Maths logical reasoning questions with Answers

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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

This reasoning problem is a box problem of 4 rows and 3 columns, where the the value of question mark in last row we have to find.

Formula :-

Sum of the squares of 1st two columns = Squares of 3rd column

1st Row

3² + 4² = 5²

➡ 9 + 16 = 25

➡ 25 = 25

L H S = R H S

2nd Row

6² + 8² = 10²

➡ 36 + 64 = 100

➡ 100 = 100

L H S = R H S

3rd Row

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

Formula :-

Five less than the product of sum of two corner's numbers in each rectangle.

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.

Problem #4

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.

Formula :- Sum of all the numbers in a square = 35

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

1st Quadrant:-

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

2nd Quadrant:-

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

3rd Quadrant:-

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

4th Quadrant:-

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

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Problem #8

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

Formula :-

Sum of the squares of numbers in 1st row = Number in 2nd row

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

In these types of reasoning questions either we can check the sum or product of all the numbers in any rows or column. If the sum or product of all the rows or columns is found to be same except the row or column in which question mark lies , then we can choose the value of question mark out of four given options to get the answer.

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

Formula :- b² - ( 1st side² - 2nd side²) = Middle number

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 suggestions 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.

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Most important reasoning questions for competitive exams using reasoning tricks

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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)

Option (2)154 is correct option

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.

These are given terms of series.
57, 56, 55, 51, 43, 34, ?
Differences of two consecutive terms can be calculated like this
57 - 56 = 1
56 - 55 = 1
55 - 51 = 4
51 - 43 = 8
43 - 34 = 9
34 - ? =
Now study the differences
1, 1, 4 , 8 , 9 , ....
This can be written as
1², 1³, 2², 2³, 3²,.....
In this pattern next difference would be 3³ and 4²
So next number would be 34 + 3³ = 34 + 27 = 61
And similarly next number would be 61 + 4² = 61 + 16 = 77
So question marks? = 61
Option (3)61 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

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

Problem #8

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

2nd quadrant

( 7 + 2 )² + 7² = 9² + 7² = 81 + 49 = 130 ( Number in the outer part of opposite sector)

1st quadrant

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

3rd quadrant

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

4th quadrant

( 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.

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Ten Latest tricky and logical Questions in box and circle reasoning for SSC CGL And SSC CHSL Exams

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Ten Most Important tricky and logical 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 .

Problem #1

This circle is divided into eight parts and each part is composed of three numbers . In each 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.

1st Method

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)

2nd Method

(8² + 6² ) ÷ 2 = ( 64 + 36 ) ÷ 2 = 100 ÷ 2 = 50 (Number in the inner part of the circle in 4th quadrant )

(5² + 7² ) ÷ 2 = ( 25 + 49 ) ÷ 2 = 74 ÷ 2 = 37 (Number in the inner part of the circle in 3rd quadrant )

(8² + 10² ) ÷ 2 = ( 64 + 100 ) ÷ 2 = 164 ÷ 2 = 82 (Number in the inner part of the circle in 2nd quadrant )

(9² + 11² ) ÷ 2 = ( 81 + 121 ) ÷ 2 = 202 ÷ 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 be equal to the number 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

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 centre of the 1st triangle)

In 2nd triangle

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

In 3rd triangle

(9 × 2 × 3 ) - ( 9 + 2 + 3 ) = 54 - 14 = 40 ( Number in centre 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)

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 )

Therefore Correct option is (4)

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Ten Most Important tricky and logical 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.

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