Reasoning questions with answers for competitive exams, Reasoning questions with answers for bank exams

Reasoning questions with answers for competitive exams,  Reasoning questions with answers for bank exams. These questions of reasoning in latest reasoning questions with answers are very very important for upcomig competitive exams  like Bank PO , SSC CGL etc . So let us start solving and understanding these Maths logical reasoning questions with answers.

Problem # 1


Reasoning questions with answers for bank exams
 
Every number in any upper row is the average of two numbers in the lower row.
Last row /Bottom Row
(9 + 7) ÷ 2  = 8 ( Number in 2nd last row ), 
(7 + 5) ÷ 2  = 6 ( Number in 2nd last row ), 
(5 + 3) ÷ 2  = 4 ( Number in 2nd last row ), 
2nd last row
(8 + 6) ÷ 2  = 7 ( Number in 3rd last row ), 
(6 + 4) ÷ 2  = 5 ( Number in 3rd last row ), 
3rd last row
(7 + 5) ÷ 2  = 6 ( Number in Top most row ), And this will be the value of quaestion mark ?
Hence option (1)6 is correct option

Problem # 2



Reasoning questions with answers for bank exams


This problem of reasoning have three figures, And every figure consist of five numbers .Out of five numbers one is at centre and four numbers are around the central number.
In every figure central number is the sum of the cube of remaining four numbers.

 33    23  +   43  +   63   = 27 + 8  + 64 + 216 =  315 (In 1st figure)
 113    73  +  83  +    63  = 1331 + 343  + 512 +216 =  2402 (In 2nd figure)
 13   53  +   ?3  +    43   = 1 + 125  +  ?3 + 64 =  1190  (In 3rd figure)
⇒   ?3  = 1190 - 64 - 125 - 1 
 ⇒   ?3  = 1190 - 190
⇒   ?3  = 1000
⇒   ? = 10
 Hence option (3)6 is correct option


Problem # 3


Reasoning questions with answers for bank exams

Take the cube of product of 1st two numbers of every row to get the 3rd number in the box.
 (2 × 3)3  =   63  = 216 (last number in 1st row)
 (3 × 4)3   =   23  = 1728 (last number in 2nd row)
 (4 × 5)3  =   203  = 8000 (last number in 3rd row)
  (5 × 6)3  =   303  = 27000 (last number in 4th row)
or 
Cube of ( 2 × 3 )  =  cube of 6 = 216 (last number in 1st row) 
Cube of ( 3 × 4) = cube of 12 = 1728 (last number in 2nd row) 
Cube of ( 4 × 5)  =  cube of 20 = 8000 (last number in 3rd row) 
Cube of ( 5 × 6)  =  cube of 30 = 27000  (last number in 4th row) 
Option (B)27000 is correct option

Problem # 4


Reasoning questions with answers for bank exams

Take the difference of every number with its preeceding number
4 - 2  = 2 ( 2nd number - 1st number ) , 
9 - 4 = 5,  ( 3rd number - 2nd number ) , 
20 - 9 = 11 , ( 4th number - 3rd number ) , 
40 -20 = 20, ( 5th number - 4th number ) , 
? - 40 ,   ( 6th number - 5th number ) 
Now this new series of difference is here
2 , 5 , 11 , 20 , ? - 40 
Again take the difference of every number with its preeceding number
5 - 2 = 3 , 
11 - 5 = 6 ,
 20 - 11 = 9 
( ? - 40 )  - 20 = 12 (If the pattern follows)
Now study the difference , this difference is multiple of 3, Hence proceeding in the same manner the next difference will be 12
So  ( ? -  40 )  - 20 = 12 
? = 12 + 20 + 40
?  = 72
Option (2)72 is correct option

Problem # 5


Reasoning questions with answers for bank exams

This problem of reasoning also have three figures, And every figure consist of four numbers .Out of four numbers one is at centre and remaining three numbers are around the central number.
In every figure central number is the sum of the number in base line and product of other two numbers . This implies 1st we have to multiply the two numbers which are in the top row then we have to add this sum to the number in bottom line to get the central number in each figure.
(5 × 2) + 10 = 10 + 10 = 20 (1st Figure)
(6 × 3) + 6 =18 + 6 = 24  ( 2nd Figure)
(7 × 5) + 6 = 35 + 6 = 41 ( 3rd Figure)
Option (1)24 is correct option


Problem # 6


Reasoning questions with answers for bank exams

This circle has been divided into four parts and each parts consist of three numbers , Two numbers are written in the outer part of the quadrant and one is at the inner part of the quadrant.
(12 - 8 )3  - 1 =  (4)3  - 1 =  64 - 1 = 63
 (4 - 3 )3  - 1  =  (1)3  - 1 =  1 - 1 =  0
 (9 - 6 )3  - 1 =  (3)3  - 1 =  27 - 1 =  26
 (12 - 10 )3  - 1 =  (2)3  - 1 =  8 - 1 =  7
                                or
Cube(12 - 8 ) -  1 = 64 - 1 =  63
Cube (4 - 3 ) -  1 = 1 - 1 =  0
Cube( 9 - 6 ) - 1 = 27 - 1 = 26
Cube(12 - 10 ) - 1 = 8 - 1 = 7
Option (1)7 is correct option


Problem # 7


Reasoning questions with answers for bank exams

This series is the difference of cube  and square of  the number starting from 2.
 23  22 )  = 8 - 4 =  4   ( 1st number in the series) 
 33  32 )  = 27 - 9 =  18   ( 2nd number in the series) 
 43  42 )  = 64 - 16 =  48   ( 3rd number in the series) 
 53  52 )  = 125 - 25 =  100   ( 4th number in the series) 
 63  62 )  = 216 - 36 =  180   ( 5th number in the series) 
 73  72 )  = 343 - 49 =  294   ( 6th number in the series) 
 83  82 )  = 512 - 64 =  448   ( 7th number in the series) 
or
Cube of 2 -  square of 2 = 8 - 4 = 4
Cube of 3 -  square of 3 = 27 - 9 = 18
Cube of 4 - square of 4 = 64 - 16 = 48
Cube of 5 -  square of 5 = 125 - 25 = 100
Cube of 6-  square of 6 = 216 - 36 = 180
Cube of 7 -  square of 7 = 343 - 49 = 294
Option (3)48 is correct option

Problem # 8


Reasoning questions with answers for bank exams


Take the sum of cube roots of two adjoining numbers from the bix box in the centre to get the number in outer part of small box .
Cube root of (343) + Cube root of (27)  = 7 + 3 = 10 ( The number in the outer small box)
Cube root of (27) + Cube root of (125)  = 3 + 5 = 8 ( The number in the outer small box)
Cube root of (125) + Cube root of (?)  = 5  +  Cube root of (?)  = 7 ( The number in the outer small box)
⇒ Cube root of (?)  = 7 - 5 
⇒ Cube root of (?)  = 2  , cubing both  sides
⇒  ?  = 8
Cube root of ( ?) + Cube root of (343)  = Cube root of (?)  + 7 = 9 ( The number in the outer small box)
⇒ ⇒ Cube root of (?)  = 9 - 7 
⇒ Cube root of (?)  = 2  , cubing both  sides
⇒  ?  = 8
Hence we get 2 in both the cases 
Hence Option (C)8 is correct option

Problem # 9


Reasoning questions with answers for bank exams 
This problem of reasoning also have three figures, And every figure consist of five numbers .Out of five numbers one is at centre and remaining four numbers are around the central number.

In this problem central number is 2 less than the sum of  all the number which are around the middle number.

(0 + 6 + 4 + 2) - 2 = 12 - 2 = 10 (  middle number in first figure) 

2nd figure
( 6 + 2 + 10 + 8) - 2 = 26 - 2 = 24 (  middle number in 2nd figure) 

3rd figure
(4 + 14 + 12 + 10 ) - 2 = 40  - 2 = 38 (  middle number in 3rd figure) 
Option (C)38 is correct option

Problem # 10


Reasoning questions with answers for bank exams

This circle has been divided into four parts and each parts consist of three numbers , Two numbers are written in the outer part of the quadrant and one is at the inner part of the quadrant.
Difference of digits of numbers in outer circle of any Quadrant - (difference of digits of numbers in inner circle of any Quadrant ) = one in every case
(12 - 8 ) - (6 - 3)  = 4 - 3 =1
(4  - 3 ) - (1 - 1)  = 1 - 0 =1
(9 - 6 ) - (6 - 4)  = 3 - 2 = 1
Similarly we have to proceed like this and choose the correct option whose difference of digits is 1. Look at the option all other option except 2nd option will not satisfies the condition required.
Hence    (12 - 10 ) - (4 - 3)  = 2 - 1 = 1
Option (2)43 is correct


In this post I discussed Ten Tricky logical reasoning questions+answers , Latest reasoning questions with answers  . Comment your valuable suggestion for further improvement.






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Ten Tricky logical reasoning questions+answers , Latest reasoning questions with answers

Ten latest and Tricky logical reasoning questions+answers are discussed in this post . These questions of reasoning in latest reasoning questions with answers are very very important for upcomig competitive exams  like Bank PO , SSC CGL etc . So let us start solving and understanding these Maths logical reasoning questions with answers.

Problem #1

Ten Tricky logical reasoning questions+answers

Magnitude of {(7 + 5) (7 - 5 ) - cube of 5 } = 125 - 12 *2 = 125 - 24 = 101
Magnitude of {(8 + 3) ( 8 - 3 ) - cube of 3 } = 11 * 5 - 27 = 55 - 27 = 28
Magnitude of {(6 + 2) ( 6 - 2 ) - cube of 2  } = 8 * 4  - 8 = 32 - 8 = 24
Magnitude of {(5 + 4) ( 5  - 4 )  - cube of 4 } =  64 -  9*1 =  64 - 9  = 55
 Note :- Magnitude of  difference of two numbers means , we have to consider the difference of two numbers irrespective of their positions, And the magnitude of difference of two numbers will always be a positive value .The magnitude of  3 - 2 and 2 - 3 will be same and will be equal to 1 ( not -1)  and  magnitude of 7 - 9  and 9 - 7 will be same and equal to  2 ( not -2 ), Similarly the magnitude of 11 - 2 and 2 - 11 will same and equal to 9 (  not -9 ).

Alternate Method 

Magnitude  { { 72  -   52  } -  (  53 ) } =  (  53  { 72  -   52  } = 125 - ( 49 - 25) =125 - 24 = 101
Magnitude  { { 82  -   32  } -  (  33 ) } = { 82  -   32  }  -  (  33  = ( 64 - 9 ) - 27 = 55 - 27 =  24
Magnitude  { { 62  -   22  } -  (  23 ) } = {62  -   22  }  -  (  23  =  ( 36  - 4) - 8 = 32 - 8 = 24
Magnitude  { { 52  -   42  } -  (  43 ) } =  (  43  { 52  -   42  } = 64 - ( 25 - 16 ) = 64  - 9 = 55

Option (2) is correct

Problem #2

Ten Tricky logical reasoning questions+answers

Difference of digits of numbers in outer circle of any Quadrant - (difference of digits of numbers in inner circle of any Quadrant ) = two  in every case
(55 - 52 ) - (9 - 8)  = 3 - 1 = 2
(29  - 21 ) - (9 - 3)  = 8 - 6 = 2
(18 - 12) - (8 - 4)  = 6 - 4 = 2
(7 - 5) - (4 - 4)  = 2 - 0 =2
Option (4) is correct

Problem #3

Ten Tricky logical reasoning questions+answers

1st quadrant 
4 ×   (7 - 3)2   = 4 ×  42  =  4  × 16 = 64
2nd quadrant 
4 ×  (5 - 5)2  = 4 × 02  =  4 × 0  = 0
3rd quadrant 
4 ×  (11 - 8)2  = 4 ×  32  =  4 × 9 = 36
4th quadrant 
4 ×   (8 - 2)2  = 4 ×  62  =  4×  36 = 144
Option  (3) is correct option

Problem #4


Ten Tricky logical reasoning questions+answers
In this figure the sum of the digits of the number obtained from the multiplication of both the numbers which are outer part of that particular quadrant.
8 × 2 = 16 = 1 + 6 = 7 digit in middle left box
6 × 5 = 30 = 3 +0 = 3 digit in middle left box
6 × 9 = 54 = 5 + 4  = 9 digit in middle right box
4 × 1 = 04 = 0  + 4 = 4 digit in middle right box
Option  (2) is correct option

Problem #5

Ten Tricky logical reasoning questions+answers
 (11 - 7 )3  =  43  = 64  (1st number in 2nd row ) 
  (14 - 11 )3 =  33 = 27  ( 2nd number in 2nd row ) 
  (64 - ?)2  =   43  = 8  ( It will be 3rd number in 2nd row ) 
So ? = 66
Alternate Method 

Add the cube root of respective number  in 2nd row to the number in 1st row to get number in 3rd row. 
7 + cube root of 64 = 7 + 4 = 11 ( 1st number in 3rd row) 
11 + cube root of 27 = 11 + 3 = 14 ( 2nd number in 3rd row) 
64 + cube root of 8 = 64 + 2 = 66 ( 3rd number in 3rd row) 
Option  (B) is correct option

Problem #6

Ten Tricky logical reasoning questions+answers
 Pick all prime number upto 11
1st prime number = 2
   Multiply it with next number
2 × 3 = 6 ( 2nd  number in the series) 
2nd prime number = 3
   Multiply it with next number
3 × 4 = 12 ( 3rd  number in the series) 
3rd prime number = 5
   Multiply it with next number
5 × 6 = 30 ( 4th  number in the series) 
4th prime number = 7
   Multiply it with next number
7 × 8 = 56 ( 5th  number in the series) 
5th prime number = 11
   Multiply it with next number
11 × 12 = 132 ( 6th   number in the series) 
Option  (D) is correct option

Problem #7

Ten Tricky logical reasoning questions+answers

Add the twice of the number in 1st column to the half of the number in 2nd column to get the 3rd number in every row. 

( 2 × 6 ) + ( 8 ÷ 2 ) = 12 + 4 = 16 ( 1st number in 3rd column) 
( 2 × 1 ) + ( 6 ÷ 2 ) = 2 + 3 = 5 ( 2nd number in 3rd column) 

( 2 × 3 ) + ( 10 ÷ 2 ) = 6 + 5 = 11 ( 3rd number in 3rd column) 
Option  (A) is correct option

Problem #8

Ten Tricky logical reasoning questions+answers
Divide with 10 the Multiplication/Product of  all the numbers which are in the outer part of triangles to get the number in the centre of the triangle. And this method will be appplicable to all the three  triangles.
1st triangle 
(5 × 6 × 4 ) ÷ 10 = 120 ÷10 =12
2nd triangle
(6 × 7 × 5 ) ÷ 10 = 210 ÷10 =21
3rd triangle
(4 × 8 × 10 ) ÷ 10 = 320 ÷10 =32
Option  (3) is correct option

Problem #9


Ten Tricky logical reasoning questions+answers
Here every figure consist of three numbers in 1st line and one number in 2nd line. And 2nd number written in every figure is the (HCF) Highest common Factor of all the three numbers in the 1st line in each figure. 
H C F ( Highest common Factor ) of 9 ,15 , 18 = 3 (1st figure )
HCF ( Highest common Factor ) of 16 ,28 , 32 = 4 ( 2nd figure )
HCF ( Highest common Factor ) of 24 ,36 , 48 = 12 (3rd figure )
Therefore option (4) is correct option .

Problem #10

Ten Tricky logical reasoning questions+answers
Take reverse of sum of both the numbers in every quadrant to get the number attached to middle of this figure. 

9 + 8  = 17 ↔️71 ( The number in the middle of  4th Quadrant)
8 + 3  = 11 ↔️11 (  The number in the middle of  1st Quadrant)
5 + 7  = 12 ↔️21 ( The number in the middle of   2nd Quadrant)
7 + 7  = 14 ↔️41 ( The number in the middle of   3rd Quadrant)
Therefore option (2) is correct option .

In this post I disscussed Ten Tricky logical reasoning questions+answers , Latest reasoning questions with answers  . Comment your valuable suggestion for further improvement.






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Ten most expected missing number series questions for SBI PO with solution , for SSC CGL , SSC CHSL , RRB NTPC and other competitive Exams

Maths    Reasoning   

Ten most expected  missing  number series questions for 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. 

Problem # 1


missing  number series questions for SBI PO with solution

Take the difference of two consecutive terms of the given series. This difference will be the pair of same odd numbers in every case.

16 - 5 = 11 (The difference of 1st and 2nd number is pair of odd number one)

49 - 16 = 33  (The difference of 2nd and 3rd number is pair of odd number three )

104 -  49 = 55  (The difference of 3rd and 4th number is pair of odd number  five)

? - 104 = 77   ( Similarly the difference of 4th and 5th number will pair of odd number seven)

⇒ ? =  77 + 104  (And in the same way the difference of 5th and 6th number is pair of odd number nine )

⇒ ? = 181

280 - ? = 99  

⇒ ? = 280 - 99 

⇒? = 181

Option (2)181 will be correct option

Problem # 2

missing  number series questions for SBI PO with solution

In this series every number is written as power of one less than the term,s position and taking as base of  5 . i.e. 1st term is written as power of 0 , 2nd term is written as power of 1, 3rd term is written as power of 2 , 4th term is written as power of 3 and so on.
5^0 - 0 = 1 - 0 = 1  (Power 0 of base 5 - same power )
5^1 - 1 = 5 - 1 = 4   (Power 1 of base 5  - same power   )
5^2 - 2 = 25 - 2 = 23   (Power 2 of base 5  - same power  )
5^3 - 3 = 125 - 3 = 123   (Power 3 of base 5  - same power )
5^4 - 4 = 625 - 4 = 621   (Power 4 of base 5  - same power  )
Option (4)621 is correct option

Problem # 3

missing  number series questions for SBI PO with solution

In this series every number is written as power of one greater than the term,s position and taking as base of  5 . i.e. 1st term is written as power of 0 , 2nd term is written as power of 1, 3rd term is written as power of 2 , 4th term is written as power of 3 and so on.
5^0 + 0 = 1 + 0 = 1  (Power 0 of base 5 + same power )
5^1 + 1 = 5 + 1 = 6   (Power 1 of base 5  + same power   )
5^2 + 2 = 25 + 2 = 27   (Power 2 of base 5  + same power  )
5^3 + 3 = 125 + 3 = 128   (Power 3 of base 5  + same power )
5^4 + 4 = 625 + 4 = 629   (Power 4 of base 5  + same power  )
Option (2)629 will be correct option.

Problem #4


missing  number series questions for SBI PO with solution

In this series every number is written as cube of  the term,s position and one number less than it . i.e. 1st term is written as cube of 1 and less than 1 , 2nd term is written as cube of 2 and less than 1, 3rd term is written as cube of 3 and one lees than it , 4t term is written as cube of 4 and one less than it and so on.
 1³ - 1 = 1 - 1 = 0
 2³ - 1 = 8 - 1 = 7
 3³ - 1 = 27 - 1 = 26
 4³ - 1 = 64 - 1 = 62
 5³ - 1 = 125 - 1 = 124
 6³ - 1 = 216 - 1 =  215
Option (1)124 will be correct option.

Problem # 5

missing  number series questions for SBI PO with solution


5  * 1  + 2 = 5  +  2 = 7 ( To get 2nd term Multiplying 1st term with 1 and add 2 to it)
7 * 2  -  4 = 14 - 4  = 10 ( To get 3rd term Multiplying 2nd term with 2 and subtract 4 to it )
10 * 3  + 6 = 30 + 6 = 36 ( To get 4th term Multiplying 3rd term with 4 and add 6 to it) 
36 * 4  - 8 = 144 - 8 = 136 ( To get 5th term Multiplying 2nd term with 2 and subtract 4 to it )
136 * 5  + 10 = 680 +10= 690 ( To get 2nd term Multiplying 1st term with 1 and add 2 to it)
Option (2)690 will be correct option.

Problem # 6


missing  number series questions for SBI PO with solution

To get 2nd term ,multiply 1st term with half and add cube of 1 to it .
12 * 0.5 + 1 = 6 + 1 =  6 +  1³ = 6 + 1 = 7
To get 3rd term ,multiply 2nd term with one ( double of half ) and add cube of 2 to it .
 7 * 1  + 8 = 7 +  2³ =  7 + 8 = 15 
To get 4th  term ,multiply 3rd term with 2 ( double of 1 ) and add cube of 3  to it .
15 * 2 +  27 =  30 +  3³ = 57
To get 5th  term ,multiply 4th term with 4 ( double of 2 ) and add cube of 4  to it .
57 * 4 + 64 =  228 +  4³ = 292
To get 6th  term ,multiply 5th term with 8 ( double of 4 ) and add cube of 3  to it .
292 * 8 + 125 = 2336 + 125 = 2461
option (4)2461 is correct option

Problem # 7

missing  number series questions for SBI PO with solution

Multiply 1st term of the series with one and half  (1.5) to get 2nd term of the series.
10 * 1.5 = 15
Multiply 2nd term of the series with three  (  double the 1.5 i.e. 2 * 1.5 = 3 ) to get 3rd term of the series.
15 * 3 = 45
Multiply 3rd term of the series with  6 (  double the 3 i.e. 3 * 2 = 6) to get 4th  term of the series.
45 * 6 = 270
Multiply 4th term of the series with 12  ( double the 6 i.e. 6 * 2 = 12 ) to get 5th term of the series.
270 *  12  = 3240
Multiply 5th term of the series with 24 (  double the 12 i.e. 12 * 2 = 24 ) to get 2nd term of the series.
3240 * 24 =  77760 
Option (2)77760 is correct option

Problem # 8

missing  number series questions for SBI PO with solution


2 * 5 - 1 = 10 -1 = 9  ( 2nd term is written as multiple of 5 with 1st term  and decrease it by 1 )
9 * 5 + 3 =  45 + 3 = 48  ( 3rd term is written as multiple of 5 with 2nd term  and increase it by 3)
48  * 5 - 5 = 240- 5 =235  ( 4th term is written as multiple of 5 with 3rd term  and decrease it by 5)
235 * 5 + 7 = 1185 + 7 = 1182  ( 5th term is written as multiple of 5 with 4th term  and decrease it by 7 )

Note carefully in this problem ,operation of decreasing and increasing used alternatively.
Option (2)1182 is correct option

Problem # 9


missing  number series questions for SBI PO with solution

In this series 1st term is 12 , To get 2nd term from it we just multiplied it with half
12 * 1/2 =  6
To get 3rd term from 2nd term we just multiplied it with double of the number what we have multiplied in last term i.e. 2 *( 1/2) =1 
6 *1 = 6 
To get 4th term from 3rd term we just multiplied it with double of the number what we have multiplied in last term i.e. 2 *( 1) =2 
6 * 2 = 12 
To get 5th term from 3rd term we just multiplied it with double of the number what we have multiplied in last term i.e. 2 *( 2 ) =4 
12 * 4 = 48 
To get 6th term from 5th term we just multiplied it with double of the number what we have multiplied in last term i.e. 2 *( 4 ) =8 
48 * 8 = 384
Hence (2)384 will be correct option

Problem # 10

missing  number series questions for SBI PO with solution

To find out the solution of  above series problem just add all the digits of the every number given , this sum will leads to the next number of the series
5 + 3 + 4 + 2 = 14 ( 2nd number in series) 
Similarly add all the digits of the 3rd number, this sum will leads to the 4th number of the series
5 + 2 + 3 + 1 = 11 ( 4th number in series) 
4 + 1 + 2 + 0 = 17 ( 6th number in series) 
6 + 7 + 3 + 2 = 18  ( last/ Missing number in series) 
Option (4)18 is correct option







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