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

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.

Ten most important problems of missing numbers in Reasoning Analogy

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 i.e. 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  (1) 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  (1)24 

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  (3)41 

Problem # 4

All the numbers in this series are consisting 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  (3)342 

Problem # 5

Every number in this series is the sum of its two  preceding 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  (3)68 

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  (3) 30

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  (1)8 

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 (6  square minus 1 ) that is 35

Writing  6² - 1 = 35

The correct option is  (2)35 

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)7 

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) 217

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Twelve Most Important 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 .

Post your answer in comments and to confirm your answer click here.

Twelve Most Important figures problems of Reasoning analogy

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) × 18 = 72  

3rd column

(16 ÷ 4)  ×  32 = 72  

2nd column

(14 ÷ ?)  ×  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 1st add 24 + 32 + 40 + 36 = 132 which  when divided by 4 gives 33 

In figure 2nd add the numbers  27 +19 +  20 + 22 = 88 Which when divided by 4 gives 22 

similarly in 3rd  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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Ten most important questions of Reasoning Analogy in missing numbers in various figures

Ten most Important questions of Reasoning Analogy of missing numbers have been discussed with their Solutions in an easy manners. These questions are very very important for competitive exams like SSC CGL , SSC CHSL And RRB NTPC Etc Because such types of problems are frequently asked in these exams.

 Problem #1  

Moving clock wise from top right number , Double the top right number to get bottom right number and add 2 to get bottom left number, now again double this number to get top left number in all the three figures .

Hence in  option (4)  24  is right answer

    Problem #2

Multiply  all the numbers which are on the corners of the given triangles to get the number which is on the centre of particular triangle.

In 1st triangle  multiply all the numbers on the corner of triangle

1 × 2 × 3 = 6

Again In 2nd  triangle  multiply all the numbers on the corner of triangle

2 × 5 × 3 = 30

Similarly In 3rd  triangle  multiply all the numbers on the corner of triangle

4 × 8 × 5 = 160

Hence in  option (3)  160  is right answer

WHAT IS THE NEXT TERMS IN THE SERIES.

Problem #3

If we add the top and bottom  numbers and the subtract from it the sum of left  and right  terms we shall get the middle number in all the figures.
in 1st figure

(3+7) - (2+5) = 10 - 7 = 3 ( Middle number in 1st figure) 
In 2nd figure

(9+9) - (7+8) =18 -15 = 3( Middle number in 2nd figure) 
In 3rd figure 

(8+8) - (6+6) = 16 -12 = 4( Middle number in 3rd figure) 

Hence in  option (2) 4 is right answer

Problem # 4

Multiply the top numbers and bottom numbers row wise the take the difference  of it .

In 1st figure 

( 5 × 6 ) - ( 3 × 8) = 30 - 24 = 6 ( Middle number in 1st figure) 

2nd figure

( 10 × 4 ) - ( 2 × 7 ) = 40 - 14 = 26 (Middle number in 2nd figure) 

3rd figure 

( 9 × 7 ) - ( 6 × 8 ) = 63 - 48 = 15 ( Middle number in 3rd figure) 

Hence in  option (2)  15  is right answer

Problem #5 

Cross multiply the numbers to get middle number in all the figures .

In 1st figure 

12 × 5 = 15 × 4 = 60 ( Middle number in 1st figure) 

In 2nd figure
3 × 14 = 6 × 7 = 42 ( Middle number in 1st figure)

In 3rd figure
26 × 3 = 13 × 6 = 78 ( Middle number in 1st figure)

Hence in  option (3)  78 is right answer

Problem #6 

Take cube roots of all the numbers which are on outer side 
 all the ellipse ( given figure ) and then add these numbers  to get middle number in the figure .

In 1st figure  cube roots of 1, 64, 27 and 8 are 1 , 4 , 3 and 2 respectively. 

Now add these numbers. i.e 1 + 4 + 3 +2 = 10

In 2nd figure  cube roots of 8, 125, 64 and 27  are 2 , 5 , 4 and 3 respectively. 

Now add these numbers. i.e 2 +5 + 4 +3 = 14

Similarly In 3rd figure  cube roots of  27 , 216 , 125 and 64 are 3 , 6 , 5 and 4 respectively. 

Now add these numbers. i.e 3 + 6 + 5 + 4 = 18

Hence in  option (4)  18  is right answer

Problem #7 

In all the figure add both numbers which are 2nd and 3rd rows then multiply this result with the number which is on the right side of given figure in the 1st row to get the number on the left side of 1st row of given figure.

In 1st figure 1 + 2 =》3 × 9 = 27

In 2nd figure  2 + 3  =》 5 × 7  = 35

In 3rd  figure  x + 4  =》 (x+4) × 4 = 36

 4x + 16 = 36

 4x = 36 - 16 = 20

x = 5

Hence in  option (3)  5  is right answer

Problem #8 

1st Figure

 From 1st figure we can see that lower number is thrice (3 times) of sum ( addition ) of both numbers which are on upper portion of figure i. e . 18 + 9 = 27 .

Now multiply it with 3 , we get 27 ×3  =81 , which is the lower number in 1st figure.

3rd Figure

And for third  figure add both the numbers 24 and 7 and then multiply  the sum with 3 like this 24 + 7 = 31 and 31 × 3 = 93

2nd  Figure

Similarly ?  in 2nd figure can be found by adding 23 and 8 and then Multiply  it with 3 as follows 

     23 + 8 = 31 × 3 = 93 , Hence 93 is the required number.

Hence in  option (3)  93  is right answer

Problem #9 

From 1st two  figures if we multiply top and bottom terms and then subtract  the result of product of left and right terms from it we shall get the number in the middle of the circles in both the given figures.

( 6 × 7 ) - ( 4 × 5 ) = 20 ( Middle Number in 1st Figure ) . 

( 9 × 7 ) - ( 6 × 3 ) = 45 ( Middle Number in 2nd Figure ). 

Similarly middle number of 3rd figure can be calculated as follows

( 8 × 8 ) - ( 4 × 6 ) = 40 ( Middle Number in 3rd Figure )

Hence in  option (3)  40  is right answer

Problem #10

If we add the left and right  numbers and the subtract from it the sum of top and bottom  terms we shall get the middle number in all the figures

( 5 + 6 ) - ( 4 + 7) = 0

( 7 + 6 ) - ( 8 + 4 ) = 1

similarly 3rd term can be calculated as follows

( 11 + 2 ) - ( 0 + 2 ) = 11

Hence in  option (3)  11  is right answer

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How to Solve linear equations of three VARIABLES BY MATRIX-METHOD

 How to find area bounded by two circles 

 How to integrate integral with square root in numerator     

How to find area of the circles which is interior to the parabola


How to find common area of two parabolas.

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  Thanks for watching and responding to this post of reasoning based on mathematics and number sequence puzzle solver.

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