Test: CAT Logical Reasoning & Data Interpretation- 5 (October 23) - CAT MCQ

Patrick visited the shop on Tuesday.
An aspect about each of the five friends is clear from (b). It is evident that a friend who made an assignment on Biology didn't visit the shop on Monday and Tuesday.
Also, from (e), Harry went to the shop after the friend who made the assignment on Biology. Therefore, the friend who made an assignment on Biology would have visited on either Wednesday or Thursday. However, Harry would have visited the shop on either Thursday or Friday.
Among the five friends, as mentioned in (b), Harry can be the only friend who printed 20 sheets. Also, from (f), Harry could not have visited on Thursday (since he printed 20 sheets). Hence, Harry must have went to the shop on Friday.
From (a), Jacob could have visited the shop on either Wednesday or Thursday. By comparing this with five friends in (b), Jacob must be the one who made an assignment on Biology.
In addition to that, Patrick did not visit the shop on Monday as evident from (f). Hence, Patrick must have visited the shop on Tuesday and Richard on Monday.
As per the information given in (d), it can be concluded that Richard must have taken a copy of 10 sheets more than that of Harry (who went to the shop on Friday). Hence, Richard must have printed 30 sheets. Moreover, from (c), Jacob must have taken a copy of 40 sheets. Furthermore, Jacob must have visited the shop on Thursday as evident from (f). Neil must have printed 50 sheets and went to the shop on Wednesday.
Since Jacob went on Thursday, the friend who visited the shop on Tuesday (Patrick) must have made an assignment on Economics (as evident from (a)). Richard did not made the assignment on Ecology (as evident from (d)). Hence, Neil must have made the assignment on Ecology and Richard must have covered the Chemistry.

The table below presents this information:

Titli produced the answer to 8 across, which had the same number of letters as the previous answer to be inserted, and one more than the subsequent answer which was produced by one of the men. So, Titli must have produced a 7-letter word, i.e. either 'Silence' or 'Rosebud'. The word next to Titli's word must be a 6-letter word, which is 'Burden'. It cannot be 'Barely' because it has been given that 'Barely' was the first word to be entered. Also, the word preceding Titli's word must be a 7-letter word.
Thus, both the 7-letter words and 'Burden' must be at consecutive places. They can be at either the second, third and fourth positions or the third, fourth and fifth positions. But, in the second case, 'Baadshah' would be at the second position, which is not possible as given in the question. So, both the 7-letter words and 'Burden' must be at the second, third and fourth positions, respectively, and 'Baadshah' would be at the fifth position. Also, 'Silence' is not at the third position. So, it must be at the second position and 'Rosebud' must be at the third position.

15 across and 15 down must start from the same letter and their length must be unequal. Also, 'Silence' is the answer to an across clue which can be either 15 across or 21 across. However, if 'Silence' is the answer to 15 across, the answer to 15 down must also start with 'S'. But there is no other word starting with 'S'; so 'Silence' must be the answer to the 21 across clue. The fifth one to be worked out was the answer to an across clue. So, 'Baadshah' should be the answer to an across clue.
Elsie's word was longer than Bineet's. So, Elsie's word must be either 'Baadshah' or 'Silence' and Bineet's word should be either 'Barely' or 'Silence'. In both the cases, 'Burden' would be Easwar's word and it would be the answer to the 15 down clue because 'Burden' has been produced by one of the men but it is not Bineet according to instruction 2 and Easwar did not solve 4 down. Sheela was neither the first nor the last to come up with an answer. So, her answer must be 'Silence'. Bineet's word must be 'Barely' and Elsie's word must be 'Baadshah'.

Easwar's number was 15 down.

For knockout format matches, bracket with respect to numbering is as shown below:

Anish Giri who is the 9^th numbered player makes it to the semifinals and Magnus Carlsen who is the 1^st numbered player reaches the quarterfinals. So, according to above tree, Anish Giri plays with Magnus Carlsen in the quarterfinals and wins to reach the semifinals.

Number of matches (in knockout format) = number of players - 1 = 32 - 1 = 31

For Row 1:
For Row 1 & Column 1:
⇒ (* + 12)/2 = 9
⇒ * = 6; then, the number of red balls kept in the three boxes is (6, 9, 12).
For Row 1 & Column 2:
⇒ (2 + 10)/2 = 6
⇒ * = 6; then, the number of red balls kept in the three boxes is (2, 6, 10)
For Row 1 & Column 3:
⇒ (4 + *)/2 = 6
⇒ * = 8; then, the number of red ball kept in the three boxes is (4, 6, 8).

For Row 2:
(a, b) is the solution of x² – 9x + 20 = 0.
⇒ x² – 4x – 5x + 20 = 0
⇒ (x – 4)(x – 5) = 0
⇒ x = 4, 5
So, a = 4, b = 5
(c, d) is the solution of x² – 11x + 24 = 0.
⇒ x² – 3x – 8x + 24 = 0
⇒ (x – 3)(x – 8) = 0
⇒ x = 3, 8
So, c = 3, d = 8
(e, f) is the solution of x² – 13x + 40 = 0.
⇒ x² – 5x – 8x + 40 = 0
⇒ (x – 5)(x – 8) = 0
⇒ x = 5, 8
So, e = 5, f = 8

For Row 3:
Total number of balls in the boxes of column 1 in row 1, row 2 and row 3 = (6 + 9 + 12) + (4 + 5) + (r + r + r) = 36 + 3r
Then, probability that the ball is black = (r + r + r)/(36 + 3r) = 1/5
⇒ 15r = 36 + 3r
⇒ 12r = 36
⇒ r = 36/12
⇒ r = 3
∴ The number of black balls in the boxes of row 3 of column 1 is (3, 3, 3).
Total number of balls in the boxes of column 3 in row 1, row 2 and row 3 = (4 + 6 + 8) + (5 + 8) + (q + q + q - 1) = 30 + 3q
Then, probability that the ball is black = (3q - 1)/(30 + 3q) = 11/42
⇒ 126q – 42 = 330 + 33q
⇒ 93q = 372
⇒ q = 4
∴ The number of black balls in the boxes of row 3 in column 3 is (4, 3, 4).
According to the given directions:
Total number of balls kept in row 3 in column 1, column 2 and column 3 = (3 + 3 + 3) + (p + p + 1 + p + 1) + (4 + 3 + 4) = 25
⇒ (3p + 22) = 25
⇒ 3p = 3
⇒ p = 1
∴ Number of black balls in row 3 of column 2 = (1, 2, 2)

Now, arranging all data in one table:

According to the question:
Required sum = (4 + 6 + 8) + (1 + 2 + 2) = 23

For Row 1:
For Row 1 & Column 1:
⇒ (* + 12)/2 = 9
⇒ * = 6; then, the number of red balls kept in the three boxes is (6, 9, 12).
For Row 1 & Column 2:
⇒ (2 + 10)/2 = 6
⇒ * = 6; then, the number of red balls kept in the three boxes is (2, 6, 10)
For Row 1 & Column 3:
⇒ (4 + *)/2 = 6
⇒ * = 8; then, the number of red ball kept in the three boxes is (4, 6, 8).

For Row 2:
(a, b) is the solution of x² – 9x + 20 = 0.
⇒ x² – 4x – 5x + 20 = 0
⇒ (x – 4)(x – 5) = 0
⇒ x = 4, 5
So, a = 4, b = 5
(c, d) is the solution of x² – 11x + 24 = 0.
⇒ x² – 3x – 8x + 24 = 0
⇒ (x – 3)(x – 8) = 0
⇒ x = 3, 8
So, c = 3, d = 8
(e, f) is the solution of x² – 13x + 40 = 0.
⇒ x² – 5x – 8x + 40 = 0
⇒ (x – 5)(x – 8) = 0
⇒ x = 5, 8
So, e = 5, f = 8

For Row 3:
Total number of balls in the boxes of column 1 in row 1, row 2 and row 3 = (6 + 9 + 12) + (4 + 5) + (r + r + r) = 36 + 3r
Then, probability that the ball is black = (r + r + r)/(36 + 3r) = 1/5
⇒ 15r = 36 + 3r
⇒ 12r = 36
⇒ r = 36/12
⇒ r = 3
∴ The number of black balls in the boxes of row 3 of column 1 is (3, 3, 3).
Total number of balls in the boxes of column 3 in row 1, row 2 and row 3 = (4 + 6 + 8) + (5 + 8) + (q + q + q - 1) = 30 + 3q
Then, probability that the ball is black = (3q - 1)/(30 + 3q) = 11/42
⇒ 126q – 42 = 330 + 33q
⇒ 93q = 372
⇒ q = 4
∴ The number of black balls in the boxes of row 3 in column 3 is (4, 3, 4).
According to the given directions:
Total number of balls kept in row 3 in column 1, column 2 and column 3 = (3 + 3 + 3) + (p + p + 1 + p + 1) + (4 + 3 + 4) = 25
⇒ (3p + 22) = 25
⇒ 3p = 3
⇒ p = 1
∴ Number of black balls in row 3 of column 2 = (1, 2, 2)

Now, arranging all data in one table:

Total number of green balls = (4 + 5) + (3 + 8) + (5 + 8) = 33
Total number of black balls = (3 + 3 + 3 + 1 + 2 + 2 + 4 + 3 + 4) = 25
∴ Required ratio = 33 : 25
Therefore, option 1 is the correct answer.

For knockout format matches, bracket with respect to numbering is as shown below:

Liren Ding is the 2^nd numbered player and reaches the quarterfinals and in the first round, no upsets are there.
So, according to the table, Liren Ding must play against the winner of match between the 7^th numbered player and the 10^th numbered player which is match no. 7 of the second round.
Match no. 7 of the second round results in upset, which means the 10^th numbered player wins this match.
Hence, the 10^th numbered player, who is Hikaru Nakamura, plays against Liren Ding in the quarterfinals.

For knockout format matches, bracket with respect to numbering is as shown below:

Magnus Carlsen, who is the 1^st numbered player, reached the semifinals and he must play with any of the 4^th, 29^th, 13^th, 20^th, 12^th, 21^st, 5^th and 28^th numbered players.
If in the first round, all even numbered matches (and none of the odd numbered ones) result in upsets, the 29^th, 13^th, 21^st, 5^th numbered players reach the second round.
There are no upsets in the second round. So, the 13^th and 5^th numbered players reach the quarterfinals.
Hence, the 13^th numbered player who is Viswanathan Anand could be the lowest seeded player facing Magnus Carlsen in the semifinals.

Patrick visited the shop on Tuesday.
An aspect about each of the five friends is clear from (b). It is evident that a friend who made an assignment on Biology didn't visit the shop on Monday and Tuesday.
Also, from (e), Harry went to the shop after the friend who made the assignment on Biology. Therefore, the friend who made an assignment on Biology would have visited on either Wednesday or Thursday. However, Harry would have visited the shop on either Thursday or Friday.
Among the five friends, as mentioned in (b), Harry can be the only friend who printed 20 sheets. Also, from (f), Harry could not have visited on Thursday (since he printed 20 sheets). Hence, Harry must have went to the shop on Friday.
From (a), Jacob could have visited the shop on either Wednesday or Thursday. By comparing this with five friends in (b), Jacob must be the one who made an assignment on Biology.
In addition to that, Patrick did not visit the shop on Monday as evident from (f). Hence, Patrick must have visited the shop on Tuesday and Richard on Monday.
As per the information given in (d), it can be concluded that Richard must have taken a copy of 10 sheets more than that of Harry (who went to the shop on Friday). Hence, Richard must have printed 30 sheets. Moreover, from (c), Jacob must have taken a copy of 40 sheets. Furthermore, Jacob must have visited the shop on Thursday as evident from (f). Neil must have printed 50 sheets and went to the shop on Wednesday.
Since Jacob went on Thursday, the friend who visited the shop on Tuesday (Patrick) must have made an assignment on Economics (as evident from (a)). Richard did not made the assignment on Ecology (as evident from (d)). Hence, Neil must have made the assignment on Ecology and Richard must have covered the Chemistry.

The table below presents this information:

Titli produced the answer to 8 across, which had the same number of letters as the previous answer to be inserted, and one more than the subsequent answer which was produced by one of the men. So, Titli must have produced a 7-letter word, i.e. either 'Silence' or 'Rosebud'. The word next to Titli's word must be a 6-letter word, which is 'Burden'. It cannot be 'Barely' because it has been given that 'Barely' was the first word to be entered. Also, the word preceding Titli's word must be a 7-letter word.
Thus, both the 7-letter words and 'Burden' must be at consecutive places. They can be at either the second, third and fourth positions or the third, fourth and fifth positions. But, in the second case, 'Baadshah' would be at the second position, which is not possible as given in the question. So, both the 7-letter words and 'Burden' must be at the second, third and fourth positions, respectively, and 'Baadshah' would be at the fifth position. Also, 'Silence' is not at the third position. So, it must be at the second position and 'Rosebud' must be at the third position.

15 across and 15 down must start from the same letter and their length must be unequal. Also, 'Silence' is the answer to an across clue which can be either 15 across or 21 across. However, if 'Silence' is the answer to 15 across, the answer to 15 down must also start with 'S'. But there is no other word starting with 'S'; so 'Silence' must be the answer to the 21 across clue. The fifth one to be worked out was the answer to an across clue. So, 'Baadshah' should be the answer to an across clue.
Elsie's word was longer than Bineet's. So, Elsie's word must be either 'Baadshah' or 'Silence' and Bineet's word should be either 'Barely' or 'Silence'. In both the cases, 'Burden' would be Easwar's word and it would be the answer to the 15 down clue because 'Burden' has been produced by one of the men but it is not Bineet according to instruction 2 and Easwar did not solve 4 down. Sheela was neither the first nor the last to come up with an answer. So, her answer must be 'Silence'. Bineet's word must be 'Barely' and Elsie's word must be 'Baadshah'.

Titli's answer was 'Rosebud'.

For Row 1:
For Row 1 & Column 1:
⇒ (* + 12)/2 = 9
⇒ * = 6; then, the number of red balls kept in the three boxes is (6, 9, 12).
For Row 1 & Column 2:
⇒ (2 + 10)/2 = 6
⇒ * = 6; then, the number of red balls kept in the three boxes is (2, 6, 10)
For Row 1 & Column 3:
⇒ (4 + *)/2 = 6
⇒ * = 8; then, the number of red ball kept in the three boxes is (4, 6, 8).

For Row 2:
(a, b) is the solution of x² – 9x + 20 = 0.
⇒ x² – 4x – 5x + 20 = 0
⇒ (x – 4)(x – 5) = 0
⇒ x = 4, 5
So, a = 4, b = 5
(c, d) is the solution of x² – 11x + 24 = 0.
⇒ x² – 3x – 8x + 24 = 0
⇒ (x – 3)(x – 8) = 0
⇒ x = 3, 8
So, c = 3, d = 8
(e, f) is the solution of x² – 13x + 40 = 0.
⇒ x² – 5x – 8x + 40 = 0
⇒ (x – 5)(x – 8) = 0
⇒ x = 5, 8
So, e = 5, f = 8

For Row 3:
Total number of balls in the boxes of column 1 in row 1, row 2 and row 3 = (6 + 9 + 12) + (4 + 5) + (r + r + r) = 36 + 3r
Then, probability that the ball is black = (r + r + r)/(36 + 3r) = 1/5
⇒ 15r = 36 + 3r
⇒ 12r = 36
⇒ r = 36/12
⇒ r = 3
∴ The number of black balls in the boxes of row 3 of column 1 is (3, 3, 3).
Total number of balls in the boxes of column 3 in row 1, row 2 and row 3 = (4 + 6 + 8) + (5 + 8) + (q + q + q - 1) = 30 + 3q
Then, probability that the ball is black = (3q - 1)/(30 + 3q) = 11/42
⇒ 126q – 42 = 330 + 33q
⇒ 93q = 372
⇒ q = 4
∴ The number of black balls in the boxes of row 3 in column 3 is (4, 3, 4).
According to the given directions:
Total number of balls kept in row 3 in column 1, column 2 and column 3 = (3 + 3 + 3) + (p + p + 1 + p + 1) + (4 + 3 + 4) = 25
⇒ (3p + 22) = 25
⇒ 3p = 3
⇒ p = 1
∴ Number of black balls in row 3 of column 2 = (1, 2, 2)

Now, arranging all data in one table:

Richard made an assignment on Chemistry.
An aspect about each of the five friends is clear from (b). It is evident that a friend who made an assignment on Biology didn't visit the shop on Monday and Tuesday.
Also, from (e), Harry went to the shop after the friend who made the assignment on Biology. Therefore, the friend who made an assignment on Biology would have visited on either Wednesday or Thursday. However, Harry would have visited the shop on either Thursday or Friday.
Among the five friends, as mentioned in (b), Harry can be the only friend who printed 20 sheets. Also, from (f), Harry could not have visited on Thursday (since he printed 20 sheets). Hence, Harry must have went to the shop on Friday.
From (a), Jacob could have visited the shop on either Wednesday or Thursday. By comparing this with five friends in (b), Jacob must be the one who made an assignment on Biology.
In addition to that, Patrick did not visit the shop on Monday as evident from (f). Hence, Patrick must have visited the shop on Tuesday and Richard on Monday.
As per the information given in (d), it can be concluded that Richard must have taken a copy of 10 sheets more than that of Harry (who went to the shop on Friday). Hence, Richard must have printed 30 sheets. Moreover, from (c), Jacob must have taken a copy of 40 sheets. Furthermore, Jacob must have visited the shop on Thursday as evident from (f). Neil must have printed 50 sheets and went to the shop on Wednesday.
Since Jacob went on Thursday, the friend who visited the shop on Tuesday (Patrick) must have made an assignment on Economics (as evident from (a)). Richard did not made the assignment on Ecology (as evident from (d)). Hence, Neil must have made the assignment on Ecology and Richard must have covered the Chemistry.

The table below presents this information:

Average daily number of sheets printed in the first three days (Monday, Tuesday, Wednesday) of the week = (30 + 10 + 50)/3 = 30
Average daily number of sheets printed in the last two days (Thursday, Friday) of the week = (40 + 20)/2 = 30
Therefore, the difference is 0.

Titli produced the answer to 8 across, which had the same number of letters as the previous answer to be inserted, and one more than the subsequent answer which was produced by one of the men. So, Titli must have produced a 7-letter word, i.e. either 'Silence' or 'Rosebud'. The word next to Titli's word must be a 6-letter word, which is 'Burden'. It cannot be 'Barely' because it has been given that 'Barely' was the first word to be entered. Also, the word preceding Titli's word must be a 7-letter word.
Thus, both the 7-letter words and 'Burden' must be at consecutive places. They can be at either the second, third and fourth positions or the third, fourth and fifth positions. But, in the second case, 'Baadshah' would be at the second position, which is not possible as given in the question. So, both the 7-letter words and 'Burden' must be at the second, third and fourth positions, respectively, and 'Baadshah' would be at the fifth position. Also, 'Silence' is not at the third position. So, it must be at the second position and 'Rosebud' must be at the third position.

15 across and 15 down must start from the same letter and their length must be unequal. Also, 'Silence' is the answer to an across clue which can be either 15 across or 21 across. However, if 'Silence' is the answer to 15 across, the answer to 15 down must also start with 'S'. But there is no other word starting with 'S'; so 'Silence' must be the answer to the 21 across clue. The fifth one to be worked out was the answer to an across clue. So, 'Baadshah' should be the answer to an across clue.
Elsie's word was longer than Bineet's. So, Elsie's word must be either 'Baadshah' or 'Silence' and Bineet's word should be either 'Barely' or 'Silence'. In both the cases, 'Burden' would be Easwar's word and it would be the answer to the 15 down clue because 'Burden' has been produced by one of the men but it is not Bineet according to instruction 2 and Easwar did not solve 4 down. Sheela was neither the first nor the last to come up with an answer. So, her answer must be 'Silence'. Bineet's word must be 'Barely' and Elsie's word must be 'Baadshah'.

Sheela's answer was 'Silence'.

Titli produced the answer to 8 across, which had the same number of letters as the previous answer to be inserted, and one more than the subsequent answer which was produced by one of the men. So, Titli must have produced a 7-letter word, i.e. either 'Silence' or 'Rosebud'. The word next to Titli's word must be a 6-letter word, which is 'Burden'. It cannot be 'Barely' because it has been given that 'Barely' was the first word to be entered. Also, the word preceding Titli's word must be a 7-letter word.
Thus, both the 7-letter words and 'Burden' must be at consecutive places. They can be at either the second, third and fourth positions or the third, fourth and fifth positions. But, in the second case, 'Baadshah' would be at the second position, which is not possible as given in the question. So, both the 7-letter words and 'Burden' must be at the second, third and fourth positions, respectively, and 'Baadshah' would be at the fifth position. Also, 'Silence' is not at the third position. So, it must be at the second position and 'Rosebud' must be at the third position.

15 across and 15 down must start from the same letter and their length must be unequal. Also, 'Silence' is the answer to an across clue which can be either 15 across or 21 across. However, if 'Silence' is the answer to 15 across, the answer to 15 down must also start with 'S'. But there is no other word starting with 'S'; so 'Silence' must be the answer to the 21 across clue. The fifth one to be worked out was the answer to an across clue. So, 'Baadshah' should be the answer to an across clue.
Elsie's word was longer than Bineet's. So, Elsie's word must be either 'Baadshah' or 'Silence' and Bineet's word should be either 'Barely' or 'Silence'. In both the cases, 'Burden' would be Easwar's word and it would be the answer to the 15 down clue because 'Burden' has been produced by one of the men but it is not Bineet according to instruction 2 and Easwar did not solve 4 down. Sheela was neither the first nor the last to come up with an answer. So, her answer must be 'Silence'. Bineet's word must be 'Barely' and Elsie's word must be 'Baadshah'.

Titli's order was third.

For Row 1:
For Row 1 & Column 1:
⇒ (* + 12)/2 = 9
⇒ * = 6; then, the number of red balls kept in the three boxes is (6, 9, 12).
For Row 1 & Column 2:
⇒ (2 + 10)/2 = 6
⇒ * = 6; then, the number of red balls kept in the three boxes is (2, 6, 10)
For Row 1 & Column 3:
⇒ (4 + *)/2 = 6
⇒ * = 8; then, the number of red ball kept in the three boxes is (4, 6, 8).

For Row 2:
(a, b) is the solution of x² – 9x + 20 = 0.
⇒ x² – 4x – 5x + 20 = 0
⇒ (x – 4)(x – 5) = 0
⇒ x = 4, 5
So, a = 4, b = 5
(c, d) is the solution of x² – 11x + 24 = 0.
⇒ x² – 3x – 8x + 24 = 0
⇒ (x – 3)(x – 8) = 0
⇒ x = 3, 8
So, c = 3, d = 8
(e, f) is the solution of x² – 13x + 40 = 0.
⇒ x² – 5x – 8x + 40 = 0
⇒ (x – 5)(x – 8) = 0
⇒ x = 5, 8
So, e = 5, f = 8

For Row 3:
Total number of balls in the boxes of column 1 in row 1, row 2 and row 3 = (6 + 9 + 12) + (4 + 5) + (r + r + r) = 36 + 3r
Then, probability that the ball is black = (r + r + r)/(36 + 3r) = 1/5
⇒ 15r = 36 + 3r
⇒ 12r = 36
⇒ r = 36/12
⇒ r = 3
∴ The number of black balls in the boxes of row 3 of column 1 is (3, 3, 3).
Total number of balls in the boxes of column 3 in row 1, row 2 and row 3 = (4 + 6 + 8) + (5 + 8) + (q + q + q - 1) = 30 + 3q
Then, probability that the ball is black = (3q - 1)/(30 + 3q) = 11/42
⇒ 126q – 42 = 330 + 33q
⇒ 93q = 372
⇒ q = 4
∴ The number of black balls in the boxes of row 3 in column 3 is (4, 3, 4).
According to the given directions:
Total number of balls kept in row 3 in column 1, column 2 and column 3 = (3 + 3 + 3) + (p + p + 1 + p + 1) + (4 + 3 + 4) = 25
⇒ (3p + 22) = 25
⇒ 3p = 3
⇒ p = 1
∴ Number of black balls in row 3 of column 2 = (1, 2, 2)

Now, arranging all data in one table:

∴ The number of slots for which the total number of balls in its boxes strictly exceeds 15 is three, viz. row 1 column 1, row 1 column 2, and row 1 column 3.

For Row 1:
For Row 1 & Column 1:
⇒ (* + 12)/2 = 9
⇒ * = 6; then, the number of red balls kept in the three boxes is (6, 9, 12).
For Row 1 & Column 2:
⇒ (2 + 10)/2 = 6
⇒ * = 6; then, the number of red balls kept in the three boxes is (2, 6, 10)
For Row 1 & Column 3:
⇒ (4 + *)/2 = 6
⇒ * = 8; then, the number of red ball kept in the three boxes is (4, 6, 8).

For Row 2:
(a, b) is the solution of x² – 9x + 20 = 0.
⇒ x² – 4x – 5x + 20 = 0
⇒ (x – 4)(x – 5) = 0
⇒ x = 4, 5
So, a = 4, b = 5
(c, d) is the solution of x² – 11x + 24 = 0.
⇒ x² – 3x – 8x + 24 = 0
⇒ (x – 3)(x – 8) = 0
⇒ x = 3, 8
So, c = 3, d = 8
(e, f) is the solution of x² – 13x + 40 = 0.
⇒ x² – 5x – 8x + 40 = 0
⇒ (x – 5)(x – 8) = 0
⇒ x = 5, 8
So, e = 5, f = 8

For Row 3:
Total number of balls in the boxes of column 1 in row 1, row 2 and row 3 = (6 + 9 + 12) + (4 + 5) + (r + r + r) = 36 + 3r
Then, probability that the ball is black = (r + r + r)/(36 + 3r) = 1/5
⇒ 15r = 36 + 3r
⇒ 12r = 36
⇒ r = 36/12
⇒ r = 3
∴ The number of black balls in the boxes of row 3 of column 1 is (3, 3, 3).
Total number of balls in the boxes of column 3 in row 1, row 2 and row 3 = (4 + 6 + 8) + (5 + 8) + (q + q + q - 1) = 30 + 3q
Then, probability that the ball is black = (3q - 1)/(30 + 3q) = 11/42
⇒ 126q – 42 = 330 + 33q
⇒ 93q = 372
⇒ q = 4
∴ The number of black balls in the boxes of row 3 in column 3 is (4, 3, 4).
According to the given directions:
Total number of balls kept in row 3 in column 1, column 2 and column 3 = (3 + 3 + 3) + (p + p + 1 + p + 1) + (4 + 3 + 4) = 25
⇒ (3p + 22) = 25
⇒ 3p = 3
⇒ p = 1
∴ Number of black balls in row 3 of column 2 = (1, 2, 2)

Now, arranging all data in one table:

∴ The total number of slots for which the total number of balls in its boxes is more than 10 and less than 20 is five.
Therefore, option 2 is the correct answer.

Titli produced the answer to 8 across, which had the same number of letters as the previous answer to be inserted, and one more than the subsequent answer which was produced by one of the men. So, Titli must have produced a 7-letter word, i.e. either 'Silence' or 'Rosebud'. The word next to Titli's word must be a 6-letter word, which is 'Burden'. It cannot be 'Barely' because it has been given that 'Barely' was the first word to be entered. Also, the word preceding Titli's word must be a 7-letter word.
Thus, both the 7-letter words and 'Burden' must be at consecutive places. They can be at either the second, third and fourth positions or the third, fourth and fifth positions. But, in the second case, 'Baadshah' would be at the second position, which is not possible as given in the question. So, both the 7-letter words and 'Burden' must be at the second, third and fourth positions, respectively, and 'Baadshah' would be at the fifth position. Also, 'Silence' is not at the third position. So, it must be at the second position and 'Rosebud' must be at the third position.

15 across and 15 down must start from the same letter and their length must be unequal. Also, 'Silence' is the answer to an across clue which can be either 15 across or 21 across. However, if 'Silence' is the answer to 15 across, the answer to 15 down must also start with 'S'. But there is no other word starting with 'S'; so 'Silence' must be the answer to the 21 across clue. The fifth one to be worked out was the answer to an across clue. So, 'Baadshah' should be the answer to an across clue.
Elsie's word was longer than Bineet's. So, Elsie's word must be either 'Baadshah' or 'Silence' and Bineet's word should be either 'Barely' or 'Silence'. In both the cases, 'Burden' would be Easwar's word and it would be the answer to the 15 down clue because 'Burden' has been produced by one of the men but it is not Bineet according to instruction 2 and Easwar did not solve 4 down. Sheela was neither the first nor the last to come up with an answer. So, her answer must be 'Silence'. Bineet's word must be 'Barely' and Elsie's word must be 'Baadshah'.

Easwar produced the answer 'Burden'.

Patrick visited the shop on Tuesday.
An aspect about each of the five friends is clear from (b). It is evident that a friend who made an assignment on Biology didn't visit the shop on Monday and Tuesday.
Also, from (e), Harry went to the shop after the friend who made the assignment on Biology. Therefore, the friend who made an assignment on Biology would have visited on either Wednesday or Thursday. However, Harry would have visited the shop on either Thursday or Friday.
Among the five friends, as mentioned in (b), Harry can be the only friend who printed 20 sheets. Also, from (f), Harry could not have visited on Thursday (since he printed 20 sheets). Hence, Harry must have went to the shop on Friday.
From (a), Jacob could have visited the shop on either Wednesday or Thursday. By comparing this with five friends in (b), Jacob must be the one who made an assignment on Biology.
In addition to that, Patrick did not visit the shop on Monday as evident from (f). Hence, Patrick must have visited the shop on Tuesday and Richard on Monday.
As per the information given in (d), it can be concluded that Richard must have taken a copy of 10 sheets more than that of Harry (who went to the shop on Friday). Hence, Richard must have printed 30 sheets. Moreover, from (c), Jacob must have taken a copy of 40 sheets. Furthermore, Jacob must have visited the shop on Thursday as evident from (f). Neil must have printed 50 sheets and went to the shop on Wednesday.
Since Jacob went on Thursday, the friend who visited the shop on Tuesday (Patrick) must have made an assignment on Economics (as evident from (a)). Richard did not made the assignment on Ecology (as evident from (d)). Hence, Neil must have made the assignment on Ecology and Richard must have covered the Chemistry.

The table below presents this information:

For knockout format matches, bracket with respect to numbering is as shown below:

If the top eight rated players make it to the quarterfinals and Carlsen who is the highest rated player makes it to final, then from the above table it is clear that if Carlsen reaches the final, then the 4th, 5th and 8th rated players can't reach the final.
From the options, Fabiano Caruana is the 4th rated player. Hence, he would definitely not play against Magnus Carlsen in the final.

Richard made an assignment on Chemistry.
An aspect about each of the five friends is clear from (b). It is evident that a friend who made an assignment on Biology didn't visit the shop on Monday and Tuesday.
Also, from (e), Harry went to the shop after the friend who made the assignment on Biology. Therefore, the friend who made an assignment on Biology would have visited on either Wednesday or Thursday. However, Harry would have visited the shop on either Thursday or Friday.
Among the five friends, as mentioned in (b), Harry can be the only friend who printed 20 sheets. Also, from (f), Harry could not have visited on Thursday (since he printed 20 sheets). Hence, Harry must have went to the shop on Friday.
From (a), Jacob could have visited the shop on either Wednesday or Thursday. By comparing this with five friends in (b), Jacob must be the one who made an assignment on Biology.
In addition to that, Patrick did not visit the shop on Monday as evident from (f). Hence, Patrick must have visited the shop on Tuesday and Richard on Monday.
As per the information given in (d), it can be concluded that Richard must have taken a copy of 10 sheets more than that of Harry (who went to the shop on Friday). Hence, Richard must have printed 30 sheets. Moreover, from (c), Jacob must have taken a copy of 40 sheets. Furthermore, Jacob must have visited the shop on Thursday as evident from (f). Neil must have printed 50 sheets and went to the shop on Wednesday.
Since Jacob went on Thursday, the friend who visited the shop on Tuesday (Patrick) must have made an assignment on Economics (as evident from (a)). Richard did not made the assignment on Ecology (as evident from (d)). Hence, Neil must have made the assignment on Ecology and Richard must have covered the Chemistry.

The table below presents this information: