Ten Logical Reasoning questions and answers for competitive exams

  No comments

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


Logical Reasoning questions and answers
In this reasoning problem 1st number (2) is associated to 10 with the help of any rule , in the same rule we have to associate 8  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 :- 

2nd Number =  a² + (2×a) + 2, where 'a' is 1st number. 

1st Number

2² + (2×2) + 2 = 4 + 4 + 2 = 10

2nd Number

8² + (2×8) + 2 = 64 + 16 + 2 = 82
Therefore option (1)2 is correct option.

Problem # 2


Logical Reasoning questions and answers
In this reasoning problem 1st number (3) is associated to 39 with the help of any rule , in the same rule we have to associate 6 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 :- 

2nd Number = a³ + (3×a) + a, where 'a' is 1st number.  

1st Number

3³ + (3×3) + 3 = 27 + 9 + 3 = 39

2nd Number

6³ + (3×6) + 3 = 216 + 18 + 6 = 240.  
Therefore option (1)240 is correct option.

Problem # 3

Logical Reasoning questions and answers
In this reasoning problem 1st number (4) is associated to 60 with the help of any rule , in the same rule we have to associate 9 to any one  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 :- 

2nd Number =  (a-1) × (a) × (a + 1), where 'a' is 1st number.  

1st Number

3 × 4 × 5 = 60

2nd Number

8 × 9 × 10 = 720
Therefore option (4)720 is correct option.

Problem # 4

Logical Reasoning questions and answers
In this problem of reasoning we have to combine two 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.

Formula : - 

(a×b) + (a+b),
where 'a' and 'b' are two given numbers.
1st Number
(27×4) + (27+4) = 108 + 31 = 139
2nd Number
(31×9) + (31+9) = 279 + 40 = 319
3rd Number
(21×6) + (21+6) = 126 + 27 = 133
Therefore option (1)153 is correct option.

Problem # 5

Logical Reasoning questions and answers
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 on the right hand side of all the four problems given above. Because in these types of reasoning problems we can change mathematical sign according to our requirements.

Formula : - 

(a×b×c)  + (a+b+c),
where 'a' , 'b' and 'c' are three given numbers.
1st Number
(5×2×3) + (5+2+3) = 30 + 10 = 40
2nd Number
(6×3×4) + (6+3+4) = 72 + 13 = 85
3rd Number
(4×1×8) + (4+1+8) = 32 + 13 = 45
Required Number
(3×2×8) + (3+2+8) = 48 + 13 = 61
Therefore option (4)61 is correct option.

Problem # 6

Logical Reasoning questions and answers

1st Method

Formula : - 

Any term = ( 2 × Previous term ) + 1
1st Term = 6
2nd Term = ( 2 × 6 ) + 1 = 12 + 1 = 13
3rd Term = ( 2 × 3 ) + 1 = 26 + 1 = 27
4th Term = ( 2 × 27 ) + 1 = 54 + 1 = 55
5th Term = ( 2 × 55 ) + 1 = 110 + 1 = 111
6th Term = ( 2 × 111 ) + 1 = 222 + 1 = 223

2nd Method

Formula : -

 Any term - Preceding Term =  Difference is multiple of 7 
13 - 6 = 7
27 - 13 = 14 = 2 × 7 = Double of 7
55 - 27 = 28 = 2 × 14 = Double of 14
111 - 55 = 56 = 2 × 28 = Double of 14
? - 111 =  It must be  2 × 56 = 112
 Hence    
? - 111 =  112
? = 112 + 111
? =  223
Therefore option (1)223 is correct option.

Problem # 7

Logical Reasoning questions and answers

Formula : - 

Any term =  one less or one more than the square of natural numbers greater than equal to 6 and smaller than equal to 11.
35 = 6² - 1
50 = 7² + 1
63 = 8² - 1
82 = 9² + 1
?  = 10² - 1 = 100 - 1 = 99
122 = 11² + 1
Therefore option (4)99 is correct option.

Problem # 8

Logical Reasoning questions and answers

 Split this number series into two series by picking alternate numbers.

13, 19 , 25  and 11, 17, 23 , ? 

1st series

13, 19 , 25 

Now considering the 2nd series. In this series every term can be found by adding 6 to its previous term.

1st Term = 13

2nd Term = 1st Term + 6 = 13  + 6 = 19

3rd Term = 2nd Term + 6 = 19  + 6 = 25

2nd series

11, 17, 23 , ?

Now considering the 1st series. In this series every term can be found by adding 6 to its previous term.

1st Term = 11

2nd Term = 1st Term + 6 = 11  + 6 = 17

3rd Term = 2nd Term + 6 = 17  + 6 = 23
4th Term = 3rd Term + 6 = 23  + 6 = 29 = ? (The value of question mark).
Therefore option (3) 29 is correct option.


Logical Reasoning questions and answers

In this reasoning problem three out of four options have been calculated by Multiplying  the 1st number with 12 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.
10 × 12 = 120
20 × 12 = 240
14 × 12 = 168(196)
12 × 12 = 120
Therefore option (3)14,196 is odd option.
Logical Reasoning questions and answers

In this problem of reasoning we have to combine both the given  numbers in such a way that after applying any mathematical operation/operations we could get the number on 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.

Formula : - 

(a×b) - (a+b),
where 'a' and 'b' are three given numbers.
1st Number
(27×4) - (27+ 4) = 108 - 31 = 77
2nd Number
(31×9) - (31+9) = 279 - 40 = 239
3rd Number
(21×6) - (21+6) = 126 - 27 = 99
Therefore option (3)99 is correct option.

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.

Share:

Missing number and reasoning problems for SSC CGL Exams

  No comments

Ten most important Missing number series questions and answers, Number Series Questions for SBI Clerk ,SBI PO with solution , for SSC GL , SSC CHSL , RRB NTPC and other competitive Exams  have been discussed in this post.  These types of series questions are asked in many other competitive exams like SI, CPO and various entrance exams. 


Ten Reasoning problems with answers for competitive exams


Problem #1



This reasoning problem have four sub problems and every sub problem have three number associated to it.
           Since last figure have question mark in it. So the solution of this problem is to find the value of question mark using three numbers associated to it . 
           But the biggest problem is how to utilised these three numbers to get the value of this question mark?
           Now we have to find or search the  formula for these three numbers in each sub problem to utilised them in any possible way to get the value of question mark. 

Formula:- Adding 1 to each digit and then combine all three numbers to get the value of question mark.

Sub Problem # 1

In this sub problem the three digits are 6 , 7 and 8, Adding 1 to each digits, we shall have
6 + 1 = 7
7 + 1 = 8
8 + 1 = 9
Now combining these three newly formed digits to get new number  as 789( the number on the right hand side).

Sub Problem # 2

In this sub problem the three digits are 2 , 3 and 4, Adding 1 to each digits, we shall have
2 + 1 = 3
3 + 1 = 4
4 + 1 = 5
Now combining these three newly formed digits to get new number  as 345( the number on the right hand side).

Sub Problem # 3

In this sub problem the three digits are 4 , 8 and 0, Adding 1 to each digits, we shall have
4 + 1 = 5
8 + 1 = 9
0 + 1 = 1
Now combining these three newly formed digits to get new number  as 591( the number on the right hand side).

Required Sub Problem # 4

In this sub problem the three digits are 8 , 0 and 7 , Adding 1 to each digits, we shall have
8 + 1 = 9
0 + 1 = 1
7 + 1 = 8
Now combining these three newly formed digits to get new number  as 918( The value of question mark).
Hence Option (4)918 is correct option.

Problem #2


This reasoning problem also have four sub problems and every sub problem have three number associated to it.
           Since last figure have question mark in it. So the solution of this problem is to find the value of question mark using three numbers associated to it . 
           But the biggest problem is how to utilised these three numbers to get the value of this question mark?
           Now we have to find or search the  formula for these three numbers in each sub problem to utilised them in any possible way to get the value of question mark. 

Formula:- Reversing the order of each digit in every number to get the value of question mark.

Sub Problem # 1

In this sub problem the three digits are 6 , 7 and 8, Reversing the order of each digit , it means we have to put the 1st digits in last place ,2nd digit in 2nd last place and 3rd digit in 3rd last place to get the number on the right hand side of each sub problem.
 6 + 5 + 7  ⇒ 7 5 6 ( the number on the right hand side).

Sub Problem # 2

In this sub problem the three digits are 2 , 3 and 4, Reversing the order of each digit , it means we have to put the 1st digits in last place ,2nd digit in 2nd last place and 3rd digit in 3rd last place to get the number on the right hand side of each sub problem.
 2 + 3 + 4  ⇒ 4 3 2 ( the number on the right hand side).

Sub Problem # 3

In this sub problem the three digits are 7 , 8 and 9, Reversing the order of each digit , it means we have to put the 1st digits in last place ,2nd digit in 2nd last place and 3rd digit in 3rd last place to get the number on the right hand side of each sub problem.
 7 + 8 + 9  ⇒ 9 8 7 ( the number on the right hand side).

Required Problem # 4

In this sub problem the three digits are 2 , 2 and 2, Reversing the order of each digit , it means we have to put the 1st digits in last place ,2nd digit in 2nd last place and 3rd digit in 3rd last place to get the number on the right hand side of each sub problem.
 2 + 2 + 2  ⇒ 2 2 2 ( The value of question mark).
But there is no option which has value 222.
Hence Option (4)None is correct option.

Problem #3

Square of 1st digit and then sum of remaining two digits

6+1+3 = 6²(1+3) = 364
8+7+2 = 8²(7+2) = 649
4+7+2 = 4²(7+2) = 169
4+2+5 = 4²(2+5) = 167 (The value of Question mark)
Hence Option (2) 167 is correct option.

Problem #4


H C F (Highest Common Factor) of three given numbers
6+2+5 = H C F of (6,2,5) = 1
4+6+8 = H C F of (4,6,8) = 2
3+6+9 = H C F of (3,6,9) = 3
4+8+9 = H C F of (4,8,9) = 1
Hence Option (4) 1 None is correct option.

Problem #5


L C M of three given numbers
L C M (Least Common Multiple) of three given numbers
6+2+5 = L C M of (6,2,5) = 30
2+3+5 = L C M of (2,3,5) = 30
6+4+2 = L C M of (6,4,2) = 12
1+3+5 = L C M of (1,3,5) = 15
Hence Option (1) 15 None is correct option.

Problem #6


6+2+5 = (6 + 2 + 5)² + 1 =  13² + 1 = 169 + 1 = 170

5+3+7 = (5 + 3 + 7)² + 1 =  15² + 1 = 225 + 1 = 226

2+4+6 = (2 + 4 + 6)² + 1 =  12² + 1 = 144 + 1 = 145

1+3+5 = (1 + 3 + 5)² + 1 =  9² + 1 = 81 + 1 = 82

Hence Option (3) 82 None is correct option.

Problem #7


2+3+5 = ( 2 + 5 ) × 3 = 7 × 3 = 21

4+3+7 = ( 4 + 7 ) × 3 = 11 × 3 = 33

4+3+9 = ( 4 + 9 ) × 3 = 13 × 3 = 39

3+1+8 = ( 3 + 8 ) × 1 = 11 × 1 = 11

Option (1)11 is correct Answer

Problem #8


2+3+5 =  2 + (5  × 3 ) = 2 + 15 = 17
4+3+7 = 4 + (3  × 7 ) = 4 + 21 = 17
4+3+9 = 4 + (3  × 9 ) = 4 + 27 = 31
3+1+8 = 3 + (1  × 8 ) = 3 + 8 = 11

Option (1)11 is correct Answer

Problem #9


5+2+3 = 5 × 2² × 3 =  5 × 4 × 3 = 60
6+3+4 = 6 × 3² × 4 =  6 × 9 × 4 = 216
4+1+8 = 4 × 1² × 8 =  4 × 1 × 8 = 32
3+2+8 =3 × 2² × 8 =  3 × 4 × 8 = 96

Option (2)96 is correct Answer


Problem #10





5+2+3=(5 × 2 × 3 ) + ( 5 + 2 + 3 ) = 30 + 10 = 40
6+3+4 = (6 × 3 × 4 ) + ( 6 + 3 + 4 ) = 72 + 13 = 85
4+1+8 = (4 × 1 × 8 ) + ( 4 + 1 + 8 ) = 32 + 13 = 45
3+2+8 = (3 × 2 × 8 ) + ( 3 + 2 + 8 ) = 48 + 13 = 61

Option (4)61 is correct Answer

Problem #11


(5 × 4) × (4 × 3 ) × ( 3 × 5) = 20 + 12 + 15 = 47
(6 × 2) × (2 × 4 ) × ( 4 × 6) = 12 + 8 + 24 = 44
(4 × 3) × (3 × 9 ) × ( 9 × 4) = 12 + 27 + 36 = 75
(3 × 1) × (1 × 7 ) × ( 7 × 3) = 3 + 7 + 21 = 31

Option (2)31 is correct Answer







<
Share:

Reasoning questions with answers for competitive exams

  No comments
Ten Most Important Reasoning questions with answers for ssc cgl and other competitive exams of 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 five rows and four columns . Here in this problem we have to find the value of question mark which is lying in 2nd column of fourth row after studying the pattern of all the numbers in this box. 
             Because after careful observation we can see that the difference of  product of 1st and 4th rows from the sum of 2nd and 3rd rows in any particular column is equal to the number in fourth row in that particular column. 

Formula:-

 5th row =  (1st  row  × 4th row ) - (2nd row + 3rd row) 
1st column

(8 × 3) - (6 + 2) = 24 - 8 = 16 (Number in last row of 1st column). 

3rd column

(7 × 2) - (3 + 4) = 14 - 7 = 7 (Number in last row of 1st column) . 

4th column

(5 × 9) - (2 + 8) = 45 - 10 = 35 (Number in last row of 1st column). 

2nd column

(6 × 5) - (4 + 1) = 30 - 5 = 25 (Number in last row of 1st column). And this number will be the value of question mark. 
The correct Answer is option (3)25

Problem # 2



This box problem of reasoning consist of four rows and five columns . And we have to find the value of question mark which is lying in fourth row of fifth column after studying the pattern of all the numbers in this box. 
 Because after careful observation we can see that the difference of  product of 1st and 4th column from the sum of 2nd and 3rd column in any particular row is equal to the number in fifth column in that particular row. 

Formula:- 


5th Column =  (1st  Column  × 4th Column ) - (2nd Column + 3rd Column) 

1st Row 
(4 × 2) - (5 + 3) = 8 - 8 = 0 (Number in last column of 1st row. 

2nd Row

(7 × 4) - (3 + 4) = 28 - 7 = 21 (Number in last column of 2nd row . 

3rd Row

(6 × 5) - (4 + 4) = 30 - 8 = 22 (Number in last column of 3rd row). 

4th Row

(9 × 5) - (6 + 5) = 45 - 11 = 34 (Number in last column of 4th row). 
The correct Answer is option (1)34

Problem # 3



Formula:- Sum of both the numbers in outer circle of same octant  + 1 = Number Opposite to both these numbers in inner circle

(2 × 4) + 1 = 8 + 1 = 9(Opposite to both these numbers in inner circle). 
(7 × 3) + 1 = 21 + 1 = 22(Opposite to both these numbers in inner circle). 
(1 × 6) + 1 = 6 + 1 = 7(Opposite to both these numbers in inner circle). 
(5 × 2) + 1 = 10 + 1 = 11(Opposite to both these numbers in inner circle). 
(3 × 4) + 1 = 12 + 1 = 13(Opposite to both these numbers in inner circle). 
(2 × 9) + 1 = 18 + 1 = 19(Opposite to both these numbers in inner circle). 
(2 × 2) + 1 = 4 + 1 = 5(Opposite to both these numbers in inner circle). 
(? × 3) + 1 = 25(Opposite to both these numbers in inner circle). 
3 × ? = 25 - 1
3 × ? = 24 
? = 24 ÷ 3
? = 8
Option  (3)8 is correct option.

Problem # 4



In this reasoning problem study all these numbers . All these numbers are Prime numbers. But if we study all these numbers row wise in all the three boxes. Then these numbers are in increasing order. So if we write all the numbers written in all the three rows in a line as follows.

1st Row 

2 , 3 , 5 , 7 , 11 , 13 

2nd Row 

17 , 19 , 23 , 29 , 31 , 37 

3rd Row 

41 , 43 , 47 , 53 , ? , 61 
So prime number after 53 will be 59 and prime number before 61 will also be 59.
Hence the value of question mark will be 59.
Option  (2)59 is correct option.

Problem # 5



In this reasoning problem study all these numbers . All these numbers are Prime numbers. But if we study all these numbers box wise in all the three boxes. Then these numbers are in increasing order. So if we write all the numbers written in all the three rows in a line as follows.

1st Box 

2 , 3 , 5 , 7 , 11 , 13 

2nd Box 

17 , 19 , 23 , 29 , ? , 37 

3rd Box 

41 , 43 , 47 , 53 , 59 , 61 
So prime number after 29 will be 31 and prime number before 37 will also be 31.
Hence the value of question mark will be 31.
Option  (3)31 is correct option.

Problem # 6


17 - 5 = 12  ⇔ 21 (Reversing the orders of both the digits)(The number opposite in the circle between two selected numbers)
17 - 12 = 5 = 05 ⇔ 50 (Reversing the orders of both the digits)
31 - 12 = 19  ⇔ 91 (Reversing the orders of both the digits)
31 - 10 = 21 ⇔ 12 (Reversing the orders of both the digits)
11 - 10 = 1 = 01 ⇔ 10 (Reversing the orders of both the digits)
11 - 6 = 5 = 05 ⇔ 50 (Reversing the orders of both the digits)
8 - 6 = 2 = 02 ⇔ 20 (Reversing the orders of both the digits)
8 - 5 = 3 = 03 ⇔ 30 (Reversing the orders of both the digits)(The number opposite in the circle between two selected numbers, and this will be the value of question mark).
Option (C)30 is correct option. 

Problem # 7


In this type of reasoning question either we have to study the pattern row wise or column wise. I mean to say that if we add or multiply all the numbers present in each row or column then its sum or product must be same. 
It is not necessary that pattern will always be found by addition or multiplication of the numbers present in rows or columns. 
Sometimes the pattern can be found of with the help of another logics like L C M or H C F of the numbers present in specific row or column.
But in this problem if we find the product of all the numbers in every column, then it is same in every column. 
So to find the value of question mark in 4th row, then we have to find the product of all the 1st three columns.
Product of all the numbers in every column is 120.
6 × 1 × 10 × 2 = 120 ( Last Column ) 
2 × 12 × 3 = 120  ( 2nd Last  Column ) 
3 × 40 = 120   ( 2nd  Column ) 
 Hence the value of  "?"  must be 120
Option (4)120 is correct option. 

Problem # 8



This box problem of reasoning consist of three rows and three columns. And we have to find the value of question mark after studying the pattern of all the letters in this box. 
If we start assigning all the letters of alphabet to numbers serial wise from 1 to 26. Then the value of  the letters present in the box above have these values .
C = 3 , E = 5  , D = 4 , K = 11 ,  F = 6,  J = 10
Now we can easily find the value of the number present in 3rd column using the value of two other numbers we just got letter,s value.

In 1st Row

 C and  E =  C + (C × E) = 3 + (3 × 5) = 3 + 15 = 18

In 2nd Row

D and K =  D + (D × K) =  4 + (4 × 11) = 4 + 44 = 48

In 3rd Row

F and J = F + (F × J) = 6 + ( 6  × 10 )  = 6 + 60 =  66
The correct Answer is option (B)66.

Problem # 9


This box problem of reasoning also consist of three rows and three columns. And we have to find the value of question mark present in 3rd row of 3rd column after studying the pattern or formula of all the letters in this box. 
                    If we start assigning all the letters of alphabet to numbers serial wise from 1 to 26. Then the value of  the letters present in the box above have these values .
C = 3 , E = 5  , D = 4 , K = 11 ,  F = 6,  J = 10
Now we can easily find the value of the number present in 3rd column using the value of two other numbers we just got letter,s value.

In 1st Row

 D  +  T = 4 + 20 = 24

In 2nd Row

G  +  S = 7 + 19 = 26

In 3rd Row

M  +  X = 13 + 24 = 37

The correct Answer is option (1)37.

Problem # 10



This box problem of reasoning consist of four rows and three columns . And we have to find the value of question mark which is lying in third row of third column after studying the pattern of all the numbers in this box. 
                      Because after careful observation we can see that  the sum  of squares of numbers in 1st and 2nd rows and cube of 3rd row is equal to the number in 4th row.

Formula:-

(Number in 1st row)² + (Number in 2nd row)² +  (Number in 3rd row)³ =  (Number in 4th row) 

1st column

 4² + 7² + 2³ = 16 + 49 + 8 = 73

2nd column

 9² + 3² + 1³ = 81 + 9 + 1 = 91

3rd column

6² + 2² + 3³ = 36 + 4 + 27 = 67

The correct Answer is option (1)67
 
Conclusion

So these were the Ten Most Important Reasoning questions with answers for competitive exams of number analogy with solutions were 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. 
Share:

Most important Twenty Number analogy questions of Reasoning with answer

  No comments

Twenty Questions of Number analogy with answer for competitive examinations  and missing number in reasoning for competitive exams like Bank PO, Bank clerk, SSC CGL, ssc chsl, RRB NTPC , group D etc have been discussed in this post.



Twenty Questions of number analogy for competitive exams





In this reasoning problem 1st number (8) is associated to 28 with the help of any rule , with the help of same rule we have to associate 27 to a number out of four given options.

Formula:- 

1st number is cube of a number and 2nd number more than one the cube of successive number of 1st number.
2³ , (2+1)³+1 ,3³ , (3+1)³ +1
2³ , 3³+1 ,3³ , 4³ +1
Hence  4³ + 1 = 64 + 1= 65
Option (4)65 is correct option.

In this reasoning problem 1st number (46) is associated to 2549 with the help of any rule , in the same rule we have to associate 23 to a number out of four given options.

Formula:-

Squares of successive digits of number on the left hand side =  Number on the right hand side

(4+1)² = 5² = 25
(6+1)² = 7² = 49
Similarly (2+1)² = 3² = 9
(3+1)² = 4² = 16
Hence 916 will be the value of question mark .
Option (3)916 is correct option.


1st Method

In this reasoning problem 1st number (37) is associated to 23 with the help of any rule , in the same rule we have to associate 19 to a number out of four given options.

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

Formula :- 


1st Number - 2nd Number = 14
37 - 23 = 14 
19 -  ? = 14 
? = 19 - 14 
? = 5
Option (4)5 is correct option.


2nd Method


All the three given numbers are prime numbers, so the value of question mark must be a prime number. And out of all the four options given only fourth option having value 5 is prime number.

Hence Option (4)5 is correct option.

 In this reasoning problem 1st number 29 is associated to 71 with the help of any rule , in the same rule we have to associate 79 to a number out of four given options.
 1st Number 
29 × 2 = 58 Now add sum of both the digits of 58 to 58 to get 2nd number.
58 + (5+8) = 58 + 13 = 71
3rd Number 
79 × 2 = 158 Now add sum of all the digits of 158 to 158 to get 4th number.
158 + (1+5+8) = 158 + 14 = 172
Hence Option (3)172 is correct option.


In this reasoning problem 1st number (79) is associated to 47 with the help of some rule , With the help of same rule we have to associate 75 to a number out of four given option. Also in this problem all the  numbers consists of two digits. These two digits can be used to find the value of question mark with the help a formula given below.

Formula :- 


( 1st digits  ×  2nd digit } - {1st digits  + 2nd digit}. 
{ 9  × 7 } - { 9  + 7 } = 63 - 16 = 47
{ 7  × 5 } - { 7  + 5 } = 35 - 12 = 23 = ? ( The value of question mark ) 
Option (2)23 is correct option.

In this reasoning problem 1st number 87 is associated to 414 with the help of some rule , With the help of same rule we have to associate 62 to a number out of four given option. Also in this problem all the  numbers consists of three digits. These three digits can be used to find the value of question mark with the help a formula given below.

Formula :- 


{(1st digit of 1st Number) ÷ 2} and {(2nd digit of 1st Number) × 2 } = 2nd Number
{ 8 ÷2 } and { 7×  2 } = 4 and 14 = 414 = 2nd Number
{ 6 ÷2 } and { 2×  2 } = 3 and 4 = 34 = 4th Number = ? ( The value of question mark ) 
Option (2)34 is correct option.

Formula :- 


{1st digit × 2nd digit } + { 3rd digit × 4th digit }  = 2nd Number.
1st Number
(2 × 5) + (7 × 3) = 10 + 21 = 31 = 2nd Number
2nd Number
(4 × 6) + (2 × 9) = 24 + 18 = 42 = 4th Number
Option (3)42 is correct option.

In this reasoning problem 1st number 144 is associated to 23 with the help of some rule , With the help of same rule we have to associate 196 to a number out of four given option. Also in this problem all the  numbers consists of three digits. These three digits can be used to find the value of question mark with the help a formula given below.

Formula :- 


{2 × Square Root (1st Number) }- 1 = 2nd Number
{2 × √144} - 1 = {2 × 12} - 1 = 24 - 1 = 23 = 2nd Number
{2 × √196} - 1 = {2 × 14} - 1 = 28 - 1 = 27 = 4th Number  = ? ( The value of question mark ) 
Option (1)27 is correct option

Also Reads these articles
 
Squaring both digits of 1st number separately, now Interchanging both digits of numbers obtained after Squaring.  Now combining these numbers to form the value of 2nd number.

1st Number = 43  

4² = 16 = 61(After Interchanging both digits)
3² = 09 = 90(After Interchanging both digits)
Now combining both the squares as 6346 = 2nd number.

3rd  Number = 68  

6² = 36 = 63(After Interchanging both digits)
8² = 64 = 46(After Interchanging both digits)
Now combining both the squares as 6346= 4th number.
Option (3)6346 is correct option

Prime factors of 6 = 2,3 and multiplication of 2 and 3 = 6

Prime factors of 15 = 3,5 and multiplication of 3 and 5 = 15

Prime factors of 35 = 5,7 and multiplication of 5 and 7 = 35

Similarly
Prime factors of ? = 7,11 and multiplication of 7 and 11 = 77.
Option (1)77 is correct option

Usually in these types of reasoning problem 1st number is associated to 2nd number with the help of some rule . Similarly 3rd number is associated to 4th number with the help of same rule.
                 But in this reasoning problem all the four numbers are written sequence wise . It Means 2nd term will be found from 1st term, 3rd term will be found from 2nd term and after applying same formula or rule 4th term will be calculated from 3rd term.

Formula :- 

nth Term = {(n+1) × Previous Term} + 4

1st term =  2
2nd term = (8 × 3) + 4 = 24 + 4 = 24
3rd term = (28 × 4) + 4 = 112 + 4 = 116
4th term = (116 × 5) + 4 = 580 + 4 = 584
Option (4) 584 is correct option

In this reasoning problem 1st number (11) is associated to 38 with the help of any rule , in the same rule we have to associate 13 to a number out of four given option.
Look carefully the given numbers consists of two digits. These three digits can be utilised with the help a formula given below.

Formula :- 

(1st Number  × 4)   - 6 = 2nd Number  
{11  ×  4 } - 6 = 44 - 6  = 44 = 2nd Number  
{13  ×  4 } - 6 = 52 - 6  = 46  - 4th Number  
Option (2)46 is correct option

Interchanging position of unit place and ten's place of given 1st  number 34 . we shall get 43, Now taking the  squares of each digit of given number 43  separately. we shall have 16 and 9. Now combining these squares to get 2nd numbers which is 169. 
     Similarly interchanging position of unit place and ten's place of 3rd number 23 . we shall get 32. Now taking the  squares of each digit of given number 32  separately. we shall have 9 and 4. Now combining these squares to get 4th numbers which is 94.  

Option (4)94 is correct option

In this reasoning problem 1st number (0) is associated to 6 with the help of some rule , With the help of same rule we have to associate 24 to a number out of four given option.

Formula :-

nth Term = n³ - n
0 = 1³ -1 = 1st number
6 = 2³ -2 = 2nd number
24 = 3³ -3  = 3rd number
? = 4³ - 4 = 64 - 4 = 60 = 4th number
Option (4)60 is correct option
In this reasoning problem 1st number (49) is associated to 23 with the help of some rule , With the help of same rule we have to associate 91 to a number out of four given option. Also in this problem all the  numbers consists of two digits. These two digits can be used to find the value of question mark with the help a formula given below.

Formula :- 

{Square Root of 1st digits  and Square Root of 2nd digit } of 1st Number = 2nd Number 
√4  and  √9 = 2 and 3  = 23 = 2nd Number
√9  and  √1 = 3 and 1  = 31 = 4th Number= ? ( The value of question mark ) 
Option (1)31 is correct option.


In this reasoning problem 1st number (48) is associated to 1664 with the help of some rule , With the help of same rule we have to associate 27 to a number out of four given option. Also in this problem all the numbers consists of two digits. These two digits can be used to find the value of question mark with the help a formula given below.

Formula :- 

{Square of 1st digits and Square of 2nd digit } of 1st Number = 2nd Number 

1st Number 48

4²  and 8² = 16 and 64 
Hence after combining we have 1664 = 2nd Number

2nd Number 27

2 = 4 and  and 7² =49.
 Hence after combining we have 449 = 4th Number= ? ( The value of question mark ) 
Option (1)449 is correct option.

Also Reads these articles
Difference of 2nd number and 1st number = 20
24 - 4 = 20
Similarly difference of 4th number and 3rd number must be 20
? - 17 = 20
? = 20 + 17
? = 37
Option (4)37 is correct option.
In this reasoning problem 1st number (49) is associated to 6 with the help of some rule , With the help of same rule we have to associate 82 to a number out of four given option. Also in this problem all the  numbers consists of two digits. These two digits can be used to find the value of question mark with the help a formula given below.

Formula :- 

Square Root of( 1st digits  ×  2nd digit } of 1st Number = 2nd Number 
√(4 *9)   = √(36)  = 6 = 2nd Number
√(8 *2)   = √(16)  = 4 = 4th Number = ? ( The value of question mark ) 
Option (2)4 is correct option.

 
 In this reasoning problem 1st number 24 is associated to 16 with the help of any rule , in the same rule we have to associate 52 to a number out of four given option.
Formula  = 2× (Sum of both digits of 1st number) = 2nd Number

 1st Number  24
2×{ 2 × 4 } = 2 × 8 = 16 = 2nd number
3rd Number  52
2×{ 5 × 2 } = 2 × 10 = 20 = 4th number
Option (4)20 is correct option.


 In this reasoning problem 1st number 2 is associated to 32 with the help of any rule , in the same rule we have to associate 3 to a number out of four given option.

1st Number 2
2⁵ = 2 × 2 × 2 × 2 × 2 = 32 = 2nd number
3rd Number 52
3⁵ = 3 × 3 × 3 × 3 × 3 = 243 = 4th number.
Option (2)243 is correct option.

Conclusion


Comment your valuable suggestion regarding the post most important Reasoning questions with answers which  includes reasoning for competitive exams, circle problems, box problems, circle problems and triangles problems  for competitive exams like SSC CGL ,SSC CHSL ,CPO ,Bank exams and RRB NTPC etc which were explained in this post.









Share: