Twelve Most Interesting figures problems of Reasoning analogy part 3

Discussed the most important  reasoning analogy problems and frequently asked picture problems in previous competitive  examinations along with their answers .
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If square root of any number is added /subtracted infinite time ,then what will be the answers .



Problem # 1


(a) 5        (b) 25            (c) 3             (d) 49


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

Since 14 is attached with 24 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  t get 12.

So 30 must be the product 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 12 . So 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 .
Multiplying 5 with 2 and subtracting 3 from result so obtained to get 8. 
Multiplying 8 with 2 and subtracting 2 from it to get 13.
 Now multiplying 13 with 2 and then subtracting 4 from  result so obtained to get 22.
At last multiplying 22 with 2 then subtracting 5 from the result so obtained to get 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 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 so obtained.

Problem # 5


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

In figure number 1 we have 0 + 6 + 4 + 2 = 12 minus 2 = 10 { middle number }
In figure second 6 + 2 + 10 + 8 = 26 - 2  = 24 .
Similarly In figure 3  we have 4 + 14 + 12 + 10 = 40 - 2 = 38 ( Question Mark )  which is are 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 in first column .
Because 40 + 24 + 35 = 99
In second column we have 30 + 35 + 34 = 99 . While in last column when we add ? + 30  +27 the sum will be 99 so desired 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 figure I  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 figure II  7 + 6 is equal to 13 an  8+ 4 which is also equal to 12 so difference of both the  sum is equal to  13  - 12 1which is the middle number

In figure III  11 + 2 is equal to 13 and  0 + 2 which is also equal to  2 so difference of both the  sum is equal to  13 - 0 = 11 which will be the middle number.
        Correct option is c(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 (4) is right option .

Problem # 9


(a) 15       (b) 12          (c) 18       

   (d) 49


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 is when divided by 4  gives 12 
,So option   ( b ) is right 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 ( d ) 15 will be right option

Problem # 12


This series 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 the 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 ) in 2nd series is 4 And  difference between third number ( 13 ) and fourth number ( 18 ) in the second series 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 right option is option (a)

answer

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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.

 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

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) = 3 
In 2nd figure
(9+9) - (7+8) =3
In 3rd figure 
(8+8) - (6+6) =6

Hence in  option (4) 6 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 - 18 =6

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

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

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 
In 2nd figure 
3  × 14 = 6 × 7 = 42
In 3rd figure 
26 × 3 = 13 × 3 = 78

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

(1)  64   (2)  31 (3)  93  (4)  27

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 in 1st Figure
9×7 - 6×3 = 45
Similarly middle number of 3rd figure can be calculated as follows
8×8 - 4×6 = 40

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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Easy Reasoning analogy methods for problems of S S C Exams

Reasoning problems with their Solutions

Problem # 1


Starting from 5 and moving clockwise  we have these numbers 5, 10,?,50,122. 
so 5 can be written  as 2^2+1 starting from 1st prime  number i. e.  Square of 1st  prime number  and plus 1

10 can be written  as 3^2+1 , Square  of 2nd prime number  and plus 1.

50 can be written as 7^2 +1,Square  of  4th prime number  and plus 1.

Similarly 122 can be written as 11^2 +1, Square  of  5th prime number  and plus 1.

Hence ? can be replaced as 5^2 +1, Square  of  3rd prime number ( 3 )  and plus 1.
So
Therefore  Required  number  is 26.

Problem # 2




In this figure largest numbers are appearing in last row. so we should search the relation column wise .
In 1st column we have to search the relationship  between  9 and 6 to give 45. so in 2nd column we have to find relationship between 12 and 7 to give 95. And same logic we shall apply in 3rd columns. 

1st Column 

In 1st columns if we add and subtract both numbers  then multiply  it then we shall have 45 as follows
we have two numbers in 1st column  9 and 6 .
(9+6)×(9-6) = (15 ) × ( 3 ) = 45

2nd Column

And again in 2nd  columns if we add and subtract both numbers   of 2nd column and then multiply  these with each others  then we shall have 45 like this 
we have two numbers in 1st column  9 and 6 .
(12+7 ) × ( 12 - 7) = (19  ) × ( 5 ) = 95

3rd Column

Same logic can be applied for 3rd column ,we add and subtract both numbers   of 3rd column and then multiply  these with each others
( 8 + 3 ) × ( 8 -3 ) = (11 × 5 ) = 55
Hence 55 shall replace  "?" in the given figure 

Test your Reasoning ability


Problem # 3



Since largest numbers  in all the three rows lie in 1st column of the given figure.
It means we have to search a relation between  20 and 7 to give 28 in 1st row. and similar relation must be between 35 and 12 to give 84 .

1st Row

So in 1st row if we multiply both numbers with each others and divide the result  with 5 we shall get 28 like this 20 × 7 =140 
Now divide 140 with 5 such that 140/5 =  28 

2nd Row

So in 2nd row if we have to multiply both numbers with each others and divide the result  with 5 we shall get 28 like this 35 × 12 = 420 .

3rd Row

Now divide 420 with 5 such that 420/5 =  84
Same logic we have to apply in 3rd row to get answer 45 . So If we  multiply  9 with 45 we shall get 275 then we have to divide 405 with 5 to give  , It will replace ? question mark as follows 
9 × 25 = 225 / 5 = 45
 So  45 shall be the right number to replace question  mark .

Problem # 4

Since largest numbers appears on 3rd columns therefore the solution must be row wise, if we treat a and b as 1st and 2nd numbers then 3rd number which is our desired numbers must be equal to (a-1)×(b) i . e. product/ multiplication  of (a-1)  and b. Hence 3rd element of 1st row must be 
(8-1)×(3) = 7×3 = 21
3rd element of 2nd row must be 
(6-1)×(5) = 5×5 = 25
So 3td element of 3rd row must be 
(12-1)×(2) = 11×2= 22
So 22 will replace "?" Question  mark.

Problem # 5


In 1st column 5 × ( 6 +7 ) = 5 × 13 = 65
In 2nd column 4 × ( 3 +2 ) = 4 × 5 = 20
So same formula will be used in 3rd column
 In 3rd  column.  9 × ( ? +4 ) = 45
                                     ( ? +4 ) = 45/9 =5
                                              ? = 5 - 4  = 1
So required answer is   " 1 ".


Problem # 6


Multiplying 1st three elements of all the columns to get 4th elements 
1 × 8 × 9 = 72
3 × 6 × 5 = 90 
2 × 7 × ? = 56
This implies ? = 4 
So ? will be replaced by 4

Problem # 7


In 1st column square of three numbers i.e. 
( 1 + 4 + 2 )^2  = 7^2  = 49
In 2nd column square of three  numbers i.e.   ( 4 + 2 +2 ) ^2 =  ( 8)^2 = 64
In 3rd column the square of  ( ? + 5 + 3 ) ^2  must be 169
this implies  ( ? + 8 )^2 = 169
( ? + 8 ) = 13
? = 5
Required answer is 5

Problem # 8

If we calculate the sum of 1st column ,2nd column , 2nd row and 3rd row ,then it is found 25 in all the cases .so total of all rows and all the columns must be 25 . It can be seen that if we put "11"  in place of question mark then total of all the rows and columns is  25 

Problem # 9


Since biggest numbers are in the fourth columns of every row. So if we mutiply 1st three numbers and then add 4th number to it ,we shall have 5th number in every row .
( 4 × 3 × 2 )  +  8 = 24 + 8 = 32
( 5 × 3 × 1 ) + 9 = 15 + 9 = 24
( 7 × 3 × 3 ) + 7 = 63 + 7 = 70
Similarly when we multiply first three numbers and then adding force number to it in the last row we shall have fourth number in the last row like this
( 2 × 9 × 4 ) + 12 = 72 + 12 = 84
So 84 will replace question mark " ?"

So required answer is 84

Problem # 10


1st Row 
multiply 3 and 2 then add one less than the 2nd number to it
( 3 × 2 ) + ( 2 - 1 ) = 6 + 1 = 1 
2nd Row 
multiply 5 and 4 then add one less than the 2nd number to it
( 5 × 4  ) + ( 4 - 1 ) = 20 + 3 = 23
3rd Row
multiply 7 and 6 then add one less than the 2nd number to it
( 7 × 6 ) + ( 6 - 1 ) = 42 + 5 = 47
4th Row
( 9 × 8 ) + ( 8 - 1) = 72 + 7 = 79
5th Row

( 10 × 9 ) + ( 9 - 1 ) = 90 + 8 = 98
So 98 will replace. " ? " 

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How to find proportion of Four numbers

If A : B = 2 : 3, B : C = 4 : 5 and      C : D = 6 : 7,    what is A  : D  ?

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