Matrix method of solving linear equations of three variables

Learn the process of solving linear equations of three variables by matrix method  .Let us understand this method with the help of an example

Matrix method of solving linear equations of 3 variables

Matrix method of solving linear equations of three variables with the help of an example. 

The system of these equations can be transformed into Matrix form as 

AX =  B  ,  ⇒ X =   A^-1 B   ->  (*)

Where A is matrix written from the coefficients of x, y and z when these equations are in symmetric form and B is the matrix written from constants from right hand sides in column form and X is matrix of all the variables in column form. 

In order to find the solution of set of these equations , first we have to find the inverse of matrix A if it exist then we can find the solution otherwise Matrix method fails to find the solution of the set of linear equations . 

Evaluation of Determinant 

|A|  = 1 (-9 - 27) -1(6 - 63) -1(6 + 21)

       =  -36 + 57 - 27

       = -63 + 57

        = -6

Since the determinant value is not equal to zero ,Therefore its inverse can be calculated.

And  formula for finding the inverse of matrix A is 

Where Adjoint A is the transpose of co factor matrix. And in order to find the co factor matrix of any matrix, we have to find co factors of all the elements present in that matrix. 

How to calculate  co factors of all the elements of the matrix A. 

 Let us calculate these cofactors. 

Co factors of 1st row are  -36 , 57 , 27

Co factors of 2nd row are  -6, 10 , 4

Co factors of 3rd row are  6 . -11 , -5

Now these co factors can be written in matrix form known as co factor Matrix. 

Co factor Matrix

Writing co factors of 1st row in 1st row of this matrix , co factors of 2nd row in 2nd row of this matrix . Similarly co factors of 3rd row in 3rd row of this matrix . 

Adjoint  Matrix

To find the Ad joint of this matrix we have to take it's transpose, Because transpose of any matrix is called Ad joint of the matrix. So writing all the elements which are in 1st row in 1st column, and  all the elements which are in 2nd row in 2nd column and  all the elements which are in 3rd row in 3rd column. 

Now we can find inverse of the matrix A by putting the value of inverse of A in equation  (4), we get

Now  putting the values  Matrix B and    A^-1  in (4) 

After simplification and using the properties of equality of two matrices  ( Two matrices of same order are equal if and only if their respective elements are equal to each other ) 

⇒  x = -54/-6 = 9

 y =  12/-6 = 2

 z  = -24/-6  = -4

Hence

 x = 9 

 y = 2

 z = -4

So this was the Matrix method of solving linear equations of three variables using inverse of matrix. Your valuables comments will be appreciated for betterment of this blog.

Also read this post for understanding inverse of matrix using elementary row transformation

 

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Ten Logical Reasoning questions and answers for competitive exams

Ten Logical Reasoning questions and answers for competitive exams  like Bank PO, Bank clerk, SSC CGL, ssc chsl, RRB NTPC , group D etc have been discussed in this post in very easy style and shortcut way .

Logical Reasoning questions and answers for competitive Exams 

Problem # 1

All the letters in each option are written from alphabet either in order or in reverse order.

From 1st option MNOP are written in  order starting from M in alphabet. 

From 2nd option TSRQ are written in reverse order starting from Q  in alphabet. i.e.  

QRST  ⇒ QRST   ⟺  TSRQ (After reversing the orders of letters).

From 3rd option  EDCB are written in reverse order starting from B  in alphabet. i.e.  

BCDE  ⇒ BCDE   ⟺  EDCB (After reversing the orders of letters).

 From 4th option  XWVU are written in reverse order starting from U in alphabet. i.e.  

UVWX  ⇒ UVWX  ⟺ XWVU ( After reversing the orders of letters).

Since  2nd,  3rd and 4th options are following the same pattern because all the letters in these options are written in order. But letters in 1st option are  written in reverse order. And this option is odd one.

Hence Option (1)MNOP is correct option.

Problem # 2

All the letters are written alternatively from alphabet either in order or in reverse order.

From 1st option MKIG are written in reverse order starting from G with a gap of one letter in alphabet. i.e.  

G H I J K L M ⇒ G H I J K L M ( Omitting blue colored letters )  ⇒ G I K M ⟺ MKIG (After reversing the orders of  letters).

From 2nd option VTRP are written in reverse order starting from P with a gap of one letter in alphabet. i.e.  

PQRSTUV ⇒  PQRSTUV (Omitting blue colored letters)  ⇒  PRTV ⟺ VTRP (After reversing the orders of letters).

From 3rd option EGIK are NOT written in reverse order starting from K with a gap of one letter in alphabet. 

From 4th option YWUS are written in reverse order starting from S with a gap of one letter in alphabet. i.e.  

STUVWXY ⇒  STUVWXY (Omitting blue colored letters)  ⇒  SUWY ⟺ YWUP (After reversing the orders of letters).

Since 1st , 2nd and 4th options are following the same pattern because all the letters in these options are written in reverse order with a gap of one letter. But letter in 3rd option are not written in reverse order.

Hence Option (3)EGIK is correct option.

Problem # 3

In this reasoning problem 1st number (5) is associated to 100 with the help of any rule , in the same rule we have to associate 7 to a number out of four given options.

Look carefully the given numbers consists of three digits. These three digits can be utilised with the help a formula given below.

Formula :-  4 × (1st number)²  = 2nd Number

 4 × (3rd number)²  = 4th Number

⇒ {5} = 4 × 5² = 4 × 25  = 100 (2nd number)

⇒{7} = 4 × 7² = 4 × 49 = 196 (4th number)

Option (4)196 is correct option.

Problem # 4

In this problem of reasoning we have to combine all the three given  numbers in such a way that after applying any mathematical operation/operations we could get the number in the right hand side of all the three problems given above. Because in these types of reasoning problems we can change mathematical sign according to our requirements.

Replacing - sign with + , + sign with × in above conditions . Now place 1st two numbers in bracket . Then apply the BODMAS rule.

(7 + 2) × 4 = 9 × 4 = 36 

(10 + 5) × 2 = 15 × 2 = 30

Similarly 

(20 + 5) × 6 = 25 × 6 = 150

Option (4)150 is correct option.

Problem # 5

In this reasoning problem 1st number (392) is associated to 28 with the help of any rule , in the same way we have to associate 722 to a number out of four given options. 

Formula :- 2 × √(1st Number ÷ 2) = 2nd Number

2 × √(392 ÷ 2) = 2 × √(196)  = 2  × 14 = 28 . 

Similarly 

2 × √(722 ÷ 2) = 2 × √(361)  = 2  × 19 = 38

Option (3)38 is correct option.

Problem # 6

 

These words are used when a book is prepared, published and read by Reader.   To publish a book , first of all it has to be written by the Author . After this book will be with the Editor for any change or editing. Then that book will be published by Publisher. After publication it will be sold by Bookseller. And at last it will be purchased by read by Reader.

The journey of the book is as follow

1. Author
2. Editor
3. Publisher
4. Bookseller
5. Reader

So according to given problem's solution the correct option will be  (D) 4 , 1 , 3 , 5 , 2

Problem # 7

Divide the sum of both the numbers with 10 to get the number in right hand side.

( 15 + 55 ) ÷ 10 = 70 ÷ 10 = 7

( 25 + 35 ) ÷ 10 = 60 ÷ 10 = 6

( 35 + 45 ) ÷ 10 = 80 ÷ 10 = 8

Similarly ( 23 + 17 ) ÷ 10 = 40 ÷ 10 = 4

Hence option (3)4 is correct.

Problem # 8

In this reasoning problem three out of four options have been calculated by Multiplying  the 1st number with its next number(successive number) to get 2nd number. This rule is applicable to only three out of four option. And one which do not follow this rule will be the correct option.

(1)   9 × ( 9 + 1 ) = 9 × 10 = 90

(2)   13 × ( 13 + 1 ) = 13 × 14 = 182

(3)   8 × ( 8 + 1 ) = 8 × 9 = 72

(4)   12 × ( 12 + 1 ) = 12 × 13 = 156, but this is given as 144

Hence option (4)12 - 144  is odd one.

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Problem # 9

Take the difference of two successive numbers, this difference is square of continuous natural numbers.

6  -  5 = 1 = 1²

10 -  6 = 4  = 2²

19 -  10 = 9  = 3²

35 - 19 = 16  = 4²

Similarly the difference of last number and 2nd last number can be calculated. This difference will be the square of some natural number.

?  - 35 = 25  = 5²

?  = 25 + 35  = 60

Hence option (2)60  is correct.

Problem # 10

Take the difference of two successive numbers, this difference is cube of continuous natural numbers.

3085 - 1357 = 1728 = 12³ 

5282 - 3085 = 2197 = 13³

8026 - 5282 = 2744 = 14³

Similarly the difference of last number and 2nd last number can be calculated. This difference will be the cube of some natural number.

? -  8026 = 3375 = 15³

? = 3375 + 8026 = 11401

Hence option (4)11401 is correct.

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Conclusion

So these were the ten  problems regarding the post Ten Logical Reasoning questions and answers pdf for competitive exams like SSC CGL ,SSC CHSL ,CPO ,Bank exams and RRB NTPC etc which were explained in this post. Feel free to comment your valuable suggestions.

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Finding inverse of 3x3 matrix using elementary transformation

We shall learn the process of finding inverse of 3x3 matrix using elementary transformation . The  method of finding inverse of the matrix 3 x 3 using elementary row transformations involves 5 to 6 steps . Let us understand this method with the help of an example.

Inverse of 3x3 matrix using elementary transformation

For finding Inverse of  3 x 3 Matrix using Elementary Row Transformations , We shall strat with this formula

  A  = I A ,

 Where I is Unit Matrix of order 3

In the 1st step we have to make 1st element of 1st row and 1st column unity. We shall start changing the given matrix written  in left hand side  step by step to unit matrix and this this process will also appliesd to the the unit matrix written on right hand side of equation (1) , Hence this change the  matrix written on the right hand side of  equation (1) to other matrix . And when the given matrix on the left hand side changed to Unit matrix , the matrix which was on the right hand side of equation of (1) will be inverse of the given matrix.

Suggested direction of elementary rows operations  are as follows

In the 1st step we have to make the 1st element of 1st column to unity by using taking suitable number common from it .

In the next step by using the above step we have to make 2nd and 3rd  element of 1st column unity by using suitable row transformations. 

 

Till now we have made two elements of 1st column zero , now we have to make 1st and 2nd elements of 3rd Row zero by using suitable row elementary transformations. Note carefully 3rd elements of 3rd row can not be changed to zero as this element will have to  be reduced to unity as in Unit Matrix. 

After this step start making the elements of 2nd Row equal to zero but using suitable row transformations.

Now we shall conclude the process of inverse of matrix  with the changing of  2nd and 3rd elements of 1st row  to zero.. 

Here matrix B is the inverse of the given matrix A. 

To verify that the matrix B obtained is correct or not. We can check it by multiplying matrix B with I, if it comes out equal to matrix A. Then our answer is correct. 

Also read this Post for solution of set of three variables equations using matrix method

So the process of finding inverse of 3x3 matrix using elementary transformation discussed in this post with the help of an example. Your comment will always be appreciated for betterment of this blog. 

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Ten most important Reasoning questions with answers for competitive exams

Ten Most Important Reasoning questions with answers for competitive exams of series, box and other type with solutions  have been discussed in this post . These types of problems are very helpful for cracking competitive exams like ssc cgl, ssc chsl and various Bank exams and many other similar exams.

10 Reasoning questions with answers for competitive exams

PROBLEM   1

This reasoning problem consists of three figures and every figure have three numbers associated to it . Two numbers are on the upper line of each box and one number is at the bottom of the dark box.  Look at last figure , it have ? in its centre . So the solution of this problem is to find the value of question mark using two numbers associated to it . 

          But the main problem is how to utilised  these two numbers to get the value of question mark?

          Now watch carefully the 1st two figures . Since these figures have some values of numbers in dark box . 

          Now we have to find or search the  formula for these three numbers in each figure to utilised them in any possible way to get number in  dark box . 

         The same formula will be applicable to third figure to find out the value of question mark.

Formula :- The Number in each dark box in every figure is H C F ( Highest common Factor) of remaining  two numbers. 

Solution:-  HCF of two numbers = 3rd number

H C F  of 6 and 10 in 1st figure is 2 (The number in 1st figure of dark square). 

H C F of 12 and 36 in 2nd figure is 12 (The number in 2nd figure of dark square).

H C F of 15 and 25  in 3rd figure is 5 (The number in 3rd figure of dark square).

Option (4)5  is correct option.

PROBLEM   2

This reasoning problem consists of three figures and every figure have three number associated to it . Two numbers are on the upper line of each box and one number is at the bottom of the dark box.  Look at last figure , it have ? in its centre . So the solution of this problem is to find the value of question mark using two numbers associated to it . 

          But the main problem is how to utilised  these two numbers to get the value of question mark?

          Now watch carefully the 1st two figures . Since these figures have some values of numbers in dark box . 

          Now we have to find or search the  formula for these three numbers in each figure to utilised them in any possible way to get number in  dark box . 

         The same formula will be applicable to third figure to find out the value of question mark.

Formula :- The Number in each dark box in every figure is L C M ( Lowest common Multiple) of remaining  two numbers. 

Solution:-  LCM of two numbers = 3rd number

L C M  of 6 and 10 in 1st figure is 30 (The number in 1st figure of dark square). 

L C M of 12 and 36 in 2nd figure is 48 (The number in 2nd figure of dark square).

L C M of  5 and 7 in 3rd figure is 35  (The number in 3rd figure of dark square).

Option (2)35  is correct option

PROBLEM   3

This box problem consist of three rows and three columns. And we have to find the value of question mark after studying the pattern of all the numbers in this box. 

Second column in every row is equal to the five time of division of 1st column with 3rd column in every row.

1st Row

5 × {28 ÷ 7} = 5 × 4 = 20 (Middle Number in 2nd column 1st row) 

2nd Row

5 × {84 ÷ 12} = 5 × 7 = 35 (Middle Number in 2nd column 2nd row) 

 3rd Row

5 × {45 ÷ 9} = 5 × 5 = 25 (Middle Number in 2nd column 3rd row) 

Option (D)25 is correct option.

PROBLEM   4

This box problem consist of three rows and four columns. And we have to find the value of question mark after studying the pattern of all the numbers in this box. 

Third row of every column is equal to the product of square root of 1st  row and cube root of 2nd row .

Formula  :- 3rd row = Square root of 1st row × Cube root of 2nd row

 1st Column

√1 ×  ∛8  = 1 × 2 = 2  

2nd Column

√4 ×  ∛27  = 2× 3 = 6  

3rd Column

√16 ×  ∛64  = 4 × 4 = 16  

4th Column

√36 ×  ∛343  = 6 × 7 = 42 = ?  (The value of question mark) 

Option (C)42 is correct option

PROBLEM   5

In Roman number system  C , CC , CCC , D , DC are number which have values in decimal system as follows.

C = 100

CC = 200

CCC = 300

D = 500

DC = 600

So 100 , 200 , 300 , 400 , 500 and 600 are the numbers in series of 100 difference. Hence in  Roman number system 400 can be written as CD.

Therefore

Option (1)CD is correct option.

PROBLEM   6

In any coding language sum of  10 and 10 has been written as 6, sum of  20 and 20 has been written as 12 . In the same coding language sum of  30 and 30 has also been written as 12. Then we have to find the value of sum of  0 and 0 ?

In the above reasoning problem it is not 10 + 10  , Actually to find its solution we have to think like this way that "each number is written in word and then its letters have been counted to find the sum of two given numbers".

TEN + TEN = 3 (Three letters in word TEN) + 3(Three letters in word TEN) = 6 (Sum of letters in both the words is SIX).

TWENTY  + TWENTY = 6(Six letters in word TWENTY)  + 6(Six letters in word TWENTY)  = 12(Sum of letters in both the words is TWELVE).

THIRTY  + THIRTY =  6(Six letters in word THIRTY) + 6(Six letters in word THIRTY)  = 12 (Sum of letters in both the words is TWELVE).

ZERO + ZERO = 4(Four letters in word ZERO) + 6(Four letters in word ZERO)  = 8 (Sum of letters in both the words is 8).

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PROBLEM   7

In these type of reasoning problems 2nd number will be obtained from 1st number either by using one mathematical operation/ set of mathematical operations or by using any mathematical operation/ set of mathematical operations to individual digits of 1st number. Same  mathematical operation/ set of mathematical operations  or formula will be used in 3rd number to get fourth number.

Formula :-  (Half of Square of 1st number)  = 2nd number  

   (Half of Square of 3rd number) = 4th number 

2nd Number

18²  ÷ 2 = 324 ÷ 2 = 162 (2nd Number)

4th Number

36²  ÷ 3 = 1296 ÷ 2 =648 ( 4th Number)

Option (4)648 is correct option.

PROBLEM   8

In these types of reasoning problems 2nd number will be obtained from 1st number either by using one mathematical operation/ set of mathematical operations or by using any mathematical operation/ set of mathematical operations to individual digits of 1st number. Same  mathematical operation/ set of mathematical operations  or formula will be used in 3rd number to get fourth number.

Formula :-   Product of two consecutive  factors  =   Any number 

1st Number  =  4  × 5 = 20

2nd Number =  6  × 7 = 42

3rd Number =  8  × 9  = 72

4th Number =  10  × 11  = 110

Option (2)110 is correct option.

PROBLEM   9

This reasoning problem consists of three figures and every figure have three numbers associated to it . So the solution of this problem is to find the value of question mark using other two numbers associated to it . 

          But the problem is how to utilised these two numbers to get the value of this question mark?

          Now we have to find or search the  formula for these two numbers in each figure to utilised them in any possible way to get third number. 

         The same formula will be applicable to third figure to find out the value of question mark.

Formula :- Half of the product of  two outer numbers in each triangle is equal to third number .

 1st Triangle

( 2 × 3 ) ÷ 2 = 6 ÷ 2 = 3 (Middle number in 1st triangle ) 
2nd Triangle

( 4 × 5 ) ÷ 2 = 20 ÷ 2 = 10 (Middle number in 2nd triangle ) 
3rd Triangle 

( 6 × 7 ) ÷ 2 = 42 ÷ 2 = 21(Middle number in 3rd triangle ) 

Option (4)21 is correct option.

PROBLEM   10

This reasoning problem consists of three figures and every figure have three numbers associated to it . So the solution of this problem is to find the value of question mark using other two numbers associated to it . 

          But the problem is how to utilised these two numbers to get the value of this question mark?

          Now we have to find or search the  formula for these two numbers in each figure to utilised them in any possible way to get third number. 

         The same formula will be applicable to third figure to find out the value of question mark.

Formula :- Sum of  two  numbers in each triangle is equal to sum of remaining number .

 1st Triangle

( 8 + 4 ) = ( 3 + 9 )  = 12

 2nd Triangle

( 4 + 7 ) = ( 6 + 5 ) = 11

3rd Triangle

( 5 + ? ) = ( 9 + 3 ) = 12

5 + ? = 12

? = 12 - 5

? = 7

Option (1)7 is correct option.

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   Ten Most Important Reasoning questions with answers for competitive exams of  series , box and other type with solutions have been discussed in this post . These types of problems are very helpful for cracking competitive exams like ssc cgl, ssc chsl and various Bank exams and many other similar exams. please feel free to comment your opinions. 

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Ten most important Reasoning questions with answers for SSC CGL Exam

Ten Most Important Reasoning questions with answers for competitive exams of circles, box and other type with solutions have been discussed in this post . These types of problems are very helpful for cracking competitive exams like ssc cgl, ssc chsl and various Bank exams and many other similar exams.

Ten Reasoning questions with answers for competitive exams

Problem # 1

This box problem consist of four rows and four columns . And we have to find the value of question mark after studying the pattern of all the numbers in this box. 

If we divide this box into two parts vertically then we can have the formula for these numbers written in this box . Because after careful observation we can see that the sum of both the numbers in left half in any particular row is equal to sum of both the numbers in right half in that particular row. 

Formula:-  (1st number + 2nd number ) = (3rd number + 4th number) 

4 + 12 = 11 + 5 = 12 (Equal sum in both the half of 1st row)
6 + 7 = 10 + 3 = 13 (Equal sum in both the half of 2nd row)
8 + 9 = 10 + 7 = 17 (Equal sum in both the half of 3rd row)
7 + 5 = ? + 4 = 12 (Equal sum in both the half of 4th row) .
Hence ? = 12 - 4 = 8
? = 8

Option (D)8 is correct option 

 Problem # 2

This circle consists of four quadrants and every  quadrant consists of two numbers . Every quadrant have one numbers in outer part and  one number in the inner part . To find the value of question mark  "?"  , we shall use two numbers which are in the inner part of adjoining sectors to calculate the value of the number which is in the outer part of every quadrant . 

Formula :-  Sum of the squares of two adjoining numbers in two inner sectors is equal to number in outer part.

1st Quadrant 

 

3² + 5² = 9 + 25 = 34

2nd Quadrant 

5² + 4²  = 25 + 16 = 41

3rd Quadrant 

4² + 2²  = 16 + 4 = 20

4th Quadrant 

2² + 3²  = 4 + 9 = 13

Option (4)13 is correct option

Problem # 3

The number in any inner sector of this circle can be broken into two parts .  1st part can be taken as base of the number and 2nd part can be taken as the power of that number. And after writing it like this, the value of same number will be equal to the number in the outer part of same sector.

2nd Quadrant 

53 ---->  5^3 = 125

3rd Quadrant 

43 ----> 4³ =  64

4th Quadrant 

25 ---->  2^5 = 32

1st Quadrant 

33 ---->  3³ = 27

 Option (1) 27 is correct option

 

Problem # 4

This circle consists of four quadrants and every  quadrant consists of two numbers . Every quadrant have one numbers in outer part and  one number in the inner part . To find the value of question mark  "?"  , we shall use two numbers which are in the inner part of adjoining sectors to calculate the value of the number which is in the outer part of every quadrant . 

3² + 5 = 9 + 5 = 14

5² + 9 = 25 + 9 = 34

9² + 2 = 81 + 2 = 83

2² + 3 = 4 + 3 = 7

Option (1)7 is correct option

Also read these posts on Reasoning Questions with Answers

1.Reasoning for competitive exams
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Problem # 5

This reasoning problem consists of three figures and every figure have five number associated to it . Four numbers are at the corner of each figure and one number is at the centre of the figure. Look at last figure ,it have ? in its centre . So the solution of this problem is to find the value of question mark using four numbers associated to it . 

          But the biggest problem is how to utilised these four numbers to get the value of this question mark?

          Now watch carefully the 1st two figures . Since these figures have some values at middle position. 

          Now we have to find or search the  formula for these four numbers in each figure to utilised them in any possible way to get middle or central number. 

         The same formula will be applicable to third figure to find out the value of question mark.

Formula :- Keep on doubling the every number to get next number starting from middle number.

Solution:- 

1st figure

4 , (Middle number)

4 × 2 = 8  

8 × 2 = 16,

16 × 2 = 32,

32 × 2 = 64

2nd figure

5 , (Middle number)

5 × 2 = 10  

10 × 2 = 20,

20 × 2 = 20,

20 × 2 =  40

40 × 2 = 80

3rd figure

? , (Middle number)

? × 2 = 5 implies ? = 2.5

2.5 × 2 = 5,

5 × 2 = 10,

10 × 2 = 20.

20 × 2 = 40.

Option (4)2.5 is correct option

Problem # 6

This box problem consist of three rows and three columns. And we have to find the value of question mark after studying the pattern of all the numbers in this box. 

Third number of every row is equal to the product of square of 1st number and square root of 2nd number.

Formula:-  Square of 1st number × Square root of 2nd number

(3)²    ×  √49 = 9 × 7 = 63 ( Number in 1st row 3rd column ) 
(4)²    × √25 = 16 × 5 = 80 ( Number in 2nd row 3rd column ) 
(5)²    ×  √25 = 25 × 5 = 125  ( Number in 3rd row 3rd column ) 
(?)²    ×  8 = 200
(?)²   = 200 ÷ 8
(?)²   = 25
?   = 5
Option (A)125 is correct option

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Problem # 7

This reasoning problem consists of three figures and every figure have five number associated to it . Four numbers are at the corner of each figure and one number is at the centre of the figure. Look at last figure ,it have ? in its centre . So the solution of this problem is to find the value of question mark using four numbers associated to it . 

          But the major problem is how to utilised these four numbers to get the value of this question mark?

          Now watch carefully the 1st two figures . Since these figures have some values at middle position. 

          Now we have to find or search the  formula for these four numbers in each figure to utilised them in any possible way to get middle or central number. 

         The same formula will be applicable to third figure to find out the value of question mark.

Formula :- Difference of Sum of three number in upper half and lower half is equal to the middle number in every figure.

( 4  + 2 +  7 ) - ( 3 + 1 ) = 13 - 4 = 9 (Middle Number in 1st star ). 

( 3  + 3 +  5 ) - ( 4 + 2 ) = 11 -  6  = 5 (Middle Number in 2nd star).

( 6  + 9 +  2 ) - ( 4 + 3 ) = 17 - 7 = 10 (Middle Number in 3rd star).

Option (3)10  is correct option

Problem # 8

This reasoning problem consists of three figures and every figure have five number associated to it . Four numbers are at the corner of each figure and one number is at the centre of the figure. Look at last figure ,it have ? in its centre . So the solution of this problem is to find the value of question mark using four numbers associated to it . 

          But the main problem is how to utilised these four numbers to get the value of this question mark?

          Now watch carefully the 1st two figures . Since these figures have some values at middle position. 

          Now we have to find or search the  formula for these four numbers in each figure to utilised them in any possible way to get middle or central number. 

         The same formula will be applicable to third figure to find out the value of question mark.

Formula :- Difference of Product of three number in upper half and lower half is equal to the middle number in every figure.

 (4 × 2 × 5 ) - ( 20 × 1 ) = 40 - 20 = 20 (Middle Number in 1st star). 

 (3 × 4 × 5 ) - ( 2 × 10 ) = 60 - 20 = 40 (Middle Number in 2nd star). 

( 8 × 6 × 1 ) - ( 14 × 3 ) = 48 - 42 = 6  (Middle Number in 3rd star). 

Option (3)6  is correct option

Problem # 9

This reasoning problem consists of three figures and every figure have five number associated to it . Four numbers are at the corner of each figure and one number is at the centre of the figure. Look at last figure ,it have ? in its centre . So the solution of this problem is to find the value of question mark using four numbers associated to it . 

          But the biggest problem is how to utilised these four numbers to get the value of this question mark?

          Now watch carefully the 1st two figures . Since these figures have some values at middle position. 

          Now we have to find or search the  formula for these four numbers in each figure to utilised them in any possible way to get middle or central number. 

         The same formula will be applicable to third figure to find out the value of question mark.

Formula :- Keep on doubling the every number to get next number starting from middle number.

3 × ( 4 + 5 + 6 ) = 3 × 15 = 45 ( Middle Number in 1st circle ) . 

1 × ( 9 + 6 + 3 ) = 1 × 18 = 18 ( Middle Number in 1st circle ) . 

7 × ( 5 + 2 + 8 ) = 7 × 15 = 105 ( Middle Number in 1st circle ) . 

Option (3)105 is correct option

Problem # 10

In this pattern 1st number which is 520 can be written as sum of  cube of 8 and 8 , 738 can be written as sum of  cube of 9 and 9, 350 have been be written as sum of  cube of 7 and 7. Similarly ? will be written as sum of cube of any number and the number itself.

8³  + 8    :   9³ + 9 :   6³  + 6    :   7³  + 7     

6³ + 6 = 216 + 6 = 222

Option (2)222 is correct option

Ten Most Important Reasoning questions with answers for competitive exams of circles, box and other type with solutions have been discussed in this post . These types of problems are very helpful for cracking competitive exams like ssc cgl, ssc chsl and various Bank exams and many other similar exams. please feel free to comment your opinions.

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