Number Analogy Reasoning questions with answers for competitive exams

Ten Most Important Number Analogy Reasoning questions with answers for competitive exams 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 about this post . 



Number Analogy Reasoning questions with answers for competitive exams




Problem # 1


Number Analogy Reasoning questions with answers for competitive exams
This box problem consist of four rows and four columns . And we have to find the value of question mark after studying the pattern of all the numbers in this box. 
        If we  add any three numbers in any row then we shall have fourth number in the same row in respective of its position.

Formula:-  


Sum of any three numbers in the same row  = 4th number in the same row

1st Row

In this row we can add 1st,  3rd and 4th number then 2nd number will be the result of these number's sum.
5 + 6 + 9 = 20

2nd Row

In this row we can add 1st, 2nd and 4th number then 3rd number will be the result of these number's sum.
4 + 8 + 3 = 15

3rd Row

In this row we can add 1st, 3rd and 4th number then 2nd number will be the result of these number's sum.
9 + 7 + 9 = 25

4th Row

In this row we can add 2nd, 3rd and 4th number then 1st number should be the result of these number's sum.
7 + 8 + ? = 22
15 + ? = 22
? = 22 - 15
? = 7
Therefore correct option is (B) 7

Problem # 2


Number Analogy Reasoning questions with answers for competitive exams 
In these types of reasoning problem where numbers are written in any circle. Then we can write these numbers in a line to study the pattern of these numbers. Either these numbers are in increasing order or in decreasing order. Sometimes these numbers are written in different pattern.
Writting all the given number in increasing order. we get
9 , 28 , ? , 126 , 217
All these numbers can be written as one more than the cube of  continuous natural numbers starting from 2.

Formula:- 

All the terms are one more than cube of some number.

1st Term = 2³ + 1 =  8 + 1 = 9
2nd Term = 3³ + 1 = 27 + 1 = 28
3rd Term = 4³ + 1 = 64 + 1 = 65(The value of Question mark)
4th Term = 5³ + 1 = 125 + 1 = 126
5th Term = 6³ + 1 =  216 + 1 = 217
Therefore correct option is (2) 65


Problem # 3


Number Analogy Reasoning questions with answers for competitive exams
In this reasoning problem 5 is related to 49 in the same way 7 will be related to ? . It  means we have to apply same mathematical operations to 7 to get ?.
    There are many method to obtain 49 from 5 , but we have to choose that method with the help of which we can obtain the value of question mark ?(out of four given option) from 7.
      So if we split 5 in to 2 , 3 and then combine the squares of these numbers(2² & 3² ⇒ 4 & 9 ⇒49). Then we shell have the the value of the number on the left side of symbol :: i. e. 49.
         Similarly if we split 7 in to 2 , 5 and then combine the squares of these numbers(2² & 5² ⇒ 4 & 25 ⇒425). Then we shell have the the value of the question mark  i. e. 425.

Therefore correct option is (1) 425

Problem # 4


Number Analogy Reasoning questions with answers for competitive exams
In this problem of reasoning we have to combine both the given  numbers in left hand side in such a way that after applying any mathematical operation/operations we could get the number on the right hand side of all the three problems given above. Because in these types of reasoning problems we can change given  mathematical sign according to our requirements.

Formula:- 


Sum of both the digits of 1st numbers ×  Product of both the digits of 2nd numbers = Number on the right hand side

1st Problem

(3+7) + (2×8) = 10 + 16 = 26 ⇔ 62(After Interchanging both the digits of number(26) just obtained).

2nd Problem

(1+9) + (1×2) = 10 + 2 = 12 ⇔ 21(After Interchanging both the digits of number(12) just obtained).

3rd Problem

(6+8)+ (4×2) = 14 + 8 = 22 ⇔ 22(After Interchanging both the digits of number(22) just obtained).
Therefore correct option is (3) 22

Problem # 5

Number Analogy Reasoning questions with answers for competitive exams
In this problem of reasoning we have to combine both the given  numbers in the left hand side in such a way that after applying any mathematical operation/operations we could get the number on the right hand side of all the three problems given above. 
                Because in these types of reasoning problems we can change given mathematical sign according to our requirements.
              In any individual problem if we add all the digits which are on the left hand side then we can get the value of the number on the right hand side.

Formula:- 


Sum of all the digits of both the numbers  on the left hand side = Number on the right hand side

1st Problem

After adding all the digits present in this problem 23@35 can be written as follows  
(2+3) + (3+5) = 13

2nd Problem

After adding all the digits present in this problem 24@13 can be written as follows 
(2+4) + (1+3) = 10

3rd Problem

After adding all the digits present in this problem 13@31 can be written  as follows  
(1+3) + (3+1) =  8
Therefore correct option is (2) 8

Problem # 6

Number Analogy Reasoning questions with answers for competitive exams
 If we multiply 1st number of the given pair with 12 then we can have 2nd number in any of the three given option except one. And that fourth option which do not follow the rule will be odd one  and will be the required answer.

Formula:- 

1st number(Any Pair)  ×  12 = 2nd number(Same Pair) 

1st Pair


10 ×  12 = 120
 

2nd Pair


20 ×  12 = 240

3rd Pair

14 ×  12 = 168(This is odd one because its multiplication is given196)

3rd Pair

12 ×  12 = 144
Therefore correct option is (3) (14, 196)

Problem # 7


Number Analogy Reasoning questions with answers for competitive exams
 
In this reasoning problem 37 is related to 23 in the same way 19 will be related to ? , It means we have to apply same mathematical operations to 19 to get the value of  ?.

1st Method

Taking the difference of both the numbers before and after :: sign ,The difference would be same in both the cases.
37 -  23 = 14
Similarly 19 - ? = 14 
- ? = 14 - 19
-? = -5
 ? = 5 

2nd Method

All the three numbers given in this problem are prime numbers so fourth will also be prime number. Therefore only one prime number out of given four options is 5.

Therefore correct option is (4) 5

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Problem # 8


Number Analogy Reasoning questions with answers for competitive exams

 This given reasoning problem having 9 numbers in it. Out of nine numbers ,one number in the middle of this figure and remaining eight numbers are in various positions. 
          To find the value of question mark (?) ,we can split this figure into four parts which includes these numbers 
1st Part consists of upper left most number and two numbers which are around it  (4 , 4 , 7)
2nd Part consists of bottom left most number and two numbers which are around it   (7 , 2 , 6) 
3rd Part consists of upper right most number and two numbers which are around it  (4 , 5 , 6)
4th Part consists of bottom right most number and two numbers which are around it  (6 , ? , 6).

Formula

 Sum of all the three numbers in each part separated above = 15

  4 + 4 + 7 = 1 (1st Part)
  7 + 2 + 6 = 15 (2nd Part)
  4 + 5 + 6 = 15 (3rd Part)
  6 + ? + 6 = 15 (4th Part)
   12 + ? = 15
   ? = 15 - 12
   ? = 3
Therefore correct option is (3) 3


Problem # 9


Number Analogy Reasoning questions with answers for competitive exams

This reasoning problem consists of three figures and every figure have four numbers associated to it . One numbers is at the centre of each figure and three numbers are at the corner of the each figure in the inner side . Look at last figure ,it have ? in its centre . 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 four numbers to get the value of this question mark?
          Now watch carefully the 1st two figures . Since these figures have some values at middle. 
          Now we have to find or search the  formula for these three numbers in each figure to utilised them in any possible way to get middle or central number. 
         The same formula will be applicable to third figure to find out the value of question mark.

Formula :-


The product of two numbers of the  right side of each triangle is equal to sum of two numbers on the left side of each triangle.

 1st Triangle 

12 × 3  = 26 + 10  = 36 

 2nd Triangle 

 6 × 9  = 43 + 11 = 54

 3rd Triangle 

9 × 5 =  ? + 15 = 30

? + 15 = 30

? = 30 - 15

? = 25

Therefore correct option is (4) 25

2nd Method

Formula :- 


 The middle Number in each figure is equal to product of two left most numbers minus number on the right side

 1st Triangle 

{( 12 × 3 )  - 10 } = ( 36 - 10) = 26

 2nd Triangle 

{( 6 × 9 )  - 11 } = ( 54 - 11) = 43

 3rd Triangle 

{( 9 × 5 )  - 15 } = ( 45 - 15) = 30

Therefore correct option is (4) 30


Problem # 10

Number Analogy Reasoning questions with answers for competitive exams


This reasoning problem also consists of three figures and every figure have four numbers associated to it . One numbers is at the centre of each figure and three numbers are at the corner of the each figure. Look at last figure , it have ? in its centre . 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 four numbers to get the value of this question mark?
          Now watch carefully the 1st two figures . Since these figures have some values at middle. 
          Now we have to find or search the  formula for these three numbers in each figure to utilised them in any possible way to get middle or central number. 
         The same formula will be applicable to third figure to find out the value of question mark.

Formula :-


 The middle Number in each figure is one tenth  of product of remaining numbers. 

Middle Number = abc/10

 1st Triangle 

(15 × 6 × 5)/10 = 450/10 = 45(Number in the middle of 1st triangle)

 2nd Triangle 

(7 × 6 × 5)/10 = 210/10 = 21(Number in the middle of 2nd triangle)

 3rd Triangle 

(20 × 30 × 1)/10 = 600/10 = 60(Number in the middle of 3rd triangle)


Therefore correct option is (2)60

Conclusion


So these were the Ten Most Important Reasoning questions with answers for competitive exams of number analogy with solutions were 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. 

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Ten Number Analogy Reasoning Questions with Answers for Competitive Exams

Ten Most Important Number Analogy Reasoning questions with answers for competitive exams 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. 



Number Analogy Reasoning questions with answers for competitive exams with solutions



Problem # 1


Separating these given numbers to two series by picking odd number and even number position
1st series 
64, 144, 324, ?  
2nd series
 96, 216 , 486 
The value of the number in even position is equal to the product of the square root of both the numbers in previous and next position of the number in even position.
2nd Term = Square root of 1st term ×   Square root of 3rd term 
                 = 64  × √144
                 =  8  ×  12
                 =  96 
4th Term = Square root of 3rd  term ×   Square root of 5th term 
                 =  √144  ×  √324
                 =  12  ×  18
                 =  216
6th Term =  Square root of 5th  term ×   Square root of 7th term 
                 =  √324  ×  √?
          486  =  18  × √?
             √?   =  486 /18
             ?   =  27, Squaring both sides, we get
              ?   =  729
Therefore correct option is (2) 726

 Problem # 2


This problem is a series problem where we have to find the value of question mark using all its previous terms by analysing the trend of all its terms whether they are in increasing order or decreasing order or in any other format.

Formula :- 

Keep on adding  1 , 2 , 3 , 4 , 5 and so on to get next number of series

1st number = -4
2nd number = 
1st number +1 = -4 +1 =  -3
3rd number =  2nd number + 2  = -3 + 2 = -1
4th number =  3rd number + 3  = -1 + 3 = 2 
5th number =  4th number + 4 = 2 +  4 = 6
6th number =  5th number + 5 = 6 + 5 = 11
Option (4)11 is correct option 

 Problem # 3


This problem is a series problem where we have to find the value of question mark using its previous term/s means by analysing the trend of all its terms whether they are in increasing order or decreasing order or in any other format.

Formula :- 

Every term is the sum of its two preceding terms.

Sum of last two terms is equal to next term
3rd term = 4 + 7 = 11
4th term = 7 + 11 = 18
5th term = 11 + 18 = 29
6th term = 18 + 29 = 47
7th term = 29 + 47 = 76 = ?
8th term = 47 + 76 = 123
9th term =  76 + 123 = 199
Therefore correct option is (4) 76

 Problem # 4

Assuming 1st three digits as single number i.e 289, Again Assuming 4th to 6th  digits as single number i.e. 324 , Similarly Assuming last three digits as single number i.e. 36?.
Now carefully analyse or study these three numbers so obtained. Since 289 is square of 17 (i.e. 17² = 289)  and  324  is square of 18 (i.e. 18² = 324) and finally 36?  must be square of  19. But 19² = 361. Hence the value of question mark will be 1.
Therefore correct option is (4) 1

             Problem # 5

In this problem of reasoning we have to combine both the given  numbers in such a way that after applying any mathematical operation/operations we could get the number in the right hand side of all the three problems given above. Because in these types of reasoning problems we can change mathematical sign according to our requirements.
Sum of all the digits is equal to the given number
(2+ 3) + (3+5) = 5 + 8 = 13
(2+ 4) + (1+3) = 6 + 4 = 10
(1+ 3) + (3+1) = 4 + 4 = 8
Therefore correct option is (2)8

 Problem # 6

 
Splitting into two series by picking alternating numbers like this
11 ,17 , 23 , ?
And 13 , 19 , 25
1st series is with a difference of 6 and 2nd series is also with a difference of 6 .
Hence if we add 6 to 23 then value of question mark can be found like this
23 + 6 = 29
Therefore correct option is (3)29

 Problem # 7



These type of series problem in reasoning can be understand by using these formulas
(1) Increasing series
(2) Decreasing series
(3) Alternate series
(4) Other series
Here in this series we can split these numbers into to series by categorising these numbers in Alternate series as follows
2 , 4 ,8 , 16 , 32  and 6 , 9 , 13 18 , ? 
Now study the 1st series , these numbers are increasing abruptly, so we can get next number from previous number by multiplication of some factor. But study the 2nd series , these numbers are not increasing abruptly, although these numbers are increasing at slower rate,  so we can not get next number from previous number by multiplication of some factor. Hence in 2nd case we can get next number from previous number by addition of some factor.
Hence 1st Series
Any Number = (2  × Previous Number)
2nd Number = 2  × 1st Number  = 2  × 2 = 4
3rd Number = 2  × 2nd Number = 2  × 4 = 8
4th Number = 2  × 3rd Number  = 2  × 8 = 16
5th Number = 2  × 4th Number  = 2  × 16 = 32
 We can not get the value of question mark from this series because question mark is at even position and all the numbers in this series are odd position.
Now 2nd Series
Any Number = { (2 + n)  +  Previous Number } , Where n is natural number staring from 1 and will be increased by 1 everytime.
2nd Number = {(2 + n) + 1st Number} = {(2 + 1) + 6} = 3 + 6 = 9
3rd Number = {(2 + n) + 2nd Number} = {(2 + 2) + 9} = 4 + 9 = 13
4th Number = {(2 + n) + 3rd Number} = {(2 + 3) + 13} = 5 + 13 = 18 
5th Number  = {(2 + n) + 4th Number} = {(2 + 4) + 18} = 6 + 18 = 24 
The correct Answer is option (1)24

 Problem # 8

In this problem of reasoning we have to combine both the given  numbers in such a way that after applying any mathematical operation/operations we could get the number in the right hand side of all the three problems given above. Because in these types of reasoning problems we can change mathematical sign according to our requirements.
(8 × 5) + (8 - 5)² = 40 + 3² = 40 + 9 = 49
(5 × 3) + (5 - 3)² = 15 + 2² = 15 + 4 = 19
(6 × 4) + (6 - 4)² = 24 + 2² =24 + 4 = 28
Therefore correct option is (2) 24

 Problem # 9

In this problem of reasoning we have to combine both the given  numbers in such a way that after applying any mathematical operation /operations we could get the number in the right hand side of all the three problems given above. Because in these types of reasoning problems we can change mathematical sign according to our requirements.
37 - 8 = 29 and 72 - 8 = 64 , Now we can combine these two results so obtained to get value of the number on the right hand side of 1st problem.
Hence 1st number = 2964.
In the same way 58 - 8 = 50 and 12 - 8 = 04,
Now we can combine these two results so obtained to get value of the number on the right hand side of 2nd problem.
Hence 2nd number 5004
Now required Number
88 - 8 = 80 and 16 - 8 = 08 ,
Now we can combine these two results so obtained to get value of the number on the right hand side of 3rd problem.
 Hence 3rd number  8008
Therefore correct option is (4) 8008

 Problem # 10

Assuming 1st four digits as single number i.e 4096,   Again assuming 5th to 8th  digits as single number i.e. 4913 , Similarly assuming last four digits as single number i.e. 5?32.
Since 4096 is cube of 16  and  4913  is cube of 17 and finally 5?32  must be cube of  18. But 18³ = 5832. Hence the value of question mark will be 8.

Therefore correct option is (1) 8


Conclusion



So these were the ten Most Important Reasoning questions with answers for competitive exams of number analogy with solutions were 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. 








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Most Important box Problems of Reasoning for different competitive Exams

Reasoning of missing number in box problems will be discussed with the help of  most important examples. Some of these examples are of 3 × 3 order and other are of different orders . 


Most Important Box Problems of Reasoning for different competitive Exams



Problem # 1


box Problems of Reasoning for different competitive Exams
This reasoning problem consists of three figures and every figure have five numbers associated to it . Four numbers are on the corner of each box and one number is in the middle of every box.  Look at last figure , it have ?(question mark) in its centre . So the solution of this problem is to find the value of question mark using other four  numbers associated to it . 

          But the main aim is how to utilised  these two numbers to get the value of question mark?.
          We have to find or search the  formula for these four numbers in each figure to utilised them in any possible way to get number in  each box . 

Formula :-

Left most number in 1st row of any box + { Product of all other numbers in same box } = Middle number in that box.

1st Box
3 + (4  × 5  × 3) = 3 + 60 = 63
2nd Box
6 + (7  × 3  × 5) = 6 + 105 = 111
3rd Box
2 + (6  × 5  × 4) = 2 + 120  = 122
Therefore option (3) 122 is correct.


Problem # 2


box Problems of Reasoning for different competitive Exams
This reasoning problem consists of two figures and every figure have five numbers associated to it . Four numbers are on the corner of each box and one number is in the middle of both the box.  Look at 2nd figure , it have ? in its centre . So the solution of this problem is to find the value of question mark using other four  numbers associated to it . 
                   We have to find  the  formula for these four numbers in both figures to utilised them in any possible way to get number in  each box . 

Formula

Product of both the numbers in 1st row - Product of both the numbers in 3rd row = Number in central row

1st Box

(11 × 12) - ( 6 × 9) = 132 - 54 = 78.

2nd  Box

(14 × 10) - ( 7 × 8) = 140 - 56 = 84
Therefore option (1) 84 is correct. 

Problem # 3


box Problems of Reasoning for different competitive Exams
This box problem consist of three rows and three columns . And we have to find the value of question mark after studying the pattern of all the numbers in this box. This question mark is in 3rd row of 3rd column. 

Formula:- 

Square of Number in 2nd row ÷ Number in 1st row = Number in 3rd row 

8² ÷ 4 = 64 ÷ 4 = 16
6² ÷ 3 = 36 ÷ 3 = 12
10² ÷ 2 = 100 ÷ 2 = 50
Therefore option (2) 50 is correct. 
 

Problem # 4


box Problems of Reasoning for different competitive Exams
This box problem consists of six rows and two columns. And we have to find the value of question mark after studying the pattern of all the numbers in this box. This question mark is in 4th row of 2nd column. 

Formula:-

The sum of both the numbers in any row = Number in 1st column of  lower row.

6 + 9 = 15 (The number in 1st column of 2nd row)

15 + 7 = 22 (The number in 1st column of 3rd row)

22 + 5 = 27  (The number in 1st column of 4th row)

27 +  ? = 36 ⇒  ? = 36 - 27⇒ = 9 (The number in 1st column of 5th row)

36 + 3 = 39 (The number in 1st column of 6th row)
Therefore option (1) 9 is correct. 

Problem # 5


box Problems of Reasoning for different competitive Exams
This box problem also consist of six rows and two columns . And we have to find the value of question mark after studying the pattern of all the numbers in this box. This question mark is in 2nd row of 2nd column. 

1st Column

Formula :-

Sum of two numbers in consecutive rows of same column = Number in Succeeding row of same column taken in order from top to bottom.

3 + 6 = 9 (Sum of the numbers in 1st and 2nd rows is equal to number in 3rd row of 1st column).

6 + 9 = 15 (Sum of the numbers in 2nd and 3rd rows is equal to number in 4th row of 1st column).

9 + 15 = 24 (Sum of the numbers in 3rd and 4th rows is equal to number in 5th  row of 1st column).

15 + 24 = 39 (Sum of the numbers in 4th and 5th rows is equal to number in 6th  row of 1st column).

2nd Column

Formula:-


Sum of two numbers in consecutive rows of same column = Number in Succeeding row of same column taken in reverse order from bottom to top.

4 + 7 = 11 (Sum of the numbers in 6th and 5th rows is equal to number in 4th row of 2nd column).

7 + 11 = 18 (Sum of the numbers in 5th and 4th rows is equal to number in 3rd row of 2nd column).

11 + 18 = 29 (Sum of the numbers in 4th and 3rd rows is equal to number in 2nd row of 2nd column).This is also the value of question mark.

18 + 29 = 47 (Sum of the numbers in 3rd and 2nd rows is equal to number in 1st row of 2nd column).
Therefore option (1) 29 is correct. 
 

Problem # 6


box Problems of Reasoning for different competitive Exams

This box problem consist of three rows and three columns . And we have to find the value of question mark after studying the pattern of all the numbers in this box. This question mark is in 2nd row of 3rd column. 
     To find the value of question mark. we  shall divide this box into two parts vertically then we can have the formula for these numbers written in this box . Because after careful observation we can see that the product of both the numbers in left half in any particular row is equal to sum of both the numbers in right half in that particular row. 
This reasoning problem can be solved by two different methods.

1st Method

Column wise

(1 × 2 ) + 1 = 2 + 1 = 3 (1st Column)
(7 × 14 ) + 7 = 98 + 7 = 105 (2nd Column)
(9 × ? ) + 9 =  117  (3rd Column)
(9 × ? )  =  117 - 9
(9 × ? )  =  108
? = 108/9
? = 12

2nd Method

(2 + 1 ) × 1 =  3
(14 + 1 ) × 7 =  15 × 7 = 105
(? + 1 ) × 9 =  117
(? + 1 )  =  117/9
(? + 1 )  =  13
?  = 13 - 1 
? = 12
Therefore option (2) 12 is correct. 

Problem # 7


box Problems of Reasoning for different competitive Exams

This box problem consist of three rows and four columns .And we have to find the value of question mark after studying the pattern of all the numbers in this box. This question mark is in 3rd row of 3rd column. 
This reasoning problem can be solved by two different methods.

1st Method

Formula:-

{R4 ÷ R3} ×  R2 = R1

(8 /4) × 3 =2 × 3 = 6
(27 /3) × 2 = 9 × 2 = 18
(9/?) × 5 = 15
(9/?)  = 15/5 = 3 
(?/9)  = 1/3
? = 9/3
? = 3

2nd Method

Multiplication of 1st and 3rd columns = Multiplication of 2nd and 4th columns.

Formula:-

{R1 × R3} = {R2 × R4}

6 × 4 = 8 × 3
18 × 3 = 2 × 27
15 × ? = 5 × 9
? = 45/15
? = 3
Therefore option (4) 3 is correct. 


Problem # 8


This reasoning problem consists of three figures and every figure have five numbers associated to it . Four numbers are on the outer side of each box and one number is in the middle of both the box.  Look at 3rd figure , it have ? in its centre . So the solution of this problem is to find the value of question mark using other four  numbers associated to it. 
                   We have to find  the  formula for these four numbers in 1st two figures to utilised them in any possible way to get number in  each box . 

1st method

Formula :- Product of all the numbers in outer parts  ÷  10 = Middle number

(5 × 3 × 4 × 2)/10 = 120/10  =  12 (1st Box)
(5 × 6 × 2 × 3)/10 = 180/10 =  18 (2nd Box)
(5 × 2 × 2 × 9)/10 = 180/10 =  18 (3rd Box)

2nd Method


Since 5 and 2 are common in all the three figures, so ignore these two numbers and check the multiplication of other two numbers to get middle number in all the three figures.
3 ×  4 = 12(1st Box)
6 ×  3 = 18(1st Box)
9 ×  2 = 18(3rd Box)

Also Reads these Articles


Reasoning of missing number in box problems  with solutions 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 regarding this post.

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