Ten most expected missing number series questions for SBI PO with solution for other competitive Exams

Maths    Reasoning   

Missing  number series questions for SBI PO with solution 

Ten most expected  missing  number series questions for SBI PO with solution , for SSC GL , SSC CHSL , RRB NTPC and other competitive Exams  have been discussed in this post.  These types of series questions are asked in many other competitive exams like SI, CPO and various entrance exams. 

Problem # 1

Take the difference of two consecutive terms of the given series. This difference will be the pair of same odd numbers in every case.

16 - 5 = 11 (The difference of 1st and 2nd number is pair of odd number one)
49 - 16 = 33  (The difference of 2nd and 3rd number is pair of odd number three )
104 -  49 = 55  (The difference of 3rd and 4th number is pair of odd number  five)
? - 104 = 77   ( Similarly the difference of 4th and 5th number will pair of odd number seven)
⇒ ? =  77 + 104  (And in the same way the difference of 5th and 6th number is pair of odd number nine )
⇒ ? = 181
280 - ? = 99  
⇒ ? = 280 - 99 
⇒? = 181

Option (2)181 will be correct option

Problem # 2

In this series every number is written as power of one less than the term,s position and taking as base of  5 . i.e. 1st term is written as power of 0 , 2nd term is written as power of 1, 3rd term is written as power of 2 , 4th term is written as power of 3 and so on.

5⁰ - 0 = 1 - 0 = 1  (Power 0 of base 5 - same power )

5¹ - 1 = 5 - 1 = 4   (Power 1 of base 5  - same power   )

5² - 2 = 25 - 2 = 23   (Power 2 of base 5  - same power  )

5³ - 3 = 125 - 3 = 123   (Power 3 of base 5  - same power )

5⁴ - 4 = 625 - 4 = 621   (Power 4 of base 5  - same power  )

Option (4)621 is correct option

Problem # 3

In this series every number is written as power of one greater than the term,s position and taking as base of  5 . i.e. 1st term is written as power of 0 , 2nd term is written as power of 1, 3rd term is written as power of 2 , 4th term is written as power of 3 and so on.

5⁰ + 0 = 1 + 0 = 1  (Power 0 of base 5 + same power )
5¹ + 1 = 5 + 1 = 6   (Power 1 of base 5  + same power   )
5² + 2 = 25 + 2 = 27   (Power 2 of base 5  + same power  )
5³ + 3 = 125 + 3 = 128   (Power 3 of base 5  + same power )
5⁴ + 4 = 625 + 4 = 629   (Power 4 of base 5  + same power  )

Option (2)629 will be correct option.

Problem # 4

In this series every number is written as cube of  the term,s position and one number less than it . i.e. 1st term is written as cube of 1 and less than 1 , 2nd term is written as cube of 2 and less than 1, 3rd term is written as cube of 3 and one lees than it , 4t term is written as cube of 4 and one less than it and so on.

 1³ - 1 = 1 - 1 = 0

 2³ - 1 = 8 - 1 = 7

 3³ - 1 = 27 - 1 = 26

 4³ - 1 = 64 - 1 = 62

 5³ - 1 = 125 - 1 = 124

 6³ - 1 = 216 - 1 =  215

Option (1)124 will be correct option.

Problem # 5

(5  × 1)  + 2 = 5  +  2 = 7 ( To get 2nd term Multiplying 1st term with 1 and add 2 to it)

(7 × 2)  -  4 = 14 - 4  = 10 ( To get 3rd term Multiplying 2nd term with 2 and subtract 4 to it ). 

(10 × 3) + 6 = 30 + 6 = 36 ( To get 4th term Multiplying 3rd term with 4 and add 6 to it) 

(36 × 4)  - 8 = 144 - 8 = 136 ( To get 5th term Multiplying 2nd term with 2 and subtract 4 to it )

(136 × 5)  + 10 = 680 +10= 690 ( To get 2nd term Multiplying 1st term with 1 and add 2 to it)

Option (2)690 will be correct option.

Problem # 6

To get 2nd term ,multiply 1st term with half and add cube of 1 to it .

(12 × 0.5 ) + 1 = 6 + 1 =  6 +  1³ = 6 + 1 = 7

To get 3rd term ,multiply 2nd term with one ( double of half ) and add cube of 2 to it .

 (7 × 1)  + 8 = 7 +  2³ =  7 + 8 = 15 

To get 4th  term ,multiply 3rd term with 2 ( double of 1 ) and add cube of 3  to it .

(15 × 2) +  27 =  30 +  3³ = 57

To get 5th  term ,multiply 4th term with 4 ( double of 2 ) and add cube of 4  to it .

( 57 × 4 ) + 64 =  228 +  4³ = 292

To get 6th  term ,multiply 5th term with 8 ( double of 4 ) and add cube of 3  to it .

( 292 × 8 ) + 125 = 2336 + 125 = 2461

option (4)2461 is correct option

Problem # 7

Multiply 1st term of the series with one and half  (1.5) to get 2nd term of the series.

10 × 1.5 = 15

Multiply 2nd term of the series with 3  (  double the 1.5 i.e.  2 × 1.5 = 3 ) to get 3rd term of the series.

15 × 3 = 45

Multiply 3rd term of the series with  6 (  double the 3 i.e. 3 × 2 = 6) to get 4th  term of the series.

45 × 6 = 270

Multiply 4th term of the series with 12  ( double the 6 i.e. 6 × 2 = 12 ) to get 5th term of the series.

270 ×  12  = 3240

Multiply 5th term of the series with 24 (  double the 12 i.e. 12 × 2 = 24 ) to get 2nd term of the series.

3240 × 24 =  77760 

Option (2)77760 is correct option

Problem # 8

(2 × 5) - 1 = 10 - 1 = 9  ( 2nd term is written as multiple of 5 with 1st term  and decrease it by 1 )

(9 × 5) + 3 =  45 + 3 = 48  ( 3rd term is written as multiple of 5 with 2nd term  and increase it by 3)

(48  × 5) - 5 = 240 - 5 =235  ( 4th term is written as multiple of 5 with 3rd term  and decrease it by 5)

( 235 × 5 ) + 7 = 1185 + 7 = 1182  ( 5th term is written as multiple of 5 with 4th term  and decrease it by 7 )

Note carefully in this problem ,operation of decreasing and increasing used alternatively.

Option (2)1182 is correct option

Also read these posts on Reasoning

1.Reasoning for competive exams
2 .Box and circle reasoning
3 . Reasoning for bank exams
4. Ten Tricky logical reasoning
5. Missing  number series questions
6. Reasoning questions with answers
7. Circle Reasoning
8.  Box Problems
9. 15 Questions Circle Problems 
10.  SSC CGL Reasoning

 

Problem # 9

In this series 1st term is 12 , To get 2nd term from it we just multiplied it with half 

12 × 1/2 =  6

To get 3rd term from 2nd term we just multiplied it with double of the number what we have multiplied in last term i.e. 2 × ( 1/2) =1 
6  × 1 = 6 
To get 4th term from 3rd term we just multiplied it with double of the number what we have multiplied in last term i.e. 2 × ( 1) =2 
6 × 2 = 12 
To get 5th term from 3rd term we just multiplied it with double of the number what we have multiplied in last term i.e. 2 × ( 2 ) =4 
12 × 4 = 48 
To get 6th term from 5th term we just multiplied it with double of the number what we have multiplied in last term i.e. 2 × ( 4 ) =8 
48 × 8 = 384
Hence (2)384 will be correct option

Problem # 10

To find out the solution of  above series problem just add all the digits of the every number given , this sum will leads to the next number of the series

5 + 3 + 4 + 2 = 14 ( 2nd number in series) 

Similarly add all the digits of the 3rd number, this sum will leads to the 4th number of the series

5 + 2 + 3 + 1 = 11 ( 4th number in series) 

4 + 1 + 2 + 0 = 17 ( 6th number in series) 

6 + 7 + 3 + 2 = 18  ( last/ Missing number in series) 

Option (4)18 is correct option

Also Reads these posts on Reasoning
ReasoningProblems #11  
ReasoningAnalogy #10
Reasoning Questions #9    
Circle Reasoning #8
TenBox-Problems #7   
TenReasoning problems #6
Ten-Important-Problem#5  
Ten--Tricky-Puzzles #4
Twelve-Figures-Problems#3  
Ten-Important-Reasoning#2  
Picture-Reasoning#1

Ten most important Missing number series questions and answers, Number Series Questions for SBI Clerk ,SBI PO with solution  for SSC GL , SSC CHSL , RRB NTPC and other competitive Exams were discussed in this post.  comment your valuable suggestions regarding this post and for further improvement.

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Ten Most Important Reasoning questions with answers for competitive exams

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

Exam cracker

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

ReasoningProblems #11          ReasoningAnalogy #10

Reasoning Questions #9         Circle Reasoning #8

TenBox-Problems #7           TenReasoning problems #6

Ten-Important-Problem#5     Ten--Tricky-Puzzles #4

Twelve-Figures-Problems#3   Ten-Important-Reasoning#2     

Picture-Reasoning#1

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 .

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Ten Most Important Reasoning Problems of Circles with Solutions, How to Solve Reasoning Circle Problems

Ten Most  Important Reasoning problems of circles with solutions. 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 Problems of Circles with Solutions.

      Problem # 1

1st circle

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

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

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 # 2

This circle consists of four quadrants and every  quadrant consists of three numbers , And every quadrant have two numbers in outer part and  1 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 

 8² ×  3⁴  =  64 × 81  = 5184

Now add all these digits  5 + 1 + 8 + 4 = 18

Now reversing the order of these digits obtained in  previous step

18   <----> 81 The number in the inner part

2nd Quadrant

  5² × 7⁴  =  25 ×  2401  = 60025

Now add all these digits  6 + 0 + 0 + 2  + 5  = 13

Now reversing the order of these digits obtained in  previous step

13   <---->  31 The number in the inner part

4th Quadrant

  8² ×  9⁴  =  64 × 6561  = 419904

Now add all these digits  4 + 1 + 9 + 9 + 0 + 4 = 27

Now reversing the order of these digits obtained in  previous step

27   <----> 72 The number in the inner part

3rd Quadrant

 7² × 7⁴  =  49 × 2401 = 117649

Now add all these digits  1 + 1 + 7 + 6 + 4 + 9  = 28

Now reversing the order of these digits obtained in  previous step

28   <----> 82 The number in the inner part

Option (1) is correct option.

Problem # 3

This  circle has been divided into eight sectors. Every sector  consist of three numbers. To solve this problem multiply both the numbers  in any sector which are in the outer part and then add 1 to it ,the result so obtained is written in the inner part of the sector which is exactly opposite to this sector .Continuing in this manner  we shall have  all the numbers placed accordingly. .  Starting from the sector immediately to the right of question mark. 

( 2 × 4 ) + 1  =  9  ( In the inner part of 5th sector ) 

( 3 × 7 ) + 1 =  22  ( In the inner part of 6th sector )

( 1 × 6 ) + 1 =  7  ( In the inner part of  7th sector )

 ( 5 × 2 ) + 1  = 11  ( In the inner part of 8th sector )

( 3 × 4 ) + 1  =  13  ( In the inner part of 1st sector )

( 2 × 9 ) + 1  =  19  ( In the inner part of 2nd sector )

( 2 × 2 ) + 1  =  5  ( In the inner part of 3rd sector )

 ( ? × 3 ) + 1  =  25  ( In the inner part of 4th sector )

Now we have to find the value of  "?" like this 

( ? × 3 ) + 1  =  25

? × 3  =  25 - 1

? × 3  =  24

? = 24/3

? = 8 

Hence option (2)  is correct option. 

Problem # 4

1st Sector

(3 + 4 )² +  4² = 7² + 4² = 49 + 16 = 65 (The number is in circle opposite to 3 and 4)

2nd Sector

(7 + 2)² + 7² =  9² +  7² = 81 + 49 = 130 (The number is in circle opposite to 7 and 2)

3rd Sector

 (6 + 9)² +  (9)² = 15² +  9² =  225 + 81 = 306  (The number is in circle opposite to 6 and 9)

4th Sector

 (5 + 8 )² +  (8)² = 13² +  8² = 169 + 64 = 233 (The number is in circle opposite to 5 and 8)

Hence option (4) is right answer.

Problem # 5

All the Numbers except one in both the circles are written in same pattern . If we add all the three digits of any number in both the circles then we shall get same result except in one number in both the circles. And that different result will be the odd one out in this problem.

In 1st circle

The sum of digits of number 319 = 3 + 1 +  9  = 13  ( Odd one out)  

The sum of digits of number 441 = 4 + 4 + 1 = 9

The sum of digits of number 243 = 2 + 4 + 3 = 9

The sum of digits of number 612 =  6 + 1 + 2 = 9

The sum of digits of number 342 = 3 + 4 + 2 =  9  

In 2nd circle

The sum of digits of number 322 =  3 + 2 + 2 = 7

The sum of digits of number 313 =  3 + 1 + 3 = 7

The sum of digits of number 142 = 1 + 4 + 2 = 7 

The sum of digits of number 304 = 3 + 0 + 4 = 7

The sum of digits of number 349 = 3+ 4 + 9 = 16 ( Odd one out)

So from both these circles two results 13 and 16 of numbers 319 and 349  are different from other.

Hence option (4) is right answer.

Problem # 6

Since this circle consists of four quadrants and every quadrant consists of three numbers ,two numbers are in outer part and one number is in the inner part  ( triangle) . The  number which is in the inner part (triangle) can be found  using both the numbers which  are in the outer part of the same quadrant by using the following formula. 

(Difference of both  numbers  in the outer part) - ( difference of  both the digits in the triangular parts ) = 1 

1st Quadrants 

(12 - 8)  - ( 6 - 3 ) = 4 - 3 = 1

2nd Quadrants 

(4 - 3)  - ( 1 - 1  ) = 1 - 0 = 1

3rd Quadrants 

(9 - 6)  - ( 6- 4 ) = 3 - 2 = 1

4th Quadrants 

(12 - 10 )  - ( ? - ? ) = 2 - 1 = 1

If we put 43 in place of question mark only then difference of its digits  will be equal to 1 and in other cases the difference of digits is not equal to 1 which is necessarily required . 

So the correct option will be  (1) . 

 

 Problem # 7

All the Numbers except one in both the circles are written in same pattern . If we add all the three digits of any number in both the circles then we shall get same result except in one number in both the circles. And that different result will be the odd one out in this problem.

In 1st circle

The sum of digits of number 245 = 2 + 4 + 5 = 11

The sum of digits of number 443 = 4 + 4 + 3 = 11

The sum of digits of number 209 = 2 + 0 + 9 = 11

The sum of digits of number 902 = 9 + 0 + 2 = 11 

The sum of digits of number 342 = 3 + 4 + 2 =  9  ( Odd one out)

In 2nd circle

The sum of digits of number 307 =  3+ 0 + 7 = 10

The sum of digits of number 343 = 3 + 4 + 3 = 10

The sum of digits of number 642 = 6 + 4 + 2 = 12( Odd one out)

The sum of digits of number 703 = 7 + 0 + 3 = 10 

The sum of digits of number 118 = 1 + 1 + 8 = 10 

So from both these circles two results 9 and 12 of numbers 342 and 642  are different from other.

Hence option (4) is right answer.

Problem # 8

Starting clockwise take the difference of two consecutive numbers. 

3 - 1 = 2

6 - 3 = 3

11 - 6 = 5

18 - 11 = 7. 

Since all these resultant numbers are prime numbers so next prime number should be 11

Therefore  ? - 18 = 11

This implies ? = 29.

Hence option (D) is correct option. 

Problem # 9

Starting clockwise from the question mark i. e.  from 3 . These numbers 3, 5 ,7 ,11 ,13 are written in a pattern of prime number series. So so after 13  the next prime number will be 17. But 17 is not in the  given option. Hence this series will start from the prime number  prior to 3 . The number prior to 3 is 2 . Because smallest prime number is 2.

Hence pattern will be 2 , 3 , 5, 7 , 11, 13.

Therefore correct option will be (D) . 

Also read these posts on Reasoning

1.Reasoning for competive exams
2 .Box and circle reasoning
3 . Reasoning for bank exams
4. Ten Tricky logical reasoning
5. Missing  number series questions
6. Reasoning questions with answers
7. Circle Reasoning
8.  Box Problems
9. 15 Questions Circle Problems 
10. SSC CGL Reasoning

 

Problem # 10

This Circle consist of four quadrant and every quadrant consists of three numbers. Here the number in the outer part of every sector contribute to the number which is in the inner part of that sector . Starting from the sector which is below the question mark , out of these two numbers written in Outer part , the the number which is greater/maximum of these two is written in the inner part. 

In 1st Sector  

Max (14,17 ) = 17 , Here greater of the two numbers will be selected

In 2nd Sector  

Max (13,18) = 18, Here greater of the two numbers will be selected

In 3rd Sector  

Max (5 , 3 ) = 5, Here greater of the two numbers will be selected

In 4th Sector

  

Max ( 4, 8 ) = 8, Here greater of the two numbers will be selected

So option (2)  will be correct option

Also Reads these posts on Reasoning

Reasoning Problems #11  
Reasoning Analogy #10
Reasoning Questions #9    
Circle Reasoning #8
Ten Box-Problems #7   
Ten Reasoning problems #6
Ten-Important-Problem#5  
Ten--Tricky-Puzzles #4
Twelve-Figures-Problems#3   
Ten-Important-Reasoning#2     
Picture-Reasoning#1

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