Showing posts with label reasoning. Show all posts
Showing posts with label reasoning. Show all posts

## Maths logical reasoning questions with Answers

Most important Reasoning questions with answers which includes circle problems, box problems, circle problems, triangles problems  for competitive exams like SSC CGL ,SSC CHSL ,CPO ,Bank exams and RRB NTPC etc have been explained in this post.

# Ten Important Reasoning questions with answers for competitive exams

## Problem #1

3² + 4² = 5²
➡ 9 + 16 = 25
➡ 25 = 25
L H S = R H S

6² + 8² = 10²
➡ 36 + 64 = 100
➡ 100 = 100
L H S = R H S

7² + 24² = 25²
➡ 49 + 576 = 625
➡ 625 = 625
L H S = R H S

### 4th Row

12² + ?² = 13²
➡ 144 + ?²  = 169
➡ ?²  = 169 -144
➡ ?²  = 25 , Taking square root on both sides
? = 5

Option (C)5 is correct option

## Problem #2

### 1st Rectangle

{(11 + 4) × ( 9 + 5 ) } - 5  = (15 × 14) - 5 = 210 - 5 = 205

### 2nd Rectangle

{(21 + 1) × ( 3 + 2 ) } - 5 = (22 × 5) - 5 = 110 - 5 = 105

### 3rd Rectangle

{(62 + 5) × ( 1 + 1 ) } - 5  = ( 67 × 2) - 5 = 134 - 5 = 129 = ?
Option (2)129 is correct option

## Problem #3

This reasoning problem consists of three triangles and every triangle have four numbers ,three around the triangle and one in the centre of the triangle.
Formula :-  Product of all the outer numbers  = Middle numbers

### 1st Figure

5 × 6 × 4 = 120 ⇒1 + 2 + 0 = 3 ( Middle number in 1st triangle )

### 2nd Figure

6 × 7 × 5 = 210  ⇒2 + 1 + 0 = 3 ( Middle number in 2nd triangle )

### 3rd Figure

4 × 8 × 10 = 320 ⇒3 + 2 + 0 = 5 ( Middle number in 3rd triangle )
Option (2)5 is correct option

This reasoning problem consists of three squares and every squares have four small squares , And each small square contains a number in it. If we add all the numbers in each small square then we shall have total of 35 in each small square.

### 1st Square

5  + 12 + 8 + 10 = 35

### 2nd Square

8 + 16 + 6 + 5 = 35

### 3rd  Square

4 + 10 + ? + 15 = 35
29 + ? = 35
? = 35 - 6
? = 6

Option (4)6 is correct option

## Problem #5

HCF ( Highest common Factor ) of 4 and 8 is 4

HCF ( Highest common Factor ) of 3 and 5 is 1

HCF ( Highest common Factor ) of  6 and 12 is 6

HCF ( Highest common Factor ) of 10 and 25 is 5
Option (2)5 is correct option

## Problem #6

All the numbers which are around the circle in any figure do not have any role to make the number in centre of the circle. But all the digits of the central number in 1st figure are reversed, So in same way all the digits of the central number in 2nd figure will be reversed to get the central/middle number in the 3rd figure. Hence 6543 will be the value of question mark.
Option (4)6543 is correct option

## Problem #7

Sum of  all the numbers in any particular column are same for all column like

### Formula :- All Numbers in any columns have equal total.

[10 + 7](1st Column) = 17 = [9 + ? + 6](2nd Column)  = [ 15 +2] = 17( 3rd Column) . Now calculating the value of ? from any two column.
→ ? = 2
Option (1)2 is correct option

*******************
Start
End
********************************

## Problem #8

Sum of square of three numbers in upper line is equal to the number in lower line in each figure.

### 1st Figure :-

2² + 5² + 6² = 4 + 25 + 36 = 65

### 2nd Figure :-

6² + 8² + 9² = 36 + 64 + 81 = 81

### 3rd Figure :-

2² + 8² + 7² = 4 + 64 + 49 = 117
Option (2)117 is correct option

## Problem #9

Sum of all the numbers in every row is 20.
2 + 3 + 7 + 8 = 20 ( Last Row )
4 + 7 + 9 = 20  ( 2nd Last Row )
11 + 9 = 20  ( 2nd Row )
Hence the value of  "?"  must be 20
Option (4)20 is correct option

## Problem #10

This Reasoning question have three triangles in it, And every triangle consists of four numbers around it , The middle number in each triangle can be written using the formula given below

### 1st Triangle :-

7² - ( 4² + 3²) = 49 - (16 + 9 ) = 49 -25 =  24  42

### 2nd Triangle :-

8² - ( 6² + 3²) = 64 - (36 + 9 ) = 64 - 45 =  19  91

### 3rd Triangle :-

9² - ( 5² + 3²) = 81 - (25 + 9 ) = 81 - 34 =  47 74
Option (3)74 is correct option

## Conclusion

comment your valuable suggestion regarding the post most important Reasoning questions with answers which includes reasoning for competitive exams, circle problems, box problems, circle problems and triangles problems  for competitive exams like SSC CGL ,SSC CHSL ,CPO ,Bank exams and RRB NTPC etc which were explained in this post.
Share:

## Most important reasoning questions for competitive exams using reasoning tricks

I have explained ten most important reasoning questions for competitive exams. These missing number Reasoning questions have been explained in simple language and easy style.

10 Most Important figure problems

## Problem #1

This reasoning problem consists of four figures and every figure have five numbers ,four around the figure and one in the centre of the figure. Watch carefully all the numbers which are around the central number are perfect squares. So we have to find out the square roots of these numbers and add these numbers then divide the sum with 5 to find the middle number ,

Middle number in every figure is 5th part of sum of roots of every the numbers

### 1st Figure

{√100 +√25 + √100 + √25 } ÷ 5 = {10 + 5 + 10 + 5 } ÷ 5  = 30 ÷ 5 = 6 ( Middle number in 1st figure )

### 2nd Figure

{√25 +√25 + √81 + √36 } ÷ 5 = {5 + 5 + 9 + 6 } ÷ 5 = 25 ÷ 5   = 5 ( Middle number in 2nd figure )

### 3rd Figure

{√25 +√25 + √25 + √25 } ÷ 5 = {5 + 5 + 5 + 5 } ÷ 5  = 20 ÷ 5 = 4 ( Middle number in 3rd figure )

### 4th Figure

{√36 +√49 + √9 + √16 } ÷ 5 = { 6  + 7 + 3 + 4 } ÷ 5  20 ÷ 5 = 4 ( Middle number in 4th figure ) .

Option (3)4 is correct option

## Problem #2

This reasoning problem consists of three figures and every figure have five numbers ,four around the figure and one in the centre of the figure. In this reasoning problem the central/middle number is the difference of sum of the numbers on the left side and right side in each figure.

### 1st Figure

( 56 + 15 ) - ( 22 + 8 ) = 71 - 30 =  41 (Middle number in 1st figure)

### 2nd Figure

( 46 + 9 ) - ( 10 + 6 ) = 55 - 16 = 39  (Middle number in 1st figure)

### 3rd Figure

( 34 + 11 ) - ( 14 + 6 ) = 45 - 20 = 25  (Middle number in 3rd figure)

Option (4)25 is correct option

## Problem #3

This problem figure also consists of three individual figures and every single figure have one number in the middle and six numbers in the upper and lower part of its circle . Since question mark of third figure is the the middle of circle , Hence this number in place of question mark shall be obtained from all other numbers which are in the lower and upper part of this circle.

## Formula :-

All the numbers in the middle of all the figure are the 7 times of sum of all the digits written around the circle.

### 1st Figure

7 [0 + 6 + 4 + 5 + 3 + 1 ] = 7 × 19 =133 (Middle number in 1st circle)

### 2nd Figure

7 [8 + 2 + 5 + 3 + 4 + 6 ] = 7 × 28 =196 (Middle number in 2nd circle)

### 3rd Figure

7 [2 + 1 + 5 + 7 + 3 + 4 ] = 7 × 22 =154 (Middle number in 3rd circle)

## Problem #4

This octagonal figure have eight numbers written on it. Look carefully these numbers are in increasing order starting from 3 and moving clockwise like this
3, 5 , 7 , 11 , 13 , 17, 19 , ? .

But these numbers do not form any pattern these they  neither have common difference nor have definite series pattern .
But if we take 1st number as ? ( question  mark ) and move clockwise like this
?, 3, 5 , 7 , 11 , 13 , 17, 19 .
Now we have a pattern of prime number, because 2 is 1st prime number and all other numbers are prime numbers. Therefore
2, 3, 5 , 7 , 11 , 13 , 17, 19 . are in prime number series pattern.

Option (4)2 is correct option

## Problem #6

This reasoning problem consists of three triangles and every triangle have four numbers associated to it. Each triangles have three numbers around it and one number is in the middle of it. So to find the middle number in each triangle multiply all the numbers around it then divide it with 10 .

### 1st triangle

(5 × 6 × 4 ) ÷ 10 = 120 ÷ 10 = 12  ( Middle number in 1st triangle )

### 2nd triangle

(6 × 7 × 5 ) ÷ 10 = 210 ÷ 10 = 21 ( Middle number in 2nd triangle )

### 3rd triangle

(4 × 8 × 10 ) ÷ 10 = 320 ÷ 10 = 32 ( Middle number in 3rd triangle ) .
Option (2)32 is correct option

## Problem #7

This reasoning problem consists of three squares and every squares have four small squares , And each small square contains a number in it. If we add all the numbers in each small square then we shall have total of 35 in each small square.
Sum of all the numbers in each box is 35.

### 1st Square

3 + 17 + 4 + 11 = 35

### 2nd Square

2 + 16 + 7 + 10 = 35

### 3rd Square

6 + 13 + ? + 15 = 35
⇒34 + ? = 35
⇒ ? = 35 - 34
⇒ ? = 1
Option (4)1 is correct option
Also read these posts on Reasoning
******************************

## Problem #8

( a + b )² + sq of any of two number when moving clockwise = Number in the outer part of opposite sector

( 7 + 2 )² +  7² = 81 + 49 = 130 ( Number in the outer part of opposite sector)

( 6 + 9 )² + 9² = 225 + 81 = 306 ( Number in the outer part of opposite sector)

( 3 + 4 )² + 4² = 49 + 16 = 65 ( Number in the outer part of opposite sector)

( 8 + 5 )² + 5² = 169 + 25 = 194 ( This number must be in  the outer part of opposite sector)

But  let us choose another option

( 8 + 5 )² +  8² = 169 + 64 = 233 ( Number in the outer part of opposite sector)
Hence option (4)306 is correct option

## Problem  #9

This reasoning problem consists of three square and every square have five numbers ,four numbers are at the corner of each square and one in the centre of the square.
In this square every middle number is written as the difference of sum of opposite numbers.

### 1st Square

( 9 + 5 ) - ( 7 + 3) = 14 - 10 = 4 ( Middle number in 1st square ) .

### 2nd Square

( 12 + 13 ) - ( 8 + 9 ) = 25 - 17 = 8 ( Middle number in 2nd square ).

### 3rd Square

( 20 + 13 ) - ( 7 + 6 ) = 33 - 13 = 20 ( Middle number in 3rd square).

Option (1)20 is correct option

## Problem #10

This octagonal figure have eight numbers written on it. Look carefully these numbers are not in any order
1, 2 , 4, 3 , 24 , 12 , 6, ?
These numbers are nor forming any pattern like this.
But if we consider these numbers pairwise opposite to each other like this
( 1 , 24 ), ( 2 , 12 ), ( 4 , 6 ), ( 3, ? ) and then check the multiplication/product of these paired numbers, we shall have same result every time  . Look at  here

1 × 24 = 2 × 12 = 4 × 6 = 3 × ? = 24
Hence 3 × ? = 24
? = 24 /3
? = 8

Option (4)8 is correct option.

These were the ten most important reasoning questions for competitive exams. I have tried to explain these missing number Reasoning questions in simple language and easy style. Please comment your valuable opinions regarding this post in comment box.
=============================================================

Share:

## Latest Questions in box and circle reasoning for SSC CGL And SSC CHSL Exams

Ten Most Important Questions of Reasoning of box and circles problems with solutions are 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 .

# This circle is divided into eight parts and each part is composed of three numbers . In each part two numbers are in outer part of the circle and one number is in the inner part of the circle. If we multiply both the numbers which are in the outer part of the circle and then add one to it then this resultant number will in the inner part of the opposite circle.

( 2 × 2 ) + 1 = 4 + 1 = 5 (In the inner part of opposite sector ).
( 9 × 2 ) + 1 = 18 + 1 = 19 (In the inner part of opposite sector ).
( 4 × 3 ) + 1 = 12 + 1 = 13 (In the inner part of opposite sector ).
( 5 × 2 ) + 1 = 10 + 1 = 11 (In the inner part of opposite sector ).
( 6 × 1 ) + 1 = 6 + 1 = 7  (In the inner part of opposite sector ).
( 3 × 7 ) + 1 = 21 + 1 = 22 (In the inner part of opposite sector ).
( 4 × 2 ) + 1 = 8 + 1 = 9 (In the inner part of opposite sector ).
( 3 × ? ) + 1 =  3? + 1 = 25
After solving
⇒3? =25 - 1
⇒ ? = 24 ÷3
⇒ ? = 8 (In the inner part of opposite sector )
Therefore Correct option is (3)

## Problem # 2

This circle is divided into four parts and each part is composed of three numbers . In each part two numbers are in outer part of the circle and one number is in the inner part of the circle. If we multiply both the numbers which are in the outer part of the circle and then add two to it then this resultant number will in the inner part of the opposite circle.
(8 × 6 ) + 2 = 48 + 2 = 50 (Number in the inner part of the circle in 4th quadrant )
(5 × 7 ) + 2 = 35 + 2 = 37 (Number in the inner part of the circle in 3rd quadrant )
(8 × 10 ) + 2 = 80 + 2 = 82 (Number in the inner part of the circle in 2nd quadrant )
(9 × 11 ) + 2 = 99 + 2 = 101 = ? (Number in the inner part of the circle in 1st quadrant )

Therefore Correct option is (4)

## Problem # 3

This circle is also divided into eight parts and  there is also one number written between  any  two numbers towards outer side of these two numbers .  If we  add these  two simultaneous  numbers then reverse the order of the result so obtained then this number will in the small circle opposite to both these numbers .
5  + 17 =  22 ↔️  22 ( Reversing the order of digits)
17 + 12 = 29 ↔️  92 ( Reversing the order of digits)
12 + 31 = 43 ↔️  34 ( Reversing the order of digits)
31 + 10 = 41 ↔️  14 ( Reversing the order of digits)
10 + 11 = 21 ↔️  12 ( Reversing the order of digits)
11+ 6 = 17 ↔️  71  ( Reversing the order of digits)
6  + 8  = 14 ↔️  41 ( Reversing the order of digits)
8  + 5 = 13 ↔️  31 = ? ( Reversing the order of digits)
Hence the value of  "? " will be 31
Therefore Correct option is (C)

## Problem # 4

Taking the difference of two consecutive terms in clockwise direction and analyse the difference , this difference always increases with the increment of 4.
10 -  4 = 6
20 - 10 = 10  with an increment of 4 as compare to last difference which is 6
34 - 20 = 14  with an increment of 4 as compare to last difference which is 10
52 - 34 = 18  with an increment of 4 as compare to last difference which is 14
? - 52 = 22  must be an increment of 4 as compare to last difference which is 18
? = 22 + 52
? = 74
100 - ? = 26  must be an increment of 4 as compare to last difference  which is 22
? = 100 - 26
? = 74
130 - 100 = 30  with an increment of 4 as compare to last difference  which is 30
130 - 100 = 30  with an increment of 4 as compare to last difference  which is 30
Therefore Correct option is (2)

## Problem # 5

This circle is divided into four parts and each part is composed of three numbers . In each part two numbers are in outer part of the circle and one number is in the inner part of the circle. If we  check the greater of the numbers which are in the outer part of the circle and  then this resultant number will in the inner part of the opposite circle.
Max (17,14) = 17  (Number in the inner part of the circle in 4th quadrant )
Max (18,13) = 18  (Number in the inner part of the circle in 3rd quadrant )
Max (5,3) = 5  (Number in the inner part of the circle in 2nd quadrant )
Max (4,8) = 8 = ? (Number in the inner part of the circle in 1st quadrant )
Therefore Correct option is (2)

## Problem # 6

Here 1st circle consist of four parts and if we divide sum of all the digits in its outer part with number of parts then result will be the middle number.
( 6 + 2 + 9 + 7  )/4 =  24/4 = 6 ( Middle number)
2nd circle consist of five parts and if we divide sum of all the digits in its outer part with number of parts then result will be the middle number.
( 7 + 2 + 5 + 8 + 3 )/5 = 25/5 = 5 ( Middle number)
3rd circle consist of six parts and if we divide sum of all the digits in its outer part with number of parts then result will be  the middle number.
(  8 + 6 + 7 + 5 + 7 + 9 )/6 = 42/6 = 7 ( Middle number)
Hence option (3) is correct answer

## Problem # 7

In all these triangles the difference of the product of  all the numbers which are in outside of triangle and sum of these numbers is equal to the numbers in the middle of each triangles.

### In 1st triangle

(5  × 3 × 2 ) - ( 5 + 3 + 2 ) = 30 - 10 = 20 ( Number in cenre of the 1st triangle)

### In 2nd triangle

(8  × 5 × 5 ) - ( 8 + 5 + 1 ) = 40 - 14 = 26 ( Number in cenre of the 2nd triangle)

### In 3rd triangle

(9  × 2 × 3 ) - ( 9 + 2 + 3 ) = 54 - 14 = 40 ( Number in cenre of the 3rd triangle)

Therefore Correct option is (4)

## Problem # 8

This figure consist of four squares around  one big square. Look carefully in this big square every number in it is perfect cube. And if we multiply the cube roots of  two adjacent numbers then we shall have the number attached to big square between these two numbers whose cube root had been multiplied.
Cube root of 512  ×  cube root of 216  = 8  × 6  =  48 ( The number at bottom line in the box)
Cube root of 216  ×  cube root of 343 = 6 × 8 = 42 ( The number at rightmost box )
Cube root of 343  ×  cube root of  64 ( Choose this number because its cube root is 4 ) = 7 × 4  = 28 ( The number at uppermost box )
Cube root of 512  ×  cube root of  64 ( Choose this number because its cube root is 4 ) = 8 × 4  = 32 ( The number at leftmost box )
Therefore Correct option is (B)
Also read these posts on Reasoning

# 1.Reasoning for competive exams

******************************

## Problem # 9

Taking the difference of two consecutive terms in clockwise direction and analyse the difference , this difference always increases with the increment of 5.
30 - 15 = 15
50 - 30 = 20 with an increment of 5 as compare to last difference which is 15
75 - 50 = 25 with an increment of 5 as compare to last difference which is 20
?  -  75 =  30 with an increment of 5 as compare to last difference which is 25
140 -  ? = 35 with an increment of 5 as compare to last difference which is 30
180 - 140 = 40.  with an increment of 5 as compare to last difference which is 35
Because in every two numbers taken in clockwise order from least number , the difference between two numbers   will be  an increment of 5 . So if we put 105 instead of question mark then we shall have complete pattern discussed above.
Therefore Correct option is (1)

## Problem # 10

This circle is divided into four parts and each part is composed of three numbers . In each part two numbers are in outer part of the circle and one number is in the inner part of the circle. If we multiply both the numbers which are in the outer part of the circle and then divide the resultant so obtained with 4  then this resultant number will in the inner part of the opposite circle.
(7 × 4 ) ÷ 4 = 28 ÷ 4 = 7 (Number in the inner part of the circle in 4th quadrant )
(5 × 8 ) ÷ 4 = 40 ÷ 4 = 10  (Number in the inner part of the circle in 3rd quadrant )
(4 × 20 ) ÷ 4 = 80 ÷ 4 = 20 (Number in the inner part of the circle in 2nd quadrant )
(6 × 8 ) ÷ 4 = 48 ÷ 4 = 12  = ? (Number in the inner part of the circle in 1st quadrant )

Ten Most Important Reasoning of box and circles problems 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 . Feel free to share your opinion regarding this post in comment box.

Thanks

Share:

## 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 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 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

(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

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

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
******************************

## 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 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

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

Also Read these posts on Reasoning

Share: