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.