Twelve Important Tricky Reasonining Problems for SSC CGL ,SSC CHSL And RRB NTPC Exams

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12 Most important problems of  tricky reasoning problems which includes box problems , triangles problems and star problems for competitive exams like SSC CGL ,SSC CHSL and RRB NTPC etc have been included in this post. 


12 Important Tricky Reasoning Problems for Reasoning Various Exams



 Problem # 1

Tricky Figure Problems for SSC CGL ,SSC CHSL And RRB NTPC Exams
Adding all the numbers which are forming triangle in corner like this

Top Right corner

10 + 13 + 9 = 32 Top Right corner

Top Left corner

15 + 10 + 7 = 32  Top left  corner

Bottom Left corner

 7 +  17 +  8 = 32  Bottom left  corner

Bottom Right corner

Similarly 8 + 9 + ? = 32 means Question mark "?"  will take the value 15. So
Option (1) is correct Option.

Problem # 2


Tricky Figure Problems for SSC CGL ,SSC CHSL And RRB NTPC Exams
Add all the numbers from corner which are forming rhombus and square respectively , because sum of all the numbers in both rhombus and square in corner position are equal to 34.

Rhombus

7 + 10 + 9 + 8 = 34

Square

15 + 5 + ? + 6 = 34 . 
26 + ? = 34
? = 34 - 26
? = 8
This means question mark ? will have  8 value. 
So Option (2) is correct Option.  

Problem # 3


Tricky Figure Problems for SSC CGL ,SSC CHSL And RRB NTPC Exams
Add all the three numbers in every small square then divide this sum with the fourth number in same square to get result every time 10.

Top Left Square 

 ( 9 + 4  + 7 )/2 = 20 /2 = 10

Top Right Square

 ( 9 + 13  + 8 )/3 = 30 /3 = 10

Bottom Left Square

( 9 + 16 + 15 )/4 = 40/4 = 10
Similarly in 

Bottom Right Square

 (6 + ? +3 )/1 = 10/1 = 10. If we take the value of question mark "?"  equal to 1 then we shall get total equal to 10 ,
Hence  Option (1) is correct Option. 

Problem # 4

 
Tricky Figure Problems for SSC CGL ,SSC CHSL And RRB NTPC Exams
In all the picture add both the numbers in 1st row and then umbers in 2nd row and left number in last row, then take the difference of these sum to get the number in right side of 3rd row.

In 1st row 9 + 8 = 17,  4 + 9 = 13  then 17 - 13 = 4 
In 2nd row  11 + 5 = 16 , 10 + 3 = 13 then 16 - 13 = 3 , Similarly
in 3rd row  7 + 16 = 23  , 6 + 12 = 18 then 23 - 18 = 5 .
Hence option (1) is correct option.

 Problem # 5

Tricky Figure Problems for SSC CGL ,SSC CHSL And RRB NTPC Exams
There are five squares in this picture. To solve this problem or to find the value of question mark , sum of all the numbers in each square when divided by the number in the corner must be equal to 6 in each four figure. In  the 

Top Left square

    (  3 + 2 +  7 ) /2  =  12/2 = 6

Top Right square

   (  9 + 5 +  4) /3  =  20/3 = 6

Bottom Left square

    (  9 + 7 +  8 )/4  =  24/4 = 6

Bottom Right square

    (  3  + ? +  1 )/1  = (4 + ?)/1= 6.
4 + ? = 6
? = 6 - 4
? = 2
This means if we put ? = 2, Only then we shall 6 as answer.
Hence option (3) is correct option. 

Problem # 6

 

Tricky Figure Problems for SSC CGL ,SSC CHSL And RRB NTPC Exams
To solve this problem divide each number in the second row with 6 , then subtract one from the answer so obtained to get the number in 1st row.
(78/6) -1 => 13 - 1 = 12 ( The  1st number in the 1st row)
(60/6 ) -1 => 10 - 1 = 9 ( The 2nd number in the 1st row)
(24/6) -1 =>  4 - 1 =  3 ( The 3rd number in the 1st row)
( ?/6 - 1 ) -1  = >10 ( The 4th number in the 1st row). that is if we put ? = 66 then upon dividing 66 with 6 we shall get 11, And after subtracting 11 from 1 , we shall get the answer.

Hence option (C) is correct option. 
 

Problem # 7

 

Tricky Figure Problems for SSC CGL ,SSC CHSL And RRB NTPC Exams
Take the square root of  difference of  21 and 12 to get middle number 3. Similarly take the square root of  difference of  15  and 6 to get middle number 3 in 1st picture.
Similarly take the square root of  difference of  29 and 4 to get middle number 5. Similarly take the square root of  difference of  36  and ? to get middle number 5 in 2nd picture.

In 1st picture

√(21 - 12 ) = √9 = 3 
√(15 - 6 ) = √9 = 3 

In 2nd picture

√(29 - 4 ) = √25 = 5 
√(36 - ? ) = √25 = 5 , If we put ? = 11 then we can solve this problem.
Hence option (4) is correct option. 

  Problem # 8


Tricky Figure Problems for SSC CGL ,SSC CHSL And RRB NTPC Exams

1st Picture

  (10)² - 16  = 100 - 16 = 84 in the same way
 (12)² - 60 = 144 - 60 = 84

2nd Picture

(9)² - 13  = 81 - 13 = 68 in the same way 
(?)² - 53  = 68
(?)² = 68 - 53
(?)² = 121
? = 11
so ? will take the value 11
Hence option (1) is correct option.

            Problem # 9

Tricky Figure Problems for SSC CGL ,SSC CHSL And RRB NTPC Exams
Sum of all the numbers except middle number in 1st picture is equal to the middle number in 2nd picture.
5 + 2 + 7 + 5 = 19 (Middle number in 2nd picture).

Sum of all the numbers except middle number in 2nd picture is equal to the middle number in 3rd picture.
6 + 11 + 2 + 4 = 19 (Middle number in 3rd picture).

Sum of all the numbers except middle number in 2nd picture is equal to the middle number in 1st  picture.
6 + 1 + 6 + ? = 15 (Middle number in 1st picture).
To make toal 15 in 1st picture ? will have to take 2 value in 3rd picture.
Hence option (4) is correct option.

      Problem # 10

Tricky Figure Problems for SSC CGL ,SSC CHSL And RRB NTPC Exams
Take the sum of squares of each number in the corner positions in each picture to get number the middle positions.
In 1st Picture 
   9² + 8² + 6² + 7² =  81 + 64 + 36 +  49 = 230
In 2nd Picture 
 6² +  7² +  4² +  3² =  36 + 49 + 16 +  9 = 110
In 3rd Picture 
9² + 6² + 4² +  5² =  81 + 36 + 16 +  25 = 158
 Therefore ? will take the value equal to 158
Hence option (3) is correct option.

   Problem # 11

Tricky Figure Problems for SSC CGL ,SSC CHSL And RRB NTPC Exams

Add all those numbers which are above central number then take the difference of this sum from the addition  of those numbers which are below the central number .
1st figure  ( 4 + 2  + 7 ) -  ( 3 + 1 ) =  13 - 4  = 9 middle number
2nd figure  ( 3 + 3  + 5 ) -  ( 4 + 2 ) =  11 - 6  = 5 middle number
3rd figure  ( 6 + 9  + 2 ) -  ( 4 + 3 ) =  17 - 7  = 10 middle number
Hence option (1) is correct option.

 Problem # 12

Tricky Figure Problems for SSC CGL ,SSC CHSL And RRB NTPC Exams
In this problem all the numbers written in 1st figure are related to 2nd figure and all the numbers written in 2nd figure are related to 3rd figure , similarly all the numbers written in 3rd figure are related to 4th figure .
i.e. number 4 in 1st figure is related to 10 in 2nd figure and number 10 in 2nd figure is related to 22 in 3rd figure. similarly number 22 in 3rd figure is related to 46 in 4th figure.
4 =  1st number in 1st figure
(4   ×  2 ) + 2 = 10 ( 1st number in 2nd figure ) 
(10  ×  2) + 2  = 22 ( 1st number in 3rd figure )
(22   ×  2) + 2 = 46 ( 1st number in 4th figure )

1 =  2nd number in 1st figure
(1  ×  2 ) + 2 = 4 ( 2nd number in 2nd figure ) 
(4  ×  2) + 2 = 10 ( 2nd number in 3rd figure ) 
(10  ×  2) + 2  = 22 ( 2nd number in 4th figure ) 

3 = Bottom number in 1st figure
 (3  ×  2) + 2 = 8  ( bottom number in 2nd figure )
(8 × 2) + 2 = 18 ( bottom number in 3rd figure )
(18 × 2) + 2 = 38  (The value of question mark)
Hence option (1) is correct option.

 Also Reads these posts on Reasoning




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Ten most important problems of Reasoning for competitive exam Part1

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Ten most important problems of Reasoning for competitive exam which includes missing numbers and missing terms and their solutions in reasoning analogy . These questions are very very important for upcoming competitive exams like SSC CGL ,SSC CHSL and RRB NTPC Etc

Ten most important problems of Reasoning for competitive exams

Reasoning for ssc cgl ,chsl, RRB NTPC

Solution 

 6 is related to 29 in the same way 24 will be related to ? , It means we have to apply same mathematical operations to 24 to get ?.
So if we multiply 6 with 5  and then subtract 1 from result obtained in previous step like this ( 6 × 5 )  - 1 which will be equal to 29.
Same operation we have to apply  to 24.
( 24 × 5 ) - 1 = 120 - 1 =  119

So Correct  option is ( c ) 119

  Problem # 2


5  :  100   ::   7  :  ?
(a) 135   (b) 91   (c) 196  (d) 49


Solution

Because in 1st case if we take Square of  5 then multiply it with 4 we shall have 100.
5² × 4 = 25 × 4 = 100
Same procedure will be applied in 3rd number by taking square of 7 then multiply it with 4
7² × 4 = 49 × 4 = 196

Correct  option is  ( c  ) 196

Problem # 3

6  :  18   ::   4  :  ?
(a)4     (b)6     (c)8    (d)10


Solution 

Divide  6² by 2
6² ÷ 2 = 36 ÷ 2 = 18
Similarly divide 4² with 2
i.e.  4² ÷ 2 = 16 ÷2 = 8
Divide the square of 1st number by 2 to get 2nd number. Similarly take square of third number and then divide it by 2 to to get the number equal to? Since square of 4 is 16 then divided by 2 to get it.
6² ÷ 2 = 36 ÷ 2 = 18

4² ÷ 2 = 16 ÷ 2 = 8 

Correct  option is (  c ) 8


Problem # 4

18  :  30   ::   36  :  ?
(a)64   (b)  62   (c)54    (d) 66


Solution  

 (18 × 2 ) - 6  = 36 - 6 = 30
 (36 × 2 ) - 6  = 72 - 6 = 66
Multiply 1st number (18) with 2 and then subtract 6 from it
18 × 2 =  36 - 6 = 30
Similarly Multiply 3rd number ( 36 ) with 2 and then subtract 6 from it ,

36 × 2 = 72 - 6 = 66

correct  option is ( d) 66


Problem # 5

12  :  20   ::   30  :  ?
(a)48  (b)42    (c)15   (d)35


Solution 
Method 1  

Split 12  =  3 × 4 
Split  20  = 4 × 5, 
Split  30 =  5 × 6,  
Study these factors (3 , 4 ) , (4 ,5 ) ,(5 ,6 ) so next pair will be  ( 6 , 7 ) , It means ? Will be replaced by the number 6 × 7 = 42

Method 2 
12 will be written as square of 3 plus 3 , 20 will be written as square of 4 plus 4 ,30 will be written as square of 5 plus 5, so next number will be written as square of 6 plus 6 which is equal to 42 , Therefore   required option will be 42.
                         Or
Make continuous factors 3 and 4  , 4 and 5 ,5 and 6 and 6 and 7 of 12, 20, 30 and 42 respectively 
12 = 3 × 4  , 
20 = 4 × 5  ,
30 = 5 × 6 so 
42 = 6 × 7

Add same number to its square to get next number 
Or multiply next number to the number
 9  = 3² + 3   ,
 20 = 4² +4 ,
 30 = 5² + 5
42 = 6² + 6
Correct  option is ( b ) 42

  Problem # 6


12  :  54   ::   15  :  ?
(a)64   (b)69   (c)56  (d)67


Solution

{(1st number ) × 5} - 6 = 2nd number
(12 × 5) - 6 = 54
Multiply 3rd number with 5 then subtract 6 from it
{(3rd number ) × 5} - 6 = 4th  number
(15 × 5) - 6 = 69
Multiply first number ( 12 ) with 5 then subtract 6 from it to get 2nd number ( 54 ) ,   Similarly   Multiply third  number ( 15 ) with 5 then subtract 6 from it.
( 1 2 × 5  ) - 6 = 60 - 6 = 54
( 1 5 × 5  ) - 6 = 75 - 6 = 69
Correct  option is ( b) 69

Problem # 7

6  :  5   ::   8  :  ?
(a) 6      (b)10        (c) 2     (d)4 


Solution
Add 4 to 1st number and divide it with 2 to get 2nd number( 6 + 4 )/2 = 5
Similarly in 2nd case Add 4 to 3rd number then divide the resultant with 2 to get 4th number
( 8 + 4 )/2 = 6
Correct  option is ( a ) 6


Problem # 8

29  :  319  ::    23 :  ?

    (a)115  (b)252 (c)151  (d)46

Solution

Add both the digits of first number  i.e. 2 and 9 then multiply it with first number to get second number. similarly in the third number add both the digits I.e  2 and 3  and multiply it with 3rd number to get fourth number.
29 × ( 2 + 9 ) = 29 × 11 = 319
23 × ( 2 + 3 ) = 23 ×  5  = 115

So 253 is the right option (a) 115

Problem # 9

6  :  64   ::   11  :  ?

(a) 127      (b) 124     (c) 144    (d) 169


Solution

Add 2 to 1st given number and take its Square to get 2nd number. Similarly add 2 to 3rd number and take it square to get fourth number.
(6 + 2 )² = 8² = 64
(11 + 2)² = 13² = 169

Correct  option is ( d ) 169

Problem # 10

36  :  50   ::   64  :  ?
(a)70   (b) 82     (c)78      (d) 72


Solution

Take Square Root of 1st number and add 1 to it then add 1 to  its Square to get 2nd number.
Similarly take square root of 64 add 1 to its square  root and take it square and add 1  to get 4th number.
36 = 6² --->  (6 + 1)² + 1 = 50 ,

 64 = 8² ------> ( 8 + 1)² + 1 = 82
√36 = 6 add 1 to it = 7 , square this number = 49 + 1 = 50
√64 = 8 add 1 to it = 9 , square this number = 81 + 1 = 82

So  Correct option is ( b ) 82



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

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

Reasoning of missing number in box problems

PROBLEM #  1


Reasoning of missing number in box problem

Step 1. 

Subtract the number in first column from the number in third column of every row. i. e. C3 - C1 = S

Step 2 . 

Multiply the the above difference with 3 to get the numbers in second column of every row i. e. C2 = 3S.

1st row    3 (48 - 28 ) = 3 × 20  = 60
2nd row 3 (7 - 5)  = 2 × 3 = 6
3rd row 3(27 - 14) = 3 × 13 = 39
4th row 3(16 - 7 ) = 3 × 9 = 27 
Hence required number is 27
Option (A) 27 is correct option

PROBLEM #  2


Reasoning of missing number in box problem

Every number in fourth row has been written as the square of all the numbers in 1st three rows in any particular column. 
Take the square of addition of  the elements of 1st three rows of 1st column like this
(1 + 4 + 2 )² = 7² = 49

Take the Take the square of addition of  the elements of 1st three rows of 1st column like this
(4 + 2 + 2 )² = 8² = 64

Take the square of addition of  the elements of 1st three rows of 1st column like this
 (? + 5 + 3 )² = (? + 8 )² = 169
(? + 8 )² = (13)² = 169 , Taking Square root
? + 8 = 13
? = 13 - 5 
? = 5
So required number will be 5
Option (C) 5 is correct option

PROBLEM #  3


Reasoning of missing number in box problem

Multiply all the numbers in the first three rows of every column and then divide the product by 2 to get the number in last row like this
(5×2×8)÷2 = 80÷2 = 40 ( Last number in 1st column ) 
(5×4×3)÷2 = 60÷2 = 30 ( Last number in 2nd column ) 
(2×1×10)÷2 = 20÷2 = 10 ( Last number in 3rd column ) 
So required number will be 10
Option (D)10 is correct option.


PROBLEM #  4


Reasoning of missing number in box problem

Adding both the numbers in second and third rows then multiply it with the number in First row to get the number in last row ,the process will be  repeated for all the three columns
( 6 + 7 ) × 5 = 65
( 3 + 2 ) × 4 = 20
( ? + 4 ) × 9 = 45
This implies ( ? + 4 ) = 45/9 = 5
? = 5 - 4 = 1
Hence required number is 1
Option (A) 1 is correct option.

PROBLEM #  5


Reasoning of missing number in box problem

Decrease the numbers in first  columns of every row by 1 then multiply it to get the number in third column of every row. 
1st Row  ( 8 - 1) ×  3  = 7 × 3 = 21
2nd Row  ( 6 - 1) ×  5 = 5 × 5 = 25
3rd Row  ( 12 - 1) ×  2 = 11 × 2 = 22
Hence required answer is 22
Option (B) 22  is correct option.

PROBLEM #  6


Reasoning of missing number in box problem

Add 1st two rows of each column to multiply with the number in 3rd column to get fourth number in every row. 
(5 + 4 ) × 2 = 9 × 2 = 18 (1st column 3rd row ) 
(6 + 3 ) × 3   = 9 × 3 = 27 ( 2nd column 3rd row ) 
(12 + 4 ) × ?  = 16 × ? = 96 (3rd column 3rd row ) 
16 × ? = 96
? = 6
Hence required number is 6
Option (D) 6  is correct option.


PROBLEM #  7




 First of all multiply  1st three numbers  in every Column then add the sum of 1st three  numbers in every row to  get the numbers in 4th row. i. e. formula for this problem is [ if a, b ,c and d are numbers in any column  then d = abc + (a+b+c) ]
[ ( 3 + 4 + 5) + (3 × 4 × 5 ) ] = 12 + 60 = 72
[ ( 2 + 5 + 6) + (2 × 5 × 6 ) ] = 13 + 60 = 73
[ ( 5 + 9 + 1) + (5 × 9 × 1 ) ] = 15 + 45 = 60
Hence  required number is 60
Option (D)60  is correct option.

Also study these missing number series questions


PROBLEM #  8



Reasoning of missing number in box problem

Decrease both the numbers in first and second columns of every row then multiply the reduced numbers to get the numbers in third column of every row
{ 8 - 1 }×{ 6 - 1 } = 7 × 5 = 35
{ 6 - 1 }×{ 4 - 1 } = 5 × 3 =  15 
{ 7 - 1 }×{ 5 - 1 } = 6 × 4 =  24
Hence required number is 24.
Option (A)24  is correct option.



PROBLEM #  9



Reasoning of missing number in box problem

Decrease the numbers in first column of every row and increase the numbers in second column of every row when multiplying both the reduced number of first and second column to get the numbers in third column of every row

{ 6 - 1 } × { 8 + 1 } = 5 × 9 = 45
{ 4 - 1 } × { 6 + 1 } = 3 × 7 = 21
{ 7 - 1 } × { 5 + 1 } = 6 × 6 = 36
Hence required number is 36
Option (C)36  is correct option.

PROBLEM #  10


Reasoning of missing number in box problem

Total of 1st row is  3 + 6 + 8 = 17
Total of 2nd row is  5 + 6 + 8 = 17
Total of 3rd row is  4 + 7 + ? = 17
Therefore replacing ? with 6 to get required answer.
Option (D)6  is correct option
So these were the most important box problems for many competitive exams

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Ten most important missing numbers questions and answers in reasoning

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Ten most important   missing numbers questions and answers in box problems in Reasoning Analogy in various figures of different order and types discussed in this post. These types of problems are very very important for the exams like SSC CGL ,SSC CHSL , RRB NTPC and many other similar competitive exams. 

Ten most important missing numbers questions and answers in reasoning


PROBLEM # 1

tricky puzzles with their Solutions

Since in 1st row we can write 3, 2 and 4  numbers to get 10 like this
( 3 × 2 ) + 4 = 6 + 4  = 10 (The number in last column  of 1st row )
And in 2nd row  ( 4 × 3 ) + 5 = 12 + 5 = 17, (The number in last column  of 2nd row )
In 3rd row ( 5 × 4 ) + 6 = 20 + 6 = 26 ,(The number in last column  of 3rd row )
Similarly ( 6 × 5 ) + 7 = 30 + 7 = 37 , (The number in last column  of 4th row )
So  " ? "   Will be replaced by 37 .
Therefore correct option will be (D)


PROBLEM # 2

   tricky puzzles with their Solutions              
Answer can be split into two parts, 1st part can be obtained by multiplying two given  numbers and second part can be obtained by adding these two numbers .
1st row 
5 × 3  = 15 and 5 + 3 = 8 so 5 , 3 = 158
2nd row 
9 × 1  = 9 and 9 + 1 = 10 so 9 , 1 = 910
3rd  row 
8 × 6  = 48 and  8 +6 = 14  so 8 , 6 = 4814
4th row 
4 × 4 = 16 and   4 + 4 = 8 so  4 , 4 = 168.
5th row
 7 × 3 = 21 and 7 + 3 = 10 so 7 , 3 = 2110
So ?  Will be replaced by 2110.
  Therefore correct option will be (A)



PROBLEM # 3


tricky puzzles with their Solutions

As the sum of 2nd and 3rd number in first and second rows in any particular column is equal to 1st element in that particular column so to get the value of " ? " . Take difference of first and second numbers column wise to get third number as follows.
44 +12 =  56, 
48 + 30 = 78 ,
14 + ? = 65 
➡️   ? = 65 - 14  = 51 
Therefore correct option will be (C)

PROBLEM # 4

tricky puzzles with their Solutions


Let us find relation between 3 and 18 , if we double the square of  3 ,we shall have 18. 
2 × (3²) = 2 × 9 = 18 (The number in last column  of 1st row )
And in 2nd row if we double the square of  4 ,we shall have 32
2 × (4²) = 2 × 16 = 32, (The number in last column  of 2nd row)
In the same way increasing the number one by one then  third ,fourth and  fifth columns can be calculated like this 

2 × (5²) = 2 × 25 = 50 (The number in last column  of 3rd row )
2 × (6²) = 2 × 36 = 72 (The number in last column  of 4th row )
2 × (7²)  = 2 × 49 = 98 (The number in last column  of 5th row )
Therefore correct option will be (C)

.

PROBLEM # 5


tricky puzzles with their Solutions

Formula a*b = (a × b) + (b - 1)

This the sum of two numbers , out of two ,1st number is product of two given  numbers and second  is the number one less than 2nd given number.

3 × 2 = (3 × 2) + (2 - 1) = 6 + 1 = 7
5 × 4 = (5 × 4) + (4 - 1) = 20 + 3 = 23
7 × 6 = (7 × 6) + (6 - 1) = 42 + 5 = 47
9 × 8 = (9 × 8) + (8 - 1) = 72 + 7 = 79
10 × 9 = (10 × 9) + (9 - 1) = 90 +8 = 98
Therefore correct option will be (C)

PROBLEM # 6



tricky puzzles with their Solutions

Suppose we have three numbers a , b and c then

Formula for this puzzle is = (a × b) + b or  b(a + 1)

Put a = 2 and  b = 6 in above formula ,we get

1st Line = (2  × 6)   +  6  = 12   +   6   = 18
Put a = 4 and  b = 20 in above formula to get 2nd line,we get
(4  × 20) + 20 = 80   +  20  = 100
Put a = 5 and  b = 21 in above formula to get 3rd line,we get
(6 × 21) + 21 = 126 +  21  = 147
So required and correct answer will be  6.


PROBLEM # 7

tricky puzzles with their Solutions

If in every row ,we add 1st and 3rd column and then multiply the sum with 2nd column, we shall have number in 4th column .

If we  consider three numbers a , b and c and start calculation by   
Formula =  (a × b) + (b × c)  or b(a + c)  --------(1)

To get 1st line , put a = 1 , b = 2 and c = 3 in (1) , we get 

1st Line =  (1×2) + (2×3) =2 + 6 = 8

To get 2nd line , put a = 2 , b = 3 and c = 4 in (1) , we get 

2nd  Line =  (2×3) + (3×4) = 6 + 12 =  18

To get 3rd line , put a = 3 , b = 4 and c = 5 in (1) , we get 

3rd Line =  (3×4) + (4×5) = 12 + 20  = 32

To get 4th line , put a = 4 , b = 5 and c = 6 in (1) , we get 

4th Line =  (4×5) + (5×6) = 20 + 30  = 50

Similarly Last line can be calculated by putting a = 5 , b = 6 and c = 7 in (1) , we get 

Last line  =  (5×6) + (6×7) = 30 + 42  = 72
Therefore correct option will be (D)

PROBLEM # 8 


tricky puzzles with their Solutions

Case 1

If we choose   1st number  as 5 then look at the pattern  opposite to given smaller number 
                              
5  ×  3 =  15
8  ×  3 =  24
12 × 3  =  36
In this pattern we can conclude 12 ×3 = 36

Case 2

If we choose   1st number  from ? then look at the opposite to given smaller number 
 ?    ×  3 = 12
5   ×  3  = 15
8   ×  3  = 24   
Therefore ? will be replaced  by 4.
Hence    4 ×  3  = 12


PROBLEM # 9


tricky puzzles with their Solutions

This circle can be divided into two parts ,  1st part containing the number 4 ,5 ,6 and 7 and second part containing 7 , 9 ,11 and ?.            Now study  the opposite number of 4  which is 7 . Then study the relation between other numbers and its opposite number ,we find difference between 4 and 7 is 3 , difference between 5 and 9 is 4 , difference between 6 and 11 is 5  . so in this pattern we can find difference between ' ? '  and  7 must be 6 . 
 7 - ? =  2 means ? = 5
  7 - 4 =  3
  9 - 5 =  4
 11 - 6 = 5 , the difference is increased by 1 every time .
  ?   - 7 = 6  means ? = 13
 Either 5 or 13 should be the required number, but 5 is not given in any option.
 So ? Will be replaced by 13 which is the required answer 
Therefore correct option will be (4)


PROBLEM # 10


tricky puzzles with their Solutions

We shall try every possible relation between different numbers given either row wise or column wise . It is found that there is no relation if we consider First row . 
 After elemination of 1st row we can develop a relation between 2nd ,3rd and 4th row which is multiplying the 2nd and 3rd numbers then double it to get 4th number column wise .
 2 × (7 × 4 ) = 2 × 28 = 56
2 × ( 15 × 6 ) = 2 × 90 = 180
2  × ( 8 ×  ?  ) = 80 
So ? = 5
These were the Ten most important  missing numbers questions and answers in reasoning . Please comment your opinion about this post. 

Therefore correct option will be (B) 

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