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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

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

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

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

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

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.

### 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

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

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)
Therefore option (3) 18 is correct.

## Problem # 9

### 1st Method

#### Formula :-

Any Term = { (Previous Term) × 2 } + n, where n is natural number less than 6

1st Term = (4 × 2) + 1 = 8 + 1 = 9
2nd Term = (9 × 2) + 2 = 18 + 2 = 20
3rd Term = (20 × 2) + 3 = 40 + 3 = 43
4th Term = (43 × 2) + 4 = 86 + 4 = 90
5th Term = (90 × 2) + 5 = 180 + 5 = 185(The value of question mark).

#### 2nd Method

Now taking difference of two consecutive terms
Again taking difference of two consecutive terms. From last row we are seeing that differences are double of previous difference , Hence next difference must be 48.
Therefore
x - 47 = 48
x = 48 + 47
x = 95
Hence differences in 2nd row would be 5 , 11 , 23 , 47, 95
And from 1st row
? - 90 = 95
? = 95 + 90
? = 185
Therefore option (3) 185 is correct.

## Problem # 10

Since question mark in this figure is in the outer circle of this figure. To find the value of question mark (?). We can combine two numbers present in the inner circle with the help of any mathematical operation/s to find the value of number in the outer circle .
Starting from 3 and moving clockwise , taking two numbers at a time in inner circle will be equal to number in outer circle.

### Formula :-

{2 × 1st Number} +  {2nd Number} =   Number in outer circle
(2 × 3) + 4 = 6 + 4 = 10
(2 × 4) + 6 = 8 + 6 = 14
(2 × 6) + 8 = 12 + 8 = 20
(2 × 8) + 3 = 16 + 3 = 19 = ?(The value of question mark)
Therefore option (1) 19 is correct.

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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## Mix reasoning practice set in english ,SSC Reasoning Practice Set PDF Download

Mix reasoning practice set in English, SSC Previous Year Reasoning Questions PDF,  Most important  Reasoning  questions with answers  which  includes circle  problems, box problems, triangles problems  for competitive exams like SSC CGL , SSC CHSL ,CPO ,Bank exams and RRB NTPC etc have been explained in this post.

# Ten Important Reasoning questions with answers for competitive exams

## Problem # 1

This box problem consist of four rows and four columns . And we have to find the value of question mark  in 4th row after studying the pattern of all the numbers in this box. This question mark is in 4th row of 2nd column.
Formula :- (R2 - R3) × R1 = R4

### Calculation

Column 1
(15 -14 ) × 10 = 10
Column 3
(8 - 6 ) × 7 = 14
Column 4
(25 -12 ) × 5 = 65
Column 2
(7 - 3 ) × 4 = 16
Option (A)16 is correct option.

## Problem # 2

This problem figure consists of four rows and every rows from top to bottom have one more row in comparison to previous row.
If we take the product of all the numbers in each row and compare the result so obtained ,then we have same result in every row.

Product of all the numbers in every row is 120.

2 × 5 × 4 × 3 = 120 ( Last Row )
4 × 15 × 2 = 120  (2nd Last Row )
10 × 12 = 120  (2nd   Row )
Hence the value of   "?"   must be 20
Option (4)120 is correct option.

## Problem # 3

This circle consists of four quadrants and every  quadrant consists of three numbers . And every quadrant have two numbers in outer part and one  number  in the inner part . To find the value of question mark  "?"  , we shall use two numbers which are in the outer part to calculate the value of the number which is in the inner part of every quadrant .

Since the question mark is in 1st quadrant ,so we have to start the calculation from 2nd quadrant.

Step 1.   Take sum of both the numbers (16 and 20 ) in outer part of  this quadrant.

Step 2. Take the square root of  the number obtained in step 1.

### Calculation

√( 20 +16) = √(36) =6

Step 1.   Take sum of both the numbers (5 and 4 ) in outer part of  this quadrant.

Step 2. Take the square root of  the number obtained in step 1.

### Calculation

√( 5 + 4 ) = √( 9) =3

Step 1.   Take sum of both the numbers (5 and 4 ) in outer part of  this quadrant.

Step 2. Take the square root of  the number obtained in step 1.

### Calculation

√(15 +10) = √(25) =5

Step 1.   Take sum of both the numbers (5 and 4 ) in outer part of  this quadrant.

Step 2. Take the square root of  the number obtained in step 1.

### Calculation

√( 30 +19) = √(49) = 7
Option  (1)7 is correct option.

## Problem # 4

Writing the 1st Number as it is and then place its cube on the right side of it. Now moving clockwise pick the next odd number for next position.
1³ = 1  ⇒ 11
3³ = 27 ⇒327
5³ = 125 ⇒5125
7³ = 343 ⇒7343
9³ = 729 ⇒9729

## Problem # 5

This box problem consists of three rows and three columns. To find the value of question marks in 3rd column of 3rd row, we have to add 1st ,2nd and 3th columns since since in every column is same in every row.

Formula :-  .
F + 4 = J  ,  J  + 4 = N
M + 4 = Q , Q + 4 = U
O + 4 = S  ,  S  + 4 = W
Option (B)W is correct option.

## Problem # 6

This box problem consists of three rows and three columns. To find the value of question marks in 3rd column of 3rd row, we have to add 1st ,2nd and 3th columns since since in every column is same in every row.

Formula :- The sum of 1st three column in every row is same .
B + 1 = D ,  D + 1  = F
N + 1 = P ,  P + 1  = R
H + 1 = J ,  J + 1  = L = ?
And 3 + 3 = 6
7 + 9 = 16
5 + 6 = 11
Hence  L11
Option (B)L11 is correct option.

## Problem # 7

Starting from 4th quadrant in anti clockwise direction ,take the sum of  both the numbers in outer part of the quadrant .Then reverse the orders of both the digits of the number so obtained to get the value of the number in the inner part of same quadrant.

9 × 8 = 72 ⟺ 27 (After reversing the orders of digits)

8 × 3 = 24 ⟺ 42 (After reversing the orders of digits)

5 × 7 = 35 ⟺ 53 (After reversing the orders of digits)

7 × 7 = 49 ⟺ 94 (After reversing the orders of digits) (The value of question mark).

Hence option (2)94  is correct.

## Problem # 8

Splitting all the numbers in the given circle into two series

## Any Term = Preceding Term + 4

### 1st term = 28

2nd term = 1st term + 4
2nd term  28  + 4 = 32
2nd term  = 32

3rd term = 2nd term + 4
3rd term  32  + 4 = 36
3rd term   = 36

4th term = 3rd term + 4
4th term 36  + 4 = 40
4th term = 40

## 2nd Series

?  , 31 , 33, 35 with a difference of 2 , So the 1st term of this series must be 2 less than the 2nd term.

## Any Term = Preceding Term - 2

4th term = 5th term - 2
4th term =  37 - 2
4th term = 35

3rd term = 4th term - 2
3rd term =  35 - 2
3rd term = 33

2nd term = 3rd term - 2
2nd term =  33 - 2
2nd term = 31

Therefore 1st term = 2nd term - 2
1st term =  31 - 2
1st term = 29

## Problem # 9

Every term can be expressed with the help of preceding term.

(2 × 1) + 2 = 2 + 2 = 4 ( 2nd number in series)
(4 × 2) - 3 = 8 - 3 = 5 ( 3rd number in series)
(5 × 3) + 4 = 15 + 4 = 19 ( 4th number in series)
(19 × 4) - 5 = 76 - 5  = 71 ( 5th number in series)
(71 × 5) - 6 = 355 - 6  = 349 ( 6th number in series)

## Problem # 10

5 = 1st term
(5 × 3) + 1 = 15 + 1 =16 = 2nd term
(16 × 3) + 2 = 48 + 2 =50 = 3rd term
(50 × 3) + 3 = 150 + 3 =153 = 4th term = value of question mark
(153 × 3) + 4 = 459 + 4 =463 = 5th term
(463 × 3) + 5 = 1389 + 5 = 1394 = 2nd term