## Thursday, 15 July 2021

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

I have explained ten most important reasoning questions of missing number for competitive exams like ssc cgl, ssc chsl ,bank clerk and rrb ntpc . These missing number Reasoning questions have been explained in simple language and easy style.

# Important reasoning questions in 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 #5

This series have seven numbers in it . And all the numbers are written in decreasing order . So we have to take the differences of two consecutive terms to proceed further.

## 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 = 9² + 7² = 81 + 49 = 130 ( Number in the outer part of opposite sector)

( 6 + 9 )² + 9² = 15² + 9² =  225 + 81 = 306 ( Number in the outer part of opposite sector)

( 3 + 4 )² + 4² = 7² + 4² =  49 + 16 = 65 ( Number in the outer part of opposite sector)

( 8 + 5 )² + 5² =13² + 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 = 13² + 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.

#### 1 comment:

1. Maths reasoning my favourite subject

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