } , { b } , {c} , {a , b}, { a , c} , { b , c } , { a , b , c } 

Equal Sets

Two sets are said to be equal to each others if they have same and same numbers of elements ,
 e. g . If  A =  {  1 , 2} and B = {x :  x ∈ N ,   x 2 } ,Then these sets A and B are equal to each other because both the sets have same numbers of elements and also same elements  1 and 2 .

Equivalent Sets 

Two sets are said to be equivalent if both the sets have same numbers of elements , And Equivalent Sets need not be equal to each other but have same cardinal numbers .

e.g  if A = { a , b , c } and B = { 1 , 2 , 3 } are Equivalent Sets

Comparable Sets

Two sets are said to be comparable if one of them is a subset of other  i.e. either A ⊆ B or B ⊆ A .
A = { a , b , c } and B = { b , c } are comparable sets.

Universal Set

A set that contains all the sets under consideration , and is denoted by U .

When we are using set containing  real numbers , then Set R is called Universal set .

If we are considering set of equilateral triangles , then Set of Triangles is called Universal set .

If A = { 2 , 4 , 6 },  B = { 1 , 2 , 3 , 4 } and C { 5 , 6 , 8 , 9 , 10 }
then Set U = { 1 , 2 , 3 , 4 ,  5 , 6 , 8 , 9 ,10 } is called Universal set.

Venn  Diagrams

Union of Sets


the union of two sets A and B is the set which contains  all those elements  which are either in set A or in from set B or in both A  and B. It is denoted by A U B .


Thus A U B = { x : x ∈ A or x ∈ B }If A = { 1 , 2 , 5 , 6  } and B = { 8 , 9 , 10 }    then  

A U B = {1 , 2 , 5 , 6 , 8 , 9 , 10 } , as this set contains  the elements either  from Set A or from Set B.


WHAT IS SET, UNION ,INTERSECTION AND VENN  DIAGRAMS


Again if A = { a , b , c , d, e , f , g } and B = { a , b , c , d  }
A U B = { a , b , c , d, e , f , g } = A
Thus  the union of a given set and its subset is always the given set 

Intersection of sets


Intersection of two sets A and B is the set which contains all those elements which are in both the set A and Set B . It is denoted by A  B .


Thus A  B = { x : x ∈ A and x ∈ B }

If A = { 1 , 2 , 5 , 6  } and B = { 5 , 6 , 10 }

then    A B = { 5 , 6  }



WHAT IS SET, UNION ,INTERSECTION AND VENN  DIAGRAMS



Again if A = { a , b , c , d, e , f , g } and B = { a , b , c , d  }
then A B = { a , b , c , d  } = B 
Thus  the intersection  of a given set and its subset is always the subset of  a given set .

Disjoint Sets


Two sets are said to be disjoint sets if they have no common element. i.e. if their intersection is a null set A B = {  } = ф
if  A = { a , b , c , d, e , f , g } and B = { h , i , j , k , l , m } then these sets A and B are disjoint as they have no common element

i. e A B = ф .


WHAT IS SET, UNION ,INTERSECTION AND VENN  DIAGRAMS

Difference of Sets


The difference of two sets A and B is the set of all those elements of  A which which are not in B . It is denoted by A - B.
Thus A - B = {x : x  A but x  B  }
If A = { 2 , 3 , 4 , 5 , 6 } and B = { 5 , 6 , 7 , 8 } then
A - B = { 2 , 3  , 4 }


WHAT IS SET, UNION ,INTERSECTION AND VENN  DIAGRAMS




Similarly  The difference of two sets B and A is the set of all those elements of  B which which are not in A. It is denoted by B - A.
Thus A - B = {x : x  B but x  A  }

If A = { 2 , 3 , 4 , 5 , 6 } and B = { 5 , 6 , 7 , 8 } then



B - A = { 7 , 8 }


WHAT IS SET, UNION ,INTERSECTION AND VENN  DIAGRAMS

Complement of a Sets



If we have a universal set U and  a given set A ⊂ U then  complement of  a set A is the set which contains all those elements of universal set U which are in not in A . It is denoted by Ac 

Thus A= { x : x  A  and x  U }


WHAT IS SET, UNION ,INTERSECTION AND VENN  DIAGRAMS

Some Important Results 

c  ф 

ф c  = 

c   ф 

( A c )  c   =    A 

U  A c   =  U

  A c   =  ф

(A U B )  c   =  A c  B c    De Morgan's Law

(A  B )  c   =  A c U c    De Morgan's Law

Practical Application of Sets

If A and  B are disjoint sets

(A B ) = n(A) + n(B) 

If A and B are not disjoint sets

n (A U B ) = n(A) + n(B) - n (A ∩ B )

Question 


In a class of 50 students , 35 opted Mathematics 25 opted Biology then how many opted both Mathematics and Biology ?

Solution


Here   n ( M U B ) represents the total ( sum ) of students in a class and n ( M ∩ B ) represents the common students who have opted both the subjects Biology and Mathematics in that class.

Here we can use this formula 

n ( M U B ) = n ( M ) + n ( B ) - n ( M ∩ B ) --- -- -- (1)
where n (M U B ) = 50 , n ( M ) = 35 , n( B ) = 25 ,
( M  B ) = ? 

so putting the above values in (1)

50 = 35 + 25 - n ( M ∩ B )
( M  B ) = ( 35 + 25 ) - 50

( M  B ) = 60- 50

( M  B ) = 10 

So There are 10 students in that class who opted both the subjects Biology and Mathematics .


Question 

In a village 350 peoples read Hindi News Paper , 235 peoples read English News Paper and 150 People read both the News Papers , How many peoples read News Papers ?

Solution

Here n ( E U H ) represents the total ( sum ) of people in a village and n ( E ∩ H ) represents the common Reader who read both the News Papers Hindi and English in that village .

Here we can use this formula 

n ( H U E ) = n ( H ) + n ( E ) - n ( H ∩ E ) --- -- -- (1)


where n (H 
  B ) = 150  , n ( H ) = 350 , n( E ) = 235 ,
 n ( H U B ) = ? 

so putting the above values in (1)
n ( H U E ) = 350 + 235 - 150( M U B ) = ( 350 + 235 ) - 150

( M B ) = 585 - 150


( M U B ) = 435 

So there are 435 Peoples who  read News Papers .


Question 

In a group of 50 teachers , 30 teachers drinks coffee and 25 drinks both coffee and tea, How many teachers drinks tea and how many drinks tea only

Answer

Here  C  and T  represents Coffee and Tea Respectively . Therefore Total numbers of teachers  in that group are 50  i.e.  n (T U C ) , and numbers of teachers who take both drinks are 30 . i . e. 
n ( C ∩ T ) = 25 And  n( C ) = 30 so n (T) = ? and Teachers drinks tea only will be represented  with n ( T - C ) ,which we have to find
n ( T U C ) = n ( T ) + n ( C ) - n ( T ∩ C ) --- -- -- (1)
              50  = n (T) + 30 - 25
              n ( T ) = 45
It means that 45 Teachers drinks tea , this implies that in these 45
Teachers there are some Teachers who  drinks both coffee as well . so by the statement  how many drinks tea only , we have to drop those Teachers who drinks both ,
 n( T - C ) = n ( T ) - n ( T ∩ C )

 n( T - C ) = 45  - 25

 n( T - C ) =  20



Conclusion

Thanks for devoting your valuable time for this post "What is Set ,union, intersection , disjoint sets , complement and difference of sets and venn diagrams " of my blog . If you liked this this blog/post, Do Follow me on my blog and share this post with your friends . We shall meet again in next post with  more interesting article  ,till then Good Bye.

Thursday, 19 July 2018

HOW TO SOLVE HARD AND IMPOSSIBLE PUZZLES PART 3


Hello Friends 
Welcome once again . Let us  discuss some of the puzzles,  which are viral in social media such as Face book  ,whats app, twitter  and Google + etc. which are based on Mathematics directly or indirectly . You also have seen such types of problems mentioned  99.9%  Fail , Solve if you are genius like this...


TODAY'S  PUZZLE FOR YOU

HOW TO  SOLVE   HARD  AND IMPOSSIBLE  PUZZLES   PART 3


I am fully hopeful that you can crack this puzzles , in case you are not able to crack/solve this , or want to confirm your answer you can click here .
Let us try to discuss and solve  some of the Puzzles given below  one by one.


PUZZLE #1 


Look at 1st Row, there is no direct connection between 9, S and 10. Also it seems no connection between 4,5,M,,1,3 in 2nd row ,Similarly 3,1,1,L,2,?,1 have no direct connection between them.

But if we recall English Alphabet A,B,C,D,E,F,.. ..Y,Z and check that which letter is at 9th and 10th position, we shall find that this is not "S" .But let us check what is the position of Letter "S" in English alphabet. We find that it is in 19th position, as 9+10= 19 , It means that "S" is equal to 19.

How to solve puzzles


So now check the position of letter "M" , which is 13th , and the  sum of all the digits in 2nd Row is 4+5+1+3 = 13, Similarly we can check that 1+2+4+5=12, 12th is the position of letter "L" and  3+2=5 ,5th is the position of letter E.

So by this trend we can add all the digits in 3rd row and put its total equal to 8, as position of the letter "I" is 9th in English Alphabet. The difference between the position of the letter "I"  and total  of all the digits in same time must be equal to each other.

Therefore 1(one) is the digit to be occupied by ? 

PUZZLE 2


In this Puzzle we have five circles in all ,and some of them are overlapping each others , Look at 1st line 15,12,24 ,It seems no relation between these numbers.But if we  add all the numbers lying in any circles , then from 1st circle of 1st line we shall have 15 + 10 = 25 and from 3rd circle of 1st line we shall have 24 + 1 = 25 .

HOW TO  SOLVE   HARD  AND IMPOSSIBLE  PUZZLES   PART 3


Similarly from 2nd circle of 2nd line we shall have 4 + 1 + 20 = 25.
It means total of all the numbers lying in any circle must be 25 . So 
from 2nd circle of 1st line we shall have 12+? + 4= 25 implies that ? must be 9 and when we consider  1st circle of 2nd  line we shall have 10 + 6 + ? = 25 this also implies that  ?  must take the value 9.


So our  Answer will be 9

PUZZLE #  3



Look at 1st line if we subtract 1 from 1 ,we get 0 , we get our 1st line's answer . Same pattern should be followed in the next and remaining lines. But when we subtract  1 from 2 in 2nd line we get 1,which is not our answer. 

How to solve picture puzzles

So let us try  new pattern right from the 1st line. If we square the 1st number and then subtract the 2nd  number from it, Let us check it from 1st line .
12  - 1 = 1 ,      It is working for 1st line
22  - 1 = 3,      It is working for 2nd line
3 - 1 = 8,       It is working for 3rd line
42  - 1 = 15,     It is working for 1st line
52  - 1 = 24,  


So following the same pattern we can calculate answer

for the line containing 6 - 1 like this 
62  - 1 = 35

So our answer is  35.


The formula for this puzzle is x-1 = 
x2  - 1 ,
where x is the 1st number





PUZZLE # 4


From the 1st line it can be seen that 21 × 2 = 42, But 39 × 3 ≠ 618 .
so if we multiply 1st digit of left side with 2 then we shall have 4,6,2 as our  digit/digits in left side of answers and when we multiply 2nd digit with 2 , we shall have 2,18,14 as right side of answer .


How to solve puzzles and quizzes


So we have to multiply with 2 on left side of our question to get left
digit and  multiply 5 with 2 to get 10 on right side of the answer.
Hence Our answer will for  45= (2×4)(2×5) = 810

The Formula for this Puzzle is  ab = (2a)(2b)


PUZZLE # 5


If we add all the digits , we can not get the answer .we have to try something new . But if we square the 1st number and add  both remaining numbers to it ,the we can get result of 1st line like this  2 + 3 + 4 = 22  +3 + 4 = 11.   
Let us check the same pattern for 2nd line 3 + 4 + 5 = 32  + 4 + 5 = 18, but 18 is not the answer . It means this will not work here.

Now we have to try one more formula for this puzzle .

If we multiply the 1st number with 2 and add both the

remaining  numbers to it.

How to solve puzzles and quizzes


Let us check all the line respectively.

1st  line  is    2×2 + 3 + 4   = 11, which 
is right answer.

2nd line  is    2×3 + 4 + 5  =  15, which 
is right answer.

3rd line   is    2×4 + 5 + 6   = 19, which 
is right answer.

So let us follow this pattern to find 
the answer of    the given puzzle as
follows

4th line  is   (2×7) + 8 + 9 = 31, which is
 right answer.

So formula we can use for this puzzle a+b+c   =   2a+b+c



PUZZLE # 6

To solve this puzzle , multiply 1st two digits with each other to get 1st two digits of the answer, After that multiply 1st and last digits to get 3rd and 4th digits of the answer, Now  add the numbers formed from 1st two steps and subtract 2nd number from it, we shall  have 5th and 6th digits of the answer.
  

How to solve puzzles and quizzes



1st line can be written as   (5×3)(5×2)(15+10-3) = 151222,
2nd line can be written as  (9×2)(9×4)(18+36-2) = 183652
3rd line can be written as  (8×6)(8×3)(48+24-6) = 482466
4th line can be written as  (5×4)(5×5)(20+25-4) = 202541
Similarly we can write last and final line 
Final line  can be written as  (7×2)(7×5)(14+35-2)=143547

Formula for this Puzzle is abc=(ab)(ac)(ab+ac-b)

PUZZLE 7


This problem is based on solution of linear equation of two variables. so it can be  solved as follows .

If we put A,B,C and D then 

we shall have following equations 
A + B = 8--------------------(1)
C - D  = 6--------------------(2)
A + C = 13-------------------(3)
B + D = 8--------------------(4)


HOW TO  SOLVE   HARD  AND IMPOSSIBLE  PUZZLES   PART 3

Now we have to solve these equations for A,B,C and D.
From eq (1) we have A = 8 - B  ,put this value of A in (3),
we get   - B + C = 13, Rewrite it 
              B - C = -5 --------------------(5)
Now  adding (2)  and (5) ,we get
C - D +  B - C = 6 - 5
B - D =  1 ---------------------(6)
Adding (4) and (6),we get 2B = 9  ⇒ B =  9/2 = 4.5
Putting the value of B in Eq (1), we get 
4.5  +  A = 8 A = 3.5 ,
Put  "A" in (3) ,we get C = 9.5
Put the value of  "B" in (4) 
4.5+D = 8 D = 3.5  So Final Answers  are 

A = 3.5,         B = 4.5,       C = 9.5,          D = 3.5


Conclusion


Thanks for devoting your valuable time for this post "How to solve various puzzles ,Quizzes , Brain Teasers and challenges " of my blog . If you liked this this blog/post, Do Follow me on my blog and share this post with your friends . We shall meet again in next post with solutions of most interesting and mind blowing puzzles ,till then Good Bye.


Let us learn Multiplication , division , arithmatic and simplification short cut ,tips and  tricks after buying this Book of Magical Mathematics .