Missing number in box Reasoning problem, How to solve various circle problems

Reasoning of missing number in box problems, circle problems  and triangle problems will be discussed with the help of 10 most important examples. Some of these examples are of  4 × 4  order and other are of  4 × 3 orders. 

Reasoning of missing number in box problems in Uniteted State

PROBLEM #  1

Formula

Every row has been written as the cube of certain number. Because 4096 is the cube of 18, 6859 is the cube of 19 , 8000 is the cube of 20  similarly 926? will also be  the cube of any number. Look carefully 1st row is the cube of 18, then 2nd row is the cube of 19, 3rd row is the cube of 20 . So in this way 4th row must be  the cube of 21.

Calculation

18³ = 4096 

19³ = 6859 

20³ = 8000

21³ = 9261

Hence value of question mark will be 1.

Option (D)1 is correct option.

PROBLEM #  2 

  

Formula

Multiply the number in 2nd row with 100 and add it to the product of numbers in 1st and 2nd rows.

Calculation

6*100 + (4*8) = 632

3*100 + (9*2) = 318

7*100 + (5*9) = 745

 Option (A)745 is correct option.

PROBLEM #  3 

This reasoning problem consists of three figures and every figure have three numbers associated to it . Two numbers are on the upper line of each box and one number is at the bottom of the dark box.  Look at last figure , it have ? in its centre . So the solution of this problem is to find the value of question mark using two numbers associated to it . 

          But the main problem is how to utilised  these two numbers to get the value of question mark?

          Now watch carefully the 1st two figures . Since these figures have some values of numbers in dark box . 

          Now we have to find or search the  formula for these three numbers in each figure to utilised them in any possible way to get number in  dark box . 

         The same formula will be applicable to third figure to find out the value of question mark.

Formula :- The DIFFERENCE of squares of both the numbers in empty box in upper line in every figure is equal to  number in dark box in lower line.

Solution:- 

9²  - 6² = 81 - 36 = 45 (1st Figure)

8²  - 2² = 64 - 4 = 60 (2nd Figure)

7²  - 3² = 49 - 9 = 40  (3rd Figure)

Option (1)40  is correct option.

PROBLEM #  4 

This reasoning problem consists of three figures and every figure have three numbers associated to it . Two numbers are on the upper line of each box and one number is at the bottom of the dark box.  Look at last figure , it have ? in its centre . So the solution of this problem is to find the value of question mark using two numbers associated to it . 

          But the main aim is how to utilised  these two numbers to get the value of question mark?

          Now watch carefully the 1st two figures . Since these figures have some values of numbers in dark box . 

          Now we have to find or search the  formula for these three numbers in each figure to utilised them in any possible way to get number in  dark box . 

         The same formula will be applicable to third figure to find out the value of question mark.

Formula :- The SUM of squares of both the numbers in empty box in upper line in every figure is equal to  number in dark box in lower line. 

Solution:- 

9²  + 6² = 81 + 36 = 117 (1st Figure)

8²  + 2² = 64 + 4 = 68 (2nd Figure)

7²  + 3² = 49 + 9 = 58 (3rd Figure)

Option (1)58 is correct option

PROBLEM #  5 

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 square roots of  two adjacent numbers then we shall have the number attached to big square between these two numbers whose square root had been multiplied.

√49  ×  √4 =  7  × 2 = 14 (The number at bottom line in the box)

√4  ×  √36 =  2  × 6 =  12 ( The number at leftmost box )

√36  ×  √? =   6  × √? = 30 ( The number at uppermost box )

√?  = 30/6 = 5, squaring both sides

? = 25 ( The number at uppermost box)

Similarly

√?  ×  √49 = 35 

√?  = 35/7

√?  = 5,  squaring both sides

? = 25  (The number at rightmost box )

 Option (A)25  is correct option

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PROBLEM # 6

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 square 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.

∛343  ×  ∛27 =  7 × 3 = 21( The number at rightmost box )

∛27  ×  ∛125 =  3  × 5 = 15 (The number at bottom line in the box)

∛125  ×  ∛? =  5  ×   ∛? = 10 ( The number at leftmost box )

5  ×  ∛? = 10 

 ∛?  = 10/5

 ∛? = 2 , cubing both sides

? = 8

Similarly

∛?  ×  ∛343 =   ∛?  × 7  = 14

 ∛? = 14/7 = 2 cubing  both sides

? = 8 ( The number at uppermost box )

Option (C)8  is correct option

PROBLEM #  7 

This reasoning problem have two figures and every figure have five number attached to it . Four numbers are at the corner of each figure and one number is at the middle of the same figure.

       Since second figure have question mark in its centre . So the solution of this problem is to find the value of question mark using four numbers associated to it . 

          But the main problem is how to utilised these four numbers to get the value of this question mark?

          Now watch carefully the 1st two figures . since these figures have some values of middle numbers. 

          Now we have to find or search the  formula for these four numbers in each figure to utilised them in any possible way to get middle or central number. 

         The same formula will be applicable to third figure to find out the value of question mark.

Formula:-   Add all the numbers in the corner of each figure  to get value of the number in the middle.

Solution:-

√4 + √16 + √9 + √25 + √4 = 2 + 4 + 3 + 5 = 14 (Middle number in 1st figure).

√9 + √49 + √36 + √36 + √1 = 3 + 7 + 6 + 1 = 17 (Middle number in 1st figure).

Option (3)17  is correct option

Reasoning of missing number in circle problems

PROBLEM #  8

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

1st Quadrant 

Step 1.   Take sum of the squares of both the numbers (8 and 4 ) in outer part of  this quadrant.

Step 2. Take the magnitude of difference of both the numbers in outer part of  this quadrant.

Step 3. Divide the result obtained in step 1 with the result obtained in step 2 to get the value of the number (20) in inner part of this quadrant.

Calculation

(8² + 4² ) ÷ ( |8 - 4 |) = ( 64 + 16 ) ÷ 4 = 80 ÷ 2 = 20

2nd Quadrant

Step 1.   Take sum of the squares of both the numbers ( 7 and 5 ) in outer part of  this quadrant.

Step 2. Take the magnitude of difference of both the numbers in outer part of  this quadrant.

Step 3. Divide the result obtained in step 1 with the result obtained in step 2 to get the value of the number (37) in inner part of this quadrant.

Calculation

(5² + 7² ) ÷ (|5 - 7 |) = ( 25 + 49 ) ÷ 2 = 74 ÷ 2 = 37

3rd Quadrant

Step 1.   Take sum of the squares of both the numbers ( 10 and 6 ) in outer part of  this quadrant.

Step 2. Take the magnitude of difference of both the numbers in outer part of  this quadrant.

Step 3. Divide the result obtained in step 1 with the result obtained in step 2 to get the value of the number (34) in inner part of this quadrant.

Calculation

(10² + 6² ) ÷ ( |10 - 6| ) = ( 100 + 36 ) ÷ 4 = 136 ÷ 4 = 34

4th Quadrant

Step 1.   Take sum of the squares of both the numbers ( 10 and 6 ) in outer part of  this quadrant.

Step 2. Take the magnitude of difference of both the numbers in outer part of  this quadrant.

Step 3.  Divide the result obtained in step 1 with the result obtained in step 2 to get the value of the number (34) in inner part of this quadrant.

Calculation

(6² + 4² ) ÷ ( |6 - 4 |) = ( 36 + 16 ) ÷ 2 = 52 ÷ 2 = 26

Option (3)26 is right option

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PROBLEM #  9 

This  big circle is  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 adjacent  numbers then reverse of order of the result so obtained then this number will be equal to the number in small circle opposite to both these numbers .

8  +  6  = 14 ⟺  41 ( Reversing the order of digits) 

6 + 11 = 17 ⟺  71  ( Reversing the order of digits)

11 + 10 =  21 ⟺ 12 ( Reversing the order of digits)

10 + 31 = 41 ⟺ 14 ( Reversing the order of digits)

31 + 12 = 43 ⟺ 34 ( Reversing the order of digits)

12 + 17 = 29 ⟺ 92 ( Reversing the order of digits)

17 + 5  = 22 ⟺ 22 ( Reversing the order of digits)

8  + 5 = 13  ⟺  31 = ? ( Reversing the order of digits)

Hence the value of  "? " will be 31 

Option (B)13 is right option

Reasoning of missing number in triangle problems

PROBLEM #  10 

This reasoning problem consists of three figures and every figure have three numbers associated to it . So the solution of this problem is to find the value of question mark using other two numbers associated to it . 

          But the problem is how to utilised these two numbers to get the value of this question mark?

          Now we have to find or search the  formula for these two numbers in each figure to utilised them in any possible way to get third number. 

         The same formula will be applicable to third figure to find out the value of question mark.

Formula :- One tenth of the product of  two outer numbers in each triangle is equal to third number .

 1st Triangle

( 5 × 4 ) ÷ 10 = 20 ÷ 10 = 2 (Middle number in 1st triangle ) 
2nd Triangle

( 6 × 5 )  ÷ 10 = 30 ÷ 10 = 3 (Middle number in 2nd triangle ) 
3rd Triangle 

( 15 × 6 )  ÷ 10 = 90 ÷ 10 = 9 (Middle number in 3rd triangle ) 

Option (3)9  is correct option

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

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Missing number Reasoning SSC CGL questions and answers

Ten most Missing number Reasoning SSC CGL questions and answers for competitive exams, reasoning questions with answers for bank exams, Reasoning tricks for competitive exams have been discussed in this article for upcoming competitive exams. 

10 Most Important Reasoning Problems for competitive Exams

Question #1

Misiing Number 

Since the question mark is in the 3rd figure. So the value of question mark will be calculated with the help of remaining four numbers in the 3rd figure in the same way, the middle numbers have been calculated in 1st and 2nd figure with the help of remaining four numbers. This question mark value can be found by this method.   

Formula:-

Sum of four numbers = Product of both the digits in middle number.

Calculation:-

7 + 4 + 7 + 12 = 30 = 5 × 6  = Multiplication  of 5 and 6 (Middle number in 1st figure ) 

15 + 7 + 12 + 8 = 42 = 6 × 7 = Multiplication  of 6 and 7 (Middle number in 2nd figure ) 

11 + 18 + 5 + ? =  ????  = 6 × 6 = Multiplication  of 6 and 6(Middle number in 3rd figure ) 

⇒ 34 + ? = 36

⇒  ? = 36 - 34

⇒  ? = 2

 Hence option (1)2 is correct.

Question #2

Since the question mark is in the 3rd figure. So the value of question mark will be calculated with the help of remaining four numbers in the  figure in the same way, the middle numbers in 1st and 2nd figure can be calculated with the help of remaining four numbers around each figure. This question mark value can be found by this method.

1st Circle

8 + 4 + 3 + 1 = 16 (Greatest number in 1st circle) 

or 

16 - (8 + 4 + 3) = The number in the middle of circle

16 - 15 = The number in the middle of circle

1  =  The number in the middle of circle

2nd Circle

4 + 3 + 5 + 8 = 20 (Greatest number in 2nd circle)

or 

20 - (4 + 3 + 5) = The number in the middle of circle

20 - 12 = The number in the middle of circle

8  =  The number in the middle of circle

3rd Circle

6 + 4 + 5 + ? = 18 

⇒ 15 + ? = 18

⇒ ? = 18 - 15

⇒ ? = 3 (Greatest number in 3rd circle)

This will be the value of question mark.

or 

18 -  (6 + 4 + 5 ) = The number in the middle of circle

18 - 15 = The number in the middle of circle

3 =  The number in the middle of circle

Hence option (1)3 is correct.

 

Question #3

Starting from 4th quadrant in anticlockwise direction , add one to the sum of squares of both the numbers in outer part of the quadrant to get the value of the number in the inner part of same quadrant.

4th Quadrant

(2 + 8)² + 1 =  10²  + 1 =  100 + 1 = 101

1st Quadrant

(3 + 8)² + 1 =  11²  + 1 = 121 + 1 = 122

2nd Quadrant

(7 + 1)²  + 1 =   8² + 1 = 64 + 1 = 65

3rd Quadrant

(5 + 4)² + 1 =   9²  + 1 = 81 + 1 = 82 (The value of question mark)

Hence option (1)82  is correct.

Question #4

Starting from 4th quadrant in anticlockwise direction , Subtract one from the sum of squares of both the numbers in outer part of the quadrant to get the value of the number in the inner part of same quadrant.

4th Quadrant

(2 + 8 )² - 1 =  10²  + 1 =  100  - 1 = 99

1st Quadrant

(3 + 8)² - 1 =  11²  + 1 = 121 - 1 = 120

2nd Quadrant

(7 + 1 )²  - 1 =   8² + 1 = 64 - 1 = 63

3rd Quadrant

(5 + 4 )² - 1 =   9²  + 1 = 81 - 1 = 80 (The value of question mark)

Hence option (2)80  is correct.

Question #5

Since the question mark is in the 3rd figure. So the value of question mark will be calculated with the help of remaining four numbers in the  figure in the same way, the middle numbers in 1st and 2nd figure can be calculated with the help of remaining four numbers around each figure. This question mark value can be found by following method.

1st Figure

63 + 34  +  8  = 105 (Greatest number in 1st figure) 

or 

105 - (63 + 34) = The number in the middle of 1st figure

105 - 97 = The number in the middle of 1st figure

8  =  The number in the middle of 1st figure

2nd Figure

30 + 10 + 5  =  45 (Greatest number in 2nd figure) 

or 

45 - (30 + 10) = The number in the middle of 2nd figure

45 - 40 = The number in the middle of 2nd figure

5  =  The number in the middle of 2nd figure

3rd Figure

31 + 62 + ? = 121 This implies ? = 28 (Greatest number in 3rd figure) 

or 

121 - (62 + 31) = The number in the middle of 3rd figure

121 - 93 = The number in the middle of 3rd figure

28  =  The number in the question mark in 3rd figure

Hence option (4)28  is correct.

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Question #6

This circle has been divided into four quadrants and each quadrant is composed of three numbers . In each quadrant two numbers are in outer quadrant of the circle and one number is in the inner quadrant of the circle. 

Starting from fourth quadrant in clockwise direction. 

Formula : - Half of sum of the squares of both the numbers in outer sector is equal to inner sector.

 4th Quadrant

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

 3rd Quadrant

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

 2nd Quadrant

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

 1st Quadrant

( 9² + 11²)/2 = ( 81 + 121)/2 =  202/2  = 101 (Number in the inner part of the circle in 1st quadrant  ).

And 101 will be the value of question mark.

Hence option (4)101  is correct.

Question #7

This circle has also been divided into four quadrants and each quadrant is composed of three numbers . In each quadrant two numbers are in outer sector of the circle and one number is in the inner sector of the circle. 

Starting from 1st quadrant in clockwise direction. 

Formula : -  Sum of the squares of both the numbers in outer sector is equal to inner sector.

  1st Quadrant

(3 + 5)² = (8)² = 64  (Number in the inner part of the circle in 1st quadrant )

 2nd Quadrant

(5 + 4)² = (9)² = 81  (Number in the inner part of the circle in 2nd quadrant )

 3rd Quadrant

(4 + 2)² = (6)² = 36  (Number in the inner part of the circle in 3rd quadrant )

 4th Quadrant

(6 + 4)² = (10)² = 100  (Number in the inner part of the circle in 4th quadrant ).

And 100 will be the value of question mark.

 

Question #8

This box problem consists of three rows and three columns. To find the value of question marks in 3rd column of 3rd row, we have to add 1st ,2nd and 3th columns since since in every column is same in every row.

Formula :- The sum of 1st three columns in every row is same .

7 + 8 + 15 = 30 ( 1st  row) 

9 + 20 + 1 = 30 (2nd row) 

14 + 2 + ? = 30 ( Because sum of all the three digits in 3rd row must be equal to the sum in other two rows)

Hence  16 + ? = 30

⇒ ? = 30 - 16 

⇒  ? = 14

Therefore the value of question mark is 14

Hence option (B)14  is correct.

Question #9

This box problem consists of four rows and four columns. To find the value of question marks in 3rd column of 3rd row, we have to divide the sum of  1st ,2nd and 3th rows with 4 to get the value of number in 4th row.

Formula :- The one fourth of  the sum of 1st three rows in every column.

5 + 8 + 7  = 20 ÷ 4 = 5 ( 1st column  in 4th row) 

9 + 6 + 13 = 28 ÷ 4 = 7 (2nd column in 4th row) 

(7 + 10 + 19) ÷ 4  = 36 ÷ 4 = 9 (4th column in 4th row) 

(8 + 9 + ?) ÷  4 =  8  (3rd column in 4th row) 

⇒ 17 + ?  =  8  ×  4

⇒ ? = 32 - 17

⇒ ? = 15

Hence option (D)15  is correct.

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Question #10

This box problem consists of four rows and four columns. To find the value of question marks in 4th column of  fourth row, we have to multiply 1st ,2nd and 3th rows  at the same time and then add 10 to the result so obtained. 

Formula :- Ten more than the product of 1st three rows in every column.

(5 × 6 × 7 )  + 10 = 210 + 10 = 220 (1st column  in 4th row) 

(6 × 5 × 4 )  + 10  = 120 + 10 = 130 (2nd column in 4th row)

(3 × 4 × 5 )  + 10  =  60 + 10 = 70 (3rd column in 4th row) 

(8 × 7 × 6 )  + 10  = 336 + 10 = 346 (4th column in 4th row) 

Hence option (B)346  is correct.

In this post I discussed Ten Missing number Reasoning SSC CGL all questions and answers . Comment your valuable suggestion for further improvement.

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Ten Logical Reasoning questions and answers for competitive exams

Ten Logical Reasoning questions and answers for competitive exams  like Bank PO, Bank clerk, SSC CGL, ssc chsl, RRB NTPC , group D etc have been discussed in this post in very easy style and shortcut way .

Logical Reasoning questions and answers for competitive Exams 

Problem # 1

All the letters in each option are written from alphabet either in order or in reverse order.

From 1st option MNOP are written in  order starting from M in alphabet. 

From 2nd option TSRQ are written in reverse order starting from Q  in alphabet. i.e.  

QRST  ⇒ QRST   ⟺  TSRQ (After reversing the orders of letters).

From 3rd option  EDCB are written in reverse order starting from B  in alphabet. i.e.  

BCDE  ⇒ BCDE   ⟺  EDCB (After reversing the orders of letters).

 From 4th option  XWVU are written in reverse order starting from U in alphabet. i.e.  

UVWX  ⇒ UVWX  ⟺ XWVU ( After reversing the orders of letters).

Since  2nd,  3rd and 4th options are following the same pattern because all the letters in these options are written in order. But letters in 1st option are  written in reverse order. And this option is odd one.

Hence Option (1)MNOP is correct option.

Problem # 2

All the letters are written alternatively from alphabet either in order or in reverse order.

From 1st option MKIG are written in reverse order starting from G with a gap of one letter in alphabet. i.e.  

G H I J K L M ⇒ G H I J K L M ( Omitting blue colored letters )  ⇒ G I K M ⟺ MKIG (After reversing the orders of  letters).

From 2nd option VTRP are written in reverse order starting from P with a gap of one letter in alphabet. i.e.  

PQRSTUV ⇒  PQRSTUV (Omitting blue colored letters)  ⇒  PRTV ⟺ VTRP (After reversing the orders of letters).

From 3rd option EGIK are NOT written in reverse order starting from K with a gap of one letter in alphabet. 

From 4th option YWUS are written in reverse order starting from S with a gap of one letter in alphabet. i.e.  

STUVWXY ⇒  STUVWXY (Omitting blue colored letters)  ⇒  SUWY ⟺ YWUP (After reversing the orders of letters).

Since 1st , 2nd and 4th options are following the same pattern because all the letters in these options are written in reverse order with a gap of one letter. But letter in 3rd option are not written in reverse order.

Hence Option (3)EGIK is correct option.

Problem # 3

In this reasoning problem 1st number (5) is associated to 100 with the help of any rule , in the same rule we have to associate 7 to a number out of four given options.

Look carefully the given numbers consists of three digits. These three digits can be utilised with the help a formula given below.

Formula :-  4 × (1st number)²  = 2nd Number

 4 × (3rd number)²  = 4th Number

⇒ {5} = 4 × 5² = 4 × 25  = 100 (2nd number)

⇒{7} = 4 × 7² = 4 × 49 = 196 (4th number)

Option (4)196 is correct option.

Problem # 4

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

Replacing - sign with + , + sign with × in above conditions . Now place 1st two numbers in bracket . Then apply the BODMAS rule.

(7 + 2) × 4 = 9 × 4 = 36 

(10 + 5) × 2 = 15 × 2 = 30

Similarly 

(20 + 5) × 6 = 25 × 6 = 150

Option (4)150 is correct option.

Problem # 5

In this reasoning problem 1st number (392) is associated to 28 with the help of any rule , in the same way we have to associate 722 to a number out of four given options. 

Formula :- 2 × √(1st Number ÷ 2) = 2nd Number

2 × √(392 ÷ 2) = 2 × √(196)  = 2  × 14 = 28 . 

Similarly 

2 × √(722 ÷ 2) = 2 × √(361)  = 2  × 19 = 38

Option (3)38 is correct option.

Problem # 6

 

These words are used when a book is prepared, published and read by Reader.   To publish a book , first of all it has to be written by the Author . After this book will be with the Editor for any change or editing. Then that book will be published by Publisher. After publication it will be sold by Bookseller. And at last it will be purchased by read by Reader.

The journey of the book is as follow

1. Author
2. Editor
3. Publisher
4. Bookseller
5. Reader

So according to given problem's solution the correct option will be  (D) 4 , 1 , 3 , 5 , 2

Problem # 7

Divide the sum of both the numbers with 10 to get the number in right hand side.

( 15 + 55 ) ÷ 10 = 70 ÷ 10 = 7

( 25 + 35 ) ÷ 10 = 60 ÷ 10 = 6

( 35 + 45 ) ÷ 10 = 80 ÷ 10 = 8

Similarly ( 23 + 17 ) ÷ 10 = 40 ÷ 10 = 4

Hence option (3)4 is correct.

Problem # 8

In this reasoning problem three out of four options have been calculated by Multiplying  the 1st number with its next number(successive number) to get 2nd number. This rule is applicable to only three out of four option. And one which do not follow this rule will be the correct option.

(1)   9 × ( 9 + 1 ) = 9 × 10 = 90

(2)   13 × ( 13 + 1 ) = 13 × 14 = 182

(3)   8 × ( 8 + 1 ) = 8 × 9 = 72

(4)   12 × ( 12 + 1 ) = 12 × 13 = 156, but this is given as 144

Hence option (4)12 - 144  is odd one.

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Problem # 9

Take the difference of two successive numbers, this difference is square of continuous natural numbers.

6  -  5 = 1 = 1²

10 -  6 = 4  = 2²

19 -  10 = 9  = 3²

35 - 19 = 16  = 4²

Similarly the difference of last number and 2nd last number can be calculated. This difference will be the square of some natural number.

?  - 35 = 25  = 5²

?  = 25 + 35  = 60

Hence option (2)60  is correct.

Problem # 10

Take the difference of two successive numbers, this difference is cube of continuous natural numbers.

3085 - 1357 = 1728 = 12³ 

5282 - 3085 = 2197 = 13³

8026 - 5282 = 2744 = 14³

Similarly the difference of last number and 2nd last number can be calculated. This difference will be the cube of some natural number.

? -  8026 = 3375 = 15³

? = 3375 + 8026 = 11401

Hence option (4)11401 is correct.

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Conclusion

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