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

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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Easy Reasoning analogy methods for problems of S S C Exams

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 


Easy Reasoning analogy methods for  problems of S S C Exams
Starting from 5 and moving clockwise  we have these numbers 5, 10,  ?, 50, 122. 
so 5 can be written  as 2²+1 starting from 1st prime  number i. e.  Square of 1st  prime number  and plus 1

10 can be written  as 3²+1 , Square  of 2nd prime number  and plus 1.


50 can be written as 7² +1, Square  of  4th prime number  and plus 1.

Similarly 122 can be written as 11² +1, Square  of  5th prime number  and plus 1.

Hence ? can be replaced as 5² +1, Square  of  3rd prime number ( 3 )  and plus 1.
So
1st Term = 5 = 2² + 1
2nd Term = 10 = 3² + 1
3rd Term = ? = 5² + 1
4th Term = 50 = 7² + 1
5th Term = 122 = 11² + 1
Therefore  Required  number  is 26.

Problem #2 


Easy Reasoning analogy methods for  problems of S S C Exams

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. similarly 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 will 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


Easy Reasoning analogy methods for  problems of S S C Exams

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

Easy Reasoning analogy methods for  problems of S S C Exams
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


Easy Reasoning analogy methods for  problems of S S C Exams
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


Easy Reasoning analogy methods for  problems of S S C Exams
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


Easy Reasoning analogy methods for  problems of S S C Exams
In 1st column square of three numbers i.e. 
( 1 + 4 + 2 )²  = 7²  = 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

Easy Reasoning analogy methods for  problems of S S C Exams
If we calculate the sum of 1st column ,2nd column or  2nd row and 3rd row ,then it is found 25 in both 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 . 

 Row Wise 

6 + 8 + ? = 9 + 3 + 13 = 10 + 14 +1 = 25 

Column Wise 

6 + 9 + 10 = 8 + 3 + 14 = ? + 13 +11 = 25 
? = 24  = 25
? = 25 - 24 
? = 1
So required answer is 1

Problem #9

Easy Reasoning analogy methods for  problems of S S C Exams

Since biggest numbers are in the fourth columns of every row. So if we multiply 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 fourth 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


Easy Reasoning analogy methods for  problems of S S C Exams

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

How to find proportion of 4 numbers, proportion of Four numbers with the help of examples


What is Proportion



Proportion is majorly based on ratio and fractions. A fraction is written in the form of a/b, while ratio a:b, then a proportion means that two ratios are equal. Here a and b are any two numbers. The ratio and proportion are key foundations to understand the various concepts in mathematics .

Proportion can be applied to solve many daily life problems based on mixture ,work and time , time and speed etc . It develops a relation between two or more quantities and thus helps in their comparison.

What is Proportion?


Proportion in general is referred to as a part, share, or number considered in comparative relation to a whole. Proportion says that when two ratios are equivalent, they are in proportion. It is an equation or statement used to depict that two ratios or fractions are equal .

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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Most Important Reasoning questions with answers for competitive exams

Most Important Reasoning questions with answers for competitive exams of number analogy with solutions have been 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. please feel free to comment your opinions. 

What is next terms in the series ?

Let us solve next number in sequence puzzle
15, 29, 56, 108, 208, ????
Here 1st term is 15, To find out 2nd term if we multiply the 1st term With 2 and subtract 1 from it, i.e (15 × 2) -1  = 30 - 1= 29
To find 2nd term we have to multiply the previous term with 2 and subtract 2 (double the previous time) from it
i.e (29 × 2) - 2 = 58 - 2 = 56
To find 3rd term we have to multiply the previous term with 4   and subtract 2  (double the previous time) from it
i.e  (56 × 2) - 4 = 112 - 4 = 108
To find 4th term we have to multiply the previous term with 8 (double the previous time) and subtract 2 from it
i.e (108 × 2) - 8 = 216 - 8 =  208
Similarly 5th term can be calculated as
multiply the previous term with 2 and subtract  subtract 2 (double the previous time) from it
i. e.  (208 × 2) - 16 = 416 - 16 =  400
Next term (400×2) - 32 = 800 - 32 =  768

Next term  (768 × 2) - 64 = 1536  - 64 = 1472
Next term  (1472 × 2) - 128 = 2944 - 128  =2816
Or we can write it in another form as follows
15 + (15–1) = 15 + 14= 29
29 + (29–2) = 29 + 27 = 56
56 + (56–4) = 56 + 52 = 108
108 + (108–8) =108 + 100 = 208
208 + (208–16) = 208 + 192= 400
400 + (400–32) = 400 + 368 = 768

768 + (768–64= 768 + 704 = 1472

1472 + (1472-128) = 1472 + 1344 = 2816

And similarly this trend follows


What’s the next number in the sequence 0, 2, 6, 12 , ?


We can get the answer in two  ways

1st Method


1 × (1–1) = 0 ( 1st term)

2 × (2–1) = 2 (2nd term)

3 × (3–1) = 6 (3rd term)

4 × (4–1) = 12 (4th term)

5 × (5–1) = 20 (5th term) 

Next term must be  20
Similarly 
6 × (6–1) = 30  (6th term)

7 × (7–1) = 42  (7th term)


2nd Method



1² – 1 = 1 - 1  = 0

2² –  2 = 4 - 2 = 2 

3² – 3 = 9 - 3 = 6

4² – 4 = 16 - 4 = 12 

5² – 5 = 25 - 5 =  20 
Next term must be  20
Similarly 

6² – 6 = 36 - 6 = 30 

7² – 7 = 49 - 7=  42

Can you solve this  Series: 30, 31, 28, 33, 26, 35, … What number should come next?


We can split this sequence into two series by picking alternate numbers from the given sequence. 
30, 31, 28, 33, 26, 35, …
1st  one is 30,  28, 26, … and 2nd one is  31, 33,  35, …
Now after careful analysing the 1st sequence ,its next terms should be  with decrease of  2 ,we get other terms as follows 
30,  28, 26, 24, 22 ,20 …
And after careful analysing the 2nd  sequence ,its next terms should be  with increase of  2 ,we get other terms as follows 
31, 33,  35, 37, 39 ,41… 

26 - 2 = 24.
So the text terms should be 30, 31, 28, 33, 26, 35, 24, 37, 22 

24 is the next number in the required sequence.

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