SBI PO Mains Mock Test - 7 - Bank Exams MCQ

We have:

• Q is the only daughter of P’s brother and is promoted in a month having 30 days.
• W’s son-in-law promoted three persons after Q.

That means, in case (1) Q get promotion in April, in case (2) Q get promotion in September.

• U promoted three persons before R, who is the mother of W.
• The number of persons promoted before U is the same as the number of persons promoted after P.
• Q is neither promoted immediately before nor after R, who is not promoted in December.

That means, in case (1) U gets promoted in February, in case (1a) U gets promoted in July, case (2) is not valid.

Based on the above given information we have:

Case (2) is not valid as the number of persons promoted before U is the same as the number of persons promoted after P.

Again, we have:

• Two persons are promoted between S and W, who is the father of V.

That means, in case (1) either S or W gets promotion in January, case (1a) is not valid.

Based on the above given information we have:

Case (1a) is not valid as two persons are promoted between S and W, who is the father of V.

Again, we have:

• Q’s father is the only son of the one who promoted four persons before T.

That means, in case (1) T must be promoted in December, and either of W or S is the father of Q, in case (1b) T must be promoted in October, and U must be the father of Q.

Since, only remaining person is V, and V is child of W, thus in case (1) V can never be the son-in-law of W, thus case (1) is not valid.

Based on the above given information we have:

Case (1) is not valid as only remaining person is V, and V can’t be W’s son-in-law.

Again, we have:

• P’s father is promoted immediately after U’s daughter-in-law.
• V is an unmarried member of the family.

Since, T is the son-in-law of W.

Thus, U must be the father of P and S must be the daughter-in-law of U.

Based on the above given information we have:

We have:

• Q is the only daughter of P’s brother and is promoted in a month having 30 days.
• W’s son-in-law promoted three persons after Q.

That means, in case (1) Q get promotion in April, in case (2) Q get promotion in September.

• U promoted three persons before R, who is the mother of W.
• The number of persons promoted before U is the same as the number of persons promoted after P.
• Q is neither promoted immediately before nor after R, who is not promoted in December.

That means, in case (1) U gets promoted in February, in case (1a) U gets promoted in July, case (2) is not valid.

Based on the above given information we have:

Case (2) is not valid as the number of persons promoted before U is the same as the number of persons promoted after P.

Again, we have:

• Two persons are promoted between S and W, who is the father of V.

That means, in case (1) either S or W gets promotion in January, case (1a) is not valid.

Based on the above given information we have:

Case (1a) is not valid as two persons are promoted between S and W, who is the father of V.

Again, we have:

• Q’s father is the only son of the one who promoted four persons before T.

That means, in case (1) T must be promoted in December, and either of W or S is the father of Q, in case (1b) T must be promoted in October, and U must be the father of Q.

Since, only remaining person is V, and V is child of W, thus in case (1) V can never be the son-in-law of W, thus case (1) is not valid.

Based on the above given information we have:

Case (1) is not valid as only remaining person is V, and V can’t be W’s son-in-law.

Again, we have:

• P’s father is promoted immediately after U’s daughter-in-law.
• V is an unmarried member of the family.

Since, T is the son-in-law of W.

Thus, U must be the father of P and S must be the daughter-in-law of U.

Based on the above given information we have:

We have:

• Q is the only daughter of P’s brother and is promoted in a month having 30 days.
• W’s son-in-law promoted three persons after Q.

That means, in case (1) Q get promotion in April, in case (2) Q get promotion in September.

• U promoted three persons before R, who is the mother of W.
• The number of persons promoted before U is the same as the number of persons promoted after P.
• Q is neither promoted immediately before nor after R, who is not promoted in December.

That means, in case (1) U gets promoted in February, in case (1a) U gets promoted in July, case (2) is not valid.

Based on the above given information we have:

Case (2) is not valid as the number of persons promoted before U is the same as the number of persons promoted after P.

Again, we have:

• Two persons are promoted between S and W, who is the father of V.

That means, in case (1) either S or W gets promotion in January, case (1a) is not valid.

Based on the above given information we have:

Case (1a) is not valid as two persons are promoted between S and W, who is the father of V.

Again, we have:

• Q’s father is the only son of the one who promoted four persons before T.

That means, in case (1) T must be promoted in December, and either of W or S is the father of Q, in case (1b) T must be promoted in October, and U must be the father of Q.

Since, only remaining person is V, and V is child of W, thus in case (1) V can never be the son-in-law of W, thus case (1) is not valid.

Based on the above given information we have:

Case (1) is not valid as only remaining person is V, and V can’t be W’s son-in-law.

Again, we have:

• P’s father is promoted immediately after U’s daughter-in-law.
• V is an unmarried member of the family.

Since, T is the son-in-law of W.

Thus, U must be the father of P and S must be the daughter-in-law of U.

Based on the above given information we have:

We have:

• Q is the only daughter of P’s brother and is promoted in a month having 30 days.
• W’s son-in-law promoted three persons after Q.

That means, in case (1) Q get promotion in April, in case (2) Q get promotion in September.

• U promoted three persons before R, who is the mother of W.
• The number of persons promoted before U is the same as the number of persons promoted after P.
• Q is neither promoted immediately before nor after R, who is not promoted in December.

That means, in case (1) U gets promoted in February, in case (1a) U gets promoted in July, case (2) is not valid.

Based on the above given information we have:

Case (2) is not valid as the number of persons promoted before U is the same as the number of persons promoted after P.

Again, we have:

• Two persons are promoted between S and W, who is the father of V.

That means, in case (1) either S or W gets promotion in January, case (1a) is not valid.

Based on the above given information we have:

Case (1a) is not valid as two persons are promoted between S and W, who is the father of V.

Again, we have:

• Q’s father is the only son of the one who promoted four persons before T.

That means, in case (1) T must be promoted in December, and either of W or S is the father of Q, in case (1b) T must be promoted in October, and U must be the father of Q.

Since, only remaining person is V, and V is child of W, thus in case (1) V can never be the son-in-law of W, thus case (1) is not valid.

Based on the above given information we have:

Case (1) is not valid as only remaining person is V, and V can’t be W’s son-in-law.

Again, we have:

• P’s father is promoted immediately after U’s daughter-in-law.
• V is an unmarried member of the family.

Since, T is the son-in-law of W.

Thus, U must be the father of P and S must be the daughter-in-law of U.

Based on the above given information we have:

1. (A) is sitting in the car, which is exactly in the middle of the row.

2. There are two cars between cars in which (A) and (C) are sitting.

3. Car at the extreme right end has ticket number 20.

Case 1:

Case 2:

4. Car immediate left of (A) is not vacant.

5. (F) is immediately left of (D).

6. Ticket number of cars of (F) and (D) are 22 and 18 respectively.

Case 1:

Case 2:

Case 2(b):

7. (E) does not sit in the car, which is adjacent to either car of (C) or car of (A).

8. Car of (E) is not parked at extreme left end. So, case 2(a) and case 2(b) would not be possible.

9. The ticket number of car (B) is 5 less than ticket number of car of (E). So, ticket number of car of (B) is 15.

10. Car of (B) is not adjacent to car of (A).

11. The ticket number of car (G) is a prime number between 21 and 26. So, the ticket number of car (G) is 23.

12. The car with ticket number 19 is immediate left of car with ticket number 16.

13. Car (A) does not have ticket number 19.

14. There are cars with ticket number 17 and 21.

Case 1:

So, car (G) is fourth car to the left of car of (D).

Hence, the correct option is (C).

1. (A) is sitting in the car, which is exactly in the middle of the row.

2. There are two cars between cars in which (A) and (C) are sitting.

3. Car at the extreme right end has ticket number 20.

Case 1:

Case 2:

4. Car immediate left of (A) is not vacant.

5. (F) is immediately left of (D).

6. Ticket number of cars of (F) and (D) are 22 and 18 respectively.

Case 1:

Case 2:

Case 2(b):

7. (E) does not sit in the car, which is adjacent to either car of (C) or car of (A).

8. Car of (E) is not parked at extreme left end. So, case 2(a) and case 2(b) would not be possible.

9. The ticket number of car (B) is 5 less than ticket number of car of (E). So, ticket number of car of (B) is 15.

10. Car of (B) is not adjacent to car of (A).

11. The ticket number of car (G) is a prime number between 21 and 26. So, the ticket number of car (G) is 23.

12. The car with ticket number 19 is immediate left of car with ticket number 16.

13. Car (A) does not have ticket number 19.

14. There are cars with ticket number 17 and 21.

Case 1:

So, four persons are to the left of car of (F).

Hence, the correct option is (B).

1. (A) is sitting in the car, which is exactly in the middle of the row.

2. There are two cars between cars in which (A) and (C) are sitting.

3. Car at the extreme right end has ticket number 20.

Case 1:

Case 2:

4. Car immediate left of (A) is not vacant.

5. (F) is immediately left of (D).

6. Ticket number of cars of (F) and (D) are 22 and 18 respectively.

Case 1:

Case 2:

Case 2(b):

7. (E) does not sit in the car, which is adjacent to either car of (C) or car of (A).

8. Car of (E) is not parked at extreme left end. So, case 2(a) and case 2(b) would not be possible.

9. The ticket number of car (B) is 5 less than ticket number of car of (E). So, ticket number of car of (B) is 15.

10. Car of (B) is not adjacent to car of (A).

11. The ticket number of car (G) is a prime number between 21 and 26. So, the ticket number of car (G) is 23.

12. The car with ticket number 19 is immediate left of car with ticket number 16.

13. Car (A) does not have ticket number 19.

14. There are cars with ticket number 17 and 21.

Case 1:

The ticket number of car of (A) is either 17 or 21.

So, ‘Cannot be determined’ is the correct answer.

Hence, the correct option is (E).

1. (A) is sitting in the car, which is exactly in the middle of the row.

2. There are two cars between cars in which (A) and (C) are sitting.

3. Car at the extreme right end has ticket number 20.

Case 1:

Case 2:

4. Car immediate left of (A) is not vacant.

5. (F) is immediately left of (D).

6. Ticket number of cars of (F) and (D) are 22 and 18 respectively.

Case 1:

Case 2:

Case 2(b):

7. (E) does not sit in the car, which is adjacent to either car of (C) or car of (A).

8. Car of (E) is not parked at extreme left end. So, case 2(a) and case 2(b) would not be possible.

9. The ticket number of car (B) is 5 less than ticket number of car of (E). So, ticket number of car of (B) is 15.

10. Car of (B) is not adjacent to car of (A).

11. The ticket number of car (G) is a prime number between 21 and 26. So, the ticket number of car (G) is 23.

12. The car with ticket number 19 is immediate left of car with ticket number 16.

13. Car (A) does not have ticket number 19.

14. There are cars with ticket number 17 and 21.

Case 1:

So, car of (B) is at extreme left end.

Hence, the correct option is (C).

For I: It is completely related to the passage, there must be awareness program to make society awake.

For II: This is must do step which should be taken by Government but it is not related to statement which is mention above.

For III: It is a myth that should be avoided or abolished but it is the part of the statement given above.

For step I:

For number: If the number is odd then add 2 else if the number is even then subtract 3.

For letter: If the letter is vowel then replace it with the second succeeding letter, else if the letter is a consonant then replace it with the second preceding letter.

For step II: the opposite letter of each letter as per alphabetical series is taken and is arranged on opposite cells whereas the numbers are moved one cell in a clockwise direction.

E.g) Opposite letters of E and I interchange their places similarly Opposite letters of W and U interchange their places and so on.

For step III: Letters are changed to the immediate succeeding letter, and numbers are changed by adding 2.

For step IV: The position of all the letters is interchanged with respect to opposite cells.

Based on the above given information we have the following result.

For step I:

For number: If the number is odd then add 2 else if the number is even then subtract 3.

For letter: If the letter is vowel then replace it with the second succeeding letter, else if the letter is a consonant then replace it with the second preceding letter.

For step II: the opposite letter of each letter as per alphabetical series is taken and is arranged on opposite cells whereas the numbers are moved one cell in a clockwise direction.

E.g) Opposite letters of E and I interchange their places similarly Opposite letters of W and U interchange their places and so on.

For step III: Letters are changed to the immediate succeeding letter, and numbers are changed by adding 2.

For step IV: The position of all the letters is interchanged with respect to opposite cells.

Based on the above given information we have the following result.

For step I:

For number: If the number is odd then add 2 else if the number is even then subtract 3.

For letter: If the letter is vowel then replace it with the second succeeding letter, else if the letter is a consonant then replace it with the second preceding letter.

For step II: the opposite letter of each letter as per alphabetical series is taken and is arranged on opposite cells whereas the numbers are moved one cell in a clockwise direction.

E.g) Opposite letters of E and I interchange their places similarly Opposite letters of W and U interchange their places and so on.

For step III: Letters are changed to the immediate succeeding letter, and numbers are changed by adding 2.

For step IV: The position of all the letters is interchanged with respect to opposite cells.

Based on the above given information we have the following result.

For step I:

For number: If the number is odd then add 2 else if the number is even then subtract 3.

For letter: If the letter is vowel then replace it with the second succeeding letter, else if the letter is a consonant then replace it with the second preceding letter.

For step II: the opposite letter of each letter as per alphabetical series is taken and is arranged on opposite cells whereas the numbers are moved one cell in a clockwise direction.

E.g) Opposite letters of E and I interchange their places similarly Opposite letters of W and U interchange their places and so on.

For step III: Letters are changed to the immediate succeeding letter, and numbers are changed by adding 2.

For step IV: The position of all the letters is interchanged with respect to opposite cells.

Based on the above given information we have the following result.

1. The difference between the ages of Ritika and Lovely is 9 years which is same as difference between the ages of Tanya and Meenakshi.

2. Nakul is 21 years old.

3. Difference between the ages of Nakul and Meenakshi is a perfect square of 4.

4. Only three persons are younger than Prateek.

So,

5. The difference between the ages of Tanya and Prateek is equal to the ages of Nakul.

Meenakshi is elder to Prateek.

6. The age of the Prateek is a perfect square.

From this case-1 gets eliminated as Prateek’s age can be either 35 years or 7 years in case-1.

Now, with case 2.

Tanya – Prateek = Nakul

Prateek = 46 – 21 = 25.

7. Sonal is 11 years younger than Prateek.

So, Sonal is 25 – 11 = 14 years.

8. Ritika is just elder than Lovely.

9. As many as persons are younger than Sonal as are elder as Lovely.

Ritika > Lovely > > > > > Prateek > Nakul > Sonal >

So, Sonal is 14 years old and is the second youngest person.

10. The sum of the ages of Ritika and Sonal is 91 years.

So, Ritika + Sonal = 91.

Ritika + 14 = 91

Ritika = 91 – 14 = 77 i.e., Ritika born = 2019 - 77 = 1942.

11. The difference between the ages of Ritika and Lovely is 9 years.

So, Lovely born = 1942 + 9 = 1951. Hence, the age of lovely is 77 – 9 = 68.

12. The difference between the ages of Lovely and Urvashi is equal to the age of Sonal.

The age of Lovely + Urvashi = Sonal

So, 68 + Urvashi = 14

Urvashi = 54 i.e. 1965 year.

Qamish is 24 years younger than Urvashi.

Qamish born = 1965 + 24 = 1989 so the age of Qamish is 30.

13. The difference between the ages of Urvashi and Qamish is same as the difference between Qamish and Opal.

Urvashi – Qamish = Qamish – opal

Urvashi + Opal = 2 Qamish

54 + Opal = 60

Opal = 6 years

So, Opal born in 2013.

The final arrangement is given below-

According to the given information-

Age of Ritika = 77 years

Age of Qamish = 30 years

Total age = 77 + 30 = 107 years

So, total age of Ritika and Qamish is 107 years.

Hence, the correct option is (B).

1. The difference between the ages of Ritika and Lovely is 9 years which is same as difference between the ages of Tanya and Meenakshi.

2. Nakul is 21 years old.

3. Difference between the ages of Nakul and Meenakshi is a perfect square of 4.

4. Only three persons are younger than Prateek.

So,

5. The difference between the ages of Tanya and Prateek is equal to the ages of Nakul.

Meenakshi is elder to Prateek.

6. The age of the Prateek is a perfect square.

From this case-1 gets eliminated as Prateek’s age can be either 35 years or 7 years in case-1.

Now, with case 2.

Tanya – Prateek = Nakul

Prateek = 46 – 21 = 25.

7. Sonal is 11 years younger than Prateek.

So, Sonal is 25 – 11 = 14 years.

8. Ritika is just elder than Lovely.

9. As many as persons are younger than Sonal as are elder as Lovely.

Ritika > Lovely > > > > > Prateek > Nakul > Sonal >

So, Sonal is 14 years old and is the second youngest person.

10. The sum of the ages of Ritika and Sonal is 91 years.

So, Ritika + Sonal = 91.

Ritika + 14 = 91

Ritika = 91 – 14 = 77 i.e., Ritika born = 2019 - 77 = 1942.

11. The difference between the ages of Ritika and Lovely is 9 years.

So, Lovely born = 1942 + 9 = 1951. Hence, the age of lovely is 77 – 9 = 68.

12. The difference between the ages of Lovely and Urvashi is equal to the age of Sonal.

The age of Lovely + Urvashi = Sonal

So, 68 + Urvashi = 14

Urvashi = 54 i.e. 1965 year.

Qamish is 24 years younger than Urvashi.

Qamish born = 1965 + 24 = 1989 so the age of Qamish is 30.

13. The difference between the ages of Urvashi and Qamish is same as the difference between Qamish and Opal.

Urvashi – Qamish = Qamish – opal

Urvashi + Opal = 2 Qamish

54 + Opal = 60

Opal = 6 years

So, Opal was born in 2013.

The final arrangement is given below-

According to the given information-

Total age of Urvashi and Sonal = 54 + 14 = 68 years

Total age of Sonal and Meenakshi = 14 + 37 = 51 years

Difference between the total age of Urvashi and Sonal and total age of Sonal and Meenakshi = 68 – 51 = 17 years.

Hence, the correct option is (D).

1. The difference between the ages of Ritika and Lovely is 9 years which is same as difference between the ages of Tanya and Meenakshi.

2. Nakul is 21 years old.

3. Difference between the ages of Nakul and Meenakshi is a perfect square of 4.

4. Only three persons are younger than Prateek.

So,

5. The difference between the ages of Tanya and Prateek is equal to the ages of Nakul.

Meenakshi is elder to Prateek.

6. The age of the Prateek is a perfect square.

From this case-1 gets eliminated as Prateek’s age can be either 35 years or 7 years in case-1.

Now, with case 2.

Tanya – Prateek = Nakul

Prateek = 46 – 21 = 25.

7. Sonal is 11 years younger than Prateek.

So, Sonal is 25 – 11 = 14 years.

8. Ritika is just elder than Lovely.

9. As many as persons are younger than Sonal as are elder as Lovely.

Ritika > Lovely > > > > > Prateek > Nakul > Sonal >

So, Sonal is 14 years old and is the second youngest person.

10. The sum of the ages of Ritika and Sonal is 91 years.

So, Ritika + Sonal = 91.

Ritika + 14 = 91

Ritika = 91 – 14 = 77 i.e., Ritika born = 2019 - 77 = 1942.

11. The difference between the ages of Ritika and Lovely is 9 years.

So, Lovely born = 1942 + 9 = 1951. Hence, the age of lovely is 77 – 9 = 68.

12. The difference between the ages of Lovely and Urvashi is equal to the age of Sonal.

The age of Lovely + Urvashi = Sonal

So, 68 + Urvashi = 14

Urvashi = 54 i.e. 1965 year.

Qamish is 24 years younger than Urvashi.

Qamish born = 1965 + 24 = 1989 so the age of Qamish is 30.

13. The difference between the ages of Urvashi and Qamish is same as the difference between Qamish and Opal.

Urvashi – Qamish = Qamish – opal

Urvashi + Opal = 2 Qamish

54 + Opal = 60

Opal = 6 years

So, Opal was born in 2013.

The final arrangement is given below-

So, two persons are born between Tanya and Prateek.

Hence, the correct option is (C).

1. The difference between the ages of Ritika and Lovely is 9 years which is same as difference between the ages of Tanya and Meenakshi.

2. Nakul is 21 years old.

3. Difference between the ages of Nakul and Meenakshi is a perfect square of 4.

4. Only three persons are younger than Prateek.

So,

5. The difference between the ages of Tanya and Prateek is equal to the ages of Nakul.

Meenakshi is elder to Prateek.

6. The age of the Prateek is a perfect square.

From this case-1 gets eliminated as Prateek’s age can be either 35 years or 7 years in case-1.

Now, with case 2.

Tanya – Prateek = Nakul

Prateek = 46 – 21 = 25.

7. Sonal is 11 years younger than Prateek.

So, Sonal is 25 – 11 = 14 years.

8. Ritika is just elder than Lovely.

9. As many as persons are younger than Sonal as are elder as Lovely.

Ritika > Lovely > > > > > Prateek > Nakul > Sonal >

So, Sonal is 14 years old and is the second youngest person.

10. The sum of the ages of Ritika and Sonal is 91 years.

So, Ritika + Sonal = 91.

Ritika + 14 = 91

Ritika = 91 – 14 = 77 i.e., Ritika born = 2019 - 77 = 1942.

11. The difference between the ages of Ritika and Lovely is 9 years.

So, Lovely born = 1942 + 9 = 1951. Hence, the age of lovely is 77 – 9 = 68.

12. The difference between the ages of Lovely and Urvashi is equal to the age of Sonal.

The age of Lovely + Urvashi = Sonal

So, 68 + Urvashi = 14

Urvashi = 54 i.e. 1965 year.

Qamish is 24 years younger than Urvashi.

Qamish born = 1965 + 24 = 1989 so the age of Qamish is 30.

13. The difference between the ages of Urvashi and Qamish is same as the difference between Qamish and Opal.

Urvashi – Qamish = Qamish – opal

Urvashi + Opal = 2 Qamish

54 + Opal = 60

Opal = 6 years

So, Opal was born in 2013.

The final arrangement is given below-

According to the given information-

Except Ritika, all persons are born in an odd year.

Hence, the correct option is (D).

1. The difference between the ages of Ritika and Lovely is 9 years which is same as difference between the ages of Tanya and Meenakshi.

2. Nakul is 21 years old.

3. Difference between the ages of Nakul and Meenakshi is a perfect square of 4.

4. Only three persons are younger than Prateek.

So,

5. The difference between the ages of Tanya and Prateek is equal to the ages of Nakul.

Meenakshi is elder to Prateek.

6. The age of the Prateek is a perfect square.

From this case-1 gets eliminated as Prateek’s age can be either 35 years or 7 years in case-1.

Now, with case 2.

Tanya – Prateek = Nakul

Prateek = 46 – 21 = 25.

7. Sonal is 11 years younger than Prateek.

So, Sonal is 25 – 11 = 14 years.

8. Ritika is just elder than Lovely.

9. As many as persons are younger than Sonal as are elder as Lovely.

Ritika > Lovely > > > > > Prateek > Nakul > Sonal >

So, Sonal is 14 years old and is the second youngest person.

10. The sum of the ages of Ritika and Sonal is 91 years.

So, Ritika + Sonal = 91.

Ritika + 14 = 91

Ritika = 91 – 14 = 77 i.e., Ritika born = 2019 - 77 = 1942.

11. The difference between the ages of Ritika and Lovely is 9 years.

So, Lovely born = 1942 + 9 = 1951. Hence, the age of lovely is 77 – 9 = 68.

12. The difference between the ages of Lovely and Urvashi is equal to the age of Sonal.

The age of Lovely + Urvashi = Sonal

So, 68 + Urvashi = 14

Urvashi = 54 i.e. 1965 year.

Qamish is 24 years younger than Urvashi.

Qamish born = 1965 + 24 = 1989 so the age of Qamish is 30.

13. The difference between the ages of Urvashi and Qamish is same as the difference between Qamish and Opal.

Urvashi – Qamish = Qamish – opal

Urvashi + Opal = 2 Qamish

54 + Opal = 60

Opal = 6 years

So, Opal was born in 2013.

The final arrangement is given below-

According to the given information-

So, only Urvashi is born just before 1973.

Hence, the correct option is (A).

Box B contains total 39 items. Only one box is placed between box B and the box which is of Piya. Box E is of Kapil. Box E does not contain 20 pens. Box B is placed below box which is of Santosh. So, case 1, case 2 and case 4 gets eliminated.

The box which is placed immediately above box A contains 6 pencils. Box A neither belongs to Geeta nor Kusum. So there will be two cases in case 3 (i.e. 3A and 3B). The box which is given to Santosh and box B contain total 56 pens.

The box which contains 15 pencils is placed immediately above the box which contains 25 pencils. So, case 3A gets eliminated. Box D and E contain total 18 pencils. Total number of pens contained by box F and A is 52. Total number of items contained by the box which is of Kapil is 10 less than the one which is of Geeta.

Total number of items contain by the box which is of Neha is 2 more than the number of pens contain by box F. So, the final arrangement is —

Box B contains total 39 items. Only one box is placed between box B and the box which is of Piya. Box E is of Kapil. Box E does not contain 20 pens. Box B is placed below box which is of Santosh. So, case 1, case 2 and case 4 gets eliminated.

The box which is placed immediately above box A contains 6 pencils. Box A neither belongs to Geeta nor Kusum. So there will be two cases in case 3 (i.e. 3A and 3B). The box which is given to Santosh and box B contain total 56 pens.

The box which contains 15 pencils is placed immediately above the box which contains 25 pencils. So, case 3A gets eliminated. Box D and E contain total 18 pencils. Total number of pens contained by box F and A is 52. Total number of items contained by the box which is of Kapil is 10 less than the one which is of Geeta.

Total number of items contain by the box which is of Neha is 2 more than the number of pens contain by box F. So, the final arrangement is —

Box B contains total 39 items. Only one box is placed between box B and the box which is of Piya. Box E is of Kapil. Box E does not contain 20 pens. Box B is placed below box which is of Santosh. So, case 1, case 2 and case 4 gets eliminated.

The box which is placed immediately above box A contains 6 pencils. Box A neither belongs to Geeta nor Kusum. So there will be two cases in case 3 (i.e. 3A and 3B). The box which is given to Santosh and box B contain total 56 pens.

The box which contains 15 pencils is placed immediately above the box which contains 25 pencils. So, case 3A gets eliminated. Box D and E contain total 18 pencils. Total number of pens contained by box F and A is 52. Total number of items contained by the box which is of Kapil is 10 less than the one which is of Geeta.

Total number of items contain by the box which is of Neha is 2 more than the number of pens contain by box F. So, the final arrangement is —

Box B contains total 39 items. Only one box is placed between box B and the box which is of Piya. Box E is of Kapil. Box E does not contain 20 pens. Box B is placed below box which is of Santosh. So, case 1, case 2 and case 4 gets eliminated.

The box which is placed immediately above box A contains 6 pencils. Box A neither belongs to Geeta nor Kusum. So there will be two cases in case 3 (i.e. 3A and 3B). The box which is given to Santosh and box B contain total 56 pens.

The box which contains 15 pencils is placed immediately above the box which contains 25 pencils. So, case 3A gets eliminated. Box D and E contain total 18 pencils. Total number of pens contained by box F and A is 52. Total number of items contained by the box which is of Kapil is 10 less than the one which is of Geeta.

Total number of items contain by the box which is of Neha is 2 more than the number of pens contain by box F. So, the final arrangement is —

Box B contains total 39 items. Only one box is placed between box B and the box which is of Piya. Box E is of Kapil. Box E does not contain 20 pens. Box B is placed below box which is of Santosh. So, case 1, case 2 and case 4 gets eliminated.

The box which is placed immediately above box A contains 6 pencils. Box A neither belongs to Geeta nor Kusum. So there will be two cases in case 3 (i.e. 3A and 3B). The box which is given to Santosh and box B contain total 56 pens.

The box which contains 15 pencils is placed immediately above the box which contains 25 pencils. So, case 3A gets eliminated. Box D and E contain total 18 pencils. Total number of pens contained by box F and A is 52. Total number of items contained by the box which is of Kapil is 10 less than the one which is of Geeta.

Total number of items contain by the box which is of Neha is 2 more than the number of pens contain by box F. So, the final arrangement is —

So, option (a) is the correct answer.

So, the best assumption that weakens the author’s contention is (c).

Input: K % 5 G > E 3 @ Y 8 9 & D 2 ^ I 4 # A 1 L £ U 6

For step I: If the symbols immediately preceded by a number and immediately followed by a consonant, then that symbol is placed immediate right of third succeeding element.

Step I: K % 5 G > E 3 Y 8 9 @ D 2 ^ & I 4 # A 1 L £ U 6

For step II: The first half of the elements are written in reverse order after the second half.

Step II: 2 ^ & I 4 # A 1 L £ U 6 D @ 9 8 Y 3 E > G 5 % K

For step III: The first six elements combination are written in reverse order after the second combination of six elements, similarly third combinations are written in reverse order after the fourth combination of six elements.

Step III: A 1 L £ U 6 # 4 I & ^ 2E > G 5 % K 3 Y 8 9 @ D

Input: @ Y K 5 % 8 > G E 3 D 2 ^ I A 1 9 & 4 # L £ U 6

Step I: @ Y K 5 % 8 G E 3 > D 2 ^ I A 1 9 & 4 L £ U # 6

Step II: ^ I A 1 9 & 4 L £ U # 6 2 D > 3 E G 8 % 5 K Y @

Thus, Seventeenth element from the right end → L

Input: K % 5 G > E 3 @ Y 8 9 & D 2 ^ I 4 # A 1 L £ U 6

For step I: If the symbols immediately preceded by a number and immediately followed by a consonant, then that symbol is placed immediate right of third succeeding element.

Step I: K % 5 G > E 3 Y 8 9 @ D 2 ^ & I 4 # A 1 L £ U 6

For step II: The first half of the elements are written in reverse order after the second half.

Step II: 2 ^ & I 4 # A 1 L £ U 6 D @ 9 8 Y 3 E > G 5 % K

For step III: The first six elements combination are written in reverse order after the second combination of six elements, similarly third combinations are written in reverse order after the fourth combination of six elements.

Step III: A 1 L £ U 6 # 4 I & ^ 2E > G 5 % K 3 Y 8 9 @ D

Input: 5 % G > 1 £ U 6 W 8 @O 9 & L E 3 ^ D 2 # K I A

Step I: 5 G > 1 % £ U 6 W 8 @O 9 L E 3 & D 2 ^ K I A #

Step II: 9 L E 3 & D 2 ^ K I A # O @ 8 W 6 U £ % 1 > G 5

Step III: 2 ^ K I A # D & 3 E L 9 £ % 1 >G 5 U 6 W 8 @ O

Thus, all such letters immediately followed by a number → L, G, U, and W