## Ten Important Reasoning questions with answers for competitive exams

## 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

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

**Formula**:- The sum of two numbers which are on the perpendicular lines is equal to third number of triangle in every figure.

**Option (2)20 is correct option**

## Problem # 2

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

**Formula**:- The sum of product of left & right and sum of lower and upper numbers in each figure is equal to middle number in every figure.

**Option (2)68 is correct option**

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

**Formula**:- The difference of product of both the numbers in upper row and product of both the numbers in lower row is equal to middle number in every figure.

**Option (1)84 is correct option**

**Formula**:- Sum of twice of any number in the inner circle and its adjoining number taken in clockwise order is equal to number in the outer cirle.

**Option (1)19 is correct option**

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

**Formula**:- The product of top three numbers in any figure is equal to product of other three numbers in every figure.

**Option (1)4 is correct option**

**Formula**:- (1st digit ×2nd digit) ÷ middle digit = 2nd number

**Option (1)8 is correct option**

## Problem # 7

**Formula**:- Reversing the digits of square root of 1st number = 2nd number

**Option (3)21 is correct option**

**Also Reads these**

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**End**

## Problem # 8

In these type of reasoning problems 2nd number will be obtained from 1st number either by using one mathematical operation/ set of mathematical operations or by using any mathematical operation/ set of mathematical operations to individual digits of 1st number. Same mathematical operation/ set of mathematical operations or formula will be used in 3rd number to get fourth number.

**Formula** :- 1st digit ^ 2nd digit = 2nd number

Taking 1st digit as base and and digit as power then after expansion we have

2^6 = 64

3 ^5 = 243

**Option (1)243 is correct option**

** **

Problem # 9

**Formula** :- ( Any number × Opposite of number already chosen ) + 1 = 61

(10 × 6) +1 = 60 +1 = 61(Middle number)

(15 × 4) +1 = 60 +1 = 61(Middle number)

(5 × ? ) +1 = 60 +1 = 61(Middle number)

(5 × ? ) +1 = 61(5 × ? ) = 60

? = 60 ÷ 5

? = 12

**Option (2)12 is correct option**

## Problem # 10

**Formula** :- ( Any number + Opposite of number already chosen ) + 13 = 71

(23 + 35 ) + 13 = 58 + 13 = 71(Middle number)

(41 + 17 ) + 13 = 58 + 13 = 71(Middle number)

** **(46 + ? ) + 13 = 71(Middle number)

46 + ? = 71 - 13

46 + ? = 58

? = 58 - 46

? = 12

**Option (2)12 is correct option**

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