Most Important circle Problems of Reasoning for different competitive Exams

Problem   1

----- Explanation   ----  

(11 × 7) - (5 × 3) = 77 - 15 = 62 

(18 × 3) - (5 × 4) = 54 - 20 = 34

(9 × 6) - (4 × 8) = 54 - 32 = 22    

 Problem  2

 ----- Explanation   ----  

2² + 5² + 3² + 7² = 4 + 25 + 9 + 49 = 87

8² + 5² + 4² + 3² = 64 + 25 + 16 + 9 = 114

9² + 4² + 1² + 6² = 81 + 16 + 1 + 36 = 134

  Problem   3

 ----- Explanation   ----  

 2 x (5 + 7 + 3) = 2 x 15 = 30 

8 x (5 + 4 + 3) = 8 x 12 = 96 

3 x (4 + 9 + 6) =  3 x 19 = 57  

 Problem   4

 ----- Explanation   ----  

 1² + (5 × 3 × 7) =1 + 105 = 106

4² + (5 × 3 × 6) =16 + 90 = 106

3² + (3 × 2 × 7) = 9 + 42 = 51 

 Problem 5

----- Explanation   ----  

   1² + (5 + 3 + 7) = 1 + 15 = 16

   4² + (5 + 3 + 6) = 16 + 14 = 30

   3² + (3 + 2 + 7)  = 9 + 12 = 21

 Problem 6

----- Explanation   ----  

(2 + 5)² + (7 - 3)² = 7² +4² =  49 + 16 = 65

(1 + 5)² + (6 - 3)² = 6² + 3² = 36 + 9 = 45 = ?

(2 + 3)² + (7 - 2)² = 5² + 5² = 25 + 25 = 50    

  Problem 7

----- Explanation   ----  

(1+2+3+4)² = 10² = 100

(2+3+4+5)² = 14² = 196

(4+5+6+7)² = 22² = 484 

 Problem 8

----- Explanation   ----  

LCM Of  1, 4, 3 and 2 = 12

LCM Of 3, 2, 5 and 4 = 60

LCM Of  4 , 5 ,6 and 2 = 60 = ? 

  Problem 9

----- Explanation   ----  

 19 + 16 + 15 + 12 = 62 ⇔26

23 + 22 + 27 + 19 = 91 ⇔19

24 + 23 + 20 + 22 = 89 ⇔98 

Problem 10

----- Explanation   ----  

√81+√16+√256+√25 = 9+4+16+5 = 34

√9+√49+√1+√64 = 3+7+1+8 = 19

√4+√36+√121+√289 = 2+6+11+7 = 26 = ?   

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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

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

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

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

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

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

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

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

 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.

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.

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.

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Missing number and reasoning problems for SSC CGL Exams

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

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Reasoning questions with answers for competitive exams

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. 

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