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
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
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
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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.
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