Missing number and reasoning problems for SSC CGL Exams

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  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. 

Ten Reasoning problems with answers for competitive exams

Problem #1

This reasoning problem have four sub problems and every sub problem have three number associated to it.

           Since last figure have question mark in it. So the solution of this problem is to find the value of question mark using three numbers associated to it . 

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

           Now we have to find or search the  formula for these three numbers in each sub problem to utilised them in any possible way to get the value of question mark. 

Formula:- Adding 1 to each digit and then combine all three numbers to get the value of question mark.

Sub Problem # 1

In this sub problem the three digits are 6 , 7 and 8, Adding 1 to each digits, we shall have

6 + 1 = 7

7 + 1 = 8

8 + 1 = 9

Now combining these three newly formed digits to get new number  as 789( the number on the right hand side).

Sub Problem # 2

In this sub problem the three digits are 2 , 3 and 4, Adding 1 to each digits, we shall have

2 + 1 = 3

3 + 1 = 4

4 + 1 = 5

Now combining these three newly formed digits to get new number  as 345( the number on the right hand side).

Sub Problem # 3

In this sub problem the three digits are 4 , 8 and 0, Adding 1 to each digits, we shall have

4 + 1 = 5

8 + 1 = 9

0 + 1 = 1

Now combining these three newly formed digits to get new number  as 591( the number on the right hand side).

Required Sub Problem # 4

In this sub problem the three digits are 8 , 0 and 7 , Adding 1 to each digits, we shall have

8 + 1 = 9

0 + 1 = 1

7 + 1 = 8

Now combining these three newly formed digits to get new number  as 918( The value of question mark).

Hence Option (4)918 is correct option.

Problem #2

This reasoning problem also have four sub problems and every sub problem have three number associated to it.

           Since last figure have question mark in it. So the solution of this problem is to find the value of question mark using three numbers associated to it . 

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

           Now we have to find or search the  formula for these three numbers in each sub problem to utilised them in any possible way to get the value of question mark. 

Formula:- Reversing the order of each digit in every number to get the value of question mark.

Sub Problem # 1

In this sub problem the three digits are 6 , 7 and 8, Reversing the order of each digit , it means we have to put the 1st digits in last place ,2nd digit in 2nd last place and 3rd digit in 3rd last place to get the number on the right hand side of each sub problem.

 6 + 5 + 7  ⇒ 7 5 6 ( the number on the right hand side).

Sub Problem # 2

In this sub problem the three digits are 2 , 3 and 4, Reversing the order of each digit , it means we have to put the 1st digits in last place ,2nd digit in 2nd last place and 3rd digit in 3rd last place to get the number on the right hand side of each sub problem.

 2 + 3 + 4  ⇒ 4 3 2 ( the number on the right hand side).

Sub Problem # 3

In this sub problem the three digits are 7 , 8 and 9, Reversing the order of each digit , it means we have to put the 1st digits in last place ,2nd digit in 2nd last place and 3rd digit in 3rd last place to get the number on the right hand side of each sub problem.

 7 + 8 + 9  ⇒ 9 8 7 ( the number on the right hand side).

Required Problem # 4

In this sub problem the three digits are 2 , 2 and 2, Reversing the order of each digit , it means we have to put the 1st digits in last place ,2nd digit in 2nd last place and 3rd digit in 3rd last place to get the number on the right hand side of each sub problem.

 2 + 2 + 2  ⇒ 2 2 2 ( The value of question mark).

But there is no option which has value 222.

Hence Option (4)None is correct option.

Problem #3

Square of 1st digit and then sum of remaining two digits

6+1+3 = 6²(1+3) = 364

8+7+2 = 8²(7+2) = 649

4+7+2 = 4²(7+2) = 169

4+2+5 = 4²(2+5) = 167 (The value of Question mark)

Hence Option (2) 167 is correct option.

Problem #4

H C F (Highest Common Factor) of three given numbers

6+2+5 = H C F of (6,2,5) = 1

4+6+8 = H C F of (4,6,8) = 2

3+6+9 = H C F of (3,6,9) = 3

4+8+9 = H C F of (4,8,9) = 1

Hence Option (4) 1 None is correct option.

Problem #5

L C M of three given numbers

L C M (Least Common Multiple) of three given numbers

6+2+5 = L C M of (6,2,5) = 30

2+3+5 = L C M of (2,3,5) = 30

6+4+2 = L C M of (6,4,2) = 12

1+3+5 = L C M of (1,3,5) = 15

Hence Option (1) 15 None is correct option.

Problem #6

6+2+5 = (6 + 2 + 5)² + 1 =  13² + 1 = 169 + 1 = 170

5+3+7 = (5 + 3 + 7)² + 1 =  15² + 1 = 225 + 1 = 226

2+4+6 = (2 + 4 + 6)² + 1 =  12² + 1 = 144 + 1 = 145

1+3+5 = (1 + 3 + 5)² + 1 =  9² + 1 = 81 + 1 = 82

Hence Option (3) 82 None is correct option.

Problem #7

2+3+5 = ( 2 + 5 ) × 3 = 7 × 3 = 21

4+3+7 = ( 4 + 7 ) × 3 = 11 × 3 = 33

4+3+9 = ( 4 + 9 ) × 3 = 13 × 3 = 39

3+1+8 = ( 3 + 8 ) × 1 = 11 × 1 = 11

Option (1)11 is correct Answer

Problem #8

2+3+5 =  2 + (5  × 3 ) = 2 + 15 = 17

4+3+7 = 4 + (3  × 7 ) = 4 + 21 = 17

4+3+9 = 4 + (3  × 9 ) = 4 + 27 = 31

3+1+8 = 3 + (1  × 8 ) = 3 + 8 = 11

Option (1)11 is correct Answer

Problem #9

5+2+3 = 5 × 2² × 3 =  5 × 4 × 3 = 60

6+3+4 = 6 × 3² × 4 =  6 × 9 × 4 = 216

4+1+8 = 4 × 1² × 8 =  4 × 1 × 8 = 32

3+2+8 =3 × 2² × 8 =  3 × 4 × 8 = 96

Option (2)96 is correct Answer

Problem #10

5+2+3=(5 × 2 × 3 ) + ( 5 + 2 + 3 ) = 30 + 10 = 40

6+3+4 = (6 × 3 × 4 ) + ( 6 + 3 + 4 ) = 72 + 13 = 85

4+1+8 = (4 × 1 × 8 ) + ( 4 + 1 + 8 ) = 32 + 13 = 45

3+2+8 = (3 × 2 × 8 ) + ( 3 + 2 + 8 ) = 48 + 13 = 61

Option (4)61 is correct Answer

Problem #11

(5 × 4) × (4 × 3 ) × ( 3 × 5) = 20 + 12 + 15 = 47

(6 × 2) × (2 × 4 ) × ( 4 × 6) = 12 + 8 + 24 = 44

(4 × 3) × (3 × 9 ) × ( 9 × 4) = 12 + 27 + 36 = 75

(3 × 1) × (1 × 7 ) × ( 7 × 3) = 3 + 7 + 21 = 31

Option (2)31 is correct Answer

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