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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