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Logical Reasoning & DI (LRDI) Test - 29

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Logical Reasoning & DI (LRDI) Test - 29
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  • Question 1
    3 / -1

    Directions For Questions

    Read the passage below and answer the questions that follow

    Following is a spread-sheet with 7 rows (numbered from 1 to 7 from top to bottom as shown) and 6 columns (columns are identified with letters of English alphabet from left to right in alphabetical order). Each of the 42 cells has to be filled with a single- digit number (including 0) so that the summation of these 42 single-digit numbers becomes 100. Among 42 cells, 30 have been filled and 12 are empty. The empty cells should be filled with either 0 or 1, only.

    The additional information which is known about the spread - sheet are as follows:

    I. Summation of the six numbers in the 4th row is 2 more than the summation of the six numbers in the 6th row

    II. Summation of the six numbers in the 4th row is 5 less than the summation of the six numbers in the 1st row

    III. Summation of the seven numbers in the 1st column (marked with ‘A’) is equal to summation of the seven numbers in the 4th column (marked with ‘D’)

    IV. Summation of the seven numbers in the 2nd column (marked with ‘B’) is 8 more than the summation of the six numbers in the 2nd row

    V. Summation of the six numbers in the 2nd row is 1 more than the summation of the six numbers in the 3rd row

    Note: columns are counted from left to right and rows are counted from top to bottom.

    ...view full instructions

    How many values of ‘n’, is the summation of the six numbers in nth row more than the summation of seven numbers in nth column (n can take any value from 1 to 6)?

    Solution

    Let the summation 6th row = X

    So, summation 4th row = X + 2 and summation 1st row = X + 7 Solving, X + 7 = 17 => X = 10 and X + 2 = 12

    Now for 6th row, sum of given values = 3 + 2 + 1 + 2 = 8 and total sum for 6th row = X = 10. So, only possible value of A6 and D6. = 1 each.

    Similarly, for 4th row, sum of given values = 3 + 2 + 3 + 2 = 10 and total sum of 4th row = X + 2 = 12. So, only possible value of A4 and E4 = 1 each

    Let summation 2nd row = Z

    So, summation 2nd column (B) = Z + 8 and summation 3rd row = Z – 1 Solving, Z + 8 = 24 => Z = 16 and Z – 1 = 15

    Now for 2nd row, sum of given values = 4 + 3 + 5 + 4 = 16 and total sum for 2nd row = Z = 16. So, only possible value of A2 and D2 = 0 each

    Now, let the summation 1st column (A) = Y = summation 4th column (D) Now for column A, sum of given values = 2 + 0 + 2 + 1 + 1 + 3 = 9 with one cell missing and for column D, sum of given values = 2+ 0 + 3 + 1 + 4 = 10 with two cells missing

    So, the maximum value of column A can be 9 + 1 (missing cell) = 10 = Y such that both the missing cells of column D = 0 each (only possibility) such that total sum remains same as column A

    Now for 3rd row, sum of given values = 2 + 5 + 4 + 0 + 3 = 14 and total sum for 3rd row = Z – 1 = 15. So, only possible value of E3 = 1 each

    So, the sum of column E = 3 + 5 + 1 + 1 + 3 + 1 + 4 = 18

    Now, the sum of solved rows = 17 + 16 + 15 + 12 + 10 + 22 = 92 So, the sum of 5th row = 100 – 92 = 8

    Sum of given values of 5th row = 1 + 2 + 0 + 3 = 6 So, the values of C5 and F5 = 1 each

    The rest of the information can be gathered as follows-

     

  • Question 2
    3 / -1

    Directions For Questions

    Read the passage below and answer the questions that follow

    Following is a spread-sheet with 7 rows (numbered from 1 to 7 from top to bottom as shown) and 6 columns (columns are identified with letters of English alphabet from left to right in alphabetical order). Each of the 42 cells has to be filled with a single- digit number (including 0) so that the summation of these 42 single-digit numbers becomes 100. Among 42 cells, 30 have been filled and 12 are empty. The empty cells should be filled with either 0 or 1, only.

    The additional information which is known about the spread - sheet are as follows:

    I. Summation of the six numbers in the 4th row is 2 more than the summation of the six numbers in the 6th row

    II. Summation of the six numbers in the 4th row is 5 less than the summation of the six numbers in the 1st row

    III. Summation of the seven numbers in the 1st column (marked with ‘A’) is equal to summation of the seven numbers in the 4th column (marked with ‘D’)

    IV. Summation of the seven numbers in the 2nd column (marked with ‘B’) is 8 more than the summation of the six numbers in the 2nd row

    V. Summation of the six numbers in the 2nd row is 1 more than the summation of the six numbers in the 3rd row

    Note: columns are counted from left to right and rows are counted from top to bottom.

    ...view full instructions

    Which of the following is the correct sequence of numbers representing totals of column A to column F?

    Solution

    Let the summation 6th row = X

    So, summation 4th row = X + 2 and summation 1st row = X + 7 Solving, X + 7 = 17 => X = 10 and X + 2 = 12

    Now for 6th row, sum of given values = 3 + 2 + 1 + 2 = 8 and total sum for 6th row = X = 10. So, only possible value of A6 and D6.

    = 1 each.

    Similarly, for 4th row, sum of given values = 3 + 2 + 3 + 2 = 10 and total sum of 4th row = X + 2 = 12. So, only possible value of A4 and E4 = 1 each

    Let summation 2nd row = Z

    So, summation 2nd column (B) = Z + 8 and summation 3rd row = Z – 1 Solving, Z + 8 = 24 => Z = 16 and Z – 1 = 15

    Now for 2nd row, sum of given values = 4 + 3 + 5 + 4 = 16 and total sum for 2nd row = Z = 16. So, only possible value of A2 and D2 = 0 each

    Now, let the summation 1st column (A) = Y = summation 4th column (D) Now for column A, sum of given values = 2 + 0 + 2 + 1 + 1 + 3 = 9 with one cell missing and for column D, sum given values = 2 + 0 + 3 + 1 + 4 = 10 with two cells missing

    So, the maximum value of column A can be 9 + 1 (missing cell) = 10 = Y such that both the missing cells of column D = 0 each (only possibility) such that total sum remains same as column A

    Now for 3rd row, sum of given values = 2 + 5 + 4 + 0 + 3 = 14 and total sum for 3rd row = Z – 1 = 15. So, only possible value of E3 = 1 each

    So, the sum of column E = 3 + 5 + 1 + 1 + 3 + 1 + 4 = 18

    Now, the sum of solved rows = 17 + 16 + 15 + 12 + 10 + 22 = 92 So, the sum of 5th row = 100 – 92 = 8

    Sum of given values of 5th row = 1 + 2 + 0 + 3 = 6 So, the values of C5 and F5 = 1 each

    The rest of the information can be gathered as follows-

    Correct order of sum of columns from A to F is 10, 24, 20, 10, 18, 18

     

  • Question 3
    3 / -1

    Directions For Questions

    Read the passage below and answer the questions that follow

    Following is a spread-sheet with 7 rows (numbered from 1 to 7 from top to bottom as shown) and 6 columns (columns are identified with letters of English alphabet from left to right in alphabetical order). Each of the 42 cells has to be filled with a single- digit number (including 0) so that the summation of these 42 single-digit numbers becomes 100. Among 42 cells, 30 have been filled and 12 are empty. The empty cells should be filled with either 0 or 1, only.

    The additional information which is known about the spread - sheet are as follows:

    I. Summation of the six numbers in the 4th row is 2 more than the summation of the six numbers in the 6th row

    II. Summation of the six numbers in the 4th row is 5 less than the summation of the six numbers in the 1st row

    III. Summation of the seven numbers in the 1st column (marked with ‘A’) is equal to summation of the seven numbers in the 4th column (marked with ‘D’)

    IV. Summation of the seven numbers in the 2nd column (marked with ‘B’) is 8 more than the summation of the six numbers in the 2nd row

    V. Summation of the six numbers in the 2nd row is 1 more than the summation of the six numbers in the 3rd row

    Note: columns are counted from left to right and rows are counted from top to bottom.

    ...view full instructions

    Which of the following is the correct order of the row numbers in decreasing order of their row summation

    Solution

    Let the summation 6th row = X

    So, summation 4th row = X + 2 and summation 1st row = X + 7 Solving, X + 7 = 17 => X = 10 and X + 2 = 12

    Now for 6th row, sum of given values = 3 + 2 + 1 + 2 = 8 and total sum for 6th row = X = 10. So, only possible value of A6 and D6. = 1 each.

    Similarly, for 4th row, sum of given values = 3 + 2 + 3 + 2 = 10 and total sum of 4th row = X + 2 = 12. So, only possible value of A4 and E4 = 1 each

    Let summation 2nd row = Z

    So, summation 2nd column (B) = Z + 8 and summation 3rd row = Z – 1 Solving, Z + 8 = 24 => Z = 16 and Z – 1 = 15

    Now for 2nd row, sum of given values = 4 + 3 + 5 + 4 = 16 and total sum for 2nd row = Z = 16. So, only possible value of A2 and D2 = 0 each

    Now, let the summation 1st column (A) = Y = summation 4th column (D) Now for column A, sum of given values = 2 + 0 + 2 + 1 + 1 + 3 = 9 with one cell missing and for column D, sum of given values = 2 + 0 + 3 + 1 + 4 = 10 with two cells missing

    So, the maximum value of column A can be 9 + 1 (missing cell) = 10 = Y such that both the missing cells of column D = 0 each (only possibility) such that total sum remains same as column A

    Now for 3rd row, sum of given values = 2 + 5 + 4 + 0 + 3 = 14 and total sum for 3rd row = Z – 1 = 15. So, only possible value of E3 = 1 each

    So, the sum of column E = 3 + 5 + 1 + 1 + 3 + 1 + 4 = 18

    Now, the sum of solved rows = 17 + 16 + 15 + 12 + 10 + 22 = 92 So, the sum of 5th row = 100 – 92 = 8

    Sum of given values of 5th row = 1 + 2 + 0 + 3 = 6 So, the values of C5 and F5 = 1 each

    The rest of the information can be gathered as follows-

    Decreasing order of their row sum is 7, 1, 2, 3, 4, 6, 5

     

  • Question 4
    3 / -1

    Directions For Questions

    Read the passage below and answer the questions that follow

    Three participants – Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants’ scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for Days 2 through 5.

    Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2.

    The following information is also known.

    I. Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day4

    II. The total score on Day 3 is the same as the total score on Day 4.

    III. Bimal’s scores are the same on Day 1 and Day 3.

    ...view full instructions

    What is Akhil's score on Day 1?

    Solution

    Let the total score of Day 1, Day 2, Day 3, Day 4 and Day 5 of all the participants be a, b, c, d and e respectively. As per the Table 1,

    a + b = 15 × 2 = 30, b + c = 15.5 × 2 = 31,

    c + d = 16 × 2 = 32 and d + e = 17 × 2 = 34

    Point 2, total score on Day 3 = total score on Day 4

    ⇒ c = d = 16 (each)

    ⇒ e = 34 – 16 = 18, b = 31 – 16 = 15 and a = 30 – 15 = 15

    Point 1, Chatur score on any day = 3, 6 or 9

    The only possibility of his Day 2 score being unique highest = 9 His only minimum score on Day 1 = 3

    ⇒ Chatur’s score on Day 3, Day 4 and Day 5 = 6 (each) Also, Akhil’s Day 4 score = Chatur’s Day 1 score = 3

    ⇒ Bimal’s Day 4 score = 16 – (3 + 6) = 7

    Now being same rank, Akhil’s Day 3 score = Bimal’s Day 3 score = 5 (each) Point 3, Bimal’s Day 1 score = Bimal’s Day 3 score = 5

    ⇒ Akhil’s Day 1 score = 15 – (5 + 3) = 7

    Now for Day 2, let the score of Akhil = p and Bimal = q Solving, p + q + 9 = 15 ⇒ p + q = 6

    Since, the rank of Akhil is 2 and Bimal is 3 Possible values of p = 4 or 5 and q = 2 or 1

    Now for Day 5, let the score of Akhil = x and Bimal = y

    Solving, x + y + 6 = 18 ⇒ x + y = 12

    Since, the rank of Akhil is 3 and Bimal is 1 Possible value of x = 5 or 4 and y = 7 or 8

    (y cannot be 9 as that being unique highest)

    The rest of the information can be gathered as follows-

    Akhil’s score on Day 1 = 7

     

  • Question 5
    3 / -1

    Directions For Questions

    Read the passage below and answer the questions that follow

    Three participants – Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants’ scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for Days 2 through 5.

    Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2.

    The following information is also known.

    I. Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day4

    II. The total score on Day 3 is the same as the total score on Day 4.

    III. Bimal’s scores are the same on Day 1 and Day 3.

    ...view full instructions

    Who attains the maximum total score?

    Solution

    Let the total score of Day 1, Day 2, Day 3, Day 4 and Day 5 of all the participants be a, b, c, d and e respectively. As per the Table 1,

    a + b = 15 × 2 = 30, b + c = 15.5 × 2 = 31, c + d = 16 × 2 = 32 and d + e = 17 × 2 = 34

    Point 2, total score on Day 3 = total score on Day 4

    ⇒ c = d = 16 (each)

    ⇒ e = 34 – 16 = 18, b = 31 – 16 = 15 and a = 30 – 15 = 15

    Point 1, Chatur score on any day = 3, 6 or 9

    The only possibility of his Day 2 score being unique highest = 9 His only minimum score on Day 1 = 3

    ⇒ Chatur’s score on Day 3, Day 4 and Day 5 = 6 (each) Also, Akhil’s Day 4 score = Chatur’s Day 1 score = 3

    ⇒ Bimal’s Day 4 score = 16 – (3 + 6) = 7

    Now being same rank, Akhil’s Day 3 score = Bimal’s Day 3 score = 5 (each) Point 3, Bimal’s Day 1 score = Bimal’s Day 3 score = 5

    ⇒ Akhil’s Day 1 score = 15 – (5 + 3) = 7

    Now for Day 2, let the score of Akhil = p and Bimal = q Solving, p + q + 9 = 15 ⇒ p + q = 6

    Since, the rank of Akhil is 2 and Bimal is 3 Possible values of p = 4 or 5 and q = 2 or 1

    Now for Day 5, let the score of Akhil = x and Bimal = y

    Solving, x + y + 6 = 18 ⇒ x + y = 12

    Since, the rank of Akhil is 3 and Bimal is 1 Possible value of x = 5 or 4 and y = 7 or 8

    (y cannot be 9 as that being unique highest)

    The rest of the information can be gathered as follows-

    Chatur attains the maximum possible score

     

  • Question 6
    3 / -1

    Directions For Questions

    Read the passage below and answer the questions that follow

    Three participants – Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants’ scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for Days 2 through 5.

    Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2.

    The following information is also known.

    I. Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day4

    II. The total score on Day 3 is the same as the total score on Day 4.

    III. Bimal’s scores are the same on Day 1 and Day 3.

    ...view full instructions

    If the total score of Bimal is a multiple of 3, what is the score of Akhil on Day 2?

    Solution

    Let the total score of Day 1, Day 2, Day 3, Day 4 and Day 5 of all the participants be a, b, c, d and e respectively. As per the Table 1,

    a + b = 15 × 2 = 30, b + c = 15.5 × 2 = 31, c + d = 16 × 2 = 32 and d + e = 17 × 2 = 34

    Point 2, total score on Day 3 = total score on Day 4

    ⇒ c = d = 16 (each)

    ⇒ e = 34 – 16 = 18, b = 31 – 16 = 15 and a = 30 – 15 = 15

    Point 1, Chatur score on any day = 3, 6 or 9

    The only possibility of his Day 2 score being unique highest = 9 His only minimum score on Day 1 = 3

    ⇒ Chatur’s score on Day 3, Day 4 and Day 5 = 6 (each) Also, Akhil’s Day 4 score = Chatur’s Day 1 score = 3

    ⇒ Bimal’s Day 4 score = 16 – (3 + 6) = 7

    Now being same rank, Akhil’s Day 3 score = Bimal’s Day 3 score = 5 (each) Point 3, Bimal’s Day 1 score = Bimal’s Day 3 score = 5

    ⇒ Akhil’s Day 1 score = 15 – (5 + 3) = 7

    Now for Day 2, let the score of Akhil = p and Bimal = q Solving, p + q + 9 = 15 ⇒ p + q = 6

    Since, the rank of Akhil is 2 and Bimal is 3 Possible values of p = 4 or 5 and q = 2 or 1

    Now for Day 5, let the score of Akhil = x and Bimal = y

    Solving, x + y + 6 = 18 ⇒ x + y = 12

    Since, the rank of Akhil is 3 and Bimal is 1 Possible value of x = 5 or 4 and y = 7 or 8 (y cannot be 9 as that being unique highest)

    The rest of the information can be gathered as follows-

    If the total score of Bimal is a multiple of 3, the only possibility = 27 ⇒ Bimal’s Day 2 score = 2

    ⇒ Akhil’s Day 2 score = 4

     

  • Question 7
    3 / -1

    Directions For Questions

    Read the passage below and answer the questions that follow

    Marvel studios has hired 360 supporting actors for their multiple projects. Marvel has started production of three movies: The Journey, Blue Notice and Wonder Women. A supporting actor is called ‘A’ grade actor if he/she is eligible to work in all the three movies. He/ She is called ‘B’ grade actor if he/she is eligible to work in at least one but not all the three movies. Otherwise he/she is called ‘C’ grade actor. Production guy Mr. George also provides following information:

    I. 2 out of every 5 supporting actors are ‘A’ grade actors.

    II. (45)th(\frac{4}{5})^{th}(54)th of the total supporting actors are ‘C’ grade actors.

    III. 88 supporting actors are not eligible to work in ‘Wonder Women’.

    IV. 26 supporting actors are not eligible to work in ‘The Journey’ only but are eligible to work in other 2 movies.

    V. 120\frac{1}{20}201th^{th}th of the total supporting actors are eligible to work only in ‘Blue Notice.

    ...view full instructions

    What percentage of supporting actors who are eligible to work in ‘Wonder Women’ are also eligible to work in at most one more movie?

    Solution

     

  • Question 8
    3 / -1

    Directions For Questions

    Read the passage below and answer the questions that follow

    Marvel studios has hired 360 supporting actors for their multiple projects. Marvel has started production of three movies: The Journey, Blue Notice and Wonder Women. A supporting actor is called ‘A’ grade actor if he/she is eligible to work in all the three movies. He/ She is called ‘B’ grade actor if he/she is eligible to work in at least one but not all the three movies. Otherwise he/she is called ‘C’ grade actor. Production guy Mr. George also provides following information:

    I. 2 out of every 5 supporting actors are ‘A’ grade actors.

    II. (45)th(\frac{4}{5})^{th}(54)th of the total supporting actors are ‘C’ grade actors.

    III. 88 supporting actors are not eligible to work in ‘Wonder Women’.

    IV. 26 supporting actors are not eligible to work in ‘The Journey’ only but are eligible to work in other 2 movies.

    V. 120\frac{1}{20}201th^{th}th of the total supporting actors are eligible to work only in ‘Blue Notice.

    ...view full instructions

    What percentage of supporting actors who are eligible to work in The Journey are also eligible to work in Blue Notice?

    Solution

     

  • Question 9
    3 / -1

    Directions For Questions

    Read the passage below and answer the questions that follow

    Marvel studios has hired 360 supporting actors for their multiple projects. Marvel has started production of three movies: The Journey, Blue Notice and Wonder Women. A supporting actor is called ‘A’ grade actor if he/she is eligible to work in all the three movies. He/ She is called ‘B’ grade actor if he/she is eligible to work in at least one but not all the three movies. Otherwise he/she is called ‘C’ grade actor. Production guy Mr. George also provides following information:

    I. 2 out of every 5 supporting actors are ‘A’ grade actors.

    II. (45)th(\frac{4}{5})^{th}(54)th of the total supporting actors are ‘C’ grade actors.

    III. 88 supporting actors are not eligible to work in ‘Wonder Women’.

    IV. 26 supporting actors are not eligible to work in ‘The Journey’ only but are eligible to work in other 2 movies.

    V. 120\frac{1}{20}201th^{th}th of the total supporting actors are eligible to work only in ‘Blue Notice.

    ...view full instructions

    Find the number of ‘B’ grade actors.

    Solution

     

  • Question 10
    3 / -1

    Directions For Questions

    Read the passage below and answer the questions that follow

    Wankhede Stadium is an oval shaped cricket ground in Mumbai. The final match of the One Day International (ODI) Cricket World Cup was once played between India and Sri Lanka at this stadium. There were four stands Kapil, Gavaskar, Tendulkar and Ganguly - to the East, West, North and South of the ground respectively. People bought tickets for these stands and watched the match at the stadium.

    Some other information is given below:

    I. The seats in each stand were in rows. Row 2 was behind row 1, row 3 was behind row 2 … and so on in each stand.

    II. The number of rows in Kapil, Gavaskar, Tendulkar and Ganguly stands were 20, 12, 15 and 10 respectively.

    III. Each row of a particular stand had an equal number of seats.

    IV. The seat numbers of row 1 of a particular stand were 1, 2, 3, ..., n, where n was the total number of seats in row 1, the seat numbers of row 2 were n + 1, n + 2, n + 3, ..., 2n, the seat numbers of row 3 were 2n + 1, 2n + 2, 2n + 3, ..., 3n, and so on.

    V. The occupancy of Kapil, Gavaskar, Tendulkar and Ganguly stands were 55.55…%, 60%, 57.14…% and 53.84…% respectively.

    VI. Due to Covid-19 pandemic no two consecutive seats could be occupied in any row of all the stands.

    ...view full instructions

    If the cost per ticket for Ganguly stand was Rs 2,500 , then what was the total revenue (in Rs. lakh) from Ganguly stand?

    Solution

    From condition (v), the occupancy of Ganguly stand was 53.84…% = 7/13 i.e., 7 seats out of 13 seats were occupied. From condition (vi), no two consecutive seats could be occupied in any row. The number of seats in each row of this stand was 13. So in row 1, seats 1, 3, 5, 7, 9, 11 and 13 were occupied.

    From condition (ii), total number of rows in Ganguly stand was 10. Hence, the revenue from Ganguly stand was = 7 × 10 × 2500 = Rs. 1,75,000

     

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