Most important Twenty Number analogy questions of Reasoning with answer

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Twenty Questions of Number analogy with answer for competitive examinations and missing number in reasoning for competitive exams like Bank PO, Bank clerk, SSC CGL, ssc chsl, RRB NTPC , group D etc have been discussed in this post.

Twenty Questions of number analogy for competitive exams

In this reasoning problem 1st number (8) is associated to 28 with the help of any rule , with the help of same rule we have to associate 27 to a number out of four given options.

Formula:-

1st number is cube of a number and 2nd number more than one the cube of successive number of 1st number.

2³ , (2+1)³+1 ,3³ , (3+1)³ +1

2³ , 3³+1 ,3³ , 4³ +1

Hence 4³ + 1 = 64 + 1= 65

Option (4)65 is correct option.

In this reasoning problem 1st number (46) is associated to 2549 with the help of any rule , in the same rule we have to associate 23 to a number out of four given options.

Formula:-

Squares of successive digits of number on the left hand side = Number on the right hand side

(4+1)² = 5² = 25

(6+1)² = 7² = 49

Similarly (2+1)² = 3² = 9

(3+1)² = 4² = 16

Hence 916 will be the value of question mark .

Option (3)916 is correct option.

1st Method

In this reasoning problem 1st number (37) is associated to 23 with the help of any rule , in the same rule we have to associate 19 to a number out of four given options.

Look carefully the given numbers consists of two digits. These two digits can be utilised with the help a formula given below.

Formula :-

1st Number - 2nd Number = 14

37 - 23 = 14

19 - ? = 14

? = 19 - 14

? = 5

Option (4)5 is correct option.

2nd Method

All the three given numbers are prime numbers, so the value of question mark must be a prime number. And out of all the four options given only fourth option having value 5 is prime number.

Hence Option (4)5 is correct option.

In this reasoning problem 1st number 29 is associated to 71 with the help of any rule , in the same rule we have to associate 79 to a number out of four given options.

1st Number

29 × 2 = 58 Now add sum of both the digits of 58 to 58 to get 2nd number.

58 + (5+8) = 58 + 13 = 71

3rd Number

79 × 2 = 158 Now add sum of all the digits of 158 to 158 to get 4th number.

158 + (1+5+8) = 158 + 14 = 172

Hence Option (3)172 is correct option.

In this reasoning problem 1st number (79) is associated to 47 with the help of some rule , With the help of same rule we have to associate 75 to a number out of four given option. Also in this problem all the numbers consists of two digits. These two digits can be used to find the value of question mark with the help a formula given below.

Formula :-

( 1st digits × 2nd digit } - {1st digits + 2nd digit}.

{ 9 × 7 } - { 9 + 7 } = 63 - 16 = 47

{ 7 × 5 } - { 7 + 5 } = 35 - 12 = 23 = ? ( The value of question mark )

Option (2)23 is correct option.

In this reasoning problem 1st number 87 is associated to 414 with the help of some rule , With the help of same rule we have to associate 62 to a number out of four given option. Also in this problem all the numbers consists of three digits. These three digits can be used to find the value of question mark with the help a formula given below.

Formula :-

{(1st digit of 1st Number) ÷ 2} and {(2nd digit of 1st Number) × 2 } = 2nd Number

{ 8 ÷2 } and { 7× 2 } = 4 and 14 = 414 = 2nd Number

{ 6 ÷2 } and { 2× 2 } = 3 and 4 = 34 = 4th Number = ? ( The value of question mark )

Option (2)34 is correct option.

Formula :-

{1st digit × 2nd digit } + { 3rd digit × 4th digit } = 2nd Number.

1st Number

(2 × 5) + (7 × 3) = 10 + 21 = 31 = 2nd Number

2nd Number

(4 × 6) + (2 × 9) = 24 + 18 = 42 = 4th Number

Option (3)42 is correct option.

In this reasoning problem 1st number 144 is associated to 23 with the help of some rule , With the help of same rule we have to associate 196 to a number out of four given option. Also in this problem all the numbers consists of three digits. These three digits can be used to find the value of question mark with the help a formula given below.

Formula :-

{2 × Square Root (1st Number) }- 1 = 2nd Number

{2 × √144} - 1 = {2 × 12} - 1 = 24 - 1 = 23 = 2nd Number

{2 × √196} - 1 = {2 × 14} - 1 = 28 - 1 = 27 = 4th Number = ? ( The value of question mark )

Option (1)27 is correct option

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Squaring both digits of 1st number separately, now Interchanging both digits of numbers obtained after Squaring. Now combining these numbers to form the value of 2nd number.

1st Number = 43

4² = 16 = 61(After Interchanging both digits)

3² = 09 = 90(After Interchanging both digits)

Now combining both the squares as 6346 = 2nd number.

3rd Number = 68

6² = 36 = 63(After Interchanging both digits)

8² = 64 = 46(After Interchanging both digits)

Now combining both the squares as 6346= 4th number.

Option (3)6346 is correct option

Prime factors of 6 = 2,3 and multiplication of 2 and 3 = 6

Prime factors of 15 = 3,5 and multiplication of 3 and 5 = 15

Prime factors of 35 = 5,7 and multiplication of 5 and 7 = 35

Similarly

Prime factors of ? = 7,11 and multiplication of 7 and 11 = 77.

Option (1)77 is correct option

Usually in these types of reasoning problem 1st number is associated to 2nd number with the help of some rule . Similarly 3rd number is associated to 4th number with the help of same rule.

But in this reasoning problem all the four numbers are written sequence wise . It Means 2nd term will be found from 1st term, 3rd term will be found from 2nd term and after applying same formula or rule 4th term will be calculated from 3rd term.

Formula :-

nth Term = {(n+1) × Previous Term} + 4

1st term = 2

2nd term = (8 × 3) + 4 = 24 + 4 = 24

3rd term = (28 × 4) + 4 = 112 + 4 = 116

4th term = (116 × 5) + 4 = 580 + 4 = 584

Option (4) 584 is correct option

In this reasoning problem 1st number (11) is associated to 38 with the help of any rule , in the same rule we have to associate 13 to a number out of four given option.

Look carefully the given numbers consists of two digits. These three digits can be utilised with the help a formula given below.

Formula :-

(1st Number × 4) - 6 = 2nd Number

{11 × 4 } - 6 = 44 - 6 = 44 = 2nd Number

{13 × 4 } - 6 = 52 - 6 = 46 - 4th Number

Option (2)46 is correct option

Interchanging position of unit place and ten's place of given 1st number 34 . we shall get 43, Now taking the squares of each digit of given number 43 separately. we shall have 16 and 9. Now combining these squares to get 2nd numbers which is 169.

Similarly interchanging position of unit place and ten's place of 3rd number 23 . we shall get 32. Now taking the squares of each digit of given number 32 separately. we shall have 9 and 4. Now combining these squares to get 4th numbers which is 94.

Option (4)94 is correct option

In this reasoning problem 1st number (0) is associated to 6 with the help of some rule , With the help of same rule we have to associate 24 to a number out of four given option.

Formula :-

nth Term = n³ - n

0 = 1³ -1 = 1st number

6 = 2³ -2 = 2nd number

24 = 3³ -3 = 3rd number

? = 4³ - 4 = 64 - 4 = 60 = 4th number

Option (4)60 is correct option

In this reasoning problem 1st number (49) is associated to 23 with the help of some rule , With the help of same rule we have to associate 91 to a number out of four given option. Also in this problem all the numbers consists of two digits. These two digits can be used to find the value of question mark with the help a formula given below.

Formula :-

{Square Root of 1st digits and Square Root of 2nd digit } of 1st Number = 2nd Number

√4 and √9 = 2 and 3 = 23 = 2nd Number

√9 and √1 = 3 and 1 = 31 = 4th Number= ? ( The value of question mark )

Option (1)31 is correct option.

In this reasoning problem 1st number (48) is associated to 1664 with the help of some rule , With the help of same rule we have to associate 27 to a number out of four given option. Also in this problem all the numbers consists of two digits. These two digits can be used to find the value of question mark with the help a formula given below.

Formula :-

{Square of 1st digits and Square of 2nd digit } of 1st Number = 2nd Number

1st Number 48

4² and 8² = 16 and 64

Hence after combining we have 1664 = 2nd Number

2nd Number 27

2 = 4 and and 7² =49.

Hence after combining we have 449 = 4th Number= ? ( The value of question mark )

Option (1)449 is correct option.

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Difference of 2nd number and 1st number = 20

24 - 4 = 20

Similarly difference of 4th number and 3rd number must be 20

? - 17 = 20

? = 20 + 17

? = 37

Option (4)37 is correct option.

In this reasoning problem 1st number (49) is associated to 6 with the help of some rule , With the help of same rule we have to associate 82 to a number out of four given option. Also in this problem all the numbers consists of two digits. These two digits can be used to find the value of question mark with the help a formula given below.

Formula :-

Square Root of( 1st digits × 2nd digit } of 1st Number = 2nd Number

√(4 *9) = √(36) = 6 = 2nd Number

√(8 *2) = √(16) = 4 = 4th Number = ? ( The value of question mark )

Option (2)4 is correct option.

In this reasoning problem 1st number 24 is associated to 16 with the help of any rule , in the same rule we have to associate 52 to a number out of four given option.

Formula = 2× (Sum of both digits of 1st number) = 2nd Number

1st Number 24

2×{ 2 × 4 } = 2 × 8 = 16 = 2nd number

3rd Number 52

2×{ 5 × 2 } = 2 × 10 = 20 = 4th number

Option (4)20 is correct option.

In this reasoning problem 1st number 2 is associated to 32 with the help of any rule , in the same rule we have to associate 3 to a number out of four given option.

1st Number 2

2⁵ = 2 × 2 × 2 × 2 × 2 = 32 = 2nd number

3rd Number 52

3⁵ = 3 × 3 × 3 × 3 × 3 = 243 = 4th number.

Option (2)243 is correct option.

Conclusion

Comment your valuable suggestion regarding the post most important Reasoning questions with answers which includes reasoning for competitive exams, circle problems, box problems, circle problems and triangles problems for competitive exams like SSC CGL ,SSC CHSL ,CPO ,Bank exams and RRB NTPC etc which were explained in this post.

Number Analogy Reasoning questions with answers for competitive exams

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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

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

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

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

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

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

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

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

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

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Problem # 9

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

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.

Ten Number Analogy Reasoning Questions with Answers for Competitive Exams

No comments

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

(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

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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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