Missing number in box Reasoning, Missing number in box puzzle

Ten most important missing number in box Reasoning, Missing number in box puzzle will be discussed with the help of ten most important examples. Some of these examples are of  3 × 3  order and other are of  4 × 4 orders.

Reasoning of missing number in box problems

 

Problem # 1

Exam Cracker

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. This question mark is in 4th row of 4th column. 

     To find the value of question mark. we  shall 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 product 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. 

Column wise :-

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

15 × 10 = 6 × 25 = 150 (Equal Product in both the half of 1st row)

4  × 16  =  8 × 8 = 64 (Equal Product in both the half of 2nd row)

6  ×  6 = 12 × 3 =  36 (Equal Product in both the half of 3rd row)

Similarly 

⇒  7 × 9 = 3 ×  ? = 63 (Equal Product in both the half of  4th row)

⇒  ? = 63

⇒  ? = 21

Option (B)21 is correct option.

 

Problem # 2

Exam Cracker

This reasoning problem consists of three figures and every figure have four numbers associated to it . Two numbers are on the upper line of each box and two number are at the bottom line of each box.  Look at last figure , it have ? in its 3rd figure . So the solution of this problem is to find the value of question mark using three numbers associated to it . 

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

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

          Now we have to find or search the  formula for these four numbers in 1st two figures to utilised them in any possible way to get number in that box . 

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

Formula:-  Product of all the numbers = 168 

1st Box 

Product of  all the numbers in 1st box is equal to 168

3 × 7 × 4 × 2 = 168 (1st figure) 

2nd Box 

Product of  all the numbers in 2nd box is equal to 168

8 × 3 × 7 × 1 = 168 (2nd figure) 

3rd Box 

Product of  all the numbers in 2nd box is equal to 168

1 × 6 × ? × 2 = 168 ( 3rd figure ) 

12 × ? = 168

? = 168 ÷ 12 

? =  14

Option (3)14 is correct option.

  

Problem # 3

Exam Cracker

This box problem consist of four 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. This question mark is in 4th row of 3rd column. 

     To find the value of question mark. we  shall divide this box into two parts horizontally then we can have the formula for these numbers written in this box . Because after careful observation we can see that the product of both the numbers in upper half in any particular column is equal to sum of both the numbers in lower half in that particular column. 

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

1st column :- 

8 + 7 = 2 + 13 = 15 (Equal sum in both the half of 1st column)

2nd column :- 

3 + 6 = 4 + 5 =  9  (Equal sum in both the half of 2nd column)

3rd column :- 

4 + 1 = 4 + ? = 5  (Equal sum in both the half of 3rd column)

⇒ 4 + ? = 5

⇒ ? = 5 - 4

⇒ ? = 1

Option (B)1 is correct option.

 

 Problem # 4

Exam Cracker

This box problem consist of three rows and three columns. And we have to find the value of question mark in 3rd row and 3rd column after studying the pattern of all the numbers in this box. 

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

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

(4)²    +  √16 = 16 + 4 = 20  (Number in 1st row 3rd column ) 

(3)²    +  √36 = 9 + 6 = 15  (Number in 2nd row 3rd column ) 

(1)²    +  √25 = 1 + 5 = 6  (Number in 3rd row 3rd column ) 

Option (B)6 is correct option.

 
Problem # 5

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.

      1st number of every row is equal to the product of square of 3rd number and square root of 2nd number.

Formula:-  Square of 3rd number  ×  Square root of  2nd number

(3)²    ×  √49 = 9 × 7 = 63 (Number in 1st row 1st column ) 

(4)²    × √25 = 16 × 5 = 80 (Number in 2nd row 1st column ) 

(?)²    ×  √64 = 200 (Number in 3rd row 1st column ) 

(?)²    ×  8 = 200

(?)²   = 200 ÷ 8

(?)²   = 25

?   = 5

Option (D)5 is correct option.

Problem # 6

Exam Cracker

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.

    Since question mark is in the 3rd column of 3rd row, so we can use any two numbers in any two particular rows  to find the value of 3rd  number in that particular row.

Formula:-  Square of 1st number  +  (2nd number ÷ 2 ) = 3rd number 

(6)²   +  ( 8 ÷ 2 ) = 36 + 4 =  40 ( Number in 1st row 3rd column ) 

(4)²   +  (  6 ÷ 2 ) = 16 + 3 =  19 ( Number in 2nd row 3rd column ) 

(7)²   +  (  2 ÷ 2 ) = 49 + 1 = 50 ( Number in 3rd row 3rd column ) 

Option (D)50 is correct option.

Problem # 7

Exam Cracker

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.

    Since question mark is in the 1st column of 3rd row, so we can use any two number in any two particular columns to find the value of 3rd  number in that particular column.

Formula:-  cube of 2nd number  ×  ( cube root of 3rd number ) = 1st number 

(4)³  ×  ∛8 = 64 × 2 = 128 ( Number in 1st row 1st column ) 

(3)³   ×  ∛64 = 27 × 4 = 108 ( Number in 2nd row 1st column ) 

(2)³   ×  ∛512 = 8 × 8 = 64 ( Number in 3rd row 1st column ) 

Option (A)64 is correct option.

Problem # 8

Exam Cracker

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

    Since question mark is in the 3rd column of 3rd row, so we can use any two number in any two particular rows to find the value of 3rd  number in that particular row.

Row wise

Formula:-  2 × (1st number - 2nd number ) = 3rd number 

( 14  - 10 ) × 2 = 4 × 2 = 8 ( Number in 3rd row 1st column) 

( 16  - 15 ) × 2 = 1 × 2 = 2 ( Number in 3rd row 1st column) 

( 19  - 13 ) × 2 = 6 × 2 =  ? = 12 ( Number in 3rd row 1st column) 

Option (D)12 is correct option.

Problem # 9

Exam Cracker

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.

    Since question mark is in the 1st column of 3rd row, so we can use any two number in any two particular columns to find the value of 3rd  number in that particular column.

Column Wise

Formula:-  cube of 1st number  +  ( cube root of 2nd number ) = 3rd number 

(2)³   +  ∛8  = 8 +  2 = 10

(3)³   +  ∛64 = 27 +  4 = 31

(1)³   +  ∛27 = 1 + 3 =  4

Option (A)4 is correct option.

Problem # 10

 

Exam Cracker

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

    Since question mark is in the 2nd column of 3rd row, so we can use any two number in any two particular columns to find the value of 3rd  number in that particular column.

Row wise

Formula:-   (3rd number - 1st number ) = Middle number 

( 3/2 ) - ( 1 ) = 1/2  (Middle number in 1st row ) 

( 8/3 ) - ( 2 ) = 2/3  (Middle number in 2nd row )

( 19/5 ) - ( 3 ) = 4/5  (Middle number in 3rd row )

Option (D)4/5 is correct option.

Ten Most Important Missing number in box Reasoning, Missing number in box puzzle 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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Simplest and shortest Matrix method to solve linear equations of 3 variables

Matrix method to solve linear equations of 3 variables

In this post we are going to understand the concept of solving linear equations of three variables with the help of matrix method .

Matrix method to solve linear equations of three variables with the help of example. 

Set of given equations are 

x - y + z = 2    

2x - y = 0  

 2y - 2 z = 1  

Rearranging these equations in symmetrical form. It means if any one of the variable in any equation is missing then write that missing variable/s with zero coefficient. As we see in this case the coefficient of z in 2nd equation and the coefficient of x in 3rd equation are missing. So we have to write coefficient of x in 2nd equation and coefficient of x in 3rd equation as zero.

x - y + z = 2 ----------------------> (1)

2x - y + 0z = 0 ------------------> (2)

0.x + 2y - 2z = 1 ----------------> (3)

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

AX = B , ⇒ X = A^-1B  -------> (*)

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| = 4 (18 - 4) -0(9 - 3) +6(12 - 18)

       = 4(14) + 0 + 6(-6) 

       = 56 - 36

        = 20

Since the determinant value of this matrix 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 this matrix. 

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

Let us calculate these cofactors.  

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

Co factors of 1st row are  (18 - 4) , -(9 - 3), (12 - 18) 

i. e. Co factors of 1st row are 14, -6 , -6

Co factors of 2nd row are -(0 -24), (12 - 18) , -(16 - 0) 

I. e. Co factors of 2nd row are  24 , -6 , -16

Co factors of 3rd row are  (0 - 36), -(4 - 18), (24 - 0) 

i.e. Co factors of 3rd row are  -36 , 14 , 24

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. 

Inverse  Matrix

Now we can find inverse of the matrix A by putting the value of inverse of A in equation  (4), 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 iff their respective elements are equal to each other ) 

Now we shall use the property of equality of two matrices ,which says that if two matrices are equal to each other then their respective elements must be  equal to each other.

⇒ x = 10

    y = 10

    z = 10

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.

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Missing number in box Reasoning problem, How to solve various circle problems

Reasoning of missing number in box problems, circle problems  and triangle problems will be discussed with the help of 10 most important examples. Some of these examples are of  4 × 4  order and other are of  4 × 3 orders. 

Reasoning of missing number in box problems in Uniteted State

PROBLEM #  1

Formula

Every row has been written as the cube of certain number. Because 4096 is the cube of 18, 6859 is the cube of 19 , 8000 is the cube of 20  similarly 926? will also be  the cube of any number. Look carefully 1st row is the cube of 18, then 2nd row is the cube of 19, 3rd row is the cube of 20 . So in this way 4th row must be  the cube of 21.

Calculation

18³ = 4096 

19³ = 6859 

20³ = 8000

21³ = 9261

Hence value of question mark will be 1.

Option (D)1 is correct option.

PROBLEM #  2 

  

Formula

Multiply the number in 2nd row with 100 and add it to the product of numbers in 1st and 2nd rows.

Calculation

6*100 + (4*8) = 632

3*100 + (9*2) = 318

7*100 + (5*9) = 745

 Option (A)745 is correct option.

PROBLEM #  3 

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 DIFFERENCE of squares of both the numbers in empty box in upper line in every figure is equal to  number in dark box in lower line.

Solution:- 

9²  - 6² = 81 - 36 = 45 (1st Figure)

8²  - 2² = 64 - 4 = 60 (2nd Figure)

7²  - 3² = 49 - 9 = 40  (3rd Figure)

Option (1)40  is correct option.

PROBLEM #  4 

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 aim 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 SUM of squares of both the numbers in empty box in upper line in every figure is equal to  number in dark box in lower line. 

Solution:- 

9²  + 6² = 81 + 36 = 117 (1st Figure)

8²  + 2² = 64 + 4 = 68 (2nd Figure)

7²  + 3² = 49 + 9 = 58 (3rd Figure)

Option (1)58 is correct option

PROBLEM #  5 

This figure consist of four squares around  one big square. Look carefully in this big square every number in it is perfect cube. And if we multiply the square roots of  two adjacent numbers then we shall have the number attached to big square between these two numbers whose square root had been multiplied.

√49  ×  √4 =  7  × 2 = 14 (The number at bottom line in the box)

√4  ×  √36 =  2  × 6 =  12 ( The number at leftmost box )

√36  ×  √? =   6  × √? = 30 ( The number at uppermost box )

√?  = 30/6 = 5, squaring both sides

? = 25 ( The number at uppermost box)

Similarly

√?  ×  √49 = 35 

√?  = 35/7

√?  = 5,  squaring both sides

? = 25  (The number at rightmost box )

 Option (A)25  is correct option

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PROBLEM # 6

This figure consist of four squares around  one big square. Look carefully in this big square every number in it is perfect cube. And if we multiply the square roots of  two adjacent numbers then we shall have the number attached to big square between these two numbers whose cube root had been multiplied.

∛343  ×  ∛27 =  7 × 3 = 21( The number at rightmost box )

∛27  ×  ∛125 =  3  × 5 = 15 (The number at bottom line in the box)

∛125  ×  ∛? =  5  ×   ∛? = 10 ( The number at leftmost box )

5  ×  ∛? = 10 

 ∛?  = 10/5

 ∛? = 2 , cubing both sides

? = 8

Similarly

∛?  ×  ∛343 =   ∛?  × 7  = 14

 ∛? = 14/7 = 2 cubing  both sides

? = 8 ( The number at uppermost box )

Option (C)8  is correct option

PROBLEM #  7 

This reasoning problem have two figures and every figure have five number attached to it . Four numbers are at the corner of each figure and one number is at the middle of the same figure.

       Since second figure have question mark 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 of middle numbers. 

          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:-   Add all the numbers in the corner of each figure  to get value of the number in the middle.

Solution:-

√4 + √16 + √9 + √25 + √4 = 2 + 4 + 3 + 5 = 14 (Middle number in 1st figure).

√9 + √49 + √36 + √36 + √1 = 3 + 7 + 6 + 1 = 17 (Middle number in 1st figure).

Option (3)17  is correct option

Reasoning of missing number in circle problems

PROBLEM #  8

This circle consists of four quadrants and every  quadrant consists of three numbers . And every quadrant have two 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 outer part to calculate the value of the number which is in the inner part of every quadrant . 

1st Quadrant 

Step 1.   Take sum of the squares of both the numbers (8 and 4 ) in outer part of  this quadrant.

Step 2. Take the magnitude of difference of both the numbers in outer part of  this quadrant.

Step 3. Divide the result obtained in step 1 with the result obtained in step 2 to get the value of the number (20) in inner part of this quadrant.

Calculation

(8² + 4² ) ÷ ( |8 - 4 |) = ( 64 + 16 ) ÷ 4 = 80 ÷ 2 = 20

2nd Quadrant

Step 1.   Take sum of the squares of both the numbers ( 7 and 5 ) in outer part of  this quadrant.

Step 2. Take the magnitude of difference of both the numbers in outer part of  this quadrant.

Step 3. Divide the result obtained in step 1 with the result obtained in step 2 to get the value of the number (37) in inner part of this quadrant.

Calculation

(5² + 7² ) ÷ (|5 - 7 |) = ( 25 + 49 ) ÷ 2 = 74 ÷ 2 = 37

3rd Quadrant

Step 1.   Take sum of the squares of both the numbers ( 10 and 6 ) in outer part of  this quadrant.

Step 2. Take the magnitude of difference of both the numbers in outer part of  this quadrant.

Step 3. Divide the result obtained in step 1 with the result obtained in step 2 to get the value of the number (34) in inner part of this quadrant.

Calculation

(10² + 6² ) ÷ ( |10 - 6| ) = ( 100 + 36 ) ÷ 4 = 136 ÷ 4 = 34

4th Quadrant

Step 1.   Take sum of the squares of both the numbers ( 10 and 6 ) in outer part of  this quadrant.

Step 2. Take the magnitude of difference of both the numbers in outer part of  this quadrant.

Step 3.  Divide the result obtained in step 1 with the result obtained in step 2 to get the value of the number (34) in inner part of this quadrant.

Calculation

(6² + 4² ) ÷ ( |6 - 4 |) = ( 36 + 16 ) ÷ 2 = 52 ÷ 2 = 26

Option (3)26 is right option

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  Box Problems 

PROBLEM #  9 

This  big circle is  divided into eight parts and  there is also one number written between  any  two numbers towards outer side of these two numbers .  If we  add these  two adjacent  numbers then reverse of order of the result so obtained then this number will be equal to the number in small circle opposite to both these numbers .

8  +  6  = 14 ⟺  41 ( Reversing the order of digits) 

6 + 11 = 17 ⟺  71  ( Reversing the order of digits)

11 + 10 =  21 ⟺ 12 ( Reversing the order of digits)

10 + 31 = 41 ⟺ 14 ( Reversing the order of digits)

31 + 12 = 43 ⟺ 34 ( Reversing the order of digits)

12 + 17 = 29 ⟺ 92 ( Reversing the order of digits)

17 + 5  = 22 ⟺ 22 ( Reversing the order of digits)

8  + 5 = 13  ⟺  31 = ? ( Reversing the order of digits)

Hence the value of  "? " will be 31 

Option (B)13 is right option

Reasoning of missing number in triangle problems

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

 1st Triangle

( 5 × 4 ) ÷ 10 = 20 ÷ 10 = 2 (Middle number in 1st triangle ) 
2nd Triangle

( 6 × 5 )  ÷ 10 = 30 ÷ 10 = 3 (Middle number in 2nd triangle ) 
3rd Triangle 

( 15 × 6 )  ÷ 10 = 90 ÷ 10 = 9 (Middle number in 3rd triangle ) 

Option (3)9  is correct option

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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Inverse method of solving linear equations of three variables

Inverse method of solving linear equations of 3 variables

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

Set of given equations are 

4x + 6z = 100     --------------------------------> (1)

3x + 6y + z = 100   -----------------------------> (2)

3x + 4y + 3z  = 100   ---------------------------> (3)

The system of these equations can be transformed into Matrix formed 

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|  = 4 (18 - 4) -0(9 - 3) +6(12 - 18)

       =  4(14) + 0 + 6(-6) 

       = 56 - 36

        = 20

Since the determinant value of this matrix  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 this matrix. 

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

Let us calculate these cofactors.  

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

Co factors of 1st row are  (18 - 4) , -(9 - 3), (12 - 18) 

i. e. Co factors of 1st row are 14, -6 , -6

Co factors of 2nd row are -(0 -24), (12 - 18) , -(16 - 0) 

I. e. Co factors of 2nd row are  24 , -6 , -16

Co factors of 3rd row are  (0 - 36), -(4 - 18), (24 - 0) 

i.e. Co factors of 3rd row are  -36 , 14 , 24

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. 

Inverse  Matrix

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 iff their respective elements are equal to each other ) 

⇒  x = 10

     y  = 10

      z = 10

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.

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