Maths Reasoning

Ten Most Important Reasoning questions with answers for competitive exams of circles,box and other type with solutions will be 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 .

Problem # 1

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Divide with 2 the sum of 1st and 3rd number in every column of each row to get the number middle row .

( 7 + 9 )/2 = 16/2 = 8

(4 + 6) /2 = 10/2 = 5

(1 + ? )/2 = 2

⇒ (1 + ? ) = 4

⇒ ? = 4 - 1

⇒ ? = 3

Therefore option (B) is correct option

Problem # 2

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 .

H C F ( Highest Common Factor ) of 4 ,8 , 12 = 4 (1st figure )

H C F ( Highest Common Factor ) of 12 ,24 , 30 = 6 ( 2nd figure )

H C F ( Highest Common Factor ) of 21 ,14 , 42 = 7 (3rd figure )

Therefore option (1) is correct option

Problem # 3

One tenth of sum of all the numbers except middle number in each box is equal to middle number in that box.

(3 * 4 * 2 * 5 )/10 = 120 / 10 = 12 ( Middle number in 1st box )

(6 * 2 * 3 * 5 )/10 = 180 / 10 = 18 ( Middle number in 2nd box )

(2 * 2 * 9 * 5 )/10 = 180 / 10 = 18 ( Middle number in 3rd box )

Therefore option (2) is correct option.

Problem # 4

Sum of the product of both the numbers in 1st and 2nd line is equal to middle numbers in each figure.

(5 * 4 ) + ( 3 * 1) = 20 + 3 = 23 ( Middle number in 1st figure )

(7 * 6 ) + ( 3 * 4) = 42 + 12 = 54 ( Middle number in 2nd figure )

(11 * 2 ) + (? * 9) = 22 + 9? = 40

⇒ 9? = 40 - 22

⇒ 9? = 18

⇒ ? = 2 ( Middle number in 3rd figure )

Therefore option (C) is correct option

Problem # 5

This problem have three figures and every figure consist of one number in the middle of the figure and four numbers around it .The product of middle number's digits is equal to the sum of the remaining numbers/digits around the figure.

7 + 4 + 7 + 12 = 30 = 5 * 6 = Product of middle number 56 in 1st figure.

15 + 7 + 12 + 8 = 42 = 6 *7 = Product of middle number 67 in 2nd figure

11 + 18 + 5 + ? = 36 = 6 * 6 = Product of middle number 66 in 3rd figure

⇒ 34 + ? = 36

⇒ ? = 36 - 34

⇒ ? = 2

Therefore option (1) is correct option

Problem # 6

This circle consists of four quadrants and every quadrant consists of three numbers , Every quadrant have two numbers in outer part and one number in its inner part . To find the value of question mark "?" , we shall use two numbers which are in the outer part of it to calculate the value of the number which is in the inner part of every quadrant .

4th Quadrant

( 7 × 4 )/4 = 28/4 = 7 ( Middle number in inner part)

3rd Quadrant

( 8 × 5 )/4 = 40/4 = 10 ( Middle number in inner part)

2nd Quadrant

( 4 × 20 )/4 = 80/4 = 20 ( Middle number in inner part)

1st Quadrant

( 6 × 8 )/4 = 48/4 = 12 ( Middle number in inner part)

Therefore option (4) is correct option

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Problem # 7

This problem have three figures and 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.

H C F ( Highest Common Factor ) of 12 ,18 , 30 = 6 (1st figure )

H C F ( Highest Common Factor ) of 16 ,32 , 40 = 8 (2nd figure )

H C F ( Highest Common Factor ) of 36 ,18 , 27 = 9 ( 3rd figure )

Therefore option (3) is correct option.

Problem # 8

Starting from 2 and moving clockwise multiplying two opposite numbers which will contribute 56 in each case.

2 * 28 = 56

14 * 4 = 56

Similarly the product of 7 and ? must be equal to 56.

7 * ? = 56

⇒ ? = 56/7 = 8

Therefore option (A) is correct option.

Problem # 9

This problem consist of three figures and every figure have three numbers out of which two numbers are smallest than 3rd number. So starting from smallest number and proceeding anticlockwise likewise.

Sq of 2 + Sq of 4 = 4 + 16 = 20 ( 1st figure)

Sq of 3 + Sq of 9 = 9 + 81 = 90 (2nd figure)

Sq of 1 + Sq of 5 = 1 + 25 = 26 (3rd figure)

Therefore option (3) is correct option.

Problem # 10

This circle have following numbers in it. 40 , 45, 25 , 40 , 30, 35 , 35 , ? .

Now after carefully studying these numbers, we find that these numbers are written in two alternate series . So starting from number 25 clockwise pick alternate numbers..

Our 1st pattern from these numbers 40 , 45 , 25 , 40 , 30, 35 , 35 , ? will be 25, 30 , 35 , 40 and these numbers are written with an increment of 5 . It means that the difference/increment between two consecutive numbers is same

25 + 5 = 30

30 + 5 = 35

35 + 5 = 40

After eliminating above numbers remaining numbers are 45 , 40 , 35 , ? .

Now note the difference between every two consecutive numbers is 5.

45 - 40 = 5

40 - 35 = 5

So 35 - ? = 5

⇒ -? = 5 - 35

⇒ -? = -30

⇒? = 30

Therefore option (1) is correct option .

Ten Most Important Reasoning questions with answers for competitive exams of circles, box and other type with solutions 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 .

Ten Most Important Reasoning questions with answers for competitive exams