## Number Analogy Reasoning questions with answers for competitive exams

Ten Most Important Number Analogy Reasoning questions with answers for competitive exams have been 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. please feel free to comment your opinions about this post .

# Number Analogy Reasoning questions with answers for competitive exams

**Problem # 1**

**Formula:- **

**Sum of any three numbers in the same row**

**=**

**4th number**

**in the same row**

**1st Row**

**2nd Row**

**3rd Row**

### 4th Row

**Therefore correct option is (B) 7**

**Problem # 2**

### Formula:-

**All the terms are one more than cube of some number.**

**Therefore correct option is (2) 65**

Problem # 3

Problem # 3

**Therefore correct option is (1) 425**

## Problem # 4

In this problem of reasoning we have to combine both the given numbers in left hand side in such a way that after applying any mathematical operation/operations we could get the number on the right hand side of all the three problems given above. Because in these types of reasoning problems we can change given mathematical sign according to our requirements.

**Formula**:-

**Sum of both the digits of 1st numbers × Product of both the digits of 2nd numbers = Number on the right hand side**

**1st Problem**

**26**⇔ 62(After Interchanging both the digits of number(26) just obtained).

### 2nd Problem

**12**⇔ 21(After Interchanging both the digits of number(12) just obtained).

### 3rd Problem

**22**⇔ 22(After Interchanging both the digits of number(22) just obtained).

**Therefore correct option is (3) 22**

**Problem # 5**

In this problem of reasoning we have to combine both the given numbers in the left hand side in such a way that after applying any mathematical operation/operations we could get the number on the right hand side of all the three problems given above. ### Formula:-

**Sum of all the digits of both the numbers on the left hand side = Number on the right hand side**

### 1st Problem

### 2nd Problem

### 3rd Problem

**Therefore correct option is (2) 8**

**Problem # 6**

**Formula:- **

**1st number(Any Pair) × 12 = 2nd number**

**(Same Pair)**

**1st Pair**

**2nd Pair**

**3rd Pair**

**3rd Pair**

**Therefore correct option is (3) (14, 196)**

**Problem # 7**

### 1st Method

### 2nd Method

### All the three numbers given in this problem are prime numbers so fourth will also be prime number. Therefore only one prime number out of given four options is 5.

**Therefore correct option is (4) 5**

**Also Read these**

**articles**

### Formula

**Sum of all the three numbers in each part separated above = 15**

**Problem # 9**

__same formula will be applicable__to third figure to find out the value of question mark.

**Formula** :-

**The product of two numbers of the**

**right side of each triangle is equal to**

**sum of two numbers on the**

**left side of each triangle**.

** 1st Triangle **

12 × 3 = 26 + 10 = 36

** 2nd Triangle **

6 × 9 = 43 + 11 = 54

** 3rd Triangle **

9 × 5 = ? + 15 = 30

? + 15 = 30

? = 30 - 15

? = 25

**Therefore correct option is (4) 25**

**2nd Method**

**Formula** :-

**The middle Number in each figure is**

**equal to product of two left most numbers minus number on the right side**.

** 1st Triangle **

{( 12 × 3 ) - 10 } = ( 36 - 10) = 26

** 2nd Triangle **

{( 6 × 9 ) - 11 } = ( 54 - 11) = 43

** 3rd Triangle **

{( 9 × 5 ) - 15 } = ( 45 - 15) = 30

**Therefore correct option is (4) 30**

**Problem # 10**

__same formula will be applicable__to third figure to find out the value of question mark.

**Formula** :-

**one tenth**of product of remaining numbers.

**Middle Number = abc/10**

** 1st Triangle **

** 2nd Triangle **

** 3rd Triangle **

**Therefore correct option is (2)60**