SBI PO Mains 30 Jan 2023 Memory Based Paper (Previous Year Paper)

There is a building in which total of ten rooms are present in which odd numbered floor has one room and even numbered floor have two rooms.

 

Arrangement 1: 

(6,5), (1,1), (6,3), (3,5), (2,3), (4,3), (1,4), (3,4), (4,2), (1,2).

The operation will be carried out from left to right on this above output.

1. First fill one room on each even numbered floor in decreasing order of the floor. i.e., numbers are filled in order of floor 6, floor 4 and floor 2. 2. For even numbered floor the digits in the output will be added. 

On taking the first three outputs and adding the digits we get,

(6,5) 6+5=11 in floor 6.

(1,1) 1+1=2 in floor 4.

(6,3) 6+3=9 in floor 2.

 

3. Then fill the odd numbered floor in increasing order. i.e., order of filling is floor 1, floor 3, floor 5 and floor 7. 4. For odd numbered floor the digits in the output will be multiplied. 

On taking the next four outputs and multiplying them, we get

(3,5) 3x5=15 in floor 1.

(2,3) 2x3=6 in floor 3.

(4,3) 4x3=12 in floor 5.

(1,4) 1x4=4 in floor 7.

 

5. The remaining will be filled in decreasing order of even numbered floor. i.e., filled in order of floor 6, floor 4 and floor 2.

(3,4) 3+4=7 in floor 6.

(4,4) 4+4=8 in floor 4.

(1,2) 1+2=3 in floor 2.

 

Arrangement 2: 

E, V, I, F, A, H, R, M, T, P.

Vowels: A, E, I.

Alphabets before letter L: F, H.

Alphabets after letter L: M, P, R, T, V

6. All the vowels will be placed with the odd prime numbers in decreasing order of the number and the increasing alphabet order of the vowels according to the English alphabet.

Decreasing order of odd prime numbers in rooms: 11, 7, 3.

Increasing order of vowels: A, E, I.

11 - A

7 - E

3 - I.

 

7.  The alphabets before the letter L according to English alphabet will be placed with the square numbers in increasing order of the number and increasing order of the letters according to the English alphabets. 

Increasing order of square numbers in rooms: 4, 9.

Increasing order of letters before L: F, H.

4 - F

9 - H.

 

8. The remaining letters will be allotted in the increasing order of the remaining numbers and the increasing order of the remaining letters according to the English alphabet.

Increasing order of the remaining numbers: 2, 6, 8, 12 and 15.

Increasing order of the remaining letters: M, P, R, T and V.

2 - M

6 - P

8 - R

12 - T

15 - V.

 

Arrangement 3: 

The letters are then placed in the linear seating arrangement in which all are facing the north direction.

9. The number of letters between I and A is equal to the number of persons between I and E. 10. No vowel sits at the end of the line. The number of letters between I and E is an even prime number. i.e., 2.

11. Three persons sit between I and the one who lives in first floor. i.e., V. 12. M sits to the immediate right of V.

13.  R sits fourth to the left of E. So, case 1 is eliminated. 14. H sits at one of the end of the line. 15. The one whose room is in 5th floor (T) sits to the immediate right of the one who has an even square number. i.e., 4 (F).

The final arrangement is,

Thus, the sum of the one who lives in floor 1 i.e., V (15) and the one who is to the immediate left of R i.e., P(6) is 15+6=21.

Thus, 21 is the answer.

There is a building in which total of ten rooms are present in which odd numbered floor has one room and even numbered floor have two rooms.

 

Arrangement 1: 

(6,5), (1,1), (6,3), (3,5), (2,3), (4,3), (1,4), (3,4), (4,2), (1,2).

The operation will be carried out from left to right on this above output.

1. First fill one room on each even numbered floor in decreasing order of the floor. i.e., numbers are filled in order of floor 6, floor 4 and floor 2. 2. For even numbered floor the digits in the output will be added. 

On taking the first three outputs and adding the digits we get,

(6,5) 6+5=11 in floor 6.

(1,1) 1+1=2 in floor 4.

(6,3) 6+3=9 in floor 2.

 

3. Then fill the odd numbered floor in increasing order. i.e., order of filling is floor 1, floor 3, floor 5 and floor 7. 4. For odd numbered floor the digits in the output will be multiplied. 

On taking the next four outputs and multiplying them, we get

(3,5) 3x5=15 in floor 1.

(2,3) 2x3=6 in floor 3.

(4,3) 4x3=12 in floor 5.

(1,4) 1x4=4 in floor 7.

 

5. The remaining will be filled in decreasing order of even numbered floor. i.e., filled in order of floor 6, floor 4 and floor 2.

(3,4) 3+4=7 in floor 6.

(4,4) 4+4=8 in floor 4.

(1,2) 1+2=3 in floor 2.

 

Arrangement 2: 

E, V, I, F, A, H, R, M, T, P.

Vowels: A, E, I.

Alphabets before letter L: F, H.

Alphabets after letter L: M, P, R, T, V

6. All the vowels will be placed with the odd prime numbers in decreasing order of the number and the increasing alphabet order of the vowels according to the English alphabet.

Decreasing order of odd prime numbers in rooms: 11, 7, 3.

Increasing order of vowels: A, E, I.

11 - A

7 - E

3 - I.

 

7.  The alphabets before the letter L according to English alphabet will be placed with the square numbers in increasing order of the number and increasing order of the letters according to the English alphabets. 

Increasing order of square numbers in rooms: 4, 9.

Increasing order of letters before L: F, H.

4 - F

9 - H.

 

8. The remaining letters will be allotted in the increasing order of the remaining numbers and the increasing order of the remaining letters according to the English alphabet.

Increasing order of the remaining numbers: 2, 6, 8, 12 and 15.

Increasing order of the remaining letters: M, P, R, T and V.

2 - M

6 - P

8 - R

12 - T

15 - V.

 

Arrangement 3: 

The letters are then placed in the linear seating arrangement in which all are facing the north direction.

9. The number of letters between I and A is equal to the number of persons between I and E. 10. No vowel sits at the end of the line. The number of letters between I and E is an even prime number. i.e., 2.

11. Three persons sit between I and the one who lives in first floor. i.e., V. 12. M sits to the immediate right of V.

13.  R sits fourth to the left of E. So, case 1 is eliminated. 14. H sits at one of the end of the line. 15. The one whose room is in 5th floor (T) sits to the immediate right of the one who has an even square number. i.e., 4 (F).

The final arrangement is,

Thus, I, III and IV is the answer.

There is a building in which total of ten rooms are present in which odd numbered floor has one room and even numbered floor have two rooms.

 

Arrangement 1: 

(6,5), (1,1), (6,3), (3,5), (2,3), (4,3), (1,4), (3,4), (4,2), (1,2).

The operation will be carried out from left to right on this above output.

1. First fill one room on each even numbered floor in decreasing order of the floor. i.e., numbers are filled in order of floor 6, floor 4 and floor 2. 2. For even numbered floor the digits in the output will be added. 

On taking the first three outputs and adding the digits we get,

(6,5) 6+5=11 in floor 6.

(1,1) 1+1=2 in floor 4.

(6,3) 6+3=9 in floor 2.

 

3. Then fill the odd numbered floor in increasing order. i.e., order of filling is floor 1, floor 3, floor 5 and floor 7. 4. For odd numbered floor the digits in the output will be multiplied. 

On taking the next four outputs and multiplying them, we get

(3,5) 3x5=15 in floor 1.

(2,3) 2x3=6 in floor 3.

(4,3) 4x3=12 in floor 5.

(1,4) 1x4=4 in floor 7.

 

5. The remaining will be filled in decreasing order of even numbered floor. i.e., filled in order of floor 6, floor 4 and floor 2.

(3,4) 3+4=7 in floor 6.

(4,4) 4+4=8 in floor 4.

(1,2) 1+2=3 in floor 2.

 

Arrangement 2: 

E, V, I, F, A, H, R, M, T, P.

Vowels: A, E, I.

Alphabets before letter L: F, H.

Alphabets after letter L: M, P, R, T, V

6. All the vowels will be placed with the odd prime numbers in decreasing order of the number and the increasing alphabet order of the vowels according to the English alphabet.

Decreasing order of odd prime numbers in rooms: 11, 7, 3.

Increasing order of vowels: A, E, I.

11 - A

7 - E

3 - I.

 

7.  The alphabets before the letter L according to English alphabet will be placed with the square numbers in increasing order of the number and increasing order of the letters according to the English alphabets. 

Increasing order of square numbers in rooms: 4, 9.

Increasing order of letters before L: F, H.

4 - F

9 - H.

 

8. The remaining letters will be allotted in the increasing order of the remaining numbers and the increasing order of the remaining letters according to the English alphabet.

Increasing order of the remaining numbers: 2, 6, 8, 12 and 15.

Increasing order of the remaining letters: M, P, R, T and V.

2 - M

6 - P

8 - R

12 - T

15 - V.

 

Arrangement 3: 

The letters are then placed in the linear seating arrangement in which all are facing the north direction.

9. The number of letters between I and A is equal to the number of persons between I and E. 10. No vowel sits at the end of the line. The number of letters between I and E is an even prime number. i.e., 2.

11. Three persons sit between I and the one who lives in first floor. i.e., V. 12. M sits to the immediate right of V.

13.  R sits fourth to the left of E. So, case 1 is eliminated. 14. H sits at one of the end of the line. 15. The one whose room is in 5th floor (T) sits to the immediate right of the one who has an even square number. i.e., 4 (F).

The final arrangement is,

The sum of numbers of the one who sits at both the ends of the line is H(9) + M(2) = 11.

Thus, 11 is the answer.

There is a building in which total of ten rooms are present in which odd numbered floor has one room and even numbered floor have two rooms.

 

Arrangement 1: 

(6,5), (1,1), (6,3), (3,5), (2,3), (4,3), (1,4), (3,4), (4,2), (1,2).

The operation will be carried out from left to right on this above output.

1. First fill one room on each even numbered floor in decreasing order of the floor. i.e., numbers are filled in order of floor 6, floor 4 and floor 2. 2. For even numbered floor the digits in the output will be added. 

On taking the first three outputs and adding the digits we get,

(6,5) 6+5=11 in floor 6.

(1,1) 1+1=2 in floor 4.

(6,3) 6+3=9 in floor 2.

 

3. Then fill the odd numbered floor in increasing order. i.e., order of filling is floor 1, floor 3, floor 5 and floor 7. 4. For odd numbered floor the digits in the output will be multiplied. 

On taking the next four outputs and multiplying them, we get

(3,5) 3x5=15 in floor 1.

(2,3) 2x3=6 in floor 3.

(4,3) 4x3=12 in floor 5.

(1,4) 1x4=4 in floor 7.

 

5. The remaining will be filled in decreasing order of even numbered floor. i.e., filled in order of floor 6, floor 4 and floor 2.

(3,4) 3+4=7 in floor 6.

(4,4) 4+4=8 in floor 4.

(1,2) 1+2=3 in floor 2.

 

Arrangement 2: 

E, V, I, F, A, H, R, M, T, P.

Vowels: A, E, I.

Alphabets before letter L: F, H.

Alphabets after letter L: M, P, R, T, V

6. All the vowels will be placed with the odd prime numbers in decreasing order of the number and the increasing alphabet order of the vowels according to the English alphabet.

Decreasing order of odd prime numbers in rooms: 11, 7, 3.

Increasing order of vowels: A, E, I.

11 - A

7 - E

3 - I.

 

7.  The alphabets before the letter L according to English alphabet will be placed with the square numbers in increasing order of the number and increasing order of the letters according to the English alphabets. 

Increasing order of square numbers in rooms: 4, 9.

Increasing order of letters before L: F, H.

4 - F

9 - H.

 

8. The remaining letters will be allotted in the increasing order of the remaining numbers and the increasing order of the remaining letters according to the English alphabet.

Increasing order of the remaining numbers: 2, 6, 8, 12 and 15.

Increasing order of the remaining letters: M, P, R, T and V.

2 - M

6 - P

8 - R

12 - T

15 - V.

 

Arrangement 3: 

The letters are then placed in the linear seating arrangement in which all are facing the north direction.

9. The number of letters between I and A is equal to the number of persons between I and E. 10. No vowel sits at the end of the line. The number of letters between I and E is an even prime number. i.e., 2.

11. Three persons sit between I and the one who lives in first floor. i.e., V. 12. M sits to the immediate right of V.

13.  R sits fourth to the left of E. So, case 1 is eliminated. 14. H sits at one of the end of the line. 15. The one whose room is in 5th floor (T) sits to the immediate right of the one who has an even square number. i.e., 4 (F).

The final arrangement is,

Thus, P - 6 - Floor 3 is the answer.

Given:-

 

Statements: 

I. K @ J → Only a few J is K.

II. J # C → All C is J.

III. C $ D → No D is C.

IV. K # B → All B is K.

Thus, the possible Venn diagram for the given statement is:

Conclusions:

I. All D is J is a possibility. → True (It is possible that All D is J as there is no negative or direct relation given between them. Hence, it is true.)

II. No J is K → False (It is definite that Some J is K. Hence, it is false)

III. Some D is not K. → False (It is possible that Some or All D is K as there is no direct relation given between them. Hence, it is false)

Hence, the correct answer is Only I follows.

1. U sits second to the left of the one whose age is 34.  2. U and the spouse of Q are sitting adjacent to each other.

3. Q sits immediate right of the one who sits opposite to the person whose age is 45. 4. Q doesn’t sit adjacent to U. 

5. P sits immediate right of the one who is the husband of U. P is not the spouse of Q. 6. In couples, the husband's age is more than the wife’s age. So, Case 1 and Case 2 is eliminated.

7. S is the male who sits opposite to the one whose age is 25. 8. The age of T is three years more than S’s spouse.

8. Neither S nor R is the spouse of Q.  10. U’s spouse age is five years more than the one who sits immediate right of T.

9. The age difference between S and Q is one year less than the age difference between R and Q. •

10. The age of Q is less than the age of P’s spouse. So, case2a gets eliminated. 

Thus, Option IV is the answer.

1. U sits second to the left of the one whose age is 34.  2. U and the spouse of Q are sitting adjacent to each other.

3. Q sits immediate right of the one who sits opposite to the person whose age is 45. 4. Q doesn’t sit adjacent to U. 

5. P sits immediate right of the one who is the husband of U. P is not the spouse of Q. 6. In couples, the husband's age is more than the wife’s age. So, Case 1 and Case 2 is eliminated.

7. S is the male who sits opposite to the one whose age is 25. 8. The age of T is three years more than S’s spouse.

8. Neither S nor R is the spouse of Q.  10. U’s spouse age is five years more than the one who sits immediate right of T.

9. The age difference between S and Q is one year less than the age difference between R and Q. •

10. The age of Q is less than the age of P’s spouse. So, case2a gets eliminated. 

Thus, no one is seated between Q’s spouse and S’s spouse when counted to the right of Q’s spouse.

1. U sits second to the left of the one whose age is 34.  2. U and the spouse of Q are sitting adjacent to each other.

3. Q sits immediate right of the one who sits opposite to the person whose age is 45. 4. Q doesn’t sit adjacent to U. 

5. P sits immediate right of the one who is the husband of U. P is not the spouse of Q. 6. In couples, the husband's age is more than the wife’s age. So, Case 1 and Case 2 is eliminated.

7. S is the male who sits opposite to the one whose age is 25. 8. The age of T is three years more than S’s spouse.

8. Neither S nor R is the spouse of Q.  10. U’s spouse age is five years more than the one who sits immediate right of T.

9. The age difference between S and Q is one year less than the age difference between R and Q. •

10. The age of Q is less than the age of P’s spouse. So, case2a gets eliminated. 

Thus, is the sum of the age of S, T, and R

1. U sits second to the left of the one whose age is 34.  2. U and the spouse of Q are sitting adjacent to each other.

3. Q sits immediate right of the one who sits opposite to the person whose age is 45. 4. Q doesn’t sit adjacent to U. 

5. P sits immediate right of the one who is the husband of U. P is not the spouse of Q. 6. In couples, the husband's age is more than the wife’s age. So, Case 1 and Case 2 is eliminated.

7. S is the male who sits opposite to the one whose age is 25. 8. The age of T is three years more than S’s spouse.

8. Neither S nor R is the spouse of Q.  10. U’s spouse age is five years more than the one who sits immediate right of T.

9. The age difference between S and Q is one year less than the age difference between R and Q. •

10. The age of Q is less than the age of P’s spouse. So, case2a gets eliminated. 

Thus, option IV is the answer.

In the given arrangement the machine arranges the numbers in each step.

Step I → Within a number, the first two digits are multiplied by each other, the middle digit is reduced by 1 and the last two digits are multiplied by each other.

Step II → All the odd digits are first arranged in descending order and then even digits are arranged in descending order within the number.

Step III → The first digit and last digit of the number are together taken as a separate number and then this number is added to the square of the addition of the middle three digits of the number.

Step IV → The digits of the number are added and their square is written. Then the number is written as it is in the end.

For example, let's take the number 67543.

Step I → First two digits are multiplied = 76 = 42, the middle digit is reduced by 1 = 5 - 1 = 4, and the last two digits are multiplied = 43 = 12. So the output of Step I is 42412.

Step II → Now first arrange the odd numbers in descending order then the even numbers in descending order. So, the output of Step II is 14422.

Step III → Add middle three digits = 4 + 4 + 2 = 10, 102 = 100. Now take the first and last digits together as a number = 12. 100 + 12 = 112.

Step IV → Now, add the digits and square them = 1 + 1 + 2 = 42 = 16. So, the output of Step IV is 16112.

Now, let's solve the given input in a similar way:

Input:    45483   45192   35654    79583    43235   55672

Step I:   20324   20018   15520    63424   12115    25514

Step II:  34220   18200    55120   36442    51112    55142

Step III:     94        110         114       228         61        152

Step IV: 16994     4110    36114    144228   4961    64152

There is a total of 3 five-digit numbers in Step IV.

Hence, the correct answer is 3.

In the given arrangement the machine arranges the numbers in each step.

Step I → Within a number, the first two digits are multiplied by each other, the middle digit is reduced by 1 and the last two digits are multiplied by each other.

Step II → All the odd digits are first arranged in descending order and then even digits are arranged in descending order within the number.

Step III → The first digit and last digit of the number are together taken as a separate number and then this number is added to the square of the addition of the middle three digits of the number.

Step IV → The digits of the number are added and their square is written. Then the number is written as it is in the end.

For example, let's take the number 67543.

Step I → First two digits are multiplied = 76 = 42, the middle digit is reduced by 1 = 5 - 1 = 4, and the last two digits are multiplied = 43 = 12. So the output of Step I is 42412.

Step II → Now first arrange the odd numbers in descending order then the even numbers in descending order. So, the output of Step II is 14422.

Step III → Add middle three digits = 4 + 4 + 2 = 10, 102 = 100. Now take the first and last digits together as a number = 12. 100 + 12 = 112.

Step IV → Now, add the digits and square them = 1 + 1 + 2 = 42 = 16. So, the output of Step IV is 16112.

Now, let's solve the given input in a similar way:

Input:    45483   45192   35654    79583    43235   55672

Step I:   20324   20018   15520    63424   12115    25514

Step II:  34220   18200    55120   36442    51112    55142

Step III:     94       110         114       228         61        152

Step IV: 16994    4110     36114   144228    4961    64152

The largest digit of the first number in Step II is 4 and the smallest digit of the last number in Step II is 1. Sum = 4 + 1 = 5.

Hence, the correct answer is 5.

In the given arrangement the machine arranges the numbers in each step.

Step I → Within a number, the first two digits are multiplied by each other, the middle digit is reduced by 1 and the last two digits are multiplied by each other.

Step II → All the odd digits are first arranged in descending order and then even digits are arranged in descending order within the number.

Step III → The first digit and last digit of the number are together taken as a separate number and then this number is added to the square of the addition of the middle three digits of the number.

Step IV → The digits of the number are added and their square is written. Then the number is written as it is in the end.

For example, let's take the number 67543.

Step I → First two digits are multiplied = 76 = 42, the middle digit is reduced by 1 = 5 - 1 = 4, and the last two digits are multiplied = 43 = 12. So the output of Step I is 42412.

Step II → Now first arrange the odd numbers in descending order then the even numbers in descending order. So, the output of Step II is 14422.

Step III → Add middle three digits = 4 + 4 + 2 = 10, 102 = 100. Now take the first and last digits together as a number = 12. 100 + 12 = 112.

Step IV → Now, add the digits and square them = 1 + 1 + 2 = 42 = 16. So, the output of Step IV is 16112.

Now, let's solve the given input in a similar way:

Input:    45483   45192   35654    79583    43235   55672

Step I:   20324   20018   15520    63424   12115    25514

Step II:  34220   18200    55120   36442    51112    55142

Step III:     94        110        114       228         61        152

Step IV: 16994     4110    36114    144228     4961    64152

The number which is third from the left in Step I is 15520. After arranging in ascending order, we get 01255. Then the digit which is third from the right is 2.

Hence, the correct answer is 2.

In the given arrangement the machine arranges the numbers in each step.

Step I → Within a number, the first two digits are multiplied by each other, the middle digit is reduced by 1 and the last two digits are multiplied by each other.

Step II → All the odd digits are first arranged in descending order and then even digits are arranged in descending order within the number.

Step III → The first digit and last digit of the number are together taken as a separate number and then this number is added to the square of the addition of the middle three digits of the number.

Step IV → The digits of the number are added and their square is written. Then the number is written as it is in the end.

For example, let's take the number 67543.

Step I → First two digits are multiplied = 76 = 42, the middle digit is reduced by 1 = 5 - 1 = 4, and the last two digits are multiplied = 43 = 12. So the output of Step I is 42412.

Step II → Now first arrange the odd numbers in descending order then the even numbers in descending order. So, the output of Step II is 14422.

Step III → Add middle three digits = 4 + 4 + 2 = 10, 10² = 100. Now take the first and last digits together as a number = 12. 100 + 12 = 112.

Step IV → Now, add the digits and square them = 1 + 1 + 2 = 4² = 16. So, the output of Step IV is 16112.

Now, let's solve the given input in a similar way:

Input:    45483   45192   35654    79583    43235   55672

Step I:   20324   20018   15520    63424   12115    25514

Step II:  34220   18200    55120   36442    51112    55142

Step III:     94        110       114       228        61        152

Step IV: 16994    4110     36114    144228     4961    64152

The second highest number in the second last step is 152 and the second smallest number is 94. Difference = 152 - 94 = 58.

Hence, the correct answer is 58.

Given:

In the given example, the conditions used are as follows:

Triangle 1 -

Input → 1, 4 → Both the numbers are perfect squares but one is an even number and the other is an odd number. So condition I will be applied and according to it, both the numbers will be increased by 1.

Output → 1 + 1 = 2, 4 + 1 = 5. 2, 5

Triangle 2 -

Input → o, k → One of the letters is a vowel and the other is a consonant smaller than N. So conditions II and IV will be applied and according to them, the vowel will be changed to the preceding alphabet and the consonant will be changed to the next alphabet.

Output → o - 1 = n, k + 1 = l. n, l

Triangle 3 -

Input → 6, 7 → The first one is an even number and the second one is an odd number and neither of them is a perfect square. So, no condition is applicable.

Output → 6, 7

Triangle 4 -

Input → i, l → One of the letters is a vowel and the other is a consonant smaller than N. So conditions II and IV will be applied and according to them, the vowel will be changed to the preceding alphabet and the consonant will be changed to the next alphabet.

Output → i - 1 = h, l + 1 = m. h, m

Triangle 5 - 

Input → s, v → Both the letters are consonant and none of them is smaller than N. So, no condition is applicable.

Output → s, v

Triangle 6 -

Input → 5, 5 → Both the numbers are odd numbers and none of them is a perfect square. So, condition III is applicable and according to it, both the numbers will be reduced by 2.

Output → 5 - 2 = 3, 5 - 2 = 3. 3, 3.

Now, the input is -

Triangle 1 -

Input → c, d → Both the letters are consonant and both of them are smaller than N. So, condition IV is applicable, and according to it, the consonants will be changed to the next alphabet.

Output → c + 1 = d, d + 1 = e. d, e

Triangle 2 -

Input → u, b → One of the letters is a vowel and the other is a consonant smaller than N. So conditions II and IV will be applied and according to them, the vowel will be changed to the preceding alphabet and the consonant will be changed to the next alphabet.

Output → u - 1 = t, b + 1 = c. t, c

Triangle 3 -

Input → 7, 7 → Both numbers are odd numbers and none of them is a perfect square. So, condition III is applicable and according to it, both the numbers will be reduced by 2.

Output → 7 - 2 = 5, 7 - 2 = 5. 5, 5.

Triangle 4 -

Input → h, f → Both the letters are consonant and both of them are smaller than N. So, condition IV is applicable, and according to it, the consonants will be changed to the next alphabet.

Output → h + 1 = i, f + 1 = g. i, g

Triangle 5 -

Input → w, q → Both the letters are consonant and none of them is smaller than N. So, no condition is applicable.

Output → w, q

Triangle 6 -

Input → 8, 3 → The first one is an even number and the second one is an odd number and neither of them is a perfect square. So, no condition is applicable.

Output → 8, 3

So, the final output is -

The second number of the output of triangle 3 = 5

The first number of the output of triangle 6 = 8

Sum = 8 + 5 = 13.

Hence, the correct answer is 13.

Given:

In the given example, the conditions used are as follows:

Triangle 1 -

Input → 1, 4 → Both the numbers are perfect squares but one is an even number and the other is an odd number. So condition I will be applied and according to it, both the numbers will be increased by 1.

Output → 1 + 1 = 2, 4 + 1 = 5. 2, 5

Triangle 2 -

Input → o, k → One of the letters is a vowel and the other is a consonant smaller than N. So conditions II and IV will be applied and according to them, the vowel will be changed to the preceding alphabet and the consonant will be changed to the next alphabet.

Output → o - 1 = n, k + 1 = l. n, l

Triangle 3 -

Input → 6, 7 → The first one is an even number and the second one is an odd number and neither of them is a perfect square. So, no condition is applicable.

Output → 6, 7

Triangle 4 -

Input → i, l → One of the letters is a vowel and the other is a consonant smaller than N. So conditions II and IV will be applied and according to them, the vowel will be changed to the preceding alphabet and the consonant will be changed to the next alphabet.

Output → i - 1 = h, l + 1 = m. h, m

Triangle 5 - 

Input → s, v → Both the letters are consonant and none of them is smaller than N. So, no condition is applicable.

Output → s, v

Triangle 6 -

Input → 5, 5 → Both the numbers are odd numbers and none of them is a perfect square. So, condition III is applicable and according to it, both the numbers will be reduced by 2.

Output → 5 - 2 = 3, 5 - 2 = 3. 3, 3.

Now, the input is -

Triangle 1 -

Input → c, d → Both the letters are consonant and both of them are smaller than N. So, condition IV is applicable, and according to it, the consonants will be changed to the next alphabet.

Output → c + 1 = d, d + 1 = e. d, e

Triangle 2 -

Input → u, b → One of the letters is a vowel and the other is a consonant smaller than N. So conditions II and IV will be applied and according to them, the vowel will be changed to the preceding alphabet and the consonant will be changed to the next alphabet.

Output → u - 1 = t, b + 1 = c. t, c

Triangle 3 -

Input → 7, 7 → Both numbers are odd numbers and none of them is a perfect square. So, condition III is applicable and according to it, both the numbers will be reduced by 2.

Output → 7 - 2 = 5, 7 - 2 = 5. 5, 5.

Triangle 4 -

Input → h, f → Both the letters are consonant and both of them are smaller than N. So, condition IV is applicable, and according to it, the consonants will be changed to the next alphabet.

Output → h + 1 = i, f + 1 = g. i, g

Triangle 5 -

Input → w, q → Both the letters are consonant and none of them is smaller than N. So, no condition is applicable.

Output → w, q

Triangle 6 -

Input → 8, 3 → The first one is an even number and the second one is an odd number and neither of them is a perfect square. So, no condition is applicable.

Output → 8, 3

So, the final output is -

The output of triangle 1 is d, e.

Hence, the correct answer is d, e.

Given:

In the given example, the conditions used are as follows:

Triangle 1 -

Input → 1, 4 → Both the numbers are perfect squares but one is an even number and the other is an odd number. So condition I will be applied and according to it, both the numbers will be increased by 1.

Output → 1 + 1 = 2, 4 + 1 = 5. 2, 5

Triangle 2 -

Input → o, k → One of the letters is a vowel and the other is a consonant smaller than N. So conditions II and IV will be applied and according to them, the vowel will be changed to the preceding alphabet and the consonant will be changed to the next alphabet.

Output → o - 1 = n, k + 1 = l. n, l

Triangle 3 -

Input → 6, 7 → The first one is an even number and the second one is an odd number and neither of them is a perfect square. So, no condition is applicable.

Output → 6, 7

Triangle 4 -

Input → i, l → One of the letters is a vowel and the other is a consonant smaller than N. So conditions II and IV will be applied and according to them, the vowel will be changed to the preceding alphabet and the consonant will be changed to the next alphabet.

Output → i - 1 = h, l + 1 = m. h, m

Triangle 5 - 

Input → s, v → Both the letters are consonant and none of them is smaller than N. So, no condition is applicable.

Output → s, v

Triangle 6 -

Input → 5, 5 → Both the numbers are odd numbers and none of them is a perfect square. So, condition III is applicable and according to it, both the numbers will be reduced by 2.

Output → 5 - 2 = 3, 5 - 2 = 3. 3, 3.

Now, the input is -

Triangle 1 -

Input → c, d → Both the letters are consonant and both of them are smaller than N. So, condition IV is applicable, and according to it, the consonants will be changed to the next alphabet.

Output → c + 1 = d, d + 1 = e. d, e

Triangle 2 -

Input → u, b → One of the letters is a vowel and the other is a consonant smaller than N. So conditions II and IV will be applied and according to them, the vowel will be changed to the preceding alphabet and the consonant will be changed to the next alphabet.

Output → u - 1 = t, b + 1 = c. t, c

Triangle 3 -

Input → 7, 7 → Both numbers are odd numbers and none of them is a perfect square. So, condition III is applicable and according to it, both the numbers will be reduced by 2.

Output → 7 - 2 = 5, 7 - 2 = 5. 5, 5.

Triangle 4 -

Input → h, f → Both the letters are consonant and both of them are smaller than N. So, condition IV is applicable, and according to it, the consonants will be changed to the next alphabet.

Output → h + 1 = i, f + 1 = g. i, g

Triangle 5 -

Input → w, q → Both the letters are consonant and none of them is smaller than N. So, no condition is applicable.

Output → w, q

Triangle 6 -

Input → 8, 3 → The first one is an even number and the second one is an odd number and neither of them is a perfect square. So, no condition is applicable.

Output → 8, 3

So, the final output is -

The second letter of the output of triangle 2 is c and the first letter of the output of triangle 5 is w.

Letters between c and w = 19.

Hence, the correct answer is 19.

Given:

In the given example, the conditions used are as follows:

Triangle 1 -

Input → 1, 4 → Both the numbers are perfect squares but one is an even number and the other is an odd number. So condition I will be applied and according to it, both the numbers will be increased by 1.

Output → 1 + 1 = 2, 4 + 1 = 5. 2, 5

Triangle 2 -

Input → o, k → One of the letters is a vowel and the other is a consonant smaller than N. So conditions II and IV will be applied and according to them, the vowel will be changed to the preceding alphabet and the consonant will be changed to the next alphabet.

Output → o - 1 = n, k + 1 = l. n, l

Triangle 3 -

Input → 6, 7 → The first one is an even number and the second one is an odd number and neither of them is a perfect square. So, no condition is applicable.

Output → 6, 7

Triangle 4 -

Input → i, l → One of the letters is a vowel and the other is a consonant smaller than N. So conditions II and IV will be applied and according to them, the vowel will be changed to the preceding alphabet and the consonant will be changed to the next alphabet.

Output → i - 1 = h, l + 1 = m. h, m

Triangle 5 - 

Input → s, v → Both the letters are consonant and none of them is smaller than N. So, no condition is applicable.

Output → s, v

Triangle 6 -

Input → 5, 5 → Both the numbers are odd numbers and none of them is a perfect square. So, condition III is applicable and according to it, both the numbers will be reduced by 2.

Output → 5 - 2 = 3, 5 - 2 = 3. 3, 3.

Now, the input is -

Triangle 1 -

Input → c, d → Both the letters are consonant and both of them are smaller than N. So, condition IV is applicable, and according to it, the consonants will be changed to the next alphabet.

Output → c + 1 = d, d + 1 = e. d, e

Triangle 2 -

Input → u, b → One of the letters is a vowel and the other is a consonant smaller than N. So conditions II and IV will be applied and according to them, the vowel will be changed to the preceding alphabet and the consonant will be changed to the next alphabet.

Output → u - 1 = t, b + 1 = c. t, c

Triangle 3 -

Input → 7, 7 → Both numbers are odd numbers and none of them is a perfect square. So, condition III is applicable and according to it, both the numbers will be reduced by 2.

Output → 7 - 2 = 5, 7 - 2 = 5. 5, 5.

Triangle 4 -

Input → h, f → Both the letters are consonant and both of them are smaller than N. So, condition IV is applicable, and according to it, the consonants will be changed to the next alphabet.

Output → h + 1 = i, f + 1 = g. i, g

Triangle 5 -

Input → w, q → Both the letters are consonant and none of them is smaller than N. So, no condition is applicable.

Output → w, q

Triangle 6 -

Input → 8, 3 → The first one is an even number and the second one is an odd number and neither of them is a perfect square. So, no condition is applicable.

Output → 8, 3

So, the final output is -

Product of numbers of the output of triangle 3 = 5*5 = 25.

Product of numbers of the output of triangle 6 = 8*3 = 24.

Sum = 25 - 24 = 1.

Hence, the correct answer is 1.

Given:-

6 persons - U, T, S, R, Q, and P

Standing in a linear row from left to right.

3 are facing North and 3 are facing South.

The consecutive distance between two persons = 8 m.

1) S travels 8 m south and then turns left and walks a few meters to reach its final position which is in the north of the final position of Q. It means S is facing South but we do not know the distance he traveled after turning left. The final position is denoted by S' in the diagram.

2) Q travels 12 m in a certain direction then turns right and travels 10m meters to reach its final position. Q is either facing North or South so will travel in either direction. 

Case I → Q travels 12m in the North direction and then turns right and travels 10m in the East direction. So the final position of Q will be in the North of the final position of S. But in condition 1), it is given that the final position of S is in the north of the final position of Q. So, this case gets canceled.

Case II → Q travels 12m in the South direction and then turns right and travels 10m in the West direction. So the final position of Q will be in the South of the final position of S. This also satisfies condition 1). Thus, Q faces South. Also, S travels 6m in the East direction to be in the North of the final position of Q.

3) P travels 16m in a certain direction then turns right and travels 24 m to reach its final position. P can travel in either the North or South direction.

4) The final position of P cannot be northeast of the final position of Q. It means P faces South and travels 16 m in the South direction and then turns right and travels 21m in the west direction. Then the final position of P will be in the Southwest of the final position of Q.

5) U travels 16m in a certain direction then turns right and travels 34m to reach its final position. U can travel in either the North or South direction. But already three persons are facing South so U faces North.

6) R walks 12m in a certain direction and then turns right and walks 10m to reach its final position which is south of the final position of U. R also walks in the North direction.

Thus, the final arrangement is -

7) T travels 6m in a certain direction and then turns right and travels 6m to reach its final position. 

The final position of U is in the Northeast direction with respect to the final position of P.

Hence, the correct answer is Northeast.

Given:-

6 persons - U, T, S, R, Q, and P

Standing in a linear row from left to right.

3 are facing North and 3 are facing South.

The consecutive distance between two persons = 8 m.

1) S travels 8 m south and then turns left and walks a few meters to reach its final position which is in the north of the final position of Q. It means S is facing South but we do not know the distance he traveled after turning left. The final position is denoted by S' in the diagram.

2) Q travels 12 m in a certain direction then turns right and travels 10m meters to reach its final position. Q is either facing North or South so will travel in either direction. 

Case I → Q travels 12m in the North direction and then turns right and travels 10m in the East direction. So the final position of Q will be in the North of the final position of S. But in condition 1), it is given that the final position of S is in the north of the final position of Q. So, this case gets canceled.

Case II → Q travels 12m in the South direction and then turns right and travels 10m in the West direction. So the final position of Q will be in the South of the final position of S. This also satisfies condition 1). Thus, Q faces South. Also, S travels 6m in the East direction to be in the North of the final position of Q.

3) P travels 16m in a certain direction then turns right and travels 24 m to reach its final position. P can travel in either the North or South direction.

4) The final position of P cannot be northeast of the final position of Q. It means P faces South and travels 16 m in the South direction and then turns right and travels 21m in the west direction. Then the final position of P will be in the Southwest of the final position of Q.

5) U travels 16m in a certain direction then turns right and travels 34m to reach its final position. U can travel in either the North or South direction. But already three persons are facing South so U faces North.

6) R walks 12m in a certain direction and then turns right and walks 10m to reach its final position which is south of the final position of U. R also walks in the North direction.

Thus, the final arrangement is -

7) T travels 6m in a certain direction and then turns right and travels 6m to reach its final position. 

U travels = 16 + 34 = 40m.

T travels = 6 + 6 = 12m.

S travels = 8 + 6 = 14m.

R travels = 12 + 10 = 22m.

Q travels = 12 + 10 = 40m.

P travels = 16 + 21 = 37m.

Thus, U travels the most.

Hence, the correct answer is U.

Given:-

6 persons - U, T, S, R, Q, and P

Standing in a linear row from left to right.

3 are facing North and 3 are facing South.

The consecutive distance between two persons = 8 m.

1) S travels 8 m south and then turns left and walks a few meters to reach its final position which is in the north of the final position of Q. It means S is facing South but we do not know the distance he traveled after turning left. The final position is denoted by S' in the diagram.

2) Q travels 12 m in a certain direction then turns right and travels 10m meters to reach its final position. Q is either facing North or South so will travel in either direction. 

Case I → Q travels 12m in the North direction and then turns right and travels 10m in the East direction. So the final position of Q will be in the North of the final position of S. But in condition 1), it is given that the final position of S is in the north of the final position of Q. So, this case gets canceled.

Case II → Q travels 12m in the South direction and then turns right and travels 10m in the West direction. So the final position of Q will be in the South of the final position of S. This also satisfies condition 1). Thus, Q faces South. Also, S travels 6m in the East direction to be in the North of the final position of Q.

3) P travels 16m in a certain direction then turns right and travels 24 m to reach its final position. P can travel in either the North or South direction.

4) The final position of P cannot be northeast of the final position of Q. It means P faces South and travels 16 m in the South direction and then turns right and travels 21m in the west direction. Then the final position of P will be in the Southwest of the final position of Q.

5) U travels 16m in a certain direction then turns right and travels 34m to reach its final position. U can travel in either the North or South direction. But already three persons are facing South so U faces North.

6) R walks 12m in a certain direction and then turns right and walks 10m to reach its final position which is south of the final position of U. R also walks in the North direction.

Thus, the final arrangement is -

7) T travels 6m in a certain direction and then turns right and travels 6m to reach its final position. 

The shortest distance T needs to travel further in order to reach the initial position of S = $\sqrt{6^2+2^2}$  = $\sqrt{36+4} = \sqrt{40} m$

Hence, the correct answer is$\sqrt{40} m$.

Given:-

6 persons - U, T, S, R, Q, and P

Standing in a linear row from left to right.

3 are facing North and 3 are facing South.

The consecutive distance between two persons = 8 m.

1) S travels 8 m south and then turns left and walks a few meters to reach its final position which is in the north of the final position of Q. It means S is facing South but we do not know the distance he traveled after turning left. The final position is denoted by S' in the diagram.

2) Q travels 12 m in a certain direction then turns right and travels 10m meters to reach its final position. Q is either facing North or South so will travel in either direction. 

Case I → Q travels 12m in the North direction and then turns right and travels 10m in the East direction. So the final position of Q will be in the North of the final position of S. But in condition 1), it is given that the final position of S is in the north of the final position of Q. So, this case gets canceled.

Case II → Q travels 12m in the South direction and then turns right and travels 10m in the West direction. So the final position of Q will be in the South of the final position of S. This also satisfies condition 1). Thus, Q faces South. Also, S travels 6m in the East direction to be in the North of the final position of Q.

3) P travels 16m in a certain direction then turns right and travels 24 m to reach its final position. P can travel in either the North or South direction.

4) The final position of P cannot be northeast of the final position of Q. It means P faces South and travels 16 m in the South direction and then turns right and travels 21m in the west direction. Then the final position of P will be in the Southwest of the final position of Q.

5) U travels 16m in a certain direction then turns right and travels 34m to reach its final position. U can travel in either the North or South direction. But already three persons are facing South so U faces North.

6) R walks 12m in a certain direction and then turns right and walks 10m to reach its final position which is south of the final position of U. R also walks in the North direction.

Thus, the final arrangement is -

7) T travels 6m in a certain direction and then turns right and travels 6m to reach its final position. 

Final position of T with respect to initial position of U. → Northeast

Final position of S with respect to initial position of Q. → Southwest

Final position of Q with respect to initial position of P. → Southwest

Final position of P with respect to initial position of R. → Southwest

Final position of T with respect to Final position of U. → Southwest

Hence, the correct answer is Final position of T with respect to initial position of U.

Given:

 

Expression:- D ^ B $ E; B % K * L; E ^ G + L; B % M

Here,

D ^ B means D is married to B.

B $ E means B is the father of E.

B % K means B is the son of K.

K * L means K is the grandmother of L.

E ^ G means E is married to G.

G + L means G is the sister-in-law of L.

B % M means B is the son of M.

Thus, the following family tree can be drawn:

Now, if K # A ^ F @ B; O $F$ C % A

then

K # A means K is the mother of A.

A ^ F means A is married to F.

F @ B means F is the brother-in-law of B.

O $ F means O is the father of F.

F $ C means F is the father of C.

C % A means C is the son of A.

The above family tree drawn can be extended as follows:

A's husband is F and he is the brother-in-law of B.

Hence, the correct answer is Brother-in-law.

Given:

 

Expression:- D ^ B $ E; B % K * L; E ^ G + L; B % M

Here,

D ^ B means D is married to B.

B $ E means B is the father of E.

B % K means B is the son of K.

K * L means K is the grandmother of L.

E ^ G means E is married to G.

G + L means G is the sister-in-law of L.

B % M means B is the son of M.

Thus, the following family tree can be drawn:

Now, D ~ H $ B

then

D ~ H means D is the daughter of H

H ! B means H is the father-in-law of B.

The above family tree drawn can be extended as follows:

H is the grandfather of E.

Hence, the correct answer is Grandfather.

Given:

 

Expression:- D ^ B $ E; B % K * L; E ^ G + L; B % M

Here,

D ^ B means D is married to B.

B $ E means B is the father of E.

B % K means B is the son of K.

K * L means K is the grandmother of L.

E ^ G means E is married to G.

G + L means G is the sister-in-law of L.

B % M means B is the son of M.

Thus, the following family tree can be drawn:

Now,  If N is the nephew of P and L = P ___ G # N ___ E

then

L = P means L is the brother of P.

P ___ G means P is either sister-in-law or brother-in-law of G. So, either the + or @ symbol will come in the blank.

G # N means G is the mother of N.

N ___ E, as N is a male (nephew of P) it means N is the son of E so here % symbol will come in the blank.

Hence, the correct answer is Either 1 or 2.

Boxes: A, B, C, D, P and Q

1. One box is placed between A and C.
2. P is added to this stack, above C below A.

 

3.  A is adjacent to D.
4. Q is placed just above D.

 

5. B is not placed at the top. So, B is placed at the bottom.

 

6. All boxes contain different number of pencils in consecutive multiple of 6 from bottom to top.
7.  A contains 30 pencils.

So, Q has 42 pencils, D has 36 pencils, P has 24 pencils, C has 18 pencils and B has 12 pencils.

Thus, the final arrangement is,

 

Thus**, Q is placed at the top.**

Boxes: A, B, C, D, P and Q

1. One box is placed between A and C.
2. P is added to this stack, above C below A.

 

3.  A is adjacent to D.
4. Q is placed just above D.

 

5. B is not placed at the top. So, B is placed at the bottom.

 

6. All boxes contain different number of pencils in consecutive multiple of 6 from bottom to top.
7. A contains 30 pencils.

So, Q has 42 pencils, D has 36 pencils, P has 24 pencils, C has 18 pencils and B has 12 pencils.

Thus, the final arrangement is,

 

The sum of pencils in box A and box P is 30 + 24 = 54.

Thus, the sum of pencils A and P is 54.

From statement I,

I. F sits third to right of E, who likes Singing.

II.  D sits immediate left of F.

III. B having Boxing as hobby sits second to right of C, who sits third to right of D.

IV.  A sit third to left of B.

V. G sits third to right of one like Dancing, who sits second to right of A. 

VI. The one who likes Football sit in-front of the one who likes Badminton, who sits third to left of E.

Clearly, we don’t know exact position of the one whose hobby is Cricket.

Hence, statement I is not sufficient alone.  

 

From statement II,

I. A sits third to left B, who like Boxing.

II. B sits third to left of D, who sits second to right of the one who likes Singing.

III. E likes Singing and sits third to left of the one who likes Cricket.

IV. The one likes Boxing sits immediate right of H.

V. The one who likes Football sit third to left of A.

VI. F sits second to left of C, who sits second to left of the one who likes Boxing.

Clearly, we don’t know hobby of the C, who sits second to right of the one who likes Cricket.

Hence, statement II is not sufficient alone.

 

From I and II:

After combining both statements we have: 

Clearly, C (Badminton) sits second to right of F (Cricket).

Thus, statement I and II together are sufficient.

Hence, option V is the correct answer.

Given six persons- B1, B2, B3, B4, B5 and B6 go on a trip on two different dates- 8 and 19.

Three different months- April, May and June.

From statement I,

 

From statement II, we can’t get any clarity or possibility because of insufficient information.

Also from statement III, we can’t get any clarity or possibility because of insufficient information. Because we can't determine on which date B4 is going. Also, we know B2 goes after B6. But we can't determine the date.

But combining statement I and III we get that case- I, III and IV are cancelled. Because only in case-II, B4 goes on an odd number date which is 19th May. Also, two persons go between B6 and B2. B2 goes after B6. So, we can find the answer,

 

B2 goes on 8th June.

Hence, only statement I and III are sufficient to find the answer.

Statement I: One person sits between T and P. T sits on one of the ends and does not face north. H sits opposite to P.

 

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C is an immediate neighbour of H.

 

Hence, statement I alone is not sufficient to answer the question.

Statement II: K an P both are immediate neighbour o each other both facing south. None of them sits at the corner. K sits opposite to C. H sits immediate right of C.

 

Hence, statement II alone is not sufficient to answer the question.

If we combine the two statements we get,

 

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Hence, H sits third from the left end.

Therefore, the correct option is 5.

Given:-

A certain number of floors 

2 flats - P and Q

Flat P is in the west of flat Q.

Immediately above or below means both persons live in the same flat.

1) C lives on an even-numbered floor below the 5th floor in flat P. It means C either lives on the 2nd floor or 4th floor.

2) D lives at a gap of two floors from C and E lives immediately below D's flat. D lives on either the 7th or 5th floor and E immediately below it in the same flat but we cannot determine the flat yet.

3) No one lives to the east of E. It means both D and E live in flat Q.

Case I -

 

Case II -

 

4) D does not live on the 5th floor and F does not live on the floor which is a multiple of 3. So case I gets eliminated.

5) One floor is between D and F. F cannot live on floor 9 as it is a multiple of 3 so F lives on floor 5.

6) G lives immediately above F. This means G and F live in the same flat. This is possible in flat P only.

Case II -

 

7) There are three floors between G and H and both lives in the same flat. H can live on either floor 2 or 10 in flat P.

8) The number of floors above C is the same as the number of floors below H. If H is placed on the 2nd floor then this condition cannot be fulfilled as there are at least 3 floors above C as of now but only 1 floor below H. So, H lives on the 10th floor. There will be 9 floors below H so there will be 9 floors above C. Thus, the building contains 4 + 9 = 13 floors.

Case II -

 

9) J lives on an even-numbered floor above the 10th floor. It means J lives on the 12th floor but we cannot determine the flat yet.

10. K lives immediately above J. It means K lives on the 13th floor and in the same flat as J.

11) No one lives to the east of K. It means K and J live in flat Q.

The final arrangement is -

 

There is a total of 26 flats in the building.

Hence, the correct answer is 26.

Given:-

A certain number of floors 

2 flats - P and Q

Flat P is in the west of flat Q.

Immediately above or below means both persons live in the same flat.

1) C lives on an even-numbered floor below the 5th floor in flat P. It means C either lives on the 2nd floor or 4th floor.

2) D lives at a gap of two floors from C and E lives immediately below D's flat. D lives on either the 7th or 5th floor and E immediately below it in the same flat but we cannot determine the flat yet.

3) No one lives to the east of E. It means both D and E live in flat Q.

Case I -

 

Case II -

 

4) D does not live on the 5th floor and F does not live on the floor which is a multiple of 3. So case I gets eliminated.

5) One floor is between D and F. F cannot live on floor 9 as it is a multiple of 3 so F lives on floor 5.

6) G lives immediately above F. This means G and F live in the same flat. This is possible in flat P only.

Case II -

 

7) There are three floors between G and H and both lives in the same flat. H can live on either floor 2 or 10 in flat P.

8) The number of floors above C is the same as the number of floors below H. If H is placed on the 2nd floor then this condition cannot be fulfilled as there are at least 3 floors above C as of now but only 1 floor below H. So, H lives on the 10th floor. There will be 9 floors below H so there will be 9 floors above C. Thus, the building contains 4 + 9 = 13 floors.

Case II -

 

9) J lives on an even-numbered floor above the 10th floor. It means J lives on the 12th floor but we cannot determine the flat yet.

10. K lives immediately above J. It means K lives on the 13th floor and in the same flat as J.

11) No one lives to the east of K. It means K and J live in flat Q.

The final arrangement is -

 

G lives on the 6th floor in flat P.

Hence, the correct answer is 6th, P.

Given:-

A certain number of floors 

2 flats - P and Q

Flat P is in the west of flat Q.

Immediately above or below means both persons live in the same flat.

1) C lives on an even-numbered floor below the 5th floor in flat P. It means C either lives on the 2nd floor or 4th floor.

2) D lives at a gap of two floors from C and E lives immediately below D's flat. D lives on either the 7th or 5th floor and E immediately below it in the same flat but we cannot determine the flat yet.

3) No one lives to the east of E. It means both D and E live in flat Q.

Case I -

 

Case II -

 

4) D does not live on the 5th floor and F does not live on the floor which is a multiple of 3. So case I gets eliminated.

5) One floor is between D and F. F cannot live on floor 9 as it is a multiple of 3 so F lives on floor 5.

6) G lives immediately above F. This means G and F live in the same flat. This is possible in flat P only.

Case II -

 

7) There are three floors between G and H and both lives in the same flat. H can live on either floor 2 or 10 in flat P.

8) The number of floors above C is the same as the number of floors below H. If H is placed on the 2nd floor then this condition cannot be fulfilled as there are at least 3 floors above C as of now but only 1 floor below H. So, H lives on the 10th floor. There will be 9 floors below H so there will be 9 floors above C. Thus, the building contains 4 + 9 = 13 floors.

Case II -

 

9) J lives on an even-numbered floor above the 10th floor. It means J lives on the 12th floor but we cannot determine the flat yet.

10. K lives immediately above J. It means K lives on the 13th floor and in the same flat as J.

11) No one lives to the east of K. It means K and J live in flat Q.

The final arrangement is -

 

There are 8 floors between C and K.

Hence, the correct answer is 8.

Given:-

A certain number of floors 

2 flats - P and Q

Flat P is in the west of flat Q.

Immediately above or below means both persons live in the same flat.

1) C lives on an even-numbered floor below the 5th floor in flat P. It means C either lives on the 2nd floor or 4th floor.

2) D lives at a gap of two floors from C and E lives immediately below D's flat. D lives on either the 7th or 5th floor and E immediately below it in the same flat but we cannot determine the flat yet.

3) No one lives to the east of E. It means both D and E live in flat Q.

Case I -

 

Case II -

 

4) D does not live on the 5th floor and F does not live on the floor which is a multiple of 3. So case I gets eliminated.

5) One floor is between D and F. F cannot live on floor 9 as it is a multiple of 3 so F lives on floor 5.

6) G lives immediately above F. This means G and F live in the same flat. This is possible in flat P only.

Case II -

 

7) There are three floors between G and H and both lives in the same flat. H can live on either floor 2 or 10 in flat P.

8) The number of floors above C is the same as the number of floors below H. If H is placed on the 2nd floor then this condition cannot be fulfilled as there are at least 3 floors above C as of now but only 1 floor below H. So, H lives on the 10th floor. There will be 9 floors below H so there will be 9 floors above C. Thus, the building contains 4 + 9 = 13 floors.

Case II -

 

9) J lives on an even-numbered floor above the 10th floor. It means J lives on the 12th floor but we cannot determine the flat yet.

10. K lives immediately above J. It means K lives on the 13th floor and in the same flat as J.

11) No one lives to the east of K. It means K and J live in flat Q.

The final arrangement is -

 

Among the given option, all except K lives on an even number of floor.

Hence, the correct answer is K.

Let's have a look at all the statements given:

I. Engaging local leaders is essential to improve education in Africa. → This statement can be concluded as the passage states that investing in local leadership should be the focus of the G20 summit and it also highlights that the expedient solutions often fail as they do not engage local leaders.

II. Economic constraints have hampered the growth of the education sector in Africa. → This statement also can be concluded as the passage states that economic constraints in Africa have adversely affected the growth of the education sector.

III. Local leadership does not affect education. → This statement cannot be concluded as the passage states that investing and engaging local leadership is essential for improving education in Africa.

Hence, the correct answer is Only I and II.

Let's have a look at all statements given:

I. Students opting for medicine and dentistry can be expected to earn more than the current average salary. → The current average salary is given £27,600 per year while the earning potential of medicine and dentistry graduates is given £47,300 per annum in the passage. It can be clearly inferred that students opting for medicine and dentistry can be expected to earn more than the current average salary.

II. The average salary of teachers in UK is less than the current average salary in UK. → In the passage it is given that law graduates can earn £25,200 and those who have teaching degrees could earn less than this. The current average UK salary is given £27,600 per year which is more than the earnings of lawyers and of teachers in UK. Thus, it can be inferred that the average salary of teachers in UK is less than the current average salary in UK.

III. Only doctors and dentists earn more than the current average salary in UK. → In the passage it is given that doctors and dentists top the list. No other professions are discussed who earn more than the current average salary in UK. That doesn't mean only these two professions earn more. It can be possible but not definite. Therefore, this statement cannot be inferred.

Hence, the correct answer is Only I and II.

As restrictions have been imposed on cash transactions and deposits, this will increase digital transactions. So, the argument I is valid.

The people living below the poverty level in the country will not face difficulties as there are no charges up to a certain extent. Not all transactions are chargeable and the limit of 25000 is justified for poor people. So, argument II is not valid.

The transaction of cash in the country increases the risk of black money in the country while digital transactions bring transparency to the system. So, argument III is valid.

Nothing is mentioned about private banks in particular so argument IV is not valid.

Hence, the correct answer is Only I and III are strong.

Given:-

6 persons - G, H, I, J, K, and L

6 days - Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday.

Each person writes a different number of letters for the essay.

1) More than three persons attend between G and J. It means G and J attend the session on Monday and Saturday but not necessarily in the same order.

2) J is not the first to attend the session. It means G attends the session on Monday and J on Saturday.

 

3) As many persons attend before H as after K.

4. K attends before L. 

Here many possibilities exist:-

 

5) K and I do not attend consecutively. This is possible only in Case I and Case III respectively. So Case II and Case IV get eliminated.

The remaining person I will be placed in the vacant position.

6. Two persons attend between I and the one who writes 430 letters. 

 

7) The one who writes the most number of letters attends the session after Wednesday but before Saturday. It means the person who writes the most letters attends the session either on Thursday or Friday.

8) The one who attends two days before J does not write the most number of letters. It means the person who writes the most letters attends the session on Friday. In Case I, the person is L and the most number of letters is 430. In Case II, the person is H.

9) The one who writes the least number of letters goes immediately after G. In Case I, the person is I and in Case II, the person is K.

 

10) G writes 60 letters less than K. G = K - 60. But in Case III, K writes the least number of letters. So, Case III gets eliminated.

11) K writes 110 letters more than J. K = J + 110.

12) The difference in the number of letters of L and J is equal to the number of letters of K. L - J = K. From 11), L - J = J + 110. In Case I, L = 430. So, 430 - 110 = 2J. J = 320/2 = 160.

K = 160 + 110 = 270.

G = 270 - 60 = 210.

The letters written by I and H cannot be determined.

Hence, the final arrangement is -

 

The smallest three-digit perfect square is 100. It means I writes 100 letters. G + I = 210 + 100 = 310.

Hence, the correct answer is 310.

Given:-

6 persons - G, H, I, J, K, and L

6 days - Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday.

Each person writes a different number of letters for the essay.

1) More than three persons attend between G and J. It means G and J attend the session on Monday and Saturday but not necessarily in the same order.

2) J is not the first to attend the session. It means G attends the session on Monday and J on Saturday.

 

3) As many persons attend before H as after K.

4. K attends before L. 

Here many possibilities exist:-

 

5) K and I do not attend consecutively. This is possible only in Case I and Case III respectively. So Case II and Case IV get eliminated.

The remaining person I will be placed in the vacant position.

6. Two persons attend between I and the one who writes 430 letters. 

 

7) The one who writes the most number of letters attends the session after Wednesday but before Saturday. It means the person who writes the most letters attends the session either on Thursday or Friday.

8) The one who attends two days before J does not write the most number of letters. It means the person who writes the most letters attends the session on Friday. In Case I, the person is L and the most number of letters is 430. In Case II, the person is H.

9) The one who writes the least number of letters goes immediately after G. In Case I, the person is I and in Case II, the person is K.

 

10) G writes 60 letters less than K. G = K - 60. But in Case III, K writes the least number of letters. So, Case III gets eliminated.

11) K writes 110 letters more than J. K = J + 110.

12) The difference in the number of letters of L and J is equal to the number of letters of K. L - J = K. From 11), L - J = J + 110. In Case I, L = 430. So, 430 - 110 = 2J. J = 320/2 = 160.

K = 160 + 110 = 270.

G = 270 - 60 = 210.

The letters written by I and H cannot be determined.

Hence, the final arrangement is -

 

L attends the session just before J

Hence, the correct answer is L.

Given:-

6 persons - G, H, I, J, K, and L

6 days - Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday.

Each person writes a different number of letters for the essay.

1) More than three persons attend between G and J. It means G and J attend the session on Monday and Saturday but not necessarily in the same order.

2) J is not the first to attend the session. It means G attends the session on Monday and J on Saturday.

 

3) As many persons attend before H as after K.

4. K attends before L. 

Here many possibilities exist:-

 

5) K and I do not attend consecutively. This is possible only in Case I and Case III respectively. So Case II and Case IV get eliminated.

The remaining person I will be placed in the vacant position.

6. Two persons attend between I and the one who writes 430 letters. 

 

7) The one who writes the most number of letters attends the session after Wednesday but before Saturday. It means the person who writes the most letters attends the session either on Thursday or Friday.

8) The one who attends two days before J does not write the most number of letters. It means the person who writes the most letters attends the session on Friday. In Case I, the person is L and the most number of letters is 430. In Case II, the person is H.

9) The one who writes the least number of letters goes immediately after G. In Case I, the person is I and in Case II, the person is K.

 

10) G writes 60 letters less than K. G = K - 60. But in Case III, K writes the least number of letters. So, Case III gets eliminated.

11) K writes 110 letters more than J. K = J + 110.

12) The difference in the number of letters of L and J is equal to the number of letters of K. L - J = K. From 11), L - J = J + 110. In Case I, L = 430. So, 430 - 110 = 2J. J = 320/2 = 160.

K = 160 + 110 = 270.

G = 270 - 60 = 210.

The letters written by I and H cannot be determined.

Hence, the final arrangement is -

 

We cannot determine the difference between the number of letters written by H and I.

Hence, the correct answer is Cannot be determined.

Given:-

6 persons - G, H, I, J, K, and L

6 days - Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday.

Each person writes a different number of letters for the essay.

1) More than three persons attend between G and J. It means G and J attend the session on Monday and Saturday but not necessarily in the same order.

2) J is not the first to attend the session. It means G attends the session on Monday and J on Saturday.

 

3) As many persons attend before H as after K.

4. K attends before L. 

Here many possibilities exist:-

 

5) K and I do not attend consecutively. This is possible only in Case I and Case III respectively. So Case II and Case IV get eliminated.

The remaining person I will be placed in the vacant position.

6. Two persons attend between I and the one who writes 430 letters. 

 

7) The one who writes the most number of letters attends the session after Wednesday but before Saturday. It means the person who writes the most letters attends the session either on Thursday or Friday.

8) The one who attends two days before J does not write the most number of letters. It means the person who writes the most letters attends the session on Friday. In Case I, the person is L and the most number of letters is 430. In Case II, the person is H.

9) The one who writes the least number of letters goes immediately after G. In Case I, the person is I and in Case II, the person is K.

 

10) G writes 60 letters less than K. G = K - 60. But in Case III, K writes the least number of letters. So, Case III gets eliminated.

11) K writes 110 letters more than J. K = J + 110.

12) The difference in the number of letters of L and J is equal to the number of letters of K. L - J = K. From 11), L - J = J + 110. In Case I, L = 430. So, 430 - 110 = 2J. J = 320/2 = 160.

K = 160 + 110 = 270.

G = 270 - 60 = 210.

The letters written by I and H cannot be determined.

Hence, the final arrangement is -

 

If H writes the second most number of letters then the order will be L > H > K > G > J > I.

G writes the fourth-highest number of letters.

Hence, the correct answer is G.

Total number of boys in class 'A' = 40

Total number of girls in class 'A' = x + a

Total number of boys in class 'B' = 20

Total number of girls in class 'B' = x + 3a

Total number of boys in class 'C' = 60

Total number of girls in class 'C' = x + 5a

If there were 10 more girls in class 'C', then the probability of selecting a girl from class 'C' would have been 40%.

Therefore, (x + 5a + 10)/(60 + x + 5a + 10) = 2/5

⇒ 5 × (x + 5a) + 50 = 140 + 2 × (x + 5a)

⇒ 3 × (x + 5a) = 90

⇒ x + 5a = 30 ...(i)

If the number of boys in class 'A' is 4 times the number of girls in the same class

Then, (x + a) × 4 = 40

⇒ x + a = 10 ...(ii)

From (i) and (ii)

⇒ 4a = 20

⇒ a = 5

And, x = 10 - a = 5

Total number of students in class 'A' = 40 + x + a = 50

Total number of students in class 'B' = 20 + x + 3a = 40

Required difference = 50 - 40 = 10

Hence, the correct answer is 10.

Total number of boys in class 'A' = 40

Total number of girls in class 'A' = x + a

Total number of boys in class 'B' = 20

Total number of girls in class 'B' = x + 3a

Total number of boys in class 'C' = 60

Total number of girls in class 'C' = x + 5a

If there were 10 more girls in class 'C', then the probability of selecting a girl from class 'C' would have been 40%.

Therefore, (x + 5a + 10)/(60 + x + 5a + 10) = 2/5

⇒ 5 × (x + 5a) + 50 = 140 + 2 × (x + 5a)

⇒ 3 × (x + 5a) = 90

⇒ x + 5a = 30 ...(i)

Total number of students in class A = 50

⇒ 40 + x + a = 50

⇒ x + a = 10 ...(ii)

Using (i) and (ii)

⇒ 4a = 20

⇒ a = 5 

And, b = 10  - a = 5

Total number of girls in class A = x + a = 10

Total number of girls in class B = x + 3a = 20

Total number of boys in class D = 40

Total number of girls in class D = 20 × 1.20 = 24

Total number of students in class D = 40 + 24 = 64

Probability of selecting 2 boys from class 'D' = 40C₂/64C₂ = (40 × 39)/(64 × 63) = 65/168

Hence, the correct answer is 65/168.

Total number of boys in class 'A' = 40

Total number of girls in class 'A' = x + a

Total number of boys in class 'B' = 20

Total number of girls in class 'B' = x + 3a

Total number of boys in class 'C' = 60

Total number of girls in class 'C' = x + 5a

If there were 10 more girls in class 'C', then the probability of selecting a girl from class 'C' would have been 40%.

Therefore, (x + 5a + 10)/(60 + x + 5a + 10) = 2/5

⇒ 5 × (x + 5a) + 50 = 140 + 2 × (x + 5a)

⇒ 3 × (x + 5a) = 90

⇒ x + 5a = 30 ...(i)

Total number of students in class B = 40

⇒ 20 + x + 3a = 40

⇒ x + 3a = 20 ...(ii)

Using (i) and (ii)

⇒ 2a = 10

⇒ a = 5 

And, b = 10  - a = 5

Total number of girls in class A = x + a = 10

Total number of girls in class B = x + 3a = 20

Total number of girls in class C = x + 5a = 30

Required average = [(40 + 20 + 60) + (10 + 20 + 30)]/3 = 60

Hence, the correct answer is 60.

Let the selling price of each 'plastic' toy that is sold by seller 'A' = P

The selling price of each 'plastic' toy that is sold by seller 'B = P (Given)

The selling price of each 'wooden' toy that is sold by seller 'A' = 2P 

The selling price of each 'wooden' toy that is sold by seller 'B' = 2P

Let the number of plastic toys sold by seller 'A' = 3x

The number of wooden toys sold by seller 'A' = (8/3) × 3x = 8x

The number of wooden toys sold by seller 'B' = (8/2) × 3x = 12x

The total number of toys sold by seller 'B' = 100

the number of plastic toys sold by seller 'B' = 100 - 12x

The number of 'plastic' toys sold by seller 'B' = The number of 'wooden' toys sold by seller 'A'.

Therefore, 100 - 12x = 8x

⇒ 20x = 100

⇒ x = 5

Putting all the values in a table we get

 

Required difference = 100 - 55 = 45

Hence, the correct answer is 45.

Let the selling price of each 'plastic' toy that is sold by seller 'A' = P

The selling price of each 'plastic' toy that is sold by seller 'B = P (Given)

The selling price of each 'wooden' toy that is sold by seller 'A' = 2P 

The selling price of each 'wooden' toy that is sold by seller 'B' = 2P

Let the number of plastic toys sold by seller 'A' = 3x

The number of wooden toys sold by seller 'A' = (8/3) × 3x = 8x

The number of wooden toys sold by seller 'B' = (8/2) × 3x = 12x

The total number of toys sold by seller 'B' = 100

the number of plastic toys sold by seller 'B' = 100 - 12x

The number of 'plastic' toys sold by seller 'B' = The number of 'wooden' toys sold by seller 'A'.

Therefore, 100 - 12x = 8x

⇒ 20x = 100

⇒ x = 5

Putting all the values in a table we get

 

Required% = [(40 - 40)/40] × 100 = 0%

Hence, the correct answer is 0%.

Let the selling price of each 'plastic' toy that is sold by seller 'A' = P

The selling price of each 'plastic' toy that is sold by seller 'B = P (Given)

The selling price of each 'wooden' toy that is sold by seller 'A' = 2P 

The selling price of each 'wooden' toy that is sold by seller 'B' = 2P

Let the number of plastic toys sold by seller 'A' = 3x

The number of wooden toys sold by seller 'A' = (8/3) × 3x = 8x

The number of wooden toys sold by seller 'B' = (8/2) × 3x = 12x

The total number of toys sold by seller 'B' = 100

the number of plastic toys sold by seller 'B' = 100 - 12x

The number of 'plastic' toys sold by seller 'B' = The number of 'wooden' toys sold by seller 'A'.

Therefore, 100 - 12x = 8x

⇒ 20x = 100

⇒ x = 5

Putting all the values in a table we get

 

The selling price of each 'plastic toy that is sold by seller 'A' = P =  130 Rs.

The selling price of each 'wooden' toy that is sold by seller 'A' = 2P  = 260 Rs.

The total revenue generated by seller 'A' from selling all the 'wooden' toys together = 260 × 40 = 10400 Rs.

Hence, the correct answer is 10400 Rs.

Shortcut Trick

Let Income of P be 240 units (since 4.166% = 1/24).

Income of Q = 240 + 10 = 250 units. Saving of Q = 68% of 250 = 170 units.

Saving of R = 1.15 × 170 = 195.5 units. Expenditure of R = 3/2 × 195.5 = 293.25 units.

Given Avg (Exp P + Exp R) = 6932 ⇒ Total = 13864. Find P's expenditure using saving ratio 7:5 and solve.

∴ The correct answer is ₹1,12,608.

Alternate Method

Given: Average Expenditure (P + R) = ₹6,932. Q Income = P Income × (1 + 1/24). Q Saving = 68% of Q Income. R Saving = 1.15 × Q Saving. R Exp : R Saving = 3 : 2.

⇒ Total Expenditure (P + R) = 6,932 × 2 = ₹13,864.

⇒ For P: Exp = 1.4 × Saving ⇒ Ratio Exp:Saving = 7:5. Income = 12 units.

⇒ Let Income P = 1200x. Exp P = 700x.

⇒ Income Q = 1200x × (25/24) = 1250x. Saving Q = 1250x × 68% = 850x.

⇒ Saving R = 850x × 1.15 = 977.5x. Exp R = 977.5x × (3/2) = 1466.25x.

⇒ Total Exp = 700x + 1466.25x = 2166.25x = 13864.

⇒ x = 13864 ÷ 2166.25 = 6.4.

⇒ Monthly Exp R = 1466.25 × 6.4 = ₹9,384.

⇒ Annual Exp R = 9,384 × 12 = ₹1,12,608.

∴ The correct answer is ₹1,12,608.

Additional Information

Percentage to Fraction Conversion

4.166% is equivalent to 1/24. This simplifies calculations involving percentage increases significantly.

Income, Expenditure and Saving

Basic identity: Income = Expenditure + Saving. All ratios must be normalized to a common variable.

Annual vs Monthly

Always check if the question asks for monthly or annual values; multiply the monthly result by 12 for the annual figure.

Shortcut Trick

Upstream speed of boat P = 160 ÷ 5 = 32 km/hr.

Downstream speed of boat Q = 32 + 8 = 40 km/hr.

Ratio of Downstream speeds P : Q is 6 : 5 ⇒ Downstream speed of P = (6/5) × 40 = 48 km/hr.

Speed of river A = (48 − 32) ÷ 2 = 8 km/hr → Speed of river B = 8 + 2 = 10 km/hr.

Upstream speed of boat Q = Downstream speed of Q − 2 × river B = 40 − 20 = 20 km/hr.

Time taken by Q = 160 ÷ 20 = 8 hr.

∴ The correct answer is 8 hr.

Alternate Method

Given:

Upstream speed of P (U_P) = 160 ÷ 5 = 32 km/hr.

Downstream speed of Q (D_Q) = U_P + 8 = 32 + 8 = 40 km/hr.

Ratio D_P : D_Q = 6 : 5.

Calculations:

⇒ D_P = (6/5) × 40 = 48 km/hr.

⇒ Speed of river A (V_A) = (D_P − U_P) ÷ 2 = (48 − 32) ÷ 2 = 8 km/hr.

⇒ Speed of river B (V_B) = V_A + 2 = 8 + 2 = 10 km/hr.

⇒ Speed of boat Q in still water (B_Q) = D_Q − V_B = 40 − 10 = 30 km/hr.

⇒ Upstream speed of boat Q (U_Q) = B_Q − V_B = 30 − 10 = 20 km/hr.

⇒ Time taken by Q for 160 km = 160 ÷ 20 = 8 hr.

∴ The correct answer is 8 hr.

Additional Information

Relative Speed in Rivers

Downstream speed = Speed of boat + Speed of stream. Upstream speed = Speed of boat − Speed of stream.

Speed of Stream Formula

Speed of stream = (Downstream speed − Upstream speed) ÷ 2.

Still Water Speed Formula

Speed of boat in still water = (Downstream speed + Upstream speed) ÷ 2.