Seating Arrangement Questions for SSC CGL

Seating Arrangement Questions for SSC CGL with Solutions

Seating arrangement is an important topic in logical reasoning for competitive exams like SSC CGL. These questions test a candidate’s ability to analyse and arrange people or objects according to given conditions. Below are 10 unique seating arrangement questions along with detailed explanations and solutions.

1. Circular Arrangement

Question: Six friends A, B, C, D, E, and F are sitting in a circle facing the center. B is to the immediate right of C. A is third to the left of B. E is between A and F. Who is sitting opposite to C?

Detailed Solution:

1. Since all are facing the center, the right and left directions remain the same.

2. B is to the immediate right of C, so placing them accordingly.

3. A is third to the left of B, which helps in fixing A’s position.

4. E is between A and F.

5. The final arrangement reveals that E is opposite to C.

Circular Arrangement Question

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2. Linear Arrangement (Single Row)

Question: Seven students P, Q, R, S, T, U, and V are sitting in a row facing north. Q is sitting third from the left. P is sitting at one of the extreme ends. U is sitting exactly between T and S. Who is sitting at the rightmost end?

Detailed Solution:

1. Placing Q in the third position from the left.

2. P is at one of the extreme ends.

3. U is between T and S, which helps in their positioning.

4. After arranging all, V is found at the rightmost end.

Linear Arrangement

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3. Linear Arrangement (Double Row)

Question: Eight persons are sitting in two parallel rows, facing each other. A, B, C, and D are sitting in row 1, facing south. P, Q, R, and S are sitting in row 2, facing north. C is sitting at one of the extreme ends. B is sitting second to the left of C. The person facing B is Q. Who is sitting opposite to A?

Detailed Solution:

1. Arranging row 1 with A, B, C, and D facing south.

2. Placing row 2’s persons opposite to them.

3. C is at the extreme end and B is second to the left.

4. The arrangement shows that R is opposite to A.

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4. Circular Arrangement (Facing Outwards)

Question: Five persons L, M, N, O, and P are sitting in a circle, facing outward. M is to the immediate left of O. P is second to the right of L. Who is sitting exactly between L and O?

Detailed Solution:

1. Since they are facing outward, left and right directions are reversed.

2. Placing M to the immediate left of O.

3. Placing P second to the right of L.

4. The final arrangement shows that N is between L and O.

Circular Seating Arrangement 

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5. Square Seating Arrangement

Question: Eight friends A, B, C, D, E, F, G, and H are sitting at the corners and middle of the sides of a square table. The ones sitting at the corners are facing the center, while the ones sitting at the sides are facing outward. E is sitting to the immediate right of H, who is sitting at a corner. Who is sitting opposite to H?

Detailed Solution:

1. Placing H at a corner.

2. E is to H’s immediate right.

3. Arranging others accordingly.

4. C is found sitting opposite to H.

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6. Mixed Arrangement

Question: Six persons J, K, L, M, N, and O are sitting in a row. Some are facing north while some are facing south. J is sitting at an extreme end. L is sitting second to the left of M. N is sitting to the immediate right of O, who is facing north. Who is sitting exactly in the middle?

Detailed Solution:

1. J is at an extreme end.

2. L is positioned second to the left of M.

3. N is immediately right of O.

4. M is found to be exactly in the middle.

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7. Linear Arrangement with Conditions

Question: Five people A, B, C, D, and E are sitting in a row facing north. A is sitting to the left of B but not next to C. D is sitting at an extreme end. Who is sitting second to the right of A?

Seating Arrangement Questions

Detailed Solution:

1. Placing D at an extreme end.

2. Arranging A to the left of B but not next to C.

3. The final arrangement shows that C is second to the right of A.

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8. Triangular Arrangement

Question: Seven people are sitting around a triangular table. Three sit at the corners, and four sit along the sides. P is sitting at a corner. R is sitting to the immediate right of Q. S is sitting opposite to P. Who is sitting to the left of S?

Detailed Solution:

1. Placing P at a corner.

2. R is immediately right of Q.

3. Arranging others accordingly.

4. T is found sitting to the left of S.

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9. Rectangular Seating Arrangement

Question: Ten people are sitting around a rectangular table, with four sitting at the corners and six sitting along the sides. X is sitting to the immediate left of Y. The person opposite to X is Z. Who is sitting diagonally opposite to Z?

Detailed Solution:

1. Arranging four people at corners and six along the sides.

2. X is to the left of Y.

3. Finding who sits opposite and diagonal to Z.

4. W is diagonally opposite to Z.

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10. Row Arrangement with Height Order

Question: Five boys A, B, C, D, and E are sitting in a row according to their heights. A is taller than C but shorter than B. D is shorter than C but taller than E. Who is the second tallest?

Detailed Solution:

1. Arranging them based on height order.

2. A is taller than C but shorter than B.

3. D is shorter than C but taller than E.

4. A is found to be the second tallest.

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Conclusion

Seating arrangement questions require careful reading and logical thinking. The best way to solve these is by drawing diagrams and arranging positions step by step. Regular practice will help in quick solving during exams like SSC CGL. Developing a habit of noting key points and testing different positions logically will boost confidence and accuracy in solving such questions efficiently.

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Logical Reasoning for SSC CGL: Important Questions and Explanations

Logical reasoning is an essential part of the SSC CGL exam, testing a candidate's ability to analyze patterns, understand relationships, and solve problems efficiently. Here, we will discuss some important logical reasoning questions along with simple explanations to help you prepare effectively.

1. Number Series

Question: Find the missing number: 2, 6, 12, 20, ?, 42
Options:
A) 30
B) 28
C) 32
D) 36

Answer: B) 30
Explanation: The pattern follows +4, +6, +8, +10... So, 20 + 10 = 30

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2. Blood Relation

Question: Pointing to a boy, Rita said, "He is the son of my grandfather’s only son." How is the boy related to Rita?
Options:
A) Brother
B) Cousin
C) Nephew
D) Uncle

Answer: A) Brother
Explanation: Grandfather’s only son is Rita’s father. His son is her brother.

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3. Odd One Out

Question: Find the odd one out:
Options:
A) Apple
B) Mango
C) Potato
D) Banana

Answer: C) Potato
Explanation: All others are fruits, but potato is a vegetable.

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4. Direction Sense

Question: A man walks 10m North, turns right and walks 5m, then turns right again and walks 10m. In which direction is he now facing?
Options:
A) East
B) West
C) North
D) South

Answer: B) West
Explanation: He initially faced North, then East, then South. Now he faces West.

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5. Coding-Decoding

Question: If CAT is coded as 3120, how will DOG be coded?
Options:
A) 4150
B) 4815
C) 4740
D) 4950

Answer: C) 4740
Explanation: Each letter’s position in the alphabet is multiplied:
C (3 × 1), A (1 × 2), T (20 × 1) → 3120
D (4 × 1), O (15 × 2), G (7 × 1) → 4740

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6. Syllogism

Question: Statements:

1. All cats are animals.

2. Some animals are wild.

Conclusion:
A) Some cats are wild.
B) All animals are cats.

Options:
A) Only A follows
B) Only B follows
C) Both follow
D) None follow

Answer: D) None follow
Explanation: The given statements do not guarantee that some cats are wild or that all animals are cats.

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7. Mirror Image

Question: What will be the mirror image of 1234 in a vertical mirror?
Options:
A) 4321
B) 2143
C) 1234
D) 3412

Answer: A) 4321
Explanation: The order of digits reverses in a mirror image.

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8. Calendar Problem

Question: If 15th August 2023 was a Tuesday, what day will 15th August 2024 be?
Options:
A) Wednesday
B) Thursday
C) Friday
D) Saturday

Answer: B) Thursday
Explanation: 2024 is a leap year, so the extra day shifts it by 2 days.

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9. Puzzle-Based Question

Question: A is the father of B, but B is not the son of A. Who is B?
Options:
A) Brother
B) Daughter
C) Uncle
D) Cousin

Answer: B) Daughter
Explanation: If B is not the son, then B must be the daughter.

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10. Clock-Based Question

Question: What is the angle between the hands of the clock at 3:30?
Options:
A) 75°
B) 90°
C) 105°
D) 120°

Answer: C) 105°
Explanation: The formula is |(30×H - 5.5×M)|
= |(30×3 - 5.5×30)| = |90 - 82.5| = 105°

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Conclusion

Logical reasoning is an important section of the SSC CGL exam, and practicing different types of questions will help improve problem-solving skills. By understanding these fundamental concepts and practicing regularly, candidates can enhance their reasoning abilities and score well in the exam. Keep practicing and stay confident!

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Most Important circle Problems of Reasoning for different competitive Exams

Problem   1

----- Explanation   ----  

(11 × 7) - (5 × 3) = 77 - 15 = 62 

(18 × 3) - (5 × 4) = 54 - 20 = 34

(9 × 6) - (4 × 8) = 54 - 32 = 22    

 Problem  2

 ----- Explanation   ----  

2² + 5² + 3² + 7² = 4 + 25 + 9 + 49 = 87

8² + 5² + 4² + 3² = 64 + 25 + 16 + 9 = 114

9² + 4² + 1² + 6² = 81 + 16 + 1 + 36 = 134

  Problem   3

 ----- Explanation   ----  

 2 x (5 + 7 + 3) = 2 x 15 = 30 

8 x (5 + 4 + 3) = 8 x 12 = 96 

3 x (4 + 9 + 6) =  3 x 19 = 57  

 Problem   4

 ----- Explanation   ----  

 1² + (5 × 3 × 7) =1 + 105 = 106

4² + (5 × 3 × 6) =16 + 90 = 106

3² + (3 × 2 × 7) = 9 + 42 = 51 

 Problem 5

----- Explanation   ----  

   1² + (5 + 3 + 7) = 1 + 15 = 16

   4² + (5 + 3 + 6) = 16 + 14 = 30

   3² + (3 + 2 + 7)  = 9 + 12 = 21

 Problem 6

----- Explanation   ----  

(2 + 5)² + (7 - 3)² = 7² +4² =  49 + 16 = 65

(1 + 5)² + (6 - 3)² = 6² + 3² = 36 + 9 = 45 = ?

(2 + 3)² + (7 - 2)² = 5² + 5² = 25 + 25 = 50    

  Problem 7

----- Explanation   ----  

(1+2+3+4)² = 10² = 100

(2+3+4+5)² = 14² = 196

(4+5+6+7)² = 22² = 484 

 Problem 8

----- Explanation   ----  

LCM Of  1, 4, 3 and 2 = 12

LCM Of 3, 2, 5 and 4 = 60

LCM Of  4 , 5 ,6 and 2 = 60 = ? 

  Problem 9

----- Explanation   ----  

 19 + 16 + 15 + 12 = 62 ⇔26

23 + 22 + 27 + 19 = 91 ⇔19

24 + 23 + 20 + 22 = 89 ⇔98 

Problem 10

----- Explanation   ----  

√81+√16+√256+√25 = 9+4+16+5 = 34

√9+√49+√1+√64 = 3+7+1+8 = 19

√4+√36+√121+√289 = 2+6+11+7 = 26 = ?   

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Ten important Questions of Reasoning series completion for competitive exams

Formula:- 

Any Term = n² × (n-1), where 'n' is natural number less than equal to 6

1st Term = 1² × (0) = 1 × 0 = 0

2nd Term = 2² × (1) = 4 × 1 = 4

3rd Term = 3² × (2) = 9 × 2 = 18

4th Term = 4² × (3) = 16 × 3 = 48

5th Term = 5² × (4) = 25 × 4 = 100

6th Term = 6² × (5) = 36 × 5 = 180

Hence Option (4)100 is correct.

Formula:-

Any Term =  Previous Term + ( sum of digits of Previous Term)

1st Term =17

2nd Term =17 + (1+7) = 17+ 8 = 25
3rd Term = 25 + (2+5) = 25 + 7 = 32

4th Term =32 + (3+2) = 32 + 5 = 37

5th Term =37 + (3+7) = 37+ 10 = 47

6th Term =47 + (4+7) = 47+ 11 = 58 = Question mark

Hence Option (2)58 is correct.

Formula:-

Difference of any two  consecutive terms = 4 +  Difference of next two consecutive terms

Difference of 2nd term and 1st term = 7 - 5 = 2
Difference of 3rd term and 2nd term = 13 - 7 = 6

Difference of 4th term and 3rd term = 23 - 13 = 10

Difference of 5th term and 4th term = 37 - 23 = 14

Difference of 6th term and 5th term = 55 - 37 = 18

Difference of 7th term and 6th term = ? - 55 = must be 4 greater than previous difference i.e. 18 + 4 = 22

? - 55 = 22

? = 22 + 55

? = 77
Hence Option (3)77 is correct.

1st term = 0.15 × 2 = 0.30 = 0.3

2nd  Term = 0.3 × 2 =  0.60 = 0.6

Also 5th Term =  1.2  × 2 = 2.4

Similarly 3rd =  0.3  × 2 = 0.6= 0.60

4th Term = 0.6 × 2 = 1.2

5th Term = 1.2 × 2 = 2.4 

Hence Option (1)0.6 is correct.

3rd term = 1 + 1 = 2

4th Term = 1 + 2 = 3

5thTerm = 2 + 3 = 5

6thTerm = 3 + 5 = 8

7th Term = 5 + 8 = 13

8th Term = 8 + 13 = 21

missing Term = 13 + 21 = ?

                           ? = 34

Hence Option 12)34 is correct.

Formula:-

Any Term = Square of Previous Term + 1

1st Term = 3

2nd  Term = (1st Term)² + 1

2nd Term =3² + 1 = 9 + 1 = 10

3rd  Term = (2nd Term)² + 1

3rd Term = 10² + 1 = 100 + 1 = 101

4th  Term = (3rd Term)² + 1

4th Term = 101² + 1 = 10201 + 1 = 10202 = The value of question mark

3² +1 , 10² + 1 , 101² + 1

Hence Option (1)19202 is correct.

Formula:-

Any Term = Twice of Previous Term + Rank of previous term

1st Term = 13

2nd Term = (13 × 2) + 1

3rd Term = (27 × 2) + 2

4th Term = (56 × 2) + 3

5th Term = (115 × 2) + 4 = 230 + 4 = 234

Hence Option (4)234 s correct.

Formula:-

Any Term = Four times of Previous Term - 1

2nd  Term = (2 × 4) - 1 = 7

3rd Term = (7 × 4) - 1 = 27

4th Term = (27 × 4) - 1 = 107

                = (107 × 4) - 1 

                = 428 - 1 = 427 

Hence Option (2)427 s correct.

Formula:-

Any Term = Twice of Previous Term  1 

2nd  Term = (2 × 2) + 1 = 4 + 1 = 5

3rd Term = (5 × 2) - 1 = 10 - 1 = 9

4th Term = (9 × 2) + 1 = 18 + 1 = 19

(19 × 2) - 1 = 38 - 1 = 37

(37 × 2) + 1 = 74 + 1 = 75

Hence Option (2)75 s correct.

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Ten Logical Reasoning questions and answers for competitive exams

Ten Logical Reasoning questions and answers for competitive  exams  like Bank PO, Bank clerk, SSC CGL, ssc chsl, RRB NTPC , group D etc have been discussed in this post in very easy style and shortcut way .

Logical Reasoning questions and answers for competitive Exams 

Problem # 1

In this reasoning problem 1st number (2) is associated to 10 with the help of any rule , in the same rule we have to associate 8  to a number out of four given options.

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

Formula :- 

2nd Number =  a² + (2×a) + 2, where 'a' is 1st number. 

1st Number

2² + (2×2) + 2 = 4 + 4 + 2 = 10

2nd Number

8² + (2×8) + 2 = 64 + 16 + 2 = 82

Therefore option (1)2 is correct option.

Problem # 2

In this reasoning problem 1st number (3) is associated to 39 with the help of any rule , in the same rule we have to associate 6 to a number out of four given options.

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

Formula :- 

2nd Number = a³ + (3×a) + a, where 'a' is 1st number.  

1st Number

3³ + (3×3) + 3 = 27 + 9 + 3 = 39

2nd Number

6³ + (3×6) + 3 = 216 + 18 + 6 = 240.  

Therefore option (1)240 is correct option.

Problem # 3

In this reasoning problem 1st number (4) is associated to 60 with the help of any rule , in the same rule we have to associate 9 to any one  number out of four given options.

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

Formula :- 

2nd Number =  (a-1) × (a) × (a + 1), where 'a' is 1st number.  

1st Number

3 × 4 × 5 = 60

2nd Number

8 × 9 × 10 = 720

Therefore option (4)720 is correct option.

Problem # 4

In this problem of reasoning we have to combine two 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.

Formula : - 

(a×b) + (a+b),

where 'a' and 'b' are two given numbers.

1st Number

(27×4) + (27+4) = 108 + 31 = 139

2nd Number

(31×9) + (31+9) = 279 + 40 = 319

3rd Number

(21×6) + (21+6) = 126 + 27 = 133

Therefore option (1)153 is correct option.

Problem # 5

In this problem of reasoning we have to combine all the three given  numbers in such a way that after applying any mathematical operation/operations we could get the number on the right hand side of all the four problems given above. Because in these types of reasoning problems we can change mathematical sign according to our requirements.

Formula : - 

(a×b×c)  + (a+b+c),

where 'a' , 'b' and 'c' are three given numbers.

1st Number

(5×2×3) + (5+2+3) = 30 + 10 = 40

2nd Number

(6×3×4) + (6+3+4) = 72 + 13 = 85

3rd Number

(4×1×8) + (4+1+8) = 32 + 13 = 45

Required Number

(3×2×8) + (3+2+8) = 48 + 13 = 61

Therefore option (4)61 is correct option.

Problem # 6

1st Method

Formula : - 

Any term = ( 2 × Previous term ) + 1

1st Term = 6

2nd Term = ( 2 × 6 ) + 1 = 12 + 1 = 13

3rd Term = ( 2 × 3 ) + 1 = 26 + 1 = 27

4th Term = ( 2 × 27 ) + 1 = 54 + 1 = 55

5th Term = ( 2 × 55 ) + 1 = 110 + 1 = 111

6th Term = ( 2 × 111 ) + 1 = 222 + 1 = 223

2nd Method

Formula : -

 Any term - Preceding Term =  Difference is multiple of 7 

13 - 6 = 7

27 - 13 = 14 = 2 × 7 = Double of 7

55 - 27 = 28 = 2 × 14 = Double of 14

111 - 55 = 56 = 2 × 28 = Double of 14

? - 111 =  It must be  2 × 56 = 112

 Hence    

? - 111 =  112

? = 112 + 111

? =  223

Therefore option (1)223 is correct option.

Problem # 7

Formula : - 

Any term =  one less or one more than the square of natural numbers greater than equal to 6 and smaller than equal to 11.

35 = 6² - 1

50 = 7² + 1

63 = 8² - 1

82 = 9² + 1

?  = 10² - 1 = 100 - 1 = 99

122 = 11² + 1

Therefore option (4)99 is correct option.

Problem # 8

 Split this number series into two series by picking alternate numbers.

13, 19 , 25  and 11, 17, 23 , ? 

1st series

13, 19 , 25 

Now considering the 2nd series. In this series every term can be found by adding 6 to its previous term.

1st Term = 13

2nd Term = 1st Term + 6 = 13  + 6 = 19

3rd Term = 2nd Term + 6 = 19  + 6 = 25

2nd series

11, 17, 23 , ?

Now considering the 1st series. In this series every term can be found by adding 6 to its previous term.

1st Term = 11

2nd Term = 1st Term + 6 = 11  + 6 = 17

3rd Term = 2nd Term + 6 = 17  + 6 = 23

4th Term = 3rd Term + 6 = 23  + 6 = 29 = ? (The value of question mark).

Therefore option (3) 29 is correct option.

In this reasoning problem three out of four options have been calculated by Multiplying  the 1st number with 12 to get 2nd number. This rule is applicable to only three out of four option. And one which do not follow this rule will be the correct option.

10 × 12 = 120

20 × 12 = 240

14 × 12 = 168(196)

12 × 12 = 120

Therefore option (3)14,196 is odd option.

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 on 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.

Formula : - 

(a×b) - (a+b),

where 'a' and 'b' are three given numbers.

1st Number

(27×4) - (27+ 4) = 108 - 31 = 77

2nd Number

(31×9) - (31+9) = 279 - 40 = 239

3rd Number

(21×6) - (21+6) = 126 - 27 = 99

Therefore option (3)99 is correct option.

Conclusion

So these were the ten  problems regarding the post Ten Logical Reasoning questions and answers pdf for competitive exams like SSC CGL ,SSC CHSL ,CPO ,Bank exams and RRB NTPC etc which were explained in this post. Feel free to comment your valuable suggestions.

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Missing number and reasoning problems for SSC CGL Exams

Ten most important Missing number series questions and answers, Number Series Questions for SBI Clerk ,SBI PO with solution , for SSC GL , SSC CHSL , RRB NTPC and other competitive Exams  have been discussed in this post.  These types of series questions are asked in many other competitive exams like SI, CPO and various entrance exams. 

Ten Reasoning problems with answers for competitive exams

Problem #1

This reasoning problem have four sub problems and every sub problem have three number associated to it.

           Since last figure have question mark in it. 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 three numbers to get the value of this question mark?

           Now we have to find or search the  formula for these three numbers in each sub problem to utilised them in any possible way to get the value of question mark. 

Formula:- Adding 1 to each digit and then combine all three numbers to get the value of question mark.

Sub Problem # 1

In this sub problem the three digits are 6 , 7 and 8, Adding 1 to each digits, we shall have

6 + 1 = 7

7 + 1 = 8

8 + 1 = 9

Now combining these three newly formed digits to get new number  as 789( the number on the right hand side).

Sub Problem # 2

In this sub problem the three digits are 2 , 3 and 4, Adding 1 to each digits, we shall have

2 + 1 = 3

3 + 1 = 4

4 + 1 = 5

Now combining these three newly formed digits to get new number  as 345( the number on the right hand side).

Sub Problem # 3

In this sub problem the three digits are 4 , 8 and 0, Adding 1 to each digits, we shall have

4 + 1 = 5

8 + 1 = 9

0 + 1 = 1

Now combining these three newly formed digits to get new number  as 591( the number on the right hand side).

Required Sub Problem # 4

In this sub problem the three digits are 8 , 0 and 7 , Adding 1 to each digits, we shall have

8 + 1 = 9

0 + 1 = 1

7 + 1 = 8

Now combining these three newly formed digits to get new number  as 918( The value of question mark).

Hence Option (4)918 is correct option.

Problem #2

This reasoning problem also have four sub problems and every sub problem have three number associated to it.

           Since last figure have question mark in it. 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 three numbers to get the value of this question mark?

           Now we have to find or search the  formula for these three numbers in each sub problem to utilised them in any possible way to get the value of question mark. 

Formula:- Reversing the order of each digit in every number to get the value of question mark.

Sub Problem # 1

In this sub problem the three digits are 6 , 7 and 8, Reversing the order of each digit , it means we have to put the 1st digits in last place ,2nd digit in 2nd last place and 3rd digit in 3rd last place to get the number on the right hand side of each sub problem.

 6 + 5 + 7  ⇒ 7 5 6 ( the number on the right hand side).

Sub Problem # 2

In this sub problem the three digits are 2 , 3 and 4, Reversing the order of each digit , it means we have to put the 1st digits in last place ,2nd digit in 2nd last place and 3rd digit in 3rd last place to get the number on the right hand side of each sub problem.

 2 + 3 + 4  ⇒ 4 3 2 ( the number on the right hand side).

Sub Problem # 3

In this sub problem the three digits are 7 , 8 and 9, Reversing the order of each digit , it means we have to put the 1st digits in last place ,2nd digit in 2nd last place and 3rd digit in 3rd last place to get the number on the right hand side of each sub problem.

 7 + 8 + 9  ⇒ 9 8 7 ( the number on the right hand side).

Required Problem # 4

In this sub problem the three digits are 2 , 2 and 2, Reversing the order of each digit , it means we have to put the 1st digits in last place ,2nd digit in 2nd last place and 3rd digit in 3rd last place to get the number on the right hand side of each sub problem.

 2 + 2 + 2  ⇒ 2 2 2 ( The value of question mark).

But there is no option which has value 222.

Hence Option (4)None is correct option.

Problem #3

Square of 1st digit and then sum of remaining two digits

6+1+3 = 6²(1+3) = 364

8+7+2 = 8²(7+2) = 649

4+7+2 = 4²(7+2) = 169

4+2+5 = 4²(2+5) = 167 (The value of Question mark)

Hence Option (2) 167 is correct option.

Problem #4

H C F (Highest Common Factor) of three given numbers

6+2+5 = H C F of (6,2,5) = 1

4+6+8 = H C F of (4,6,8) = 2

3+6+9 = H C F of (3,6,9) = 3

4+8+9 = H C F of (4,8,9) = 1

Hence Option (4) 1 None is correct option.

Problem #5

L C M of three given numbers

L C M (Least Common Multiple) of three given numbers

6+2+5 = L C M of (6,2,5) = 30

2+3+5 = L C M of (2,3,5) = 30

6+4+2 = L C M of (6,4,2) = 12

1+3+5 = L C M of (1,3,5) = 15

Hence Option (1) 15 None is correct option.

Problem #6

6+2+5 = (6 + 2 + 5)² + 1 =  13² + 1 = 169 + 1 = 170

5+3+7 = (5 + 3 + 7)² + 1 =  15² + 1 = 225 + 1 = 226

2+4+6 = (2 + 4 + 6)² + 1 =  12² + 1 = 144 + 1 = 145

1+3+5 = (1 + 3 + 5)² + 1 =  9² + 1 = 81 + 1 = 82

Hence Option (3) 82 None is correct option.

Problem #7

2+3+5 = ( 2 + 5 ) × 3 = 7 × 3 = 21

4+3+7 = ( 4 + 7 ) × 3 = 11 × 3 = 33

4+3+9 = ( 4 + 9 ) × 3 = 13 × 3 = 39

3+1+8 = ( 3 + 8 ) × 1 = 11 × 1 = 11

Option (1)11 is correct Answer

Problem #8

2+3+5 =  2 + (5  × 3 ) = 2 + 15 = 17

4+3+7 = 4 + (3  × 7 ) = 4 + 21 = 17

4+3+9 = 4 + (3  × 9 ) = 4 + 27 = 31

3+1+8 = 3 + (1  × 8 ) = 3 + 8 = 11

Option (1)11 is correct Answer

Problem #9

5+2+3 = 5 × 2² × 3 =  5 × 4 × 3 = 60

6+3+4 = 6 × 3² × 4 =  6 × 9 × 4 = 216

4+1+8 = 4 × 1² × 8 =  4 × 1 × 8 = 32

3+2+8 =3 × 2² × 8 =  3 × 4 × 8 = 96

Option (2)96 is correct Answer

Problem #10

5+2+3=(5 × 2 × 3 ) + ( 5 + 2 + 3 ) = 30 + 10 = 40

6+3+4 = (6 × 3 × 4 ) + ( 6 + 3 + 4 ) = 72 + 13 = 85

4+1+8 = (4 × 1 × 8 ) + ( 4 + 1 + 8 ) = 32 + 13 = 45

3+2+8 = (3 × 2 × 8 ) + ( 3 + 2 + 8 ) = 48 + 13 = 61

Option (4)61 is correct Answer

Problem #11

(5 × 4) × (4 × 3 ) × ( 3 × 5) = 20 + 12 + 15 = 47

(6 × 2) × (2 × 4 ) × ( 4 × 6) = 12 + 8 + 24 = 44

(4 × 3) × (3 × 9 ) × ( 9 × 4) = 12 + 27 + 36 = 75

(3 × 1) × (1 × 7 ) × ( 7 × 3) = 3 + 7 + 21 = 31

Option (2)31 is correct Answer

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