SBI PO Mains Memory Based Paper (Held On: 13th September 2025) (Previous Year Paper)

Given: Eight people of a three-generation family were born on the same date and same month of different years, 1944, 1954, 1961, 1979, 1993, 1998, 2007, and 2014. They are of different ages and like different fruits, viz., Apple, Mango, Banana, Guava, Plum, Watermelon, Kiwi, and Orange. Their age is calculated with respect to the base year 2025.

1. The age of the one who likes Watermelon is the cube of a number. 

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

2. Four people were born between P and the one who likes Watermelon.

 

3. The age difference between M and the one who likes Apple is 17 years. 

 

4. The age digit of M’s child is the reverse of the age digit of M’s father. (There is only one pair: 81 years and 18 years)

 

5. The age of R’s daughter is equal to the sum of the ages of T and the one who likes Kiwi. (There is only one pair: 64 = 46 + 18 years, so, R's daughter = 64 years)

6. S is the sister-in-law of M and has two children. 

7. The age difference between P and the one who likes Guava is the same as the age difference between R’s only granddaughter and the one who likes Guava. (There is only one pair: 81 - 46 = 35 and 46 - 11 = 35 years, So, T likes Guava)

Now, we can eliminate Case 1. (Because P and R's only granddaughter are not the same)

 

So, P is M's father and R is M's mother.

Now, we can eliminate Case 1.

8. Three people were born between N and the one who likes Mango.  9. N’s mother is immediately older than N, who does not like Kiwi. 

 

10. R is three people older than the one who likes Plum. 

 

11. The age of the one who likes Orange is an odd number. (So, S likes Banana)

 

12. Q does not like Kiwi and Plum.  

 

13. One of the parents of O is not alive. (So, O is M's child, so that we can eliminate Case II of blood relation.)

This is our final arrangement,

 

Total age of males who belong to the first generation = O(18) + N(27) = 45 years.

Hence, the correct answer is 45 years.

Given: Eight people of a three-generation family were born on the same date and same month of different years, 1944, 1954, 1961, 1979, 1993, 1998, 2007, and 2014. They are of different ages and like different fruits, viz., Apple, Mango, Banana, Guava, Plum, Watermelon, Kiwi, and Orange. Their age is calculated with respect to the base year 2025.

1. The age of the one who likes Watermelon is the cube of a number. 

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

2. Four people were born between P and the one who likes Watermelon.

 

3. The age difference between M and the one who likes Apple is 17 years. 

 

4. The age digit of M’s child is the reverse of the age digit of M’s father. (There is only one pair: 81 years and 18 years)

 

5. The age of R’s daughter is equal to the sum of the ages of T and the one who likes Kiwi. (There is only one pair: 64 = 46 + 18 years, so, R's daughter = 64 years)

6. S is the sister-in-law of M and has two children. 

7. The age difference between P and the one who likes Guava is the same as the age difference between R’s only granddaughter and the one who likes Guava. (There is only one pair: 81 - 46 = 35 and 46 - 11 = 35 years, So, T likes Guava)

Now, we can eliminate Case 1. (Because P and R's only granddaughter are not the same)

 

So, P is M's father and R is M's mother.

Now, we can eliminate Case 1.

8. Three people were born between N and the one who likes Mango.  9. N’s mother is immediately older than N, who does not like Kiwi. 

 

10. R is three people older than the one who likes Plum. 

 

11. The age of the one who likes Orange is an odd number. (So, S likes Banana)

 

12. Q does not like Kiwi and Plum.  

 

13. One of the parents of O is not alive. (So, O is M's child, so that we can eliminate Case II of blood relation.)

This is our final arrangement,

 

Hence, the correct answer is 1993.

Given: Eight people of a three-generation family were born on the same date and same month of different years, 1944, 1954, 1961, 1979, 1993, 1998, 2007, and 2014. They are of different ages and like different fruits, viz., Apple, Mango, Banana, Guava, Plum, Watermelon, Kiwi, and Orange. Their age is calculated with respect to the base year 2025.

1. The age of the one who likes Watermelon is the cube of a number. 

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

2. Four people were born between P and the one who likes Watermelon.

 

3. The age difference between M and the one who likes Apple is 17 years. 

 

4. The age digit of M’s child is the reverse of the age digit of M’s father. (There is only one pair: 81 years and 18 years)

 

5. The age of R’s daughter is equal to the sum of the ages of T and the one who likes Kiwi. (There is only one pair: 64 = 46 + 18 years, so, R's daughter = 64 years)

6. S is the sister-in-law of M and has two children. 

7. The age difference between P and the one who likes Guava is the same as the age difference between R’s only granddaughter and the one who likes Guava. (There is only one pair: 81 - 46 = 35 and 46 - 11 = 35 years, So, T likes Guava)

Now, we can eliminate Case 1. (Because P and R's only granddaughter are not the same)

 

So, P is M's father and R is M's mother.

Now, we can eliminate Case 1.

8. Three people were born between N and the one who likes Mango.  9. N’s mother is immediately older than N, who does not like Kiwi. 

 

10. R is three people older than the one who likes Plum. 

 

11. The age of the one who likes Orange is an odd number. (So, S likes Banana)

 

12. Q does not like Kiwi and Plum.  

 

13. One of the parents of O is not alive. (So, O is M's child, so that we can eliminate Case II of blood relation.)

This is our final arrangement,

 

Hence, the correct answer is Watermelon.

Given: Eight people of a three-generation family were born on the same date and same month of different years, 1944, 1954, 1961, 1979, 1993, 1998, 2007, and 2014. They are of different ages and like different fruits, viz., Apple, Mango, Banana, Guava, Plum, Watermelon, Kiwi, and Orange. Their age is calculated with respect to the base year 2025.

1. The age of the one who likes Watermelon is the cube of a number. 

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

2. Four people were born between P and the one who likes Watermelon.

 

3. The age difference between M and the one who likes Apple is 17 years. 

 

4. The age digit of M’s child is the reverse of the age digit of M’s father. (There is only one pair: 81 years and 18 years)

 

5. The age of R’s daughter is equal to the sum of the ages of T and the one who likes Kiwi. (There is only one pair: 64 = 46 + 18 years, so, R's daughter = 64 years)

6. S is the sister-in-law of M and has two children. 

7. The age difference between P and the one who likes Guava is the same as the age difference between R’s only granddaughter and the one who likes Guava. (There is only one pair: 81 - 46 = 35 and 46 - 11 = 35 years, So, T likes Guava)

Now, we can eliminate Case 1. (Because P and R's only granddaughter are not the same)

 

So, P is M's father and R is M's mother.

Now, we can eliminate Case 1.

8. Three people were born between N and the one who likes Mango.  9. N’s mother is immediately older than N, who does not like Kiwi. 

 

10. R is three people older than the one who likes Plum. 

 

11. The age of the one who likes Orange is an odd number. (So, S likes Banana)

 

12. Q does not like Kiwi and Plum.  

 

13. One of the parents of O is not alive. (So, O is M's child, so that we can eliminate Case II of blood relation.)

This is our final arrangement,

 

Hence, the correct answer is O - Kiwi.

Given: Eight people of a three-generation family were born on the same date and same month of different years, 1944, 1954, 1961, 1979, 1993, 1998, 2007, and 2014. They are of different ages and like different fruits, viz., Apple, Mango, Banana, Guava, Plum, Watermelon, Kiwi, and Orange. Their age is calculated with respect to the base year 2025.

1. The age of the one who likes Watermelon is the cube of a number. 

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

2. Four people were born between P and the one who likes Watermelon.

 

3. The age difference between M and the one who likes Apple is 17 years. 

 

4. The age digit of M’s child is the reverse of the age digit of M’s father. (There is only one pair: 81 years and 18 years)

 

5. The age of R’s daughter is equal to the sum of the ages of T and the one who likes Kiwi. (There is only one pair: 64 = 46 + 18 years, so, R's daughter = 64 years)

6. S is the sister-in-law of M and has two children. 

7. The age difference between P and the one who likes Guava is the same as the age difference between R’s only granddaughter and the one who likes Guava. (There is only one pair: 81 - 46 = 35 and 46 - 11 = 35 years, So, T likes Guava)

Now, we can eliminate Case 1. (Because P and R's only granddaughter are not the same)

 

So, P is M's father and R is M's mother.

Now, we can eliminate Case 1.

8. Three people were born between N and the one who likes Mango.  9. N’s mother is immediately older than N, who does not like Kiwi. 

 

10. R is three people older than the one who likes Plum. 

 

11. The age of the one who likes Orange is an odd number. (So, S likes Banana)

 

12. Q does not like Kiwi and Plum.  

 

13. One of the parents of O is not alive. (So, O is M's child, so that we can eliminate Case II of blood relation.)

This is our final arrangement,

 

Hence, the correct answer is Total age of Q and M is a multiple of 5.

Logic:

Step I: Reverse the letters of the word and combine all letters

Step II:

Letters: Last 2 letters of the word

Number: If the number of letters in a word is even, then subtract 1, and if the number of letters in a word is odd, then add 1.

Step III: 

Letters: If the letter is a vowel, change it to the previous letter, and if the letter is a consonant, change it to the next letter.

Symbol: If the number is odd, change it to *, and if the number is even, change it to %.

Step IV: Pick the second letter of * and pick the first letter of %.

Phrase: Digital innovation strategy execution

Step I: latigidnoitavonniygetartsnoitucexe

Step II: di8 in9 st7 ex10

Step III: eh% ho* tu* dy%

Step IV: ed% ou*

The second letter from the left in Step I = a

and the third letter from the right in Step I = e

Letters between a and e are b, c, and d, i.e., 3 letters.

Hence, the correct answer is Three.

Logic:

Step I: Reverse the letters of the word and combine all letters

Step II:

Letters: Last 2 letters of the word

Number: If the number of letters in a word is even, then subtract 1, and if the number of letters in a word is odd, then add 1.

Step III: 

Letters: If the letter is a vowel, change it to the previous letter, and if the letter is a consonant, change it to the next letter.

Symbol: If the number is odd, change it to *, and if the number is even, change it to %.

Step IV: Pick the second letter of * and pick the first letter of %.

Phrase: Digital innovation strategy execution

Step I: latigidnoitavonniygetartsnoitucexe

Step II: di8 in9 st7 ex10

Step III: eh% ho* tu* dy%

Step IV: ed% ou*

Hence, the correct answer is ed% ou*.

Logic:

Step I: Reverse the letters of the word and combine all letters

Step II:

Letters: Last 2 letters of the word

Number: If the number of letters in a word is even, then subtract 1, and if the number of letters in a word is odd, then add 1.

Step III: 

Letters: If the letter is a vowel, change it to the previous letter, and if the letter is a consonant, change it to the next letter.

Symbol: If the number is odd, change it to *, and if the number is even, change it to %.

Step IV: Pick the second letter of * and pick the first letter of %.

Phrase: Digital innovation strategy execution

Step I: latigidnoitavonniygetartsnoitucexe

Step II: di8 in9 st7 ex10

Step III: eh% ho* tu* dy%

Step IV: ed% ou*

Hence, the correct answer is ho.*

Given: There is a four-floor residential building such that Floor 2 is above Floor 1, Floor 3 is above Floor 2, and Floor 4 is above Floor 3. Each floor has two flats, named Flat X and Flat Y, where Flat X is to the west of Flat Y. The total area of each flat is 2880 sq ft. Each flat consists of a certain number of rooms, and the area of every room within a flat is equal. No two flats have the same number of rooms.

1. The flat in which each room has an area of 360 sq ft is immediately above the flat having 5 rooms, which is located to the west of the flat in which each room has an area of 480 sq ft, which is situated on an odd-numbered floor. (2880 ÷ 360 = 8 rooms, 2880 ÷ 480 = 6 rooms, 2880 ÷ 5 = 576 sq ft)

Here, we have two cases, i.e., Case 1 and Case 2.

 

2. The flat having 4 rooms is on an even-numbered floor and is immediately above the flat in which each room has an area of 180 sq ft, which is located to the east of the flat in which each room has an area of 320 sq ft. (2880 ÷ 4 = 720 sq ft, 2880 ÷ 180 = 16 rooms, 2880 ÷ 320 = 9 rooms)

Now, we can eliminate Case 2. (Because there is no vacant place for a 320 sq ft area)

 

3. The total number of rooms on Floor 2 is 16, and none of the flats on this floor have rooms with an area less than 160 sq ft. (i.e., Floor 2 Flat A = 12 rooms, 2880 ÷ 12 = 240 sq ft)

 

4. The area of each room of one of the flats on Floor 4 is 20 sq ft less than the area of each room of Flat Y on Floor 1. (i.e., Area of each room on floor 4 is 180 - 20 = 160 sq ft, so rooms = 2880 ÷ 160 = 18 rooms)

This is our final arrangement,

 

Total rooms on floor 3 = 5 + 6 = 11 rooms.

Hence, the correct answer is 11 rooms.

Given: There is a four-floor residential building such that Floor 2 is above Floor 1, Floor 3 is above Floor 2, and Floor 4 is above Floor 3. Each floor has two flats, named Flat X and Flat Y, where Flat X is to the west of Flat Y. The total area of each flat is 2880 sq ft. Each flat consists of a certain number of rooms, and the area of every room within a flat is equal. No two flats have the same number of rooms.

1. The flat in which each room has an area of 360 sq ft is immediately above the flat having 5 rooms, which is located to the west of the flat in which each room has an area of 480 sq ft, which is situated on an odd-numbered floor. (2880 ÷ 360 = 8 rooms, 2880 ÷ 480 = 6 rooms, 2880 ÷ 5 = 576 sq ft)

Here, we have two cases, i.e., Case 1 and Case 2.

 

2. The flat having 4 rooms is on an even-numbered floor and is immediately above the flat in which each room has an area of 180 sq ft, which is located to the east of the flat in which each room has an area of 320 sq ft. (2880 ÷ 4 = 720 sq ft, 2880 ÷ 180 = 16 rooms, 2880 ÷ 320 = 9 rooms)

Now, we can eliminate Case 2. (Because there is no vacant place for a 320 sq ft area)

 

3. The total number of rooms on Floor 2 is 16, and none of the flats on this floor have rooms with an area less than 160 sq ft. (i.e., Floor 2 Flat A = 12 rooms, 2880 ÷ 12 = 240 sq ft)

 

4. The area of each room of one of the flats on Floor 4 is 20 sq ft less than the area of each room of Flat Y on Floor 1. (i.e., Area of each room on floor 4 is 180 - 20 = 160 sq ft, so rooms = 2880 ÷ 160 = 18 rooms)

This is our final arrangement,

 

Hence, the correct answer is 12 rooms.

Given: There is a four-floor residential building such that Floor 2 is above Floor 1, Floor 3 is above Floor 2, and Floor 4 is above Floor 3. Each floor has two flats, named Flat X and Flat Y, where Flat X is to the west of Flat Y. The total area of each flat is 2880 sq ft. Each flat consists of a certain number of rooms, and the area of every room within a flat is equal. No two flats have the same number of rooms.

1. The flat in which each room has an area of 360 sq ft is immediately above the flat having 5 rooms, which is located to the west of the flat in which each room has an area of 480 sq ft, which is situated on an odd-numbered floor. (2880 ÷ 360 = 8 rooms, 2880 ÷ 480 = 6 rooms, 2880 ÷ 5 = 576 sq ft)

Here, we have two cases, i.e., Case 1 and Case 2.

 

2. The flat having 4 rooms is on an even-numbered floor and is immediately above the flat in which each room has an area of 180 sq ft, which is located to the east of the flat in which each room has an area of 320 sq ft. (2880 ÷ 4 = 720 sq ft, 2880 ÷ 180 = 16 rooms, 2880 ÷ 320 = 9 rooms)

Now, we can eliminate Case 2. (Because there is no vacant place for a 320 sq ft area)

 

3. The total number of rooms on Floor 2 is 16, and none of the flats on this floor have rooms with an area less than 160 sq ft. (i.e., Floor 2 Flat A = 12 rooms, 2880 ÷ 12 = 240 sq ft)

 

4. The area of each room of one of the flats on Floor 4 is 20 sq ft less than the area of each room of Flat Y on Floor 1. (i.e., Area of each room on floor 4 is 180 - 20 = 160 sq ft, so rooms = 2880 ÷ 160 = 18 rooms)

This is our final arrangement,

 

Hence, the correct answer is The highest number of rooms is in Flat Y on Floor 4.

Given: There is a four-floor residential building such that Floor 2 is above Floor 1, Floor 3 is above Floor 2, and Floor 4 is above Floor 3. Each floor has two flats, named Flat X and Flat Y, where Flat X is to the west of Flat Y. The total area of each flat is 2880 sq ft. Each flat consists of a certain number of rooms, and the area of every room within a flat is equal. No two flats have the same number of rooms.

1. The flat in which each room has an area of 360 sq ft is immediately above the flat having 5 rooms, which is located to the west of the flat in which each room has an area of 480 sq ft, which is situated on an odd-numbered floor. (2880 ÷ 360 = 8 rooms, 2880 ÷ 480 = 6 rooms, 2880 ÷ 5 = 576 sq ft)

Here, we have two cases, i.e., Case 1 and Case 2.

 

2. The flat having 4 rooms is on an even-numbered floor and is immediately above the flat in which each room has an area of 180 sq ft, which is located to the east of the flat in which each room has an area of 320 sq ft. (2880 ÷ 4 = 720 sq ft, 2880 ÷ 180 = 16 rooms, 2880 ÷ 320 = 9 rooms)

Now, we can eliminate Case 2. (Because there is no vacant place for a 320 sq ft area)

 

3. The total number of rooms on Floor 2 is 16, and none of the flats on this floor have rooms with an area less than 160 sq ft. (i.e., Floor 2 Flat A = 12 rooms, 2880 ÷ 12 = 240 sq ft)

 

4. The area of each room of one of the flats on Floor 4 is 20 sq ft less than the area of each room of Flat Y on Floor 1. (i.e., Area of each room on floor 4 is 180 - 20 = 160 sq ft, so rooms = 2880 ÷ 160 = 18 rooms)

This is our final arrangement,

 

Total rooms in Flat X = 8 + 5 + 12 + 9 = 34 rooms.

Hence, the correct answer is 34 rooms.

1. L is placed at the extreme right end of the arrangement.

_ _ _ _ _ _ _ _ L

2. Letter I is placed immediately to the left of D.

I D

3. Exactly two letters are placed between D and T. 

I D _ _ T

4. L is placed fourth to the right of T.

I D _ _ T _ _ _ L

5. E is placed immediately left of N. 6. The second I is placed somewhere to the right of N but is not adjacent to N. 

Here, we have 3 Cases, i.e., Case I, Case II, and Case III.

Case I: I D E N T I _ _ L

Case II: I D E N T _ I _ L

Case III: I D E N T _ _ I L

7. A is placed immediately to the right of C. 

Now, we can eliminate Case II.

Case I: I D E N T I C A L

Case III: I D E N T C A I L

In case III, it's not a meaningful word.

So, the correct word is IDENTICAL.

Hence, the correct answer is N.

Given: Twelve people are sitting at an equal distance in two parallel rows. A, B, C, D, E, and F are sitting in Row 1 facing north, and M, N, O, P, Q, and R are sitting in Row 2 facing south, such that the persons sitting in Row 1 face the persons sitting in Row 2. Each person celebrates his/her birthday on a different date in the month of 30 days. A, B, C, D, E, and F are born on odd-numbered dates, while M, N, O, P, Q, and R are born on even-numbered dates.

1. D sits third to the right of B, who celebrates his birthday on the 15th, and one of them sits at the end of Row 1. 

Here, we have 2 Cases, i.e., Case 1 and Case 2.

2. There are three people between the person who celebrates his birthday on the 15th and the person who celebrates his birthday on an odd-numbered date greater than 3 but less than 6.

Now, we can eliminate Case 2.

3. N faces the person who sits third to the left of F, who celebrates his birthday on the 1st. 

4. M sits immediately right of P, who faces E. 

5. Q sits exactly between N and R, but R does not face A.

6. The person who faces E celebrates his birthday on double the date of C’s birthday.

7. The person who celebrates his birthday on 11 is an immediate neighbour of C. 8. The person who faces A celebrates his birthday on the cube of 2. 

9. The person who celebrates his birthday on the 25th sits immediately right of the person who celebrates his birthday on the 13th. (Both dates are odd, i.e., the persons are in Row 1.)

10. The person who sits diagonally opposite B celebrates his birthday on the square of an even prime number.

11. The person sitting at the right end of Row 2 celebrates his birthday on the 6th.
12. The person who faces C celebrates his birthday on double the birthday date of M, while the person who sits immediate right of R celebrates his birthday on half the birth date of O.

This is our final arrangement,

Hence, the correct answer is Q.

Given: Twelve people are sitting at an equal distance in two parallel rows. A, B, C, D, E, and F are sitting in Row 1 facing north, and M, N, O, P, Q, and R are sitting in Row 2 facing south, such that the persons sitting in Row 1 face the persons sitting in Row 2. Each person celebrates his/her birthday on a different date in the month of 30 days. A, B, C, D, E, and F are born on odd-numbered dates, while M, N, O, P, Q, and R are born on even-numbered dates.

1. D sits third to the right of B, who celebrates his birthday on the 15th, and one of them sits at the end of Row 1. 

Here, we have 2 Cases, i.e., Case 1 and Case 2.

2. There are three people between the person who celebrates his birthday on the 15th and the person who celebrates his birthday on an odd-numbered date greater than 3 but less than 6.

Now, we can eliminate Case 2.

3. N faces the person who sits third to the left of F, who celebrates his birthday on the 1st. 

4. M sits immediately right of P, who faces E. 

5. Q sits exactly between N and R, but R does not face A.

6. The person who faces E celebrates his birthday on double the date of C’s birthday.

7. The person who celebrates his birthday on 11 is an immediate neighbour of C. 8. The person who faces A celebrates his birthday on the cube of 2. 

9. The person who celebrates his birthday on the 25th sits immediately right of the person who celebrates his birthday on the 13th. (Both dates are odd, i.e., the persons are in Row 1.)

10. The person who sits diagonally opposite B celebrates his birthday on the square of an even prime number.

11. The person sitting at the right end of Row 2 celebrates his birthday on the 6th.
12. The person who faces C celebrates his birthday on double the birthday date of M, while the person who sits immediate right of R celebrates his birthday on half the birth date of O.

This is our final arrangement,

Hence, the correct answer is A - N.

Given: Twelve people are sitting at an equal distance in two parallel rows. A, B, C, D, E, and F are sitting in Row 1 facing north, and M, N, O, P, Q, and R are sitting in Row 2 facing south, such that the persons sitting in Row 1 face the persons sitting in Row 2. Each person celebrates his/her birthday on a different date in the month of 30 days. A, B, C, D, E, and F are born on odd-numbered dates, while M, N, O, P, Q, and R are born on even-numbered dates.

1. D sits third to the right of B, who celebrates his birthday on the 15th, and one of them sits at the end of Row 1. 

Here, we have 2 Cases, i.e., Case 1 and Case 2.

2. There are three people between the person who celebrates his birthday on the 15th and the person who celebrates his birthday on an odd-numbered date greater than 3 but less than 6.

Now, we can eliminate Case 2.

3. N faces the person who sits third to the left of F, who celebrates his birthday on the 1st. 

4. M sits immediately right of P, who faces E. 

5. Q sits exactly between N and R, but R does not face A.

6. The person who faces E celebrates his birthday on double the date of C’s birthday.

7. The person who celebrates his birthday on 11 is an immediate neighbour of C. 8. The person who faces A celebrates his birthday on the cube of 2. 

9. The person who celebrates his birthday on the 25th sits immediately right of the person who celebrates his birthday on the 13th. (Both dates are odd, i.e., the persons are in Row 1.)

10. The person who sits diagonally opposite B celebrates his birthday on the square of an even prime number.

11. The person sitting at the right end of Row 2 celebrates his birthday on the 6th.
12. The person who faces C celebrates his birthday on double the birthday date of M, while the person who sits immediate right of R celebrates his birthday on half the birth date of O.

This is our final arrangement,

Hence, the correct answer is M.

Given: Twelve people are sitting at an equal distance in two parallel rows. A, B, C, D, E, and F are sitting in Row 1 facing north, and M, N, O, P, Q, and R are sitting in Row 2 facing south, such that the persons sitting in Row 1 face the persons sitting in Row 2. Each person celebrates his/her birthday on a different date in the month of 30 days. A, B, C, D, E, and F are born on odd-numbered dates, while M, N, O, P, Q, and R are born on even-numbered dates.

1. D sits third to the right of B, who celebrates his birthday on the 15th, and one of them sits at the end of Row 1. 

Here, we have 2 Cases, i.e., Case 1 and Case 2.

2. There are three people between the person who celebrates his birthday on the 15th and the person who celebrates his birthday on an odd-numbered date greater than 3 but less than 6.

Now, we can eliminate Case 2.

3. N faces the person who sits third to the left of F, who celebrates his birthday on the 1st. 

4. M sits immediately right of P, who faces E. 

5. Q sits exactly between N and R, but R does not face A.

6. The person who faces E celebrates his birthday on double the date of C’s birthday.

7. The person who celebrates his birthday on 11 is an immediate neighbour of C. 8. The person who faces A celebrates his birthday on the cube of 2. 

9. The person who celebrates his birthday on the 25th sits immediately right of the person who celebrates his birthday on the 13th. (Both dates are odd, i.e., the persons are in Row 1.)

10. The person who sits diagonally opposite B celebrates his birthday on the square of an even prime number.

11. The person sitting at the right end of Row 2 celebrates his birthday on the 6th.
12. The person who faces C celebrates his birthday on double the birthday date of M, while the person who sits immediate right of R celebrates his birthday on half the birth date of O.

This is our final arrangement,

Hence, the correct answer is 25.

Statement I: Box L is kept immediately above box M. Box J is kept two boxes above box M. Box N is adjacent to box M but below box K.

 

Statement II: Box K is kept above box N. Box O is kept below box J. More than one box is placed between box K and box O. Box M is not kept at the top or bottom.

From I and II,

 

Hence, the correct answer is Data given in both statements I and II together are sufficient to answer.

Statement I: C delivers the presentation immediately after A, but before B. Exactly two people deliver presentations between C and E. D delivers the presentation immediately before A. F delivers the presentation after D but before B.

 

Statement II: More than two people deliver presentations after A. E delivers the presentation two days before B. C delivers the presentation after D. An equal number of people deliver presentations before F and after A. D delivers the presentation before A.

 

Hence, the correct answer is Data given in statement II alone is sufficient to answer.

Statement I: S sits second to the right of W. U sits immediately to the right of S. V sits somewhere to the left of W. Exactly three persons sit between X and U.

Statement II: One person sits between T and S. X sits immediately to the right of T. V sits second to the right of T. W is not an immediate neighbour of S.

Hence, the correct answer is Data given in either statement I or statement II alone is sufficient to answer.

Option 1: Incorrect

The passage does not claim that digital infrastructure has completely eliminated inequality. It only states that service delivery and inclusion have improved, not that disparities no longer exist.

Option 2: Incorrect

The paragraph clearly highlights environmental and sustainability concerns, indicating that infrastructure growth alone is not sufficient for long-term progress.

Option 3: Correct

The passage discusses both the benefits of rapid development (growth, inclusion, innovation) and the challenges it brings (environmental impact, data privacy, unequal adaptation). This supports the inference that development requires careful management.

Option 4: Incorrect

Although traditional sectors face challenges, the passage does not suggest their inevitable closure. It merely indicates difficulty in adapting.

Option 5: Incorrect

Environmental concerns are explicitly mentioned, making this option contradictory to the passage.

Hence, the correct answer is Rapid development brings both opportunities and challenges that need careful management.

Option 1: Incorrect

The passage does not argue for stopping development; instead, it stresses managing growth responsibly.

Option 2: Correct

The final lines of the paragraph explicitly emphasize balancing growth with inclusivity, innovation with regulation, and development with sustainability. This directly leads to this conclusion.

Option 3: Incorrect

There is no suggestion that digital platforms should entirely replace traditional sectors.

Option 4: Incorrect

The passage acknowledges global economic uncertainties, indicating that they do affect growth strategies.

Option 5: Incorrect

The paragraph implies that sustainability and growth must coexist, not that they are mutually exclusive.

Hence, the correct answer is Policymakers must strike a balance between development, regulation, and inclusivity.

Option 1: Incorrect

The passage does not assume resistance from citizens toward technology adoption.

Option 2: Correct

The government’s emphasis on digital infrastructure and large projects assumes that technology and infrastructure can drive economic growth, efficiency, and inclusion.

Option 3: Incorrect

Environmental concerns are treated as important challenges, not as secondary or ignorable.

Option 4: Incorrect

The passage highlights benefits in semi-urban and rural areas, contradicting this assumption.

Option 5: Incorrect

There is no assumption that regulations inherently discourage innovation; rather, the need for balance is emphasized.

Hence, the correct answer is Economic growth can be enhanced through technology and infrastructure development.

Logic:

Step 1: Find the highest digit in the number and write its square. Rearrange the letters in alphabetical order.

For each vowel → next alphabet

For each consonant → previous alphabet

Write the positive difference of the remaining two digits.

Step 2: Reverse the letters obtained in Step 1. Replace the number with the sum of the number obtained in Step 1.

Step 3: Write the 1st and 2nd elements in small letters. Replace the number with multiply by 2.

Step 4: Each letter is replaced by its previous alphabet. The number is multiplied by 3.

Step 5: Replace consonants by (%) and vowels by (#).

In the number:

Odd Digit → -1

Even Digit → +1, and 0 has no change.

Step 5 is the final step of the rearrangement.

Input: DE584 AX739 PM261 WS948 CN375 UO682

Step I: CF641 BW814 LO361 RV814 BM492 PV644

Step II: FC11 WB13 OL10 VR13 MB15 VP14

Step III: fc22 wb26 ol20 vr26 mb30 vp28

Step IV: eb66 va78 nk60 uq78 la90 uo84

Step V: #%77 %#69 %%70 #%69 %#80 ##95

This is the final arrangement.

Hence, the correct answer is Two.

Logic:

Step 1: Find the highest digit in the number and write its square. Rearrange the letters in alphabetical order.

For each vowel → next alphabet

For each consonant → previous alphabet

Write the positive difference of the remaining two digits.

Step 2: Reverse the letters obtained in Step 1. Replace the number with the sum of the number obtained in Step 1.

Step 3: Write the 1st and 2nd elements in small letters. Replace the number with multiply by 2.

Step 4: Each letter is replaced by its previous alphabet. The number is multiplied by 3.

Step 5: Replace consonants by (%) and vowels by (#).

In the number:

Odd Digit → -1

Even Digit → +1, and 0 has no change.

Step 5 is the final step of the rearrangement.

Input: DE584 AX739 PM261 WS948 CN375 UO682

Step I: CF641 BW814 LO361 RV814 BM492 PV644

Step II: FC11 WB13 OL10 VR13 MB15 VP14

Step III: fc22 wb26 ol20 vr26 mb30 vp28

Step IV: eb66 va78 nk60 uq78 la90 uo84

Step V: #%77 %#69 %%70 #%69 %#80 ##95

This is the final arrangement.

Hence, the correct answer is Step III.

Logic:

Step 1: Find the highest digit in the number and write its square. Rearrange the letters in alphabetical order.

For each vowel → next alphabet

For each consonant → previous alphabet

Write the positive difference of the remaining two digits.

Step 2: Reverse the letters obtained in Step 1. Replace the number with the sum of the number obtained in Step 1.

Step 3: Write the 1st and 2nd elements in small letters. Replace the number with multiply by 2.

Step 4: Each letter is replaced by its previous alphabet. The number is multiplied by 3.

Step 5: Replace consonants by (%) and vowels by (#).

In the number:

Odd Digit → -1

Even Digit → +1, and 0 has no change.

Step 5 is the final step of the rearrangement.

Input: DE584 AX739 PM261 WS948 CN375 UO682

Step I: CF641 BW814 LO361 RV814 BM492 PV644

Step II: FC11 WB13 OL10 VR13 MB15 VP14

Step III: fc22 wb26 ol20 vr26 mb30 vp28

Step IV: eb66 va78 nk60 uq78 la90 uo84

Step V: #%77 %#69 %%70 #%69 %#80 ##95

This is the final arrangement.

Hence, the correct answer is WB13.

Logic:

Step 1: Find the highest digit in the number and write its square. Rearrange the letters in alphabetical order.

For each vowel → next alphabet

For each consonant → previous alphabet

Write the positive difference of the remaining two digits.

Step 2: Reverse the letters obtained in Step 1. Replace the number with the sum of the number obtained in Step 1.

Step 3: Write the 1st and 2nd elements in small letters. Replace the number with multiply by 2.

Step 4: Each letter is replaced by its previous alphabet. The number is multiplied by 3.

Step 5: Replace consonants by (%) and vowels by (#).

In the number:

Odd Digit → -1

Even Digit → +1, and 0 has no change.

Step 5 is the final step of the rearrangement.

Input: DE584 AX739 PM261 WS948 CN375 UO682

Step I: CF641 BW814 LO361 RV814 BM492 PV644

Step II: FC11 WB13 OL10 VR13 MB15 VP14

Step III: fc22 wb26 ol20 vr26 mb30 vp28

Step IV: eb66 va78 nk60 uq78 la90 uo84

Step V: #%77 %#69 %%70 #%69 %#80 ##95

This is the final arrangement.

Hence, the correct answer is eb.

Logic:

Step 1: Find the highest digit in the number and write its square. Rearrange the letters in alphabetical order.

For each vowel → next alphabet

For each consonant → previous alphabet

Write the positive difference of the remaining two digits.

Step 2: Reverse the letters obtained in Step 1. Replace the number with the sum of the number obtained in Step 1.

Step 3: Write the 1st and 2nd elements in small letters. Replace the number with multiply by 2.

Step 4: Each letter is replaced by its previous alphabet. The number is multiplied by 3.

Step 5: Replace consonants by (%) and vowels by (#).

In the number:

Odd Digit → -1

Even Digit → +1, and 0 has no change.

Step 5 is the final step of the rearrangement.

Input: DE584 AX739 PM261 WS948 CN375 UO682

Step I: CF641 BW814 LO361 RV814 BM492 PV644

Step II: FC11 WB13 OL10 VR13 MB15 VP14

Step III: fc22 wb26 ol20 vr26 mb30 vp28

Step IV: eb66 va78 nk60 uq78 la90 uo84

Step V: #%77 %#69 %%70 #%69 %#80 ##95

This is the final arrangement.

Hence, the correct answer is 76.

Option 1: Incorrect

The passage does not assume electricity generation to be completely pollution-free.

Option 2: Incorrect

Public transport usage is not discussed in the passage.

Option 3: Correct

The promotion of electric vehicles assumes that addressing issues like charging infrastructure and supply will lead to wider adoption, which is essential for the policy to succeed.

Option 4: Incorrect

There is no assumption regarding fossil fuel price stability.

Option 5: Incorrect

The passage does not restrict electric vehicles to urban areas only.

Hence, the correct answer is Improved infrastructure will encourage wider adoption of electric vehicles.

Option 1: Incorrect

Banning social media is an extreme and impractical step not suggested by the passage.

Option 2: Correct

The passage highlights declining reading habits and limited support for libraries, making increased funding and engaging programs the most practical and relevant action.

Option 3: Incorrect

Replacing textbooks with summaries would further weaken reading habits.

Option 4: Incorrect

Ignoring the issue contradicts the concerns raised by educators.

Option 5: Incorrect

Reducing workload does not directly address the decline in reading habits.

Hence, the correct answer is Increase funding for libraries and introduce engaging reading programs for youth.

Option 1: Incorrect

The cost of AI systems is not the central concern raised in the passage.

Option 2: Incorrect

The speed of human recruiters is not presented as a drawback or a strong argument.

Option 3: Correct

The passage clearly highlights the risk of reinforcing hidden biases and the lack of transparency in AI-based decision-making, making this the strongest argument against excessive reliance on such tools.

Option 4: Incorrect

Applicant preference is not discussed in the passage.

Option 5: Incorrect

Lack of technical expertise is not mentioned as a major concern.

Hence, the correct answer is AI tools may perpetuate hidden biases and lack transparency in decision-making.

Given: Ten different books - History, Science, Geography, Economics, Literature, Mathematics, Psychology, Philosophy, Biology, and Politics - are kept in an almirah having four vertical stacks: Stack 1, Stack 2, Stack 3, and Stack 4. Stack 2 is to the east of Stack 1, Stack 3 is to the east of Stack 2, and Stack 4 is to the east of Stack 3. Each stack contains a different number of books. The number of books decreases from west to east, i.e., Stack 1 has more books than Stack 2, Stack 2 has more books than Stack 3, and Stack 3 has more books than Stack 4. At least one book is kept in each stack, and the top of all four stacks contains a book.

 

1. Philosophy is kept at the top of Stack 3. 
2. Science is kept immediately above History in the same stack.
3. History is kept two positions above Biology in the same stack.

 

4. The psychology book is kept to the east of the Politics book, which is not kept in Stack 3.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

Case 1:

 

Case 2: 

 

5. The geography book is kept two books above the Mathematics book in the same stack.

Now, we can eliminate Case 2.

Case 1:

 

6. The literature book is not kept in Stack 4.

This is our final arrangement,

 

Hence, the correct answer is Science.

Given: Ten different books - History, Science, Geography, Economics, Literature, Mathematics, Psychology, Philosophy, Biology, and Politics - are kept in an almirah having four vertical stacks: Stack 1, Stack 2, Stack 3, and Stack 4. Stack 2 is to the east of Stack 1, Stack 3 is to the east of Stack 2, and Stack 4 is to the east of Stack 3. Each stack contains a different number of books. The number of books decreases from west to east, i.e., Stack 1 has more books than Stack 2, Stack 2 has more books than Stack 3, and Stack 3 has more books than Stack 4. At least one book is kept in each stack, and the top of all four stacks contains a book.

 

1. Philosophy is kept at the top of Stack 3.  2. Science is kept immediately above History in the same stack. 3. History is kept two positions above Biology in the same stack.

 

4. The psychology book is kept to the east of the Politics book, which is not kept in Stack 3.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

Case 1:

 

Case 2: 

 

5. The geography book is kept two books above the Mathematics book in the same stack.

Now, we can eliminate Case 2.

Case 1:

 

6. The literature book is not kept in Stack 4.

This is our final arrangement,

 

Hence, the correct answer is 4.

Given: Ten different books - History, Science, Geography, Economics, Literature, Mathematics, Psychology, Philosophy, Biology, and Politics - are kept in an almirah having four vertical stacks: Stack 1, Stack 2, Stack 3, and Stack 4. Stack 2 is to the east of Stack 1, Stack 3 is to the east of Stack 2, and Stack 4 is to the east of Stack 3. Each stack contains a different number of books. The number of books decreases from west to east, i.e., Stack 1 has more books than Stack 2, Stack 2 has more books than Stack 3, and Stack 3 has more books than Stack 4. At least one book is kept in each stack, and the top of all four stacks contains a book.

 

1. Philosophy is kept at the top of Stack 3.  2. Science is kept immediately above History in the same stack. 3. History is kept two positions above Biology in the same stack.

 

4. The psychology book is kept to the east of the Politics book, which is not kept in Stack 3.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

Case 1:

 

Case 2: 

 

5. The geography book is kept two books above the Mathematics book in the same stack.

Now, we can eliminate Case 2.

Case 1:

 

6. The literature book is not kept in Stack 4.

This is our final arrangement,

 

Hence, the correct answer is Geography - Psychology.

Given: Ten different books - History, Science, Geography, Economics, Literature, Mathematics, Psychology, Philosophy, Biology, and Politics - are kept in an almirah having four vertical stacks: Stack 1, Stack 2, Stack 3, and Stack 4. Stack 2 is to the east of Stack 1, Stack 3 is to the east of Stack 2, and Stack 4 is to the east of Stack 3. Each stack contains a different number of books. The number of books decreases from west to east, i.e., Stack 1 has more books than Stack 2, Stack 2 has more books than Stack 3, and Stack 3 has more books than Stack 4. At least one book is kept in each stack, and the top of all four stacks contains a book.

 

1. Philosophy is kept at the top of Stack 3.  2. Science is kept immediately above History in the same stack. 3. History is kept two positions above Biology in the same stack.

 

4. The psychology book is kept to the east of the Politics book, which is not kept in Stack 3.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

Case 1:

 

Case 2: 

 

5. The geography book is kept two books above the Mathematics book in the same stack.

Now, we can eliminate Case 2.

Case 1:

 

6. The literature book is not kept in Stack 4.

This is our final arrangement,

 

Hence, the correct answer is History.

Option 1: Incorrect

A complete and permanent ban on private vehicles is extreme and impractical. The passage does not suggest such a drastic measure.

Option 2: Correct

The passage highlights inadequate public transport and weak enforcement of emission norms. Strengthening public transport and enforcing regulations directly addresses the core problems identified.

Option 3: Incorrect

Ignoring short-term pollution spikes contradicts the health concerns emphasized in the passage.

Option 4: Incorrect

Shifting administrative offices to rural areas is neither discussed nor logically connected to controlling urban air pollution.

Option 5: Incorrect

The passage notes that awareness campaigns alone have had a limited impact. Hence, relying only on them would be insufficient.

Hence, the correct answer is Strengthen public transport infrastructure and ensure strict enforcement of emission regulations.

Option 1: Incorrect

The passage identifies multiple causes of air pollution, including vehicular traffic and construction, not just industrial activity.

Option 2: Correct

Despite emission norms and restrictions, pollution levels remain high, indicating that existing measures have not been fully effective.

Option 3: Incorrect

The passage mentions awareness campaigns, implying that people are not completely unaware of pollution-related risks.

Option 4: Incorrect

Construction activity is explicitly mentioned as a contributing factor, making this option factually incorrect.

Option 5: Incorrect

Health experts warn about the consequences of pollution, but do not oppose urban development itself.

Hence, the correct answer is Existing pollution control measures have been partially ineffective.

A * B (18m) → A is 8m north of B

A % B (24m) → A is 14m south of B

A & B (28m) → A is 18m east of B

A @ B (35m) → A is 25m west of B

then,

M * N (16m) → M is 6m north of N

O @ P (32m) → O is 22m west of P

N & O (36m) → N is 26m east of O

Q & P (42m) → Q is 32m east of P

S @ R (58m) → S is 48m west of R

S % T (27m) → S is 17m south of T

U % Q (20m) → U is 10m south of Q

R * U (22m) → R is 12m north of U

X * R (30m) → X is 20m north of R

The possible diagram is as follows,

Hence, the correct answer is North-east.

A * B (18m) → A is 8m north of B

A % B (24m) → A is 14m south of B

A & B (28m) → A is 18m east of B

A @ B (35m) → A is 25m west of B

then,

M * N (16m) → M is 6m north of N

O @ P (32m) → O is 22m west of P

N & O (36m) → N is 26m east of O

Q & P (42m) → Q is 32m east of P

S @ R (58m) → S is 48m west of R

S % T (27m) → S is 17m south of T

U % Q (20m) → U is 10m south of Q

R * U (22m) → R is 12m north of U

X * R (30m) → X is 20m north of R

The possible diagram is as follows,

The distance between U and X = 20 + 2 + 10 = 32 m.

Hence, the correct answer is 32 m.

A * B (18m) → A is 8m north of B

A % B (24m) → A is 14m south of B

A & B (28m) → A is 18m east of B

A @ B (35m) → A is 25m west of B

then,

M * N (16m) → M is 6m north of N

O @ P (32m) → O is 22m west of P

N & O (36m) → N is 26m east of O

Q & P (42m) → Q is 32m east of P

S @ R (58m) → S is 48m west of R

S % T (27m) → S is 17m south of T

U % Q (20m) → U is 10m south of Q

R * U (22m) → R is 12m north of U

X * R (30m) → X is 20m north of R

The possible diagram is as follows,

Hence, the correct answer is X is northeast of P.

General Solution:

First, we solve for 'x' and 'y' using the percentage distributions and given conditions. Let 'B' be the total number of bats available and 'F' be the total number of footballs available.

Solve for x and y:

⇒ Total % of bats sold = 100%

⇒ 10 + (3x - 2y) + 45 + (2x + y) = 100

⇒ 5x - y = 45 ... (Equation 1)

⇒ Total % of footballs sold = 100%

⇒ (2x + 4y) + 15 + 2y + 35 = 100

⇒ 2x + 6y = 50 ⇒ x + 3y = 25 ... (Equation 2)

⇒ Multiply Eq. 1 by 3: 15x - 3y = 135. Add to Eq. 2: 16x = 160 ⇒ x = 10.

⇒ Substitute x=10 in Eq. 2: 10 + 3y = 25 ⇒ 3y = 15 ⇒ y = 5.

Thus,

Find Total Bats (B) and Total Footballs (F):

⇒ Condition for B: Bats sold by B = Footballs sold by B

⇒ 20% of B = 15% of F ⇒ 0.2B = 0.15F ⇒ B = 0.75F.

⇒ Condition for A: Difference between bats and footballs sold is 780.

⇒ |10% of B - 40% of F| = 780

⇒ |0.1(0.75F) - 0.4F| = 780 ⇒ |0.075F - 0.4F| = 780

⇒ 0.325F = 780 ⇒ F = 780 / 0.325 = 2400.

⇒ B = 0.75 × 2400 = 1800.

Calculations:

⇒ Items sold by A: Bats (180) + Footballs (0.4 × 2400 = 960) = 1140.

⇒ Items sold by C: Bats (0.45 × 1800 = 810) + Footballs (240) = 1050.

⇒ Average (A, C) = (1140 + 1050) / 2 = 1095.

⇒ Items sold by B: Bats (0.2 × 1800 = 360) + Footballs (360) = 720.

⇒ Items sold by D: Bats (0.25 × 1800 = 450) + Footballs (840) = 1290.

⇒ Average (B, D) = (720 + 1290) / 2 = 1005.

⇒ Difference = 1095 - 1005 = 90.

⇒ Percentage = (90 / 1005) × 100 ≈ 8.955%.

∴ The percentage is 8.96%.

General Solution:

First, we solve for 'x' and 'y' using the percentage distributions and given conditions. Let 'B' be the total number of bats available and 'F' be the total number of footballs available.

Solve for x and y:

⇒ Total % of bats sold = 100%

⇒ 10 + (3x - 2y) + 45 + (2x + y) = 100

⇒ 5x - y = 45 ... (Equation 1)

⇒ Total % of footballs sold = 100%

⇒ (2x + 4y) + 15 + 2y + 35 = 100

⇒ 2x + 6y = 50 ⇒ x + 3y = 25 ... (Equation 2)

⇒ Multiply Eq. 1 by 3: 15x - 3y = 135. Add to Eq. 2: 16x = 160 ⇒ x = 10.

⇒ Substitute x=10 in Eq. 2: 10 + 3y = 25 ⇒ 3y = 15 ⇒ y = 5.

Thus,

Find Total Bats (B) and Total Footballs (F):

⇒ Condition for B: Bats sold by B = Footballs sold by B

⇒ 20% of B = 15% of F ⇒ 0.2B = 0.15F ⇒ B = 0.75F.

⇒ Condition for A: Difference between bats and footballs sold is 780.

⇒ |10% of B - 40% of F| = 780

⇒ |0.1(0.75F) - 0.4F| = 780 ⇒ |0.075F - 0.4F| = 780

⇒ 0.325F = 780 ⇒ F = 780 / 0.325 = 2400.

⇒ B = 0.75 × 2400 = 1800.

Calculations:

⇒ Bats sold by A = 10% of 1800 = 180. Returns = 25% of 180 = 45.

⇒ Footballs sold by B = 15% of 2400 = 360. Returns = 30.

⇒ Initial Stock: Bats = 1800, Footballs = 2400. Total = 4200.

⇒ Total bats sold (all shopkeepers) = 1800. Total footballs sold (all shopkeepers) = 2400.

⇒ Net sales after returns: Bats = 1800 - 45 = 1755. Footballs = 2400 - 30 = 2370.

⇒ Remaining Stock: Bats = 1800 - 1755 = 45. Footballs = 2400 - 2370 = 30.

⇒ Total remaining stock = 45 + 30 = 75.

⇒ Stock for next cycle = 75 × 1.2 = 90.

∴ The total number of items available is 90.

General Solution:

First, we solve for 'x' and 'y' using the percentage distributions and given conditions. Let 'B' be the total number of bats available and 'F' be the total number of footballs available.

Solve for x and y:

⇒ Total % of bats sold = 100%

⇒ 10 + (3x - 2y) + 45 + (2x + y) = 100

⇒ 5x - y = 45 ... (Equation 1)

⇒ Total % of footballs sold = 100%

⇒ (2x + 4y) + 15 + 2y + 35 = 100

⇒ 2x + 6y = 50 ⇒ x + 3y = 25 ... (Equation 2)

⇒ Multiply Eq. 1 by 3: 15x - 3y = 135. Add to Eq. 2: 16x = 160 ⇒ x = 10.

⇒ Substitute x=10 in Eq. 2: 10 + 3y = 25 ⇒ 3y = 15 ⇒ y = 5.

Thus,

Find Total Bats (B) and Total Footballs (F):

⇒ Condition for B: Bats sold by B = Footballs sold by B

⇒ 20% of B = 15% of F ⇒ 0.2B = 0.15F ⇒ B = 0.75F.

⇒ Condition for A: Difference between bats and footballs sold is 780.

⇒ |10% of B - 40% of F| = 780

⇒ |0.1(0.75F) - 0.4F| = 780 ⇒ |0.075F - 0.4F| = 780

⇒ 0.325F = 780 ⇒ F = 780 / 0.325 = 2400.

⇒ B = 0.75 × 2400 = 1800.

Calculations:

⇒ Footballs sold by D = 35% of 2400 = 840.

⇒ Cost Price (CP) of one football = ₹300.

⇒ Profit of D = 20%. Selling Price (SP) of D = 300 × 1.2 = ₹360 per football.

⇒ Footballs sold by C = 10% of 2400 = 240.

⇒ Shopkeeper C sells at the same SP (₹360).

⇒ Profit per football for C = SP - CP = 360 - 300 = ₹60.

⇒ Total Profit for C = 240 × 60 = ₹14,400.

∴ The total profit earned by C is ₹14,400.

General Solution:

First, we need to determine the actual number of boys and girls in each school by combining the percentage of boys with the given ratio of girls.

Express total students and boys in terms of the number of girls:

⇒ Let the number of girls in schools P, Q, R, and S be 6k, 3k, 5k, and 11k respectively.

⇒ School P: Boys = 60%, so Girls = 40%. ⇒ 40% of Total = 6k ⇒ Total = 15k, Boys = 9k.

⇒ School Q: Boys = 70%, so Girls = 30%. ⇒ 30% of Total = 3k ⇒ Total = 10k, Boys = 7k.

⇒ School R: Boys = 20%, so Girls = 80%. ⇒ 80% of Total = 5k ⇒ Total = 6.25k, Boys = 1.25k.

⇒ School S: Boys = 45%, so Girls = 55%. ⇒ 55% of Total = 11k ⇒ Total = 20k, Boys = 9k.

Solve for 'k' using the additional information:

⇒ Difference between boys in S and the combined boys in Q and R is 15.

⇒ |BoysS - (BoysQ + BoysR)| = 15

⇒ |9k - (7k + 1.25k)| = 15

⇒ |9k - 8.25k| = 15

⇒ 0.75k = 15 ⇒ k = 20.

Final Counts for each school:

⇒ School P: Girls = 120, Boys = 180, Total = 300.

⇒ School Q: Girls = 60, Boys = 140, Total = 200.

⇒ School R: Girls = 100, Boys = 25, Total = 125.

⇒ School S: Girls = 220, Boys = 180, Total = 400.

Calculations:

⇒ Total boys in P and S together = 180 + 180 = 360.

⇒ Girls in school T = 125% of Girls in Q = 1.25 × 60 = 75.

⇒ Ratio in school T is Boys : Girls = 5 : 3.

⇒ 3 parts = 75 

⇒ Boys in T = (5/3) × 75 = 125. Total students in T = 75 + 125 = 200.

Total boys (360) / Total students in T (200) = 180%.

∴ The correct percentage is 180%.

General Solution:

First, we need to determine the actual number of boys and girls in each school by combining the percentage of boys with the given ratio of girls.

Express total students and boys in terms of the number of girls:

⇒ Let the number of girls in schools P, Q, R, and S be 6k, 3k, 5k, and 11k respectively.

⇒ School P: Boys = 60%, so Girls = 40%. ⇒ 40% of Total = 6k ⇒ Total = 15k, Boys = 9k.

⇒ School Q: Boys = 70%, so Girls = 30%. ⇒ 30% of Total = 3k ⇒ Total = 10k, Boys = 7k.

⇒ School R: Boys = 20%, so Girls = 80%. ⇒ 80% of Total = 5k ⇒ Total = 6.25k, Boys = 1.25k.

⇒ School S: Boys = 45%, so Girls = 55%. ⇒ 55% of Total = 11k ⇒ Total = 20k, Boys = 9k.

Solve for 'k' using the additional information:

⇒ Difference between boys in S and the combined boys in Q and R is 15.

⇒ |BoysS - (BoysQ + BoysR)| = 15

⇒ |9k - (7k + 1.25k)| = 15

⇒ |9k - 8.25k| = 15

⇒ 0.75k = 15 ⇒ k = 20.

Final Counts for each school:

⇒ School P: Girls = 120, Boys = 180, Total = 300.

⇒ School Q: Girls = 60, Boys = 140, Total = 200.

⇒ School R: Girls = 100, Boys = 25, Total = 125.

⇒ School S: Girls = 220, Boys = 180, Total = 400.

Calculations:

⇒ Boys in P who fail = (100% - 75%) of 180 = 25% of 180 = 45.

⇒ Girls in R who fail = (100% - 80%) of 100 = 20% of 100 = 20.

⇒ Total failed students = 45 + 20 = 65.

∴ 65 students failed to pass.

General Solution:

First, we need to determine the actual number of boys and girls in each school by combining the percentage of boys with the given ratio of girls.

Express total students and boys in terms of the number of girls:

⇒ Let the number of girls in schools P, Q, R, and S be 6k, 3k, 5k, and 11k respectively.

⇒ School P: Boys = 60%, so Girls = 40%. ⇒ 40% of Total = 6k ⇒ Total = 15k, Boys = 9k.

⇒ School Q: Boys = 70%, so Girls = 30%. ⇒ 30% of Total = 3k ⇒ Total = 10k, Boys = 7k.

⇒ School R: Boys = 20%, so Girls = 80%. ⇒ 80% of Total = 5k ⇒ Total = 6.25k, Boys = 1.25k.

⇒ School S: Boys = 45%, so Girls = 55%. ⇒ 55% of Total = 11k ⇒ Total = 20k, Boys = 9k.

Solve for 'k' using the additional information:

⇒ Difference between boys in S and the combined boys in Q and R is 15.

⇒ |BoysS - (BoysQ + BoysR)| = 15

⇒ |9k - (7k + 1.25k)| = 15

⇒ |9k - 8.25k| = 15

⇒ 0.75k = 15 ⇒ k = 20.

Final Counts for each school:

⇒ School P: Girls = 120, Boys = 180, Total = 300.

⇒ School Q: Girls = 60, Boys = 140, Total = 200.

⇒ School R: Girls = 100, Boys = 25, Total = 125.

⇒ School S: Girls = 220, Boys = 180, Total = 400.

Calculations:

⇒ School Q in 2026: Boys increase by 20% = 140 × 1.2 = 168. Girls remain 60. Total = 228.

⇒ School S in 2026: Girls decrease by 15% = 220 × 0.85 = 187. Boys remain 180. Total = 367.

⇒ Difference = 367 - 228 = 139.

∴ The difference is 139.

General Solution:

First, we need to determine the actual number of boys and girls in each school by combining the percentage of boys with the given ratio of girls.

Express total students and boys in terms of the number of girls:

⇒ Let the number of girls in schools P, Q, R, and S be 6k, 3k, 5k, and 11k respectively.

⇒ School P: Boys = 60%, so Girls = 40%. ⇒ 40% of Total = 6k ⇒ Total = 15k, Boys = 9k.

⇒ School Q: Boys = 70%, so Girls = 30%. ⇒ 30% of Total = 3k ⇒ Total = 10k, Boys = 7k.

⇒ School R: Boys = 20%, so Girls = 80%. ⇒ 80% of Total = 5k ⇒ Total = 6.25k, Boys = 1.25k.

⇒ School S: Boys = 45%, so Girls = 55%. ⇒ 55% of Total = 11k ⇒ Total = 20k, Boys = 9k.

Solve for 'k' using the additional information:

⇒ Difference between boys in S and the combined boys in Q and R is 15.

⇒ |Boys_S - (Boys_Q + Boys_R)| = 15

⇒ |9k - (7k + 1.25k)| = 15

⇒ |9k - 8.25k| = 15

⇒ 0.75k = 15 ⇒ k = 20.

Final Counts for each school:

⇒ School P: Girls = 120, Boys = 180, Total = 300.

⇒ School Q: Girls = 60, Boys = 140, Total = 200.

⇒ School R: Girls = 100, Boys = 25, Total = 125.

⇒ School S: Girls = 220, Boys = 180, Total = 400.

Calculations:

⇒ Total students in P and R together = 300 + 125 = 425.

⇒ Average number of girls in all schools = (120 + 60 + 100 + 220) / 4 = 500 / 4 = 125.

⇒ Required Ratio = 425 : 125.

⇒ Dividing both sides by 25 = 17 : 5.

∴ The correct option is 17:5.

Shortcut Trick

Let Volume of Cylinder = x and Volume of Sphere = y.

Initial liquid volume = 0.5x. Total capacity filled by (liquid + 5 spheres) = x.

Volume of 1 sphere (y) corresponds to a height of 2 cm. Volume of 5 spheres = 5y.

Let total height of cylinder be H. Initial height = 0.5H. Rise for 5 spheres = 5 × 2 = 10 cm.

Since 0.5H + 10 cm = H, then 0.5H = 10 cm, so H = 20 cm. 1 cm height = x/20 volume.

Rise of 2 cm = (x/20) × 2 = x/10. Therefore, y = x/10 ⇒ x = 10y.

Target: 0.5x = 5y. From options, 3/10(10y) + 2y = 3y + 2y = 5y.

∴ The correct answer is 3/10 × (Volume of Cylinder) + 2 × (Volume of Sphere).

Alternate Method

Given: Volume of cylinder = x, Volume of sphere = y. Initial liquid = 0.5x.

Formula: Volume = Base Area × Height. Volume displaced = Volume of submerged object.

⇒ Let A be the base area of the cylinder. 1 sphere causes 2 cm rise ⇒ y = A × 2.

⇒ Total spheres dropped = 1 + 4 = 5 spheres. Total volume of spheres = 5y.

⇒ Total volume (x) = Initial Liquid (0.5x) + Volume of 5 spheres (5y).

⇒ x − 0.5x = 5y ⇒ 0.5x = 5y.

⇒ Since y = 2A and x = H × A, then 5(2A) = 0.5(H × A) ⇒ 10A = 0.5HA ⇒ H = 20 cm.

⇒ Relationship: x = 10y (since 10 cm height is 0.5x and corresponds to 5y).

⇒ 50% of Volume = 0.5x = 5y.

⇒ Check Option 4: (3/10)x + 2y = (3/10)(10y) + 2y = 3y + 2y = 5y.

∴ The correct answer is 3/10 × (Volume of Cylinder) + 2 × (Volume of Sphere).

Additional Information

Archimedes Principle in Cylinders

When a solid is immersed in a liquid-filled cylinder, the volume of the displaced liquid equals the volume of the submerged part of the solid: V_solid = πr²h_rise.

Cylinder and Sphere Volumes

Volume of Cylinder = πr²h and Volume of Sphere = (4/3)πR³. In level-rise problems, the base area πr² remains constant.

Shortcut Trick

Identify N: The smallest even prime number is 2.

Substitute N = 2 into the given expression: √[(2 + 576) ÷ 2].

⇒ √[578 ÷ 2] = √289.

⇒ 17.

∴ The correct answer is 17.

Alternate Method

Given:

The value of N is defined as the smallest even prime number.

Expression to evaluate: √[(N + 576) / 2]

Formula Used:

Definition of prime numbers: N = 2.

Calculations:

⇒ Smallest even prime number (N) = 2

⇒ Expression = √[(2 + 576) ÷ 2]

⇒ Expression = √[578 ÷ 2]

⇒ Expression = √289

⇒ Expression = 17

∴ The correct answer is 17.

Additional Information

Prime Number Properties

A prime number has exactly two factors: 1 and itself. 2 is the only even prime number and the smallest prime number.

Perfect Squares

Understanding squares of numbers up to 30 is crucial: 17² = 289, 18² = 324, 19² = 361.

Arithmetic Progression of Squares

The series follows the pattern (N + k)². If N = 2, the series is 4, 9, 16, 25, 36, ... representing consecutive squares starting from 2².

Shortcut Trick

Identify the smallest even prime number: N = 2.

The series pattern is given as (N + k − 1)² for the k^th term.

For the 7^th term (k = 7): (2 + 7 − 1)² = 8².

8² = 64.

∴ The correct answer is 64.

Alternate Method

Given:

Pattern: N², (N + 1)², (N + 2)², (N + 3)², (N + 4)², ...

N = Smallest even prime number ⇒ 2

Formula Used:

The k^th term of the series T_k = (N + k − 1)²

2² T1 → 3² T2 → 4² T3 → 5² T4 → 6² T5 → 7² T6 → 8²

Calculations:

⇒ According to the passage, N = 2 (the only even prime number).

⇒ The 7^th term corresponds to k = 7.

⇒ T₇ = (2 + 7 − 1)²

⇒ T₇ = (2 + 6)²

⇒ T₇ = 8²

⇒ T₇ = 64

∴ The correct answer is 64.

Additional Information

Prime Number Properties

A prime number has exactly two factors. 2 is unique as it is the only even prime number and the smallest prime number.

Square Series Pattern

A series following (N+x)² logic is a Perfect Square Series where the base values follow an Arithmetic Progression.

General Solution:

First, we need to find the value of x and the specific number of trucks and buses manufactured and sold for each month.

Relation between Trucks and Buses:

⇒ Let Buses manufactured in a month be B. Then Trucks manufactured (T) = (2/3)B.

⇒ Total vehicles (V) = T + B = (2/3)B + B = (5/3)B.

⇒ From this, B = 3/5 of Total Vehicles and T = 2/5 of Total Vehicles.

Solving for x using May data:

⇒ For May: Total vehicles = x. Buses manufactured = (3/5)x.

⇒ Buses sold in May = 84% of (3/5)x = 63 (given in table).

⇒ 0.84 × 0.6x = 63 ⇒ 0.504x = 63 ⇒ x = 63 / 0.504 = 125.

Calculating inventories for all months (x = 125):

April (2x = 250):

⇒ Buses Manufactured = (3/5) × 250 = 150; Trucks Manufactured = (2/5) × 250 = 100.

⇒ Buses Sold = 20% of 150 = 30; Trucks Sold = 75% of 100 = 75.

May (x = 125):

⇒ Buses Manufactured = (3/5) × 125 = 75; Trucks Manufactured = (2/5) × 125 = 50.

⇒ Buses Sold = 84% of 75 = 63; Trucks Sold = 80% of 50 = 40.

June (2.4x = 300):

⇒ Buses Manufactured = (3/5) × 300 = 180; Trucks Manufactured = (2/5) × 300 = 120.

⇒ Buses Sold = 50% of 180 = 90; Trucks Sold = 45% of 120 = 54.

Calculations:

⇒ Trucks manufactured (July) = 200% of Trucks sold (April) = 2 × 75 = 150.

⇒ Buses manufactured (July) = 150% of Buses sold (June) = 1.5 × 90 = 135.

⇒ Unsold Trucks (July) = 40% of 150 = 60.

⇒ Unsold Buses (July) = 20% of 135 = 27.

⇒ Total unsold = 60 + 27 = 87.

∴ The correct answer is 87.

General Solution:

First, we need to find the value of x and the specific number of trucks and buses manufactured and sold for each month.

Relation between Trucks and Buses:

⇒ Let Buses manufactured in a month be B. Then Trucks manufactured (T) = (2/3)B.

⇒ Total vehicles (V) = T + B = (2/3)B + B = (5/3)B.

⇒ From this, B = 3/5 of Total Vehicles and T = 2/5 of Total Vehicles.

Solving for x using May data:

⇒ For May: Total vehicles = x. Buses manufactured = (3/5)x.

⇒ Buses sold in May = 84% of (3/5)x = 63 (given in table).

⇒ 0.84 × 0.6x = 63 ⇒ 0.504x = 63 ⇒ x = 63 / 0.504 = 125.

Calculating inventories for all months (x = 125):

April (2x = 250):

⇒ Buses Manufactured = (3/5) × 250 = 150; Trucks Manufactured = (2/5) × 250 = 100.

⇒ Buses Sold = 20% of 150 = 30; Trucks Sold = 75% of 100 = 75.

May (x = 125):

⇒ Buses Manufactured = (3/5) × 125 = 75; Trucks Manufactured = (2/5) × 125 = 50.

⇒ Buses Sold = 84% of 75 = 63; Trucks Sold = 80% of 50 = 40.

June (2.4x = 300):

⇒ Buses Manufactured = (3/5) × 300 = 180; Trucks Manufactured = (2/5) × 300 = 120.

⇒ Buses Sold = 50% of 180 = 90; Trucks Sold = 45% of 120 = 54.

Calculations:

⇒ Unsold trucks in May = 10. Defective = 20% of 10 = 2. Non-defective = 10 - 2 = 8.

⇒ Unsold buses in June = 90. Defective = 30% of 90 = 27. Non-defective = 90 - 27 = 63.

⇒ Total non-defective = 8 + 63 = 71.

∴ The correct answer is 71.

General Solution:

First, we need to find the value of x and the specific number of trucks and buses manufactured and sold for each month.

Relation between Trucks and Buses:

⇒ Let Buses manufactured in a month be B. Then Trucks manufactured (T) = (2/3)B.

⇒ Total vehicles (V) = T + B = (2/3)B + B = (5/3)B.

⇒ From this, B = 3/5 of Total Vehicles and T = 2/5 of Total Vehicles.

Solving for x using May data:

⇒ For May: Total vehicles = x. Buses manufactured = (3/5)x.

⇒ Buses sold in May = 84% of (3/5)x = 63 (given in table).

⇒ 0.84 × 0.6x = 63 ⇒ 0.504x = 63 ⇒ x = 63 / 0.504 = 125.

Calculating inventories for all months (x = 125):

April (2x = 250):

⇒ Buses Manufactured = (3/5) × 250 = 150; Trucks Manufactured = (2/5) × 250 = 100.

⇒ Buses Sold = 20% of 150 = 30; Trucks Sold = 75% of 100 = 75.

May (x = 125):

⇒ Buses Manufactured = (3/5) × 125 = 75; Trucks Manufactured = (2/5) × 125 = 50.

⇒ Buses Sold = 84% of 75 = 63; Trucks Sold = 80% of 50 = 40.

June (2.4x = 300):

⇒ Buses Manufactured = (3/5) × 300 = 180; Trucks Manufactured = (2/5) × 300 = 120.

⇒ Buses Sold = 50% of 180 = 90; Trucks Sold = 45% of 120 = 54.

Calculations:

⇒ Unsold in May = (75 - 63) + (50 - 40) = 12 + 10 = 22.

⇒ Unsold in June = (180 - 90) + (120 - 54) = 90 + 66 = 156.

⇒ Difference = 156 - 22 = 134.

∴ The correct answer is 134.

General Solution:

First, we need to find the value of x and the specific number of trucks and buses manufactured and sold for each month.

Relation between Trucks and Buses:

⇒ Let Buses manufactured in a month be B. Then Trucks manufactured (T) = (2/3)B.

⇒ Total vehicles (V) = T + B = (2/3)B + B = (5/3)B.

⇒ From this, B = 3/5 of Total Vehicles and T = 2/5 of Total Vehicles.

Solving for x using May data:

⇒ For May: Total vehicles = x. Buses manufactured = (3/5)x.

⇒ Buses sold in May = 84% of (3/5)x = 63 (given in table).

⇒ 0.84 × 0.6x = 63 ⇒ 0.504x = 63 ⇒ x = 63 / 0.504 = 125.

Calculating inventories for all months (x = 125):

April (2x = 250):

⇒ Buses Manufactured = (3/5) × 250 = 150; Trucks Manufactured = (2/5) × 250 = 100.

⇒ Buses Sold = 20% of 150 = 30; Trucks Sold = 75% of 100 = 75.

May (x = 125):

⇒ Buses Manufactured = (3/5) × 125 = 75; Trucks Manufactured = (2/5) × 125 = 50.

⇒ Buses Sold = 84% of 75 = 63; Trucks Sold = 80% of 50 = 40.

June (2.4x = 300):

⇒ Buses Manufactured = (3/5) × 300 = 180; Trucks Manufactured = (2/5) × 300 = 120.

⇒ Buses Sold = 50% of 180 = 90; Trucks Sold = 45% of 120 = 54.

Calculations:

⇒ k = (Buses sold in May) + (Trucks sold in May) = 63 + 40 = 103.

⇒ Percentage = (Buses sold / k) × 100 = (63 / 103) × 100 = 61.165...% ⇒ 61.17%.

∴ The correct answer is 61.17%.

General Solution:

First, we need to find the value of x and the specific number of trucks and buses manufactured and sold for each month.

Relation between Trucks and Buses:

⇒ Let Buses manufactured in a month be B. Then Trucks manufactured (T) = (2/3)B.

⇒ Total vehicles (V) = T + B = (2/3)B + B = (5/3)B.

⇒ From this, B = 3/5 of Total Vehicles and T = 2/5 of Total Vehicles.

Solving for x using May data:

⇒ For May: Total vehicles = x. Buses manufactured = (3/5)x.

⇒ Buses sold in May = 84% of (3/5)x = 63 (given in table).

⇒ 0.84 × 0.6x = 63 ⇒ 0.504x = 63 ⇒ x = 63 / 0.504 = 125.

Calculating inventories for all months (x = 125):

April (2x = 250):

⇒ Buses Manufactured = (3/5) × 250 = 150; Trucks Manufactured = (2/5) × 250 = 100.

⇒ Buses Sold = 20% of 150 = 30; Trucks Sold = 75% of 100 = 75.

May (x = 125):

⇒ Buses Manufactured = (3/5) × 125 = 75; Trucks Manufactured = (2/5) × 125 = 50.

⇒ Buses Sold = 84% of 75 = 63; Trucks Sold = 80% of 50 = 40.

June (2.4x = 300):

⇒ Buses Manufactured = (3/5) × 300 = 180; Trucks Manufactured = (2/5) × 300 = 120.

⇒ Buses Sold = 50% of 180 = 90; Trucks Sold = 45% of 120 = 54.

Calculations:

⇒ Trucks manufactured in June = 120.

⇒ Total buses manufactured = 150 (April) + 75 (May) + 180 (June) = 405.

⇒ Percentage = (120 / 405) × 100 = 29.629...% ⇒ 29.63%.

∴ The correct answer is 29.63%.

General Solution:

We need to determine the count of students who registered, appeared, qualified, and did not qualify for each year based on the provided conditions.

Year 2020:

⇒ Appeared = Qualified + Not Qualified = 162 + 288 = 450.

⇒ Total Registered = Appeared / 0.90 = 450 / 0.90 = 500.

⇒ Not Appeared = 500 - 450 = 50.

Year 2021:

⇒ Ratio of Appeared (2020 : 2021) = 5 : 4.

⇒ 5 units = 450 ⇒ 1 unit = 90 ⇒ Appeared (2021) = 4 × 90 = 360.

⇒ Total Registered = Appeared / 0.80 = 360 / 0.80 = 450.

⇒ Not Appeared = 450 - 360 = 90.

Year 2022:

⇒ Appeared (2022) = Appeared (2020) - 100 = 450 - 100 = 350.

⇒ Total Registered = Appeared / 0.70 = 350 / 0.70 = 500.

⇒ Not Appeared = 500 - 350 = 150.

Totals:

⇒ Total Appeared = 450 + 360 + 350 = 1160.

⇒ Total Qualified = 686 (Given).

⇒ Total Registered = 500 + 450 + 500 = 1450.

⇒ p% = (Total Qualified / Total Appeared) × 100 = (686 / 1160) × 100 ≈ 59.13%.

Calculations:

⇒ I. Not appeared 2021 = 90. (True).

⇒ II. Registered 2021 (450) + Appeared 2022 (350) = 800. (True).

⇒ III. Qualified 2020 (162) is 30% of Registered 2020 (500). 500 × 0.30 = 150. (162 ≠ 150). (NOT TRUE).

∴ The correct option is Only III.

General Solution:

We need to determine the count of students who registered, appeared, qualified, and did not qualify for each year based on the provided conditions.

Year 2020:

⇒ Appeared = Qualified + Not Qualified = 162 + 288 = 450.

⇒ Total Registered = Appeared / 0.90 = 450 / 0.90 = 500.

⇒ Not Appeared = 500 - 450 = 50.

Year 2021:

⇒ Ratio of Appeared (2020 : 2021) = 5 : 4.

⇒ 5 units = 450 ⇒ 1 unit = 90 ⇒ Appeared (2021) = 4 × 90 = 360.

⇒ Total Registered = Appeared / 0.80 = 360 / 0.80 = 450.

⇒ Not Appeared = 450 - 360 = 90.

Year 2022:

⇒ Appeared (2022) = Appeared (2020) - 100 = 450 - 100 = 350.

⇒ Total Registered = Appeared / 0.70 = 350 / 0.70 = 500.

⇒ Not Appeared = 500 - 350 = 150.

Totals:

⇒ Total Appeared = 450 + 360 + 350 = 1160.

⇒ Total Qualified = 686 (Given).

⇒ Total Registered = 500 + 450 + 500 = 1450.

⇒ p% = (Total Qualified / Total Appeared) × 100 = (686 / 1160) × 100 ≈ 59.13%.

Calculations:

⇒ Qualified (2021) = 0.75 × 360 = 270.

⇒ Total Qualified = Qualified(2020) + Qualified(2021) + Qualified(2022).

⇒ 686 = 162 + 270 + Qualified(2022).

⇒ 686 = 432 + Qualified(2022) ⇒ Qualified(2022) = 254.

⇒ Not Qualified (2022) = Appeared(2022) - Qualified(2022).

⇒ Not Qualified (2022) = 350 - 254 = 96.

∴ The correct option is 96.