CAT — Tables

Step 1:

The average number of coins per sack in each box is given to be a distinct integer. Hence, the nine boxes must have average values 1, 2, 3, 4, 5, 6, 7, 8, and 9, in some order.

Since the sum of these averages is

(9 × 10) / 2 = 45,

and every row and column has the same total number of coins, each row and each column must also contain a total of 45 coins.

Further, because the average number of coins in a box is an integer, the sum of the three sacks in every box must be divisible by 3.

An average of 1 is possible only with the combination (1, 1, 1).

This combination can be placed only in the 3rd row, 3rd column, as it satisfies both conditions:

• the minimum value is 1, and
• the median is also 1.

Similarly, an average of 9 is possible only with (9, 9, 9).

Since the box in 3rd row, 1st column has a median of 8, it cannot have an average of 9.

The box in 3rd row, 1st column satisfies condition (iii). Hence, the only valid combination is (7, 8, 9).

The box in 3rd row, 2nd column must satisfy conditions (i) and (ii). Therefore, its minimum and maximum values are 1 and 9, respectively.

The possible combinations are (1, 5, 9) and (1, 8, 9). However, only (1, 8, 9) allows the total of the third row to become 45.

Thus, the third row is:

• 3rd row, 1st column: (7, 8, 9) → Average = 8
• 3rd row, 2nd column: (1, 8, 9) → Average = 6
• 3rd row, 3rd column: (1, 1, 1) → Average = 1

Step 2:

If the average of a box is 2, then the total number of coins in that box must be 6.

Since each sack must contain fewer than 5 coins, this is possible only for the box in the 2nd row, 2nd column.

Among the possible combinations, only (1, 2, 3) satisfies the required conditions.

(The combination (1, 1, 4) satisfies only two conditions and therefore cannot be used.)

Hence:

• 2nd row, 1st column: (1, 2, 9) → Average = 4
• 2nd row, 2nd column: (1, 2, 3) → Average = 2
• 2nd row, 3rd column: (9, 9, 9) → Average = 9

Similarly, the first row is determined as:

• 1st row, 1st column: (1, 1, 7) → Average = 3
• 1st row, 2nd column: (3, 9, 9) → Average = 7
• 1st row, 3rd column: (1, 6, 8) → Average = 5

The total number of coins in all the boxes in the 3rd row is 45.

Step 1:

The average number of coins per sack in each box is given to be a distinct integer. Hence, the nine boxes must have average values 1, 2, 3, 4, 5, 6, 7, 8, and 9, in some order.

Since the sum of these averages is

(9 × 10) / 2 = 45,

and every row and column has the same total number of coins, each row and each column must also contain a total of 45 coins.

Further, because the average number of coins in a box is an integer, the sum of the three sacks in every box must be divisible by 3.

An average of 1 is possible only with the combination (1, 1, 1).

This combination can be placed only in the 3rd row, 3rd column, as it satisfies both conditions:

• the minimum value is 1, and
• the median is also 1.

Similarly, an average of 9 is possible only with (9, 9, 9).

Since the box in 3rd row, 1st column has a median of 8, it cannot have an average of 9.

The box in 3rd row, 1st column satisfies condition (iii). Hence, the only valid combination is (7, 8, 9).

The box in 3rd row, 2nd column must satisfy conditions (i) and (ii). Therefore, its minimum and maximum values are 1 and 9, respectively.

The possible combinations are (1, 5, 9) and (1, 8, 9). However, only (1, 8, 9) allows the total of the third row to become 45.

Thus, the third row is:

• 3rd row, 1st column: (7, 8, 9) → Average = 8
• 3rd row, 2nd column: (1, 8, 9) → Average = 6
• 3rd row, 3rd column: (1, 1, 1) → Average = 1

Step 2:

If the average of a box is 2, then the total number of coins in that box must be 6.

Since each sack must contain fewer than 5 coins, this is possible only for the box in the 2nd row, 2nd column.

Among the possible combinations, only (1, 2, 3) satisfies the required conditions.

(The combination (1, 1, 4) satisfies only two conditions and therefore cannot be used.)

Hence:

• 2nd row, 1st column: (1, 2, 9) → Average = 4
• 2nd row, 2nd column: (1, 2, 3) → Average = 2
• 2nd row, 3rd column: (9, 9, 9) → Average = 9

Similarly, the first row is determined as:

• 1st row, 1st column: (1, 1, 7) → Average = 3
• 1st row, 2nd column: (3, 9, 9) → Average = 7
• 1st row, 3rd column: (1, 6, 8) → Average = 5

5 boxes have atleast one sack containing 9 coins.

Step 1:

The average number of coins per sack in each box is given to be a distinct integer. Hence, the nine boxes must have average values 1, 2, 3, 4, 5, 6, 7, 8, and 9, in some order.

Since the sum of these averages is

(9 × 10) / 2 = 45,

and every row and column has the same total number of coins, each row and each column must also contain a total of 45 coins.

Further, because the average number of coins in a box is an integer, the sum of the three sacks in every box must be divisible by 3.

An average of 1 is possible only with the combination (1, 1, 1).

This combination can be placed only in the 3rd row, 3rd column, as it satisfies both conditions:

• the minimum value is 1, and
• the median is also 1.

Similarly, an average of 9 is possible only with (9, 9, 9).

Since the box in 3rd row, 1st column has a median of 8, it cannot have an average of 9.

The box in 3rd row, 1st column satisfies condition (iii). Hence, the only valid combination is (7, 8, 9).

The box in 3rd row, 2nd column must satisfy conditions (i) and (ii). Therefore, its minimum and maximum values are 1 and 9, respectively.

The possible combinations are (1, 5, 9) and (1, 8, 9). However, only (1, 8, 9) allows the total of the third row to become 45.

Thus, the third row is:

• 3rd row, 1st column: (7, 8, 9) → Average = 8
• 3rd row, 2nd column: (1, 8, 9) → Average = 6
• 3rd row, 3rd column: (1, 1, 1) → Average = 1

Step 2:

If the average of a box is 2, then the total number of coins in that box must be 6.

Since each sack must contain fewer than 5 coins, this is possible only for the box in the 2nd row, 2nd column.

Among the possible combinations, only (1, 2, 3) satisfies the required conditions.

(The combination (1, 1, 4) satisfies only two conditions and therefore cannot be used.)

Hence:

• 2nd row, 1st column: (1, 2, 9) → Average = 4
• 2nd row, 2nd column: (1, 2, 3) → Average = 2
• 2nd row, 3rd column: (9, 9, 9) → Average = 9

Similarly, the first row is determined as:

• 1st row, 1st column: (1, 1, 7) → Average = 3
• 1st row, 2nd column: (3, 9, 9) → Average = 7
• 1st row, 3rd column: (1, 6, 8) → Average = 5

Step 1:

The average number of coins per sack in each box is given to be a distinct integer. Hence, the nine boxes must have average values 1, 2, 3, 4, 5, 6, 7, 8, and 9, in some order.

Since the sum of these averages is

(9 × 10) / 2 = 45,

and every row and column has the same total number of coins, each row and each column must also contain a total of 45 coins.

Further, because the average number of coins in a box is an integer, the sum of the three sacks in every box must be divisible by 3.

An average of 1 is possible only with the combination (1, 1, 1).

This combination can be placed only in the 3rd row, 3rd column, as it satisfies both conditions:

• the minimum value is 1, and
• the median is also 1.

Similarly, an average of 9 is possible only with (9, 9, 9).

Since the box in 3rd row, 1st column has a median of 8, it cannot have an average of 9.

The box in 3rd row, 1st column satisfies condition (iii). Hence, the only valid combination is (7, 8, 9).

The box in 3rd row, 2nd column must satisfy conditions (i) and (ii). Therefore, its minimum and maximum values are 1 and 9, respectively.

The possible combinations are (1, 5, 9) and (1, 8, 9). However, only (1, 8, 9) allows the total of the third row to become 45.

Thus, the third row is:

• 3rd row, 1st column: (7, 8, 9) → Average = 8
• 3rd row, 2nd column: (1, 8, 9) → Average = 6
• 3rd row, 3rd column: (1, 1, 1) → Average = 1

Step 2:

If the average of a box is 2, then the total number of coins in that box must be 6.

Since each sack must contain fewer than 5 coins, this is possible only for the box in the 2nd row, 2nd column.

Among the possible combinations, only (1, 2, 3) satisfies the required conditions.

(The combination (1, 1, 4) satisfies only two conditions and therefore cannot be used.)

Hence:

• 2nd row, 1st column: (1, 2, 9) → Average = 4
• 2nd row, 2nd column: (1, 2, 3) → Average = 2
• 2nd row, 3rd column: (9, 9, 9) → Average = 9

Similarly, the first row is determined as:

• 1st row, 1st column: (1, 1, 7) → Average = 3
• 1st row, 2nd column: (3, 9, 9) → Average = 7
• 1st row, 3rd column: (1, 6, 8) → Average = 5

Step 1:

The average number of coins per sack in each box is given to be a distinct integer. Hence, the nine boxes must have average values 1, 2, 3, 4, 5, 6, 7, 8, and 9, in some order.

Since the sum of these averages is

(9 × 10) / 2 = 45,

and every row and column has the same total number of coins, each row and each column must also contain a total of 45 coins.

Further, because the average number of coins in a box is an integer, the sum of the three sacks in every box must be divisible by 3.

An average of 1 is possible only with the combination (1, 1, 1).

This combination can be placed only in the 3rd row, 3rd column, as it satisfies both conditions:

• the minimum value is 1, and
• the median is also 1.

Similarly, an average of 9 is possible only with (9, 9, 9).

Since the box in 3rd row, 1st column has a median of 8, it cannot have an average of 9.

The box in 3rd row, 1st column satisfies condition (iii). Hence, the only valid combination is (7, 8, 9).

The box in 3rd row, 2nd column must satisfy conditions (i) and (ii). Therefore, its minimum and maximum values are 1 and 9, respectively.

The possible combinations are (1, 5, 9) and (1, 8, 9). However, only (1, 8, 9) allows the total of the third row to become 45.

Thus, the third row is:

• 3rd row, 1st column: (7, 8, 9) → Average = 8
• 3rd row, 2nd column: (1, 8, 9) → Average = 6
• 3rd row, 3rd column: (1, 1, 1) → Average = 1

Step 2:

If the average of a box is 2, then the total number of coins in that box must be 6.

Since each sack must contain fewer than 5 coins, this is possible only for the box in the 2nd row, 2nd column.

Among the possible combinations, only (1, 2, 3) satisfies the required conditions.

(The combination (1, 1, 4) satisfies only two conditions and therefore cannot be used.)

Hence:

• 2nd row, 1st column: (1, 2, 9) → Average = 4
• 2nd row, 2nd column: (1, 2, 3) → Average = 2
• 2nd row, 3rd column: (9, 9, 9) → Average = 9

Similarly, the first row is determined as:

• 1st row, 1st column: (1, 1, 7) → Average = 3
• 1st row, 2nd column: (3, 9, 9) → Average = 7
• 1st row, 3rd column: (1, 6, 8) → Average = 5

Step 1:

From condition (5), the protein content in P1 is twice that of M3. Let the protein content in M3 be x grams. Therefore, P1 contains 2x grams of protein.

According to condition (4), the values of protein, fat, and other nutrients are all non-zero multiples of 4.

From condition (1), both pseudo-cereals contain more carbohydrates and more protein than any millet.

If 2x = 24, then the fat content in P1 becomes:

100 − 66 − 24 − 10 = 0,

which is not a valid non-zero multiple of 4.

Similarly, if 2x = 16, then the fat content in P1 becomes:

100 − 66 − 16 − 10 = 8,

forcing x = 8 and leaving the remaining nutrients in M3 as:

100 − 56 − 8 − 12 = 24,

which is feasible.

Hence, x = 8.

Step 2:

From condition (3), all unknown carbohydrate values are non-zero multiples of 5.

Using condition (1), M2 must contain 65 g of carbohydrates, which gives it 12 g of protein.

(Note that the carbohydrate content cannot be below 65 g, otherwise the protein content would exceed 14 g.)

For P1, the carbohydrate content is 66 g. The possible multiples of 5 greater than 66 are 70, 75, 80, 85, 90, and 95.

However, 95, 90, and 85 are not possible because the remaining nutrients would exceed the available 15 g.

From condition (2), C1 and C2 have higher carbohydrate content than either pseudo-cereal. Therefore, their carbohydrate values must be 75 g and 80 g.

Since P2 has 70 g of carbohydrates, comparing C1 and C2:

• C1 must have 80 g of carbohydrates, giving 8 g of protein.
• C2 must have 75 g of carbohydrates, giving 12 g of protein.

The remaining 28 g in M1 (fat + other nutrients) must be split into non-zero multiples of 4.

The possible pairs are:

• (4, 24)
• (8, 20)
• (12, 16)

in any order.

Hence, the final values are:

• C1: Carbohydrates = 80, Protein = 8, Fat = 0, Other = 12
• C2: Carbohydrates = 75, Protein = 12, Fat = 3, Other = 10
• M1: Carbohydrates = 62, Protein = 10, Fat = 4/8/12, Other = 24/20/16
• M2: Carbohydrates = 65, Protein = 12, Fat = 7, Other = 16
• M3: Carbohydrates = 56, Protein = 8, Fat = 12, Other = 24
• P1: Carbohydrates = 66, Protein = 16, Fat = 8, Other = 10
• P2: Carbohydrates = 70, Protein = 14, Fat = 8, Other = 8

From the completed table, M1 contains 62 g of carbohydrates per 100 g.

The food grains with a carbohydrate content greater than 62 g per 100 g are:

• C1
• C2
• M2
• P1
• P2

Thus, 5 food grains have a higher carbohydrate content than M1.

Hence, the correct answer is 5.

Step 1:

From condition (5), the protein content in P1 is twice that of M3. Let the protein content in M3 be x grams. Therefore, P1 contains 2x grams of protein.

According to condition (4), the values of protein, fat, and other nutrients are all non-zero multiples of 4.

From condition (1), both pseudo-cereals contain more carbohydrates and more protein than any millet.

If 2x = 24, then the fat content in P1 becomes:

100 − 66 − 24 − 10 = 0,

which is not a valid non-zero multiple of 4.

Similarly, if 2x = 16, then the fat content in P1 becomes:

100 − 66 − 16 − 10 = 8,

forcing x = 8 and leaving the remaining nutrients in M3 as:

100 − 56 − 8 − 12 = 24,

which is feasible.

Hence, x = 8.

Step 2:

From condition (3), all unknown carbohydrate values are non-zero multiples of 5.

Using condition (1), M2 must contain 65 g of carbohydrates, which gives it 12 g of protein.

(Note that the carbohydrate content cannot be below 65 g, otherwise the protein content would exceed 14 g.)

For P1, the carbohydrate content is 66 g. The possible multiples of 5 greater than 66 are 70, 75, 80, 85, 90, and 95.

However, 95, 90, and 85 are not possible because the remaining nutrients would exceed the available 15 g.

From condition (2), C1 and C2 have higher carbohydrate content than either pseudo-cereal. Therefore, their carbohydrate values must be 75 g and 80 g.

Since P2 has 70 g of carbohydrates, comparing C1 and C2:

• C1 must have 80 g of carbohydrates, giving 8 g of protein.
• C2 must have 75 g of carbohydrates, giving 12 g of protein.

The remaining 28 g in M1 (fat + other nutrients) must be split into non-zero multiples of 4.

The possible pairs are:

• (4, 24)
• (8, 20)
• (12, 16)

in any order.

Hence, the final values are:

• C1: Carbohydrates = 80, Protein = 8, Fat = 0, Other = 12
• C2: Carbohydrates = 75, Protein = 12, Fat = 3, Other = 10
• M1: Carbohydrates = 62, Protein = 10, Fat = 4/8/12, Other = 24/20/16
• M2: Carbohydrates = 65, Protein = 12, Fat = 7, Other = 16
• M3: Carbohydrates = 56, Protein = 8, Fat = 12, Other = 24
• P1: Carbohydrates = 66, Protein = 16, Fat = 8, Other = 10
• P2: Carbohydrates = 70, Protein = 14, Fat = 8, Other = 8

In 100 g of M2, there were 12 g of proteins.

Step 1:

From condition (5), the protein content in P1 is twice that of M3. Let the protein content in M3 be x grams. Therefore, P1 contains 2x grams of protein.

According to condition (4), the values of protein, fat, and other nutrients are all non-zero multiples of 4.

From condition (1), both pseudo-cereals contain more carbohydrates and more protein than any millet.

If 2x = 24, then the fat content in P1 becomes:

100 − 66 − 24 − 10 = 0,

which is not a valid non-zero multiple of 4.

Similarly, if 2x = 16, then the fat content in P1 becomes:

100 − 66 − 16 − 10 = 8,

forcing x = 8 and leaving the remaining nutrients in M3 as:

100 − 56 − 8 − 12 = 24,

which is feasible.

Hence, x = 8.

Step 2:

From condition (3), all unknown carbohydrate values are non-zero multiples of 5.

Using condition (1), M2 must contain 65 g of carbohydrates, which gives it 12 g of protein.

(Note that the carbohydrate content cannot be below 65 g, otherwise the protein content would exceed 14 g.)

For P1, the carbohydrate content is 66 g. The possible multiples of 5 greater than 66 are 70, 75, 80, 85, 90, and 95.

However, 95, 90, and 85 are not possible because the remaining nutrients would exceed the available 15 g.

From condition (2), C1 and C2 have higher carbohydrate content than either pseudo-cereal. Therefore, their carbohydrate values must be 75 g and 80 g.

Since P2 has 70 g of carbohydrates, comparing C1 and C2:

• C1 must have 80 g of carbohydrates, giving 8 g of protein.
• C2 must have 75 g of carbohydrates, giving 12 g of protein.

The remaining 28 g in M1 (fat + other nutrients) must be split into non-zero multiples of 4.

The possible pairs are:

• (4, 24)
• (8, 20)
• (12, 16)

in any order.

Hence, the final values are:

• C1: Carbohydrates = 80, Protein = 8, Fat = 0, Other = 12
• C2: Carbohydrates = 75, Protein = 12, Fat = 3, Other = 10
• M1: Carbohydrates = 62, Protein = 10, Fat = 4/8/12, Other = 24/20/16
• M2: Carbohydrates = 65, Protein = 12, Fat = 7, Other = 16
• M3: Carbohydrates = 56, Protein = 8, Fat = 12, Other = 24
• P1: Carbohydrates = 66, Protein = 16, Fat = 8, Other = 10
• P2: Carbohydrates = 70, Protein = 14, Fat = 8, Other = 8

In 100 g of M3 there were 24 g of other nutrients.

Step 1:

From condition (5), the protein content in P1 is twice that of M3. Let the protein content in M3 be x grams. Therefore, P1 contains 2x grams of protein.

According to condition (4), the values of protein, fat, and other nutrients are all non-zero multiples of 4.

From condition (1), both pseudo-cereals contain more carbohydrates and more protein than any millet.

If 2x = 24, then the fat content in P1 becomes:

100 − 66 − 24 − 10 = 0,

which is not a valid non-zero multiple of 4.

Similarly, if 2x = 16, then the fat content in P1 becomes:

100 − 66 − 16 − 10 = 8,

forcing x = 8 and leaving the remaining nutrients in M3 as:

100 − 56 − 8 − 12 = 24,

which is feasible.

Hence, x = 8.

Step 2:

From condition (3), all unknown carbohydrate values are non-zero multiples of 5.

Using condition (1), M2 must contain 65 g of carbohydrates, which gives it 12 g of protein.

(Note that the carbohydrate content cannot be below 65 g, otherwise the protein content would exceed 14 g.)

For P1, the carbohydrate content is 66 g. The possible multiples of 5 greater than 66 are 70, 75, 80, 85, 90, and 95.

However, 95, 90, and 85 are not possible because the remaining nutrients would exceed the available 15 g.

From condition (2), C1 and C2 have higher carbohydrate content than either pseudo-cereal. Therefore, their carbohydrate values must be 75 g and 80 g.

Since P2 has 70 g of carbohydrates, comparing C1 and C2:

• C1 must have 80 g of carbohydrates, giving 8 g of protein.
• C2 must have 75 g of carbohydrates, giving 12 g of protein.

The remaining 28 g in M1 (fat + other nutrients) must be split into non-zero multiples of 4.

The possible pairs are:

• (4, 24)
• (8, 20)
• (12, 16)

in any order.

Hence, the final values are:

• C1: Carbohydrates = 80, Protein = 8, Fat = 0, Other = 12
• C2: Carbohydrates = 75, Protein = 12, Fat = 3, Other = 10
• M1: Carbohydrates = 62, Protein = 10, Fat = 4/8/12, Other = 24/20/16
• M2: Carbohydrates = 65, Protein = 12, Fat = 7, Other = 16
• M3: Carbohydrates = 56, Protein = 8, Fat = 12, Other = 24
• P1: Carbohydrates = 66, Protein = 16, Fat = 8, Other = 10
• P2: Carbohydrates = 70, Protein = 14, Fat = 8, Other = 8

The median is the value that lies in the middle when all observations are arranged in ascending or descending order.

The protein content (in grams per 100 g) of the seven food grains is:

8, 8, 10, 12, 12, 14, 16

The middle (4th) observation is 12.

Hence, the median protein content is 12 g.

Step 1:

Let the scores on Day 1, Day 2, Day 3, Day 4, and Day 5 be D1, D2, D3, D4, and D5, respectively.

From the given information,

• D1 + D2 = 30
• D2 + D3 = 31
• D3 + D4 = 32
• D4 + D5 = 34

Using condition (2),

D3 = D4 = 16

Therefore,

• D1 = 15
• D2 = 15
• D3 = 16
• D4 = 16
• D5 = 18

From conditions (1) and (3), let the unknown scores be represented as a, b, and c.

Since the total score on Day 3 is 16,

2a + c = 16

Also, c > a.

The possible pairs are (4, 8) and (5, 6).

Since all of Chatur's scores are multiples of 3, the only valid choice is:

a = 5, c = 6

This also implies that Chatur's unique highest score occurs on Day 2, so

b = 3.

Step 2:

The known scores are:

• Akhil: Day 1 = 7, Day 2 = x, Day 3 = 5, Day 4 = 3, Day 5 = p
• Bimal: Day 1 = 5, Day 2 = y, Day 3 = 5, Day 4 = 7, Day 5 = q
• Chatur: Day 1 = 3, Day 2 = 9, Day 3 = 6, Day 4 = 6, Day 5 = 6

Since Chatur's score of 9 is the unique highest score,

9 > x > y

and

q > 6 > p.

Using the daily totals,

x + y = 6

and

p + q = 12.

The possible values are:

• (x, y) = (5, 1) or (4, 2)
• (p, q) = (5, 7) or (4, 8)

Step 1:

Let the scores on Day 1, Day 2, Day 3, Day 4, and Day 5 be D1, D2, D3, D4, and D5, respectively.

From the given information,

• D1 + D2 = 30
• D2 + D3 = 31
• D3 + D4 = 32
• D4 + D5 = 34

Using condition (2),

D3 = D4 = 16

Therefore,

• D1 = 15
• D2 = 15
• D3 = 16
• D4 = 16
• D5 = 18

From conditions (1) and (3), let the unknown scores be represented as a, b, and c.

Since the total score on Day 3 is 16,

2a + c = 16

Also, c > a.

The possible pairs are (4, 8) and (5, 6).

Since all of Chatur's scores are multiples of 3, the only valid choice is:

a = 5, c = 6

This also implies that Chatur's unique highest score occurs on Day 2, so

b = 3.

Step 2:

The known scores are:

• Akhil: Day 1 = 7, Day 2 = x, Day 3 = 5, Day 4 = 3, Day 5 = p
• Bimal: Day 1 = 5, Day 2 = y, Day 3 = 5, Day 4 = 7, Day 5 = q
• Chatur: Day 1 = 3, Day 2 = 9, Day 3 = 6, Day 4 = 6, Day 5 = 6

Since Chatur's score of 9 is the unique highest score,

9 > x > y

and

q > 6 > p.

Using the daily totals,

x + y = 6

and

p + q = 12.

The possible values are:

• (x, y) = (5, 1) or (4, 2)
• (p, q) = (5, 7) or (4, 8)

Step 1:

Let the scores on Day 1, Day 2, Day 3, Day 4, and Day 5 be D1, D2, D3, D4, and D5, respectively.

From the given information,

• D1 + D2 = 30
• D2 + D3 = 31
• D3 + D4 = 32
• D4 + D5 = 34

Using condition (2),

D3 = D4 = 16

Therefore,

• D1 = 15
• D2 = 15
• D3 = 16
• D4 = 16
• D5 = 18

From conditions (1) and (3), let the unknown scores be represented as a, b, and c.

Since the total score on Day 3 is 16,

2a + c = 16

Also, c > a.

The possible pairs are (4, 8) and (5, 6).

Since all of Chatur's scores are multiples of 3, the only valid choice is:

a = 5, c = 6

This also implies that Chatur's unique highest score occurs on Day 2, so

b = 3.

Step 2:

The known scores are:

• Akhil: Day 1 = 7, Day 2 = x, Day 3 = 5, Day 4 = 3, Day 5 = p
• Bimal: Day 1 = 5, Day 2 = y, Day 3 = 5, Day 4 = 7, Day 5 = q
• Chatur: Day 1 = 3, Day 2 = 9, Day 3 = 6, Day 4 = 6, Day 5 = 6

Since Chatur's score of 9 is the unique highest score,

9 > x > y

and

q > 6 > p.

Using the daily totals,

x + y = 6

and

p + q = 12.

The possible values are:

• (x, y) = (5, 1) or (4, 2)
• (p, q) = (5, 7) or (4, 8)

25 is the minimum possible total score of Bimal when y = 1 and q = 7.

Step 1:

Let the scores on Day 1, Day 2, Day 3, Day 4, and Day 5 be D1, D2, D3, D4, and D5, respectively.

From the given information,

• D1 + D2 = 30
• D2 + D3 = 31
• D3 + D4 = 32
• D4 + D5 = 34

Using condition (2),

D3 = D4 = 16

Therefore,

• D1 = 15
• D2 = 15
• D3 = 16
• D4 = 16
• D5 = 18

From conditions (1) and (3), let the unknown scores be represented as a, b, and c.

Since the total score on Day 3 is 16,

2a + c = 16

Also, c > a.

The possible pairs are (4, 8) and (5, 6).

Since all of Chatur's scores are multiples of 3, the only valid choice is:

a = 5, c = 6

This also implies that Chatur's unique highest score occurs on Day 2, so

b = 3.

Step 2:

The known scores are:

• Akhil: Day 1 = 7, Day 2 = x, Day 3 = 5, Day 4 = 3, Day 5 = p
• Bimal: Day 1 = 5, Day 2 = y, Day 3 = 5, Day 4 = 7, Day 5 = q
• Chatur: Day 1 = 3, Day 2 = 9, Day 3 = 6, Day 4 = 6, Day 5 = 6

Since Chatur's score of 9 is the unique highest score,

9 > x > y

and

q > 6 > p.

Using the daily totals,

x + y = 6

and

p + q = 12.

The possible values are:

• (x, y) = (5, 1) or (4, 2)
• (p, q) = (5, 7) or (4, 8)

Total score of Bimal is a multiple of 3 when y = 2 and q = 8.

Then, score of Akhil on Day 2 will be 4.

Step 1:

Let the scores on Day 1, Day 2, Day 3, Day 4, and Day 5 be D1, D2, D3, D4, and D5, respectively.

From the given information,

• D1 + D2 = 30
• D2 + D3 = 31
• D3 + D4 = 32
• D4 + D5 = 34

Using condition (2),

D3 = D4 = 16

Therefore,

• D1 = 15
• D2 = 15
• D3 = 16
• D4 = 16
• D5 = 18

From conditions (1) and (3), let the unknown scores be represented as a, b, and c.

Since the total score on Day 3 is 16,

2a + c = 16

Also, c > a.

The possible pairs are (4, 8) and (5, 6).

Since all of Chatur's scores are multiples of 3, the only valid choice is:

a = 5, c = 6

This also implies that Chatur's unique highest score occurs on Day 2, so

b = 3.

Step 2:

The known scores are:

• Akhil: Day 1 = 7, Day 2 = x, Day 3 = 5, Day 4 = 3, Day 5 = p
• Bimal: Day 1 = 5, Day 2 = y, Day 3 = 5, Day 4 = 7, Day 5 = q
• Chatur: Day 1 = 3, Day 2 = 9, Day 3 = 6, Day 4 = 6, Day 5 = 6

Since Chatur's score of 9 is the unique highest score,

9 > x > y

and

q > 6 > p.

Using the daily totals,

x + y = 6

and

p + q = 12.

The possible values are:

• (x, y) = (5, 1) or (4, 2)
• (p, q) = (5, 7) or (4, 8)

Step 1:

Let us determine the annual amount of money raised by each firm.

Brzygoo:

There are two possible distributions of the funds raised (in crores):

Case 1

• 2012: 1
• 2013: 2
• 2014: 3
• 2015: 1

Case 2

• 2012: 1
• 2013: 3
• 2014: 2
• 2015: 1

Czechy:

The amounts raised are:

• 2013: 1
• 2014: a
• 2015: b
• 2016: c
• 2017: 1

Since the total amount raised is 9 crores,

a + b + c = 9 − 2 = 7

The only possible values satisfying this condition are:

(a, b, c) = (2, 3, 2)

(No other arrangement is possible, as changing the last year of existence would not satisfy the total amount of 7.)

Drjbna:

The yearly amounts raised (in crores) are:

• 2011: 1
• 2012: 2
• 2013: 4
• 2014: 2
• 2015: 1



Hence, the exact yearly amounts can be uniquely determined only for Czechy and Drjbna.

Step 1:

Let us determine the possible annual amounts raised by Alfloo and Elavalaki.

Alfloo:

Assume the yearly amounts raised (in crores) are:

• 2009: 1
• 2010: 3
• 2011: 4
• 2012: 5
• 2013: 4
• 2014: 3
• 2015: 2
• 2016: 1

This gives the minimum possible amount in 2010, assuming 2015 = 2 crores.

However, the total comes to 23 crores, which is not feasible.

Therefore, the amount raised in 2010 must increase by 1 crore.

The possible distributions are:

Case 1

• 2009: 1
• 2010: 2
• 2011: 3
• 2012: 4
• 2013: 5
• 2014: 3
• 2015: 2
• 2016: 1

Case 2

• 2009: 1
• 2010: 2
• 2011: 3
• 2012: 5
• 2013: 4
• 2014: 3
• 2015: 2
• 2016: 1

Note: If the amount raised in 2011 were 4 crores, the minimum possible total would be 22 crores (sequence 1-2-4-5-4-3-2-1), which is also not feasible.

Elavalaki:

The possible yearly amounts are:

Case 1

Sub-case 1

• 2010: 1
• 2011: 2
• 2012: 3
• 2013: 4
• 2014: 2
• 2015: 1

Sub-case 2

• 2010: 1
• 2011: 2
• 2012: 4
• 2013: 3
• 2014: 2
• 2015: 1

Case 2

• 2010: 1
• 2011: 3
• 2012: 5
• 2013: 3
• 2014: 1

Hence, the amount of money raised in 2015 could be:

= 2 + 1 + 3 + 1 + 1 or 0

= 7 crores or 8 crores.

The largest possible amount of money raised in 2013 (in Rs. crores) by each firm is:

• Alfloo: 5
• Brzygoo: 3
• Czechy: 1
• Drjbna: 4
• Elavalaki: 4

Therefore, the maximum total amount raised in 2013 is:

= 5 + 3 + 1 + 4 + 4

= 17 crores

Hence, the required answer is 17 crores.

If Elavalaki raised Rs. 3 crore in 2013, the following two cases are possible:

Case 1

• 2010: 1
• 2011: 3
• 2012: 5
• 2013: 3
• 2014: 1
• 2015: 0

Case 2

• 2010: 1
• 2011: 2
• 2012: 4
• 2013: 3
• 2014: 2
• 2015: 1



Accordingly, the minimum possible amount of money raised by all five companies in 2012 is:

= 4 + 1 + 0 + 2 + 4

= 11 crores.

If the total amount of money raised in 2014 is Rs. 12 crores, the possible amounts raised by the firms in 2014 are:

• Alfloo: 3
• Brzygoo: 2 or 3
• Czechy: 2
• Drjbna: 2
• Elavalaki: 2 or 1

These values add up to the required total of 12 crores.

Under this arrangement, the statement "Brzygoo raised the same amount of money as Elavalaki in 2013" cannot be satisfied.

Hence, the given statement is not possible.

Step 1:

Ullas has an average rating of 2.2, so his total score is:

2.2 × 5 = 11

Since his median rating is 2, the middle rating must be 2.

As his mode is also 2, another rating must be 2.

Further, the range is 3, so the remaining two ratings must be 1 and 4 or 2 and 5. The only valid set is:

(1, 2, 2, 2, 4)

Similarly, Vasu has an average rating of 3.8, giving a total score of:

3.8 × 5 = 19

Since the median and mode are both 4, and the range is 3, the only possible ratings are:

(2, 4, 4, 4, 5)

Applying the same logic:

• Waman: (1, 2, 4, 5, 5)
• Xavier: (1, 3, 4, 5, 5)
• Yusuf: (1, 1, 3, 4, 4)

Step 2:

The average ratings awarded by R1, R2, R3, R4, and R5 are 3.4, 2.2, 3.8, 2.8, and 3.4, respectively.

Hence, their total ratings are:

• R1 = 17
• R2 = 11
• R3 = 19
• R4 = 14
• R5 = 17

Using conditions (a) and (b), the partial assignments are:

• Ullas: R1 = 1
• Vasu: R4 = 5
• Waman: R1 = 5, R2 = 1, R3 = 5
• Xavier: R2 = 5, R3 = 5
• Yusuf: R2 = 1, R3 = 1

Since the total rating awarded by R2 is 11, the ratings given by R2 to Ullas and Vasu must be 2 and 2, respectively.

Similarly, the total rating awarded by R3 is 19, so the ratings given by R3 to Ullas and Vasu must be 4 and 4, respectively.

Filling in the remaining values gives:

• Ullas: R1 = 1, R2 = 2, R3 = 4, R4 = 2, R5 = 2
• Vasu: R1 = 4, R2 = 2, R3 = 4, R4 = 5, R5 = 4
• Waman: R1 = 5, R2 = 1, R3 = 5, R4 = 4, R5 = 2
• Xavier: R1 = 3, R2 = 5, R3 = 5, R4 = 1, R5 = 4
• Yusuf: R1 = 4, R2 = 1, R3 = 1, R4 = 3, R5 = 4

From the completed rating assignment, every individual rating can be determined uniquely using the given information.

Therefore, no individual rating remains undetermined.

Hence, the correct answer is 0.

Step 1:

Ullas has an average rating of 2.2, so his total score is:

2.2 × 5 = 11

Since his median rating is 2, the middle rating must be 2.

As his mode is also 2, another rating must be 2.

Further, the range is 3, so the remaining two ratings must be 1 and 4 or 2 and 5. The only valid set is:

(1, 2, 2, 2, 4)

Similarly, Vasu has an average rating of 3.8, giving a total score of:

3.8 × 5 = 19

Since the median and mode are both 4, and the range is 3, the only possible ratings are:

(2, 4, 4, 4, 5)

Applying the same logic:

• Waman: (1, 2, 4, 5, 5)
• Xavier: (1, 3, 4, 5, 5)
• Yusuf: (1, 1, 3, 4, 4)

Step 2:

The average ratings awarded by R1, R2, R3, R4, and R5 are 3.4, 2.2, 3.8, 2.8, and 3.4, respectively.

Hence, their total ratings are:

• R1 = 17
• R2 = 11
• R3 = 19
• R4 = 14
• R5 = 17

Using conditions (a) and (b), the partial assignments are:

• Ullas: R1 = 1
• Vasu: R4 = 5
• Waman: R1 = 5, R2 = 1, R3 = 5
• Xavier: R2 = 5, R3 = 5
• Yusuf: R2 = 1, R3 = 1

Since the total rating awarded by R2 is 11, the ratings given by R2 to Ullas and Vasu must be 2 and 2, respectively.

Similarly, the total rating awarded by R3 is 19, so the ratings given by R3 to Ullas and Vasu must be 4 and 4, respectively.

Filling in the remaining values gives:

• Ullas: R1 = 1, R2 = 2, R3 = 4, R4 = 2, R5 = 2
• Vasu: R1 = 4, R2 = 2, R3 = 4, R4 = 5, R5 = 4
• Waman: R1 = 5, R2 = 1, R3 = 5, R4 = 4, R5 = 2
• Xavier: R1 = 3, R2 = 5, R3 = 5, R4 = 1, R5 = 4
• Yusuf: R1 = 4, R2 = 1, R3 = 1, R4 = 3, R5 = 4

R2 did not give a rating of 4 to any workers.

Step 1:

Ullas has an average rating of 2.2, so his total score is:

2.2 × 5 = 11

Since his median rating is 2, the middle rating must be 2.

As his mode is also 2, another rating must be 2.

Further, the range is 3, so the remaining two ratings must be 1 and 4 or 2 and 5. The only valid set is:

(1, 2, 2, 2, 4)

Similarly, Vasu has an average rating of 3.8, giving a total score of:

3.8 × 5 = 19

Since the median and mode are both 4, and the range is 3, the only possible ratings are:

(2, 4, 4, 4, 5)

Applying the same logic:

• Waman: (1, 2, 4, 5, 5)
• Xavier: (1, 3, 4, 5, 5)
• Yusuf: (1, 1, 3, 4, 4)

Step 2:

The average ratings awarded by R1, R2, R3, R4, and R5 are 3.4, 2.2, 3.8, 2.8, and 3.4, respectively.

Hence, their total ratings are:

• R1 = 17
• R2 = 11
• R3 = 19
• R4 = 14
• R5 = 17

Using conditions (a) and (b), the partial assignments are:

• Ullas: R1 = 1
• Vasu: R4 = 5
• Waman: R1 = 5, R2 = 1, R3 = 5
• Xavier: R2 = 5, R3 = 5
• Yusuf: R2 = 1, R3 = 1

Since the total rating awarded by R2 is 11, the ratings given by R2 to Ullas and Vasu must be 2 and 2, respectively.

Similarly, the total rating awarded by R3 is 19, so the ratings given by R3 to Ullas and Vasu must be 4 and 4, respectively.

Filling in the remaining values gives:

• Ullas: R1 = 1, R2 = 2, R3 = 4, R4 = 2, R5 = 2
• Vasu: R1 = 4, R2 = 2, R3 = 4, R4 = 5, R5 = 4
• Waman: R1 = 5, R2 = 1, R3 = 5, R4 = 4, R5 = 2
• Xavier: R1 = 3, R2 = 5, R3 = 5, R4 = 1, R5 = 4
• Yusuf: R1 = 4, R2 = 1, R3 = 1, R4 = 3, R5 = 4

R1 gave a rating of 3 to Xavier.

Step 1:

Ullas has an average rating of 2.2, so his total score is:

2.2 × 5 = 11

Since his median rating is 2, the middle rating must be 2.

As his mode is also 2, another rating must be 2.

Further, the range is 3, so the remaining two ratings must be 1 and 4 or 2 and 5. The only valid set is:

(1, 2, 2, 2, 4)

Similarly, Vasu has an average rating of 3.8, giving a total score of:

3.8 × 5 = 19

Since the median and mode are both 4, and the range is 3, the only possible ratings are:

(2, 4, 4, 4, 5)

Applying the same logic:

• Waman: (1, 2, 4, 5, 5)
• Xavier: (1, 3, 4, 5, 5)
• Yusuf: (1, 1, 3, 4, 4)

Step 2:

The average ratings awarded by R1, R2, R3, R4, and R5 are 3.4, 2.2, 3.8, 2.8, and 3.4, respectively.

Hence, their total ratings are:

• R1 = 17
• R2 = 11
• R3 = 19
• R4 = 14
• R5 = 17

Using conditions (a) and (b), the partial assignments are:

• Ullas: R1 = 1
• Vasu: R4 = 5
• Waman: R1 = 5, R2 = 1, R3 = 5
• Xavier: R2 = 5, R3 = 5
• Yusuf: R2 = 1, R3 = 1

Since the total rating awarded by R2 is 11, the ratings given by R2 to Ullas and Vasu must be 2 and 2, respectively.

Similarly, the total rating awarded by R3 is 19, so the ratings given by R3 to Ullas and Vasu must be 4 and 4, respectively.

Filling in the remaining values gives:

• Ullas: R1 = 1, R2 = 2, R3 = 4, R4 = 2, R5 = 2
• Vasu: R1 = 4, R2 = 2, R3 = 4, R4 = 5, R5 = 4
• Waman: R1 = 5, R2 = 1, R3 = 5, R4 = 4, R5 = 2
• Xavier: R1 = 3, R2 = 5, R3 = 5, R4 = 1, R5 = 4
• Yusuf: R1 = 4, R2 = 1, R3 = 1, R4 = 3, R5 = 4

The median of the ratings given by R3 to the five workers is 4.

Step 1:

Ullas has an average rating of 2.2, so his total score is:

2.2 × 5 = 11

Since his median rating is 2, the middle rating must be 2.

As his mode is also 2, another rating must be 2.

Further, the range is 3, so the remaining two ratings must be 1 and 4 or 2 and 5. The only valid set is:

(1, 2, 2, 2, 4)

Similarly, Vasu has an average rating of 3.8, giving a total score of:

3.8 × 5 = 19

Since the median and mode are both 4, and the range is 3, the only possible ratings are:

(2, 4, 4, 4, 5)

Applying the same logic:

• Waman: (1, 2, 4, 5, 5)
• Xavier: (1, 3, 4, 5, 5)
• Yusuf: (1, 1, 3, 4, 4)

Step 2:

The average ratings awarded by R1, R2, R3, R4, and R5 are 3.4, 2.2, 3.8, 2.8, and 3.4, respectively.

Hence, their total ratings are:

• R1 = 17
• R2 = 11
• R3 = 19
• R4 = 14
• R5 = 17

Using conditions (a) and (b), the partial assignments are:

• Ullas: R1 = 1
• Vasu: R4 = 5
• Waman: R1 = 5, R2 = 1, R3 = 5
• Xavier: R2 = 5, R3 = 5
• Yusuf: R2 = 1, R3 = 1

Since the total rating awarded by R2 is 11, the ratings given by R2 to Ullas and Vasu must be 2 and 2, respectively.

Similarly, the total rating awarded by R3 is 19, so the ratings given by R3 to Ullas and Vasu must be 4 and 4, respectively.

Filling in the remaining values gives:

• Ullas: R1 = 1, R2 = 2, R3 = 4, R4 = 2, R5 = 2
• Vasu: R1 = 4, R2 = 2, R3 = 4, R4 = 5, R5 = 4
• Waman: R1 = 5, R2 = 1, R3 = 5, R4 = 4, R5 = 2
• Xavier: R1 = 3, R2 = 5, R3 = 5, R4 = 1, R5 = 4
• Yusuf: R1 = 4, R2 = 1, R3 = 1, R4 = 3, R5 = 4

Restaurant R4 gave its median rating to exactly one of the workers.

Let the GDP of Country 9 be x and its per capita GDP be y.

Then, the population of each country can be expressed as a multiple of x/y:

• Country 5: 10/36
• Country 8: 7/41
• Country 3: 13/20
• Country 7: 8/30

Comparing these fractions, 7/41 is the smallest.

Hence, Country 8 had the smallest population in 2024.

Let the GDP of Country 9 be x and the per capita GDP of Country 10 be y.

The ratio of the GDPs of Country 4 and Country 5 in 2026 is:

= (0.12 × 1.005²) / (0.10 × 1.007²)

≈ 1.195

Hence, the required ratio is 1.195.

Let the GDP of Country 9 be x and the per capita GDP of Country 10 be y.

The population of each of the given countries can be expressed as a multiple of x/y:

• Country 4: (12/38) × 1.005³
• Country 1: (15/41) × 0.999³
• Country 7: (8/30) × 0.999³
• Country 5: (10/36) × 1.003¹³

Comparing these values, Country 1 has the largest population.

Hence, the correct answer is Country 1.

Let the GDP of Country 9 be x and the per capita GDP of Country 10 be y.

For each country listed, the GDP growth rate exceeds the population growth rate.

Since the growth in GDP is higher than the growth in population, the per capita GDP of every listed country will be higher in 2027 than it was in 2024.

Hence, none of the given countries will have a lower per capita GDP in 2027 compared to 2024.