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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

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

Difference of both the 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

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

## Problem #4 In this figure the sum of the digits of the numbers 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

(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 +  64⅓ = 7 + 4 = 11 ( 1st number in 3rd row)
11 +  27⅓ = 11 + 3 = 14 ( 2nd number in 3rd row)
64 +  8⅓ = 64 + 2 = 66 ( 3rd number in 3rd row)
Option  (B) is correct option

## Problem #6

Pick all prime number up to 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

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
Also read these posts on Reasoning
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## Problem #8

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 applicable 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

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

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

Also Read these posts on Reasoning

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