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Showing posts with label Others. Show all posts

10 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 .

Problem # 1


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  is 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 

Square of 8 × Taking Power 4 of  3  =  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

Square of 5 ×  Taking Power 4 of  7  =  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

Square of 8 × Taking Power 4 of  9  =  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

Square of 7 × Taking Power 4 of  7  =  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 # 2


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 # 3


Sq (3 + 4 ) + Sq (4 ) = Sq 7 + Sq 4 = 49 + 16 = 65 (The number is in circle opposite to 3 and 4)

Sq (7 + 2 ) + Sq (7 ) = Sq 9 + Sq 7 = 81 + 49 = 130 (The number is in circle opposite to 7 and 2)

 Sq (6 + 9 ) + Sq (9 ) = Sq 15 + Sq 9 =  225 + 81 = 306  (The number is in circle opposite to 6 and 9)

Sq (5 + 8 ) + Sq (8 ) = Sq 13 + Sq 8 = 169 + 64 = 233 (The number is in circle opposite to 5 and 8)
Hence option (4) is right answer.

Problem # 4


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 # 5



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 # 6


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 # 7



Starting clockwise take the difference of two consective 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 # 8



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 # 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) . 

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



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Ten most important missing numbers questions and answers in reasoning

Ten most important   missing numbers questions and answers in box problems in Reasoning Analogy in various figures of different order and types discussed in this post. These types of problems are very very important for the exams like SSC CGL ,SSC CHSL , RRB NTPC and many other similar competitive exams.

PROBLEM # 1

Since in 1st row we can write 3, 2 and 4  numbers to get 10 like this
( 3 × 2 ) + 4 = 6 + 4  = 10 (The number in last column  of 1st row )
And in 2nd row  ( 4 × 3 ) + 5 = 12 + 5 = 17, (The number in last column  of 2nd row )
In 3rd row ( 5 × 4 ) + 6 = 20 + 6 = 26 ,(The number in last column  of 3rd row )
Similarly ( 6 × 5 ) + 7 = 30 + 7 = 37 , (The number in last column  of 4th row )
So  " ? "   Will be replaced by 37 .
Therefore correct option will be (D)


PROBLEM # 2
                 
Let us find relation between 3 and 18 , if we double the square of  3 ,we shall have 18. 
2 × ( 3^2) = 2 × 9 = 18 (The number in last column  of 1st row )
And in 2nd row if we double the square of  4 ,we shall have 32
2 × ( 4^2) = 2 × 16 = 32, (The number in last column  of 2nd row)
In the same way increasing the number one by one then  third ,fourth and  fifth columns can be calculated like this 

          2 × ( 5^2) = 2 × 25 = 50 (The number in last column  of 3rd row )
          2 × ( 6^2) = 2 × 36 = 72 (The number in last column  of 4th row )
          2 × (7^2)  = 2 × 49 = 98 (The number in last column  of 5th row )
Therefore correct option will be (C)


PROBLEM # 3

As the sum of first and second number in first and second rows in any particular column is equal to third element in that particular column so to get the value of " ? " . Differentiate  first and second numbers column wise to get third number as follows.
56 - 12 =  44, 
78 - 30 = 48 ,
65 - ? = 14 ➡️  ? = ? = 65 - 14  = 51 
Therefore correct option will be (C)

PROBLEM # 4


Answer can be split into two parts, 1st part can be obtained by multiplying two given  numbers and second part can be obtained by adding these two numbers .
1st row 5 × 3  = 15 and 5 + 3 = 8 so 5 , 3 = 158
2nd row 9 × 1  = 9 and 9 + 1 = 10 so 9 , 1 = 910
3rd  row 8 × 6  = 48 and  8 +6 = 14  so 8 , 6 = 4814
4th row 4 × 4 = 16 and   4 + 4 = 8 so  4 , 4 = 168.
5th row 7 × 3 = 21 and 7 + 3 = 10 so 7 , 3 = 2110
So ?  Will be replaced by 2110.
  Therefore correct option will be (A)

 .

PROBLEM # 5



Formula a*b = (a × b) + (b - 1)

This the sum of two numbers , out of two ,1st number is product of two given  numbers and second  is the number one less than 2nd given number.

3*2 = (3 × 2) + (2 - 1) = 6 + 1 = 7
5*4 = (5 × 4) + (4 - 1) = 20 + 3 = 23
7*6 = (7 × 6) + (6 - 1) = 42 + 5 = 47
9*8 = (9 × 8) + (8 - 1) = 72 + 7 = 79
10*9 = (10 × 9) + (9 - 1) = 90 +8 = 98
Therefore correct option will be (C)

PROBLEM # 6


Find the missing number ?
Suppose we have three numbers a , b and c then

Formula for this puzzle is = (a × b) + b or  b(a + 1)

Put a = 2 and  b = 6 in above formula ,we get

1st Line = (2  × 6)   +  6  = 12   +   6   = 18
Put a = 4 and  b = 20 in above formula to get 2nd line,we get
(4  × 20) + 20 = 80   +  20  = 100
Put a = 5 and  b = 21 in above formula to get 3rd line,we get
(6 × 21) + 21 = 126 +  21  = 147
So required and correct answer will be  6


PROBLEM # 7


If in every row ,we add 1st and 3rd column and then multiply the sum with 2nd column, we shall have number in 4th column .

If we  consider three numbers a , b and c and start calculation by   Formula =  (a × b) + (b × c)  or b(a + c)  --------(1)

To get 1st line , put a = 1 , b = 2 and c = 3 in (1) , we get 

1st Line =  (1×2) + (2×3) =2 + 6 = 8

To get 2nd line , put a = 2 , b = 3 and c = 4 in (1) , we get 

2nd  Line =  (2×3) + (3×4) = 6 + 12 =  18

To get 3rd line , put a = 3 , b = 4 and c = 5 in (1) , we get 

3rd Line =  (3×4) + (4×5) = 12 + 20  = 32

To get 4th line , put a = 4 , b = 5 and c = 6 in (1) , we get 

4th Line =  (4×5) + (5×6) = 20 + 30  = 50

Similarly Last line can be calculated by putting a = 5 , b = 6 and c = 7 in (1) , we get 

Last line  =  (5×6) + (6×7) = 30 + 42  = 72
Therefore correct option will be (D)

PROBLEM # 8 

Find the missing number ?

Case 1


If we choose   1st number  from 5 then look at the pattern  opposite to given smaller number 
                              
5  ×  3 =  15
8  ×  3 =  24
12 × 3  =  36

In this pattern we can conclude 12 ×3 = 36

Case 2

If we choose   1st number  from ? then look at the opposite to given smaller number 

 ?    ×  3 = 12
5   ×  3  = 15
8   ×  3  = 24   
Therefore ? will be replaced  by 4.
Hence    4 ×  3  = 12

PROBLEM # 9


This circle can be divided into two parts ,  1st part containing the number 4 ,5 ,6 and 7 and second part containing 7 , 9 ,11 and ?.            Now study  the opposite number of 4  which is 7 . Then study the relation between other numbers and its opposite number ,we find difference between 4 and 7 is 3 , difference between 5 and 9 is 4 , difference between 6 and 11 is 5  . so in this pattern we can find difference between ' ? '  and  7 must be 6 . 
 7 - ? =  2 means ? = 5
 7 - 4 =  3
 9 - 5 =  4
11 - 6 = 5 , the difference is increased by 1 every time .
 ?   - 7 = 6  means ? = 13
Either 5 or 13 should be the required number, but 5 is not given in any option.
So ? Will be replaced by 13 which is the required answer 
Therefore correct option will be (4)

PROBLEM # 10

We shall try every possible relation between different numbers given either row wise or column wise . It is found that there is no relation if we consider First row . 
 After elemination of 1st row we can develop a relation between 2nd ,3rd and 4th row which is multiplying the 2nd and 3rd numbers then double it to get 4th number column wise .
 2 × (7 × 4 ) = 2 × 28 = 56
2 × ( 15 × 6 ) = 2 × 90 = 180
2  × ( 8 ×  ?  ) = 80 
So ? = 5
These were the Ten most important  missing numbers questions and answers in reasoning . Please comment your opinion about this post. 

Therefore correct option will be (B) 

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Ten most important problems of missing numbers in Reasoning Analogy part 2



Let us discuss series of missing numbers and missing terms and their  solutions in reasoning analogy . These questions are very very important for upcoming competitive exams like SSC CGL ,SSC CHSL and RRB NTPC Etc

Problem # 1


This series consists of two alternate series 1st series consist of number 1,3, 5 and 7 and second series consist of  the series 2 ,5 ,8,?

So in 1st series there is difference of 2, so next number in the series must be 2 greater than 7 that is 9 .
In the 2nd series  we observed a difference of 3 show the next number must be 11 that is 3 greater than 8 the last term therefore the last two terms of the series 11 and 9

The correct option is  ( a ) 11,9

Problem # 2


This series is also the combination of two series one with the number  2, 48 ,16 ,32 and second series consist of these numbers 6, 9 13 ,18 , ? . So ? Will be filled out using 2nd series not from 1st series.

Observing 2nd series carefully ,In  second series numbers are increased by  3 ,4 and 5 ,so next Increment should be of 6. This implies next number should be  24 with an increment of 6.

The correct option is  ( a ) 

Problem # 3

Observe carefully all the numbers in the series are prime number , so  next prime number after 37 will be 41 .This implies that  ?  will be replaced by 41.

The correct option is  ( c ) 

Problem # 4


All the numbers in this series are consist of three digits numbers and Middle digit is the sum of first and last digit in all the numbers except one that is 342 in this case 3 + 2 is equal to 5 not equal to 4 .So  correct option is option is  (c) ,this is the wrong number in the series .

The correct option is  ( C ) 

Problem # 5

Every number in this series is the sum of its two  preceeding number plus 3.

14  = ( 7 + 4 ) + 3
24 = ( 14 + 7 ) + 3
41 = ( 24  +14 ) + 3
?   = ( 41 + 24 ) + 3 = 65+3 = 68

The correct option is  (  C ) 

Problem # 6

This series consists of triplet ,that is every third number is product  its two preceding numbers

  As third  number 10 is the product of 2 and 5, sixth number  18 is the product of fourth (3) and fifth (6 ) numbers .

So ninth number of the series  must be the product of seventh and  eighth numbers. 4 × 7 = 28 instead of 30

The correct option is  ( C ) 

Problem # 7


Here second  number is the sum of the digits of first number the forth number is the sum of digits of third number similarly every even positioned number is  the sum of its preceding number's digits question  mark will be replaced by the sum of 5 + 3 that is 8

The correct option is  ( a ) 

Problem # 8


Writing 3 = 2² - 1

Writing 8 = 3² - 1

Writing  15 = 4² - 1

Writing 24 = 5² - 1


Show next number should be  6² - 1 ( square minus 1 ) that is 35

Writing  6² - 1 = 35

The correct option is  (2 ) 

Problem # 9

Every number in this series is the double of its preceding number .So double of 1 is   2 , double of 2 is 4 ,double of 4 should be 8 ( instead of 7 ) , double of 8 is 16 and double of 16 should be 32 .  So 7 is the wrong number in this series.

The correct option is  ( 1 ) 

Problem # 10

In the series every number is the the cube of its  position plus 1 .

Since

  2 =  1³ + 1

   9 =  2³ + 1

  28 =  3³ + 1

  65  =  4³ + 1

  126 =  5³ + 1

  In the same way next number of the series must be  6³ + 1 = 216 + 1 = 217

The correct option is  ( 1 ) 

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Twelve Most Interesting figures problems of Reasoning analogy part 3

Discussed most Important  Reasoning analogy problems and frequently asked picture problems in previous competitive  examinations along with their answers .
Twelve Most Interesting figures problems of Reasoning analogy part 3
Post your answer in comments and to confirm your answer click here.

Problem # 1



Outermost numbers are the  product of Square roots of two  numbers attached to it.

Since 14 is attached with 4 and 49 , the square root of 4 and 49 are 2 and 7 respectively now multiplying 2 and 7 to get 14.

Since 12 is attached 
with 36 and 4 , the square root of 36 and 4 are 6 and 2 respectively now multiplying 6 and  2  to get 12.

So 30 must be the product of Square Root  of 36  and ? , so if we put 25 instead of Question Mark   ? .Then product of square root of 25 and the square root of 36 which will be  equal to 30.

Similarly product of square root of  49 and Square Root of ? must be equal to 35 . But we had already put 25 in place of ? to give product 35 .

Hence  (b) 25  is right option.

Problem # 2


Starting from 3 and moving clockwise multiply 3 with 2 and subtracting 1 from it  to get 5 .
( 3 х 2 ) - 1 =  6 - 1  = 5
Multiplying 5 with 2 and subtracting 2 from result so obtained to get 8. 
( 5 х 2 ) - 2 =  10 - 2 = 8
Multiplying 8 with 2 and subtracting 3 from it to get 13.
( 8 х 2 ) - 3 = 16 - 3 = 13
Now multiplying 13 with 2 and then subtracting 4 from  result so obtained to get 22.
( 13 х 2 ) - 4 = 26 - 4 = 22
At last multiplying 22 with 2 then subtracting 5 from the result so obtained to get 39 .
( 22 х 2 ) - 5 = 44 - 5 =  39.
so (C) 39 is the right answer

Problem # 3

1st Step 

In  1st Row B ,D are F are written in a  gap of one letter .
In 3rd Row N ,P and R are written in a gap of one letter
In 2nd Row H and J are also written in a gap of one letter ,So J and ? must be written with a gap of one letter .This means after J the Lett with one gap will be L.

2nd Step 

In 1st row  digits associated are 3 ,3 and 6 means last digit in the Row is sum of 1st two digits 3 + 3 = 6.
In 3rd row  digits associated are 7 ,9  and 16 means last digit/number in this Row is sum of 1st two digits 7 + 9 = 16.
In 2nd  row  digits associated are 5 ,6 and ? will be replaced by sum of other two digits 5 + 6 = 11 .
So (a) L11 is the right option
          

Problem # 4

 Here in this Figure  result will  be calculated column wise in any particular column. In first column the fourth element is calculated by dividing second element with third  element and then multiplying first and the results will be  obtained.

1st column

(12 ÷ 3) x 18 = 72  

3rd column

(16 ÷ 4) x 32 = 72  

2nd column

(14 ÷ ?) x 16 = 112 
⇒ 14 ÷ ?  = 112 ÷ 16
⇒ 14 ÷ ?  = 7
 ?  = 2
Therefore option ( A )2 is  right  answer .

Problem # 5


Add all the numbers in any particular figure and then subtract 2 from it to get the middle number .

In 1st figure we have 0 + 6 + 4 + 2 = 12 minus 2 = 10 { middle number in 1st figure}
In 2nd figure  6 + 2 + 10 + 8 = 26 - 2  = 24 { middle number in 2nd figure} .
Similarly In 3rd figure  we have 4 + 14 + 12 + 10 = 40 - 2 = 38 ( Question Mark )  which is our required number. 
Therefore option ( C ) is  right  answer .


             

Want to check more reasoning problems with their solutions ,click here,. 

Problem # 6


If we add all the numbers column wise then we get 99 in all the three columns so we shall  have 42 in place of question mark to  have total   99 in all the three  columns .
Because 40 + 24 + 35 = 99 ( Last number in 1st column ) 
In second column we have 30 + 35 + 34 = 99 ,(Last number in 2nd column ) . While in last column when we add ? + 30  + 27 ,then the sum of all the numbers in last column will be 99 so required number  will be 42.
The right option is ( a) 42
                   

Problem # 7



Find the sum of numbers which are in the Row ( horizontal ) then subtract this total from the sum which are in column ( vertical ) to find the number which is in the middle of the given figure .
In 1st figure   5 + 6 is equal to 11 and  4 + 7  which is also equal to 11 so difference of both the  sum is equal to zero ,which is the middle number in 1st figure. 
(5 + 6) - ( 4 +7 ) = 11 - 11 = 0
In 2nd figure  7 + 6 is equal to 13 and  8+ 4 which is equal to 12 so difference of both the  sum is equal to 1,which is the middle number
(7 + 6) - ( 8 +4 ) = 13 - 12 = 1
In 3rd figure  11 + 2 is equal to 13 and  0 + 2 which is  equal to  2 so difference of both the  sum is equal to  13 - 0 = 11 which will be in the middle number.
(11 + 2) - ( 0 +2 ) = 13 - 2 = 11
 Correct option is 3(11)

Problem # 8


Multiplying all the elements in 2nd row with 2 then add elements of 1st and 2nd row downward to get the elements in  3rd row
1st Column    13  + ( 7×2 ) = 13 + 14 = 27
2nd Column 54 + ( 45 × 2 )  = 54 + 90 = 144
3rd Column   ? + ( 2× 32 ) = 68
This implies ? = 68 - 64 = 4 
So the correct option is  A(4) 

Problem # 9

To find the middle number in each figure add all the numbers except middle number then divide it with 4 to  get middle number .

In figure 1 add 24 + 32 + 40 + 36 = 132 which  when divided by 4 gives 33 

In figure II add the numbers  27 +19 +  20 + 22 = 88 Which when divided by 4 gives 22 

similarly in III  figure when we add all the numbers 6 + 16 + 14 + 1 2 = 48 which will  after division  by 4  gives 12 .
So option   ( 2 )12  is correct option

Problem # 10


Sum of all the digits of 1st and 2nd rows are 17 .Similarly sum of all digits of 3rd row must be 4 +7 + ? = 17 .This implies  the value of  ? should be 6.
Hence the correct  option will be  ( a ) 6

Problem # 11



Spilt this circle into two parts ,one above x axis and other below x axis , the sum of digits in both the semi circle must be equal to each  other . 
This means ? + 4 + 5 + 6 = 11 + 9 + 3+7
? +15  = 30
? = 15
So option ( 4 ) 15 will be right option

Problem # 12


This pattern consists of two series of alternative numbers ,first one consists of 2, 4, 8, 16,32 so on and second series consists of the number 6, 9, 13, 18, so on .
 So our next number in from given series will be derived from the second series 6, 9, 13, 18, so on .
Look at the difference between first two number 6 and 9 which is equal to 3 ,And  difference between second number (9) and third number (13 ) is 4 And  difference between third number ( 13 ) and fourth number ( 18 ) is 5 so difference between 18 and next number must be 6 it  means next number must be 24 because 24 - 18 = 6
Show the correct option is  (1)24

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If square root of any number is added /subtracted infinite time ,then what will be the answers .

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