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Logical Reasoning & DI (LRDI) Test - 7
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Logical Reasoning & DI (LRDI) Test - 7

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  • Question 1/20
    3 / -1

    Directions For Questions

    The hiring process for Capnigent company is going on in a college. For selection in various positions in a company, it conducts a printed test with four different divisions, each with a maximum of 100 marks.

    The following table gives the gross as well as the cut-off marks for each division fixed for different positions.

    If there is no mark given for a particular division, then it means that for that role, there is no individual cutoff in that division.

    A student will get the job only if he gets marks greater than or equal to the cut-off marks in each of the division and his aggregate marks are at least equal to the aggregate cut-off marks as specified by the company.

    ...view full instructions

    If a student is selected for exactly five roles what could be the minimum aggregate marks obtained by him? 

    Solutions

    The highest cutoff for each section is represented in the table below:

    We can calculate the minimum score he should get to clear the cutoff of each Division.

    The minimum marks he should get to clear the cut-off of each division = 98+97+95+96=386.

    This will automatically clear the cutoff for the aggregate marks. i.e 385 marks.

    Now, if a student is elected for exactly five roles, then he must have been failed to clear the cut-off of the individual division.

    Suppose the student scored 97 in Division 1, thus he failed to qualify for the role of Purchasing Manager.

    While he will qualify for all the other roles: Thus, the aggregate score will be 97+97+95+96=385.

    We cannot reduce the score in Division 1 as then he will not qualify for Quality control also. 

    Similarly, he can score 1 marks less than the cutoff for any division.

    Minimum score possible= 385.

  • Question 2/20
    3 / -1

    Directions For Questions

    The hiring process for Capnigent company is going on in a college. For selection in various positions in a company, it conducts a printed test with four different divisions, each with a maximum of 100 marks.

    The following table gives the gross as well as the cut-off marks for each division fixed for different positions.

    If there is no mark given for a particular division, then it means that for that role, there is no individual cutoff in that division.

    A student will get the job only if he gets marks greater than or equal to the cut-off marks in each of the division and his aggregate marks are at least equal to the aggregate cut-off marks as specified by the company.

    ...view full instructions

    If a student is not selected for any role what is the maximum aggregate score he could get?

    Solutions

    The highest cutoff for each section is represented in the table below:

    We can calculate the minimum score he should get to clear the cutoff of each Division.

    The minimum marks he should get to clear the cut-off of each division = 98+97+95+96=386.

    If the student failed to clear the highest cutoff of each division by 1 marks then he will score: 97+96+94+95=382.

    Thus he will fail to qualify the following roles: Operation Manager, Marketing Manager, Purchasing Manager, Data Analyst and even Quality Control.

    The only role he will qualify for will be of the Office Manager, thus he should get the aggregate score of 381.

  • Question 3/20
    3 / -1

    Directions For Questions

    The hiring process for Capnigent company is going on in a college. For selection in various positions in a company, it conducts a printed test with four different divisions, each with a maximum of 100 marks.

    The following table gives the gross as well as the cut-off marks for each division fixed for different positions.

    If there is no mark given for a particular division, then it means that for that role, there is no individual cutoff in that division.

    A student will get the job only if he gets marks greater than or equal to the cut-off marks in each of the division and his aggregate marks are at least equal to the aggregate cut-off marks as specified by the company.

    ...view full instructions

    If a student is selected for exactly three roles what could be the maximum aggregate marks obtained by him?

    Solutions

    The highest cutoff for each section is represented in the table below:

    We can calculate the minimum score he should get to clear the cutoff of each Division.

    The minimum marks he should get to clear the cut-off of each division = 98+97+95+96=386.

    If the student gets 100 in each division then he will qualify for each role.

    A student is elected for exactly three roles, thus he should fail to qualify for the divisional cutoff scores in order to get maximum aggregate marks.

    He should have scored: 97+96+100+95= 388 marks in order to qualify for the roles of Quality Control, Data Analyst and Office Manager.

  • Question 4/20
    3 / -1

    Directions For Questions

    The hiring process for Capnigent company is going on in a college. For selection in various positions in a company, it conducts a printed test with four different divisions, each with a maximum of 100 marks.

    The following table gives the gross as well as the cut-off marks for each division fixed for different positions.

    If there is no mark given for a particular division, then it means that for that role, there is no individual cutoff in that division.

    A student will get the job only if he gets marks greater than or equal to the cut-off marks in each of the division and his aggregate marks are at least equal to the aggregate cut-off marks as specified by the company.

    ...view full instructions

    If a student is selected for the role of Purchasing manager, while he failed to qualify for other roles. What can be the maximum overall score he could get?

    Solutions

    The highest cutoff for each section is represented in the table below:

    Suppose the student just qualifies for the role of Purchasing Manager:

    Then he should score: 98, 90, 92 in the first three divisions.

    The roles Operation manager, Quality Control, Office manager and data analyst are already excluded for him.

    For marketing, he should score less than 96 in division 4.

    Suppose he scores 95 in division 4, the overall score will be= 98+90+92+95= 375.

    Now we can increase the scores in the other divisions, such that he doesn't qualify for other roles, in order to maximize his aggregate score.

    Case, 100+ 94+92+95=381, In this case, he scored the maximum aggregate marks while got selected only for the role of Purchasing manager.

    The answer is option D.

  • Question 5/20
    3 / -1

    Directions For Questions

    The hiring process for Capnigent company is going on in a college. For selection in various positions in a company, it conducts a printed test with four different divisions, each with a maximum of 100 marks.

    The following table gives the gross as well as the cut-off marks for each division fixed for different positions.

    If there is no mark given for a particular division, then it means that for that role, there is no individual cutoff in that division.

    A student will get the job only if he gets marks greater than or equal to the cut-off marks in each of the division and his aggregate marks are at least equal to the aggregate cut-off marks as specified by the company.

    ...view full instructions

    If a student gets selected, only for Operation Manager & Purchasing Manager, find the maximum possible score in Division 4?

    Solutions

    For a student to get selected for Operations ad Purchasing, he has to clear minimum cut-offs in all divisions and overall cut-off.

    i.e 98 + 97 + 92 + 87 = 374. The overall score should be more than or equal to 380 for Operations. 

    Student cleared divisional cutoffs of Quality control and Office manager. Therefore his total score should be below 382 (which is less than 385-Quality).

    Student's scores in Divisions 3 & 4 should be less than 95 and 90 so as not to get qualified for Data analyst.

    Student's scores in Division 4 should be less than 96 so as not to get qualified for the marketing manager.

    We need the maximum values of Division 3 ad Division 4. The highest possible overall marks of the student are 381.

    a+b+c+d=381 where a≥ 98, b≥ 97, d< 96a+b+c+d=381\ where\ a\ge\ 98,\ b\ge\ 97,\ d<\ 96

    The condition for C varies.

    Case 1

    If   96>d≥ 90\ \ 96>d\ge\ 90, then 92≤ c<9592\le\ c<95

    98 + 97 + c + d = 381 

    If c = 94. d = 92

    98 + 97 + c + d = 381

    If c = 92. d = 94

    Case 2

    If   90>d ≥ 87\ \ 90>d\ \ge\ 87, then 92≤ c≤ 10092\le\ c\le\ 100

    98 + 97 + c + d = 381

    If d = 87, then c = 99

    Maximum possible score in division 4 is 94.

  • Question 6/20
    3 / -1

    Directions For Questions

    The CEE (Common Entrance Exam) in Philippines has 4 levels. A candidate clears a level and proceeds to the next level only if he gets at least one question correct in that level. Candidates have to attempt all the questions in each level. Each level consists of 3 questions but the number of questions in subsequent levels can be increased based on the performance of a candidate in this level as follows:

     

    1. If a candidate answers all the questions correctly, the paper in the next level has two questions more than the number of questions that the candidate had in this level.
    2. If a candidate answers at least half of the number of questions in a particular level correctly, then in the next level, that candidate will have one question more than what he had in this level.
    3. If a candidate answers less than half the number of questions in a particular level correctly, then in the next level, that candidate will have the same number of questions as in this level.

     

    Each question is awarded 3 marks if answered correctly and -1 mark if answered incorrectly.

     

    4 friends - A, B, C, D who took the CEE are comparing their experiences.

    ...view full instructions

    A scored a net total of 7 marks in the level 2 exam. On the whole, he attempted not more than 20 questions in all the levels put together. What is the maximum possible net score of A in level 3 exam?

    Solutions

    A net score of 7 marks is possible if A attempts 3 questions correctly and 2 questions incorrectly. So, level 2 exam should have 5 questions. So, level 3 exam would have 6 questions. The total number of question in level 1, 2 and 3 put together is 3+5+6 = 14. Since A should not attempt more than 20 questions in all the level put together, he would have 6 questions in level 4. So, he would have to answer a maximum of 2 questions correctly in level 3. So, 4 incorrect questions. The score in level 3 is therefore, 2*3 - 4*1 = 2 marks.

  • Question 7/20
    3 / -1

    Directions For Questions

    The CEE (Common Entrance Exam) in Philippines has 4 levels. A candidate clears a level and proceeds to the next level only if he gets at least one question correct in that level. Candidates have to attempt all the questions in each level. Each level consists of 3 questions but the number of questions in subsequent levels can be increased based on the performance of a candidate in this level as follows:

     

    1. If a candidate answers all the questions correctly, the paper in the next level has two questions more than the number of questions that the candidate had in this level.
    2. If a candidate answers at least half of the number of questions in a particular level correctly, then in the next level, that candidate will have one question more than what he had in this level.
    3. If a candidate answers less than half the number of questions in a particular level correctly, then in the next level, that candidate will have the same number of questions as in this level.

     

    Each question is awarded 3 marks if answered correctly and -1 mark if answered incorrectly.

     

    4 friends - A, B, C, D who took the CEE are comparing their experiences.

    ...view full instructions

    If B’s net score in level 3 exam is 10 marks, then his score in level 2 exams cannot be

    Solutions

    A net score of 10 marks means 4 correct questions and 2 incorrect questions.
    So, the level 3 exam has 6 questions.
    Level 2 exam can have either 4 questions or 5 questions.
    Case1: When the level 2 exam has 4 questions.
    If he answers 2 correctly, he will score 4.In level 3 there will be 5 questions. Hence it can be negated.
    If he answers 3 correctly, he will score 8. In level 3, there will be 5 questions. Hence it can be negated.
    If he answers 4 correctly, he will score 12. In level 3, there will be 6 questions. Valid scenario.
    Case2: When the level 2 exam has 5 questions.
    If he answers 2 correctly, then there will be 5 questions in level3. Hence this can be negated.
    If he answers 3 correctly, he will score 7. In level 3, there will be 6 questions. Valid scenario
    If he answers 4 correctly, he will score 11. In level 3, there will be 6 questions. Valid scenario.
    If he answers all 5 correctly, there will be 7 questions. Hence negated.
    He can score either 7 or 11 or 12.
    Hence B is the correct answer.

  • Question 8/20
    3 / -1

    Directions For Questions

    The CEE (Common Entrance Exam) in Philippines has 4 levels. A candidate clears a level and proceeds to the next level only if he gets at least one question correct in that level. Candidates have to attempt all the questions in each level. Each level consists of 3 questions but the number of questions in subsequent levels can be increased based on the performance of a candidate in this level as follows:

     

    1. If a candidate answers all the questions correctly, the paper in the next level has two questions more than the number of questions that the candidate had in this level.
    2. If a candidate answers at least half of the number of questions in a particular level correctly, then in the next level, that candidate will have one question more than what he had in this level.
    3. If a candidate answers less than half the number of questions in a particular level correctly, then in the next level, that candidate will have the same number of questions as in this level.

     

    Each question is awarded 3 marks if answered correctly and -1 mark if answered incorrectly.

     

    4 friends - A, B, C, D who took the CEE are comparing their experiences.

    ...view full instructions

    If the number of correct attempts of C in level 1 is the same as that of D in level 3, and C scored a net of 5 marks in level 2, then what is the maximum possible net score of D in level 4 exam?

    Solutions

    To get a net score of 5 marks, one should either score 2 correct + 1 incorrect or 3 correct + 4 incorrect and so on. However, the max possible number of questions in level 2 is only 5 (when a student gets all questions correct in level 1). So the higher order cases where number of questions > 7 are impossible. Hence, C had 3 questions in level 2 of which he answered 2 correctly and 1 incorrectly. As there has been no increase in questions from level 1 to level 2, he must not have answered even half of his questions correctly in level 1. So he must have answered only 1 question out of 3 correctly in level 1. So D must also have attempted only one question correctly out of 3 in level 3. However, to get the max number of questions in level 4, we can assume he answered all questions correctly in level 1 and level 2. Hence, he would have had 3+2= 5 questions in level 2 and 5+2= 7 questions in level 3. So D's score would be 3/3 correct in level 1, 5/5 correct in level 2 and 1/7 correct in level 3. As he did not get more than half of the questions correct in level 3, the number of questions will not increase. So the number of questions in level 4 would still be 7. If he answers all these questions correctly, he should get 7*3 = 21 marks.

  • Question 9/20
    3 / -1

    Directions For Questions

    The CEE (Common Entrance Exam) in Philippines has 4 levels. A candidate clears a level and proceeds to the next level only if he gets at least one question correct in that level. Candidates have to attempt all the questions in each level. Each level consists of 3 questions but the number of questions in subsequent levels can be increased based on the performance of a candidate in this level as follows:

     

    1. If a candidate answers all the questions correctly, the paper in the next level has two questions more than the number of questions that the candidate had in this level.
    2. If a candidate answers at least half of the number of questions in a particular level correctly, then in the next level, that candidate will have one question more than what he had in this level.
    3. If a candidate answers less than half the number of questions in a particular level correctly, then in the next level, that candidate will have the same number of questions as in this level.

     

    Each question is awarded 3 marks if answered correctly and -1 mark if answered incorrectly.

     

    4 friends - A, B, C, D who took the CEE are comparing their experiences.

    ...view full instructions

    In each of the levels 2, 3 and 4, no two candidates have the same number of questions. Then, what is the maximum score that any candidate can score in all the levels put together?

    Solutions

    One possible set of questions in the four levels for each of the four candidates is

    A - 3, 5, 7, 9
    B - 3, 3, 5, 7
    C - 3, 4, 6, 8
    D - 3, 0, 0, 0

    In this case, D does not pass the first level and hence does not move on to the second level.

    So, the maximum score can be (3+5+7+9)*3 = 72 marks

  • Question 10/20
    3 / -1

    Directions For Questions

    The CEE (Common Entrance Exam) in Philippines has 4 levels. A candidate clears a level and proceeds to the next level only if he gets at least one question correct in that level. Candidates have to attempt all the questions in each level. Each level consists of 3 questions but the number of questions in subsequent levels can be increased based on the performance of a candidate in this level as follows:

     

    1. If a candidate answers all the questions correctly, the paper in the next level has two questions more than the number of questions that the candidate had in this level.
    2. If a candidate answers at least half of the number of questions in a particular level correctly, then in the next level, that candidate will have one question more than what he had in this level.
    3. If a candidate answers less than half the number of questions in a particular level correctly, then in the next level, that candidate will have the same number of questions as in this level.

     

    Each question is awarded 3 marks if answered correctly and -1 mark if answered incorrectly.

     

    4 friends - A, B, C, D who took the CEE are comparing their experiences.

    ...view full instructions

    What is the second least total score, given that he/she gets atleast one question correct in all the rounds?

    Solutions

    It is given that student appears in all rounds. Hence he/she will get at least one question correct in each round so as to move to the next round.

    Hence, the 2nd least minimum score is 5.

  • Question 11/20
    3 / -1

    Directions For Questions

    Eight persons - A, B, C, D, E, F, G, H are seated around a circular table facing inwards not necessarily in the same order. These persons are married to P, Q, R, S, T, U, V, W not necessarily in the same order.
    1)B sits second to the right of the person who is married to R.
    2)Three persons sit between D and the person who is married to V, who is neither B nor seated adjacent to B
    3)H sits three places away from the person who is married to R and he is not married to Q.
    4)The person who is married to Q is adjacent to B and the person who is married to W.
    5)A sits three places away from the person who is married to R.
    6)The person who is married to T and C are adjacent to each other.
    7)E sits opposite the person who is married to P.
    8)C is married to neither R nor S.

    ...view full instructions

    How many different arrangements can be drawn from these statements?

    Solutions


    Based on statement 1, 3, and 4, we can draw the following figure

    As per the statement , three persons sit between D and the person who is married to V, who is neither B nor seated adjacent to B. So, the only possibility is when D is on the left of B, and this means that V is married to H. 

    It is given that the person who is married to T and C are adjacent to each other, thus T and C are between R and V. Also, E is opposite the person who is married to P, and only B can marry P based on these conditions. So, we can say that E is married to T and C is next to R. C is married to neither R nor S, thus only U can be married to C, and only S can be married to D. 

    Now, the only possibility left for R and W is G or F in any order. 

    There are 2 arrangements which can be drawn. F and G can interchange their places.

  • Question 12/20
    3 / -1

    Directions For Questions

    Eight persons - A, B, C, D, E, F, G, H are seated around a circular table facing inwards not necessarily in the same order. These persons are married to P, Q, R, S, T, U, V, W not necessarily in the same order.
    1)B sits second to the right of the person who is married to R.
    2)Three persons sit between D and the person who is married to V, who is neither B nor seated adjacent to B
    3)H sits three places away from the person who is married to R and he is not married to Q.
    4)The person who is married to Q is adjacent to B and the person who is married to W.
    5)A sits three places away from the person who is married to R.
    6)The person who is married to T and C are adjacent to each other.
    7)E sits opposite the person who is married to P.
    8)C is married to neither R nor S.

    ...view full instructions

    Who is sitting opposite to F?

    Solutions

    The given arrangements can be shown in the diagram.
    F is sitting opposite to G.

  • Question 13/20
    3 / -1

    Directions For Questions

    Eight persons - A, B, C, D, E, F, G, H are seated around a circular table facing inwards not necessarily in the same order. These persons are married to P, Q, R, S, T, U, V, W not necessarily in the same order.
    1)B sits second to the right of the person who is married to R.
    2)Three persons sit between D and the person who is married to V, who is neither B nor seated adjacent to B
    3)H sits three places away from the person who is married to R and he is not married to Q.
    4)The person who is married to Q is adjacent to B and the person who is married to W.
    5)A sits three places away from the person who is married to R.
    6)The person who is married to T and C are adjacent to each other.
    7)E sits opposite the person who is married to P.
    8)C is married to neither R nor S.

    ...view full instructions

    Who is married to C?

    Solutions

    The given arrangements can be shown in the diagram.

    C is married to U.

  • Question 14/20
    3 / -1

    Directions For Questions

    Eight persons - A, B, C, D, E, F, G, H are seated around a circular table facing inwards not necessarily in the same order. These persons are married to P, Q, R, S, T, U, V, W not necessarily in the same order.
    1)B sits second to the right of the person who is married to R.
    2)Three persons sit between D and the person who is married to V, who is neither B nor seated adjacent to B
    3)H sits three places away from the person who is married to R and he is not married to Q.
    4)The person who is married to Q is adjacent to B and the person who is married to W.
    5)A sits three places away from the person who is married to R.
    6)The person who is married to T and C are adjacent to each other.
    7)E sits opposite the person who is married to P.
    8)C is married to neither R nor S.

    ...view full instructions

    Who is married to S?

    Solutions

    The given arrangements can be shown in the diagram.

    D is married to S.

  • Question 15/20
    3 / -1

    Directions For Questions

    Eight persons - A, B, C, D, E, F, G, H are seated around a circular table facing inwards not necessarily in the same order. These persons are married to P, Q, R, S, T, U, V, W not necessarily in the same order.
    1)B sits second to the right of the person who is married to R.
    2)Three persons sit between D and the person who is married to V, who is neither B nor seated adjacent to B
    3)H sits three places away from the person who is married to R and he is not married to Q.
    4)The person who is married to Q is adjacent to B and the person who is married to W.
    5)A sits three places away from the person who is married to R.
    6)The person who is married to T and C are adjacent to each other.
    7)E sits opposite the person who is married to P.
    8)C is married to neither R nor S.

    ...view full instructions

    Who is married to 'W'?

    Solutions

    The given arrangements can be shown in the diagram.

    W is married to either G or F. Hence cannot be determined.

    The answer is Option (D).

  • Question 16/20
    3 / -1

    Directions For Questions

    The two tables show the number of students studying in a class and the number of new students joining a particular class in each of the academic years. Admissions in the school start from class 6 and students who fail in a particular class in a particular academic year are dropped for the next academic year, except for class 6, where they are retained. Answer the questions based on the given data.

    ...view full instructions

    In the year 2011 - 2012, which class had the highest number of failures?

    Solutions

    In 2009-10, the total number of students in class 6 was 58. These students were supposed to advance to class 7 in 2010-11.
    It is given that the total number of new joinees in 2010-11 in class 7 is 3, and the total strength of class 7 is 59.
    If all 58 students had passed the exam in 2009-10, the total strength of class 7 in 2010-11 would have been 58 + 3 = 61.

    Thus, the number of students who failed in class 6 in 2009-10 = 61 - 59 = 2.

    Similarly, the number of students who failed in other classes in all the years can be found.

    The table consisting of all the failed students is as follows:

    From the above table, it can be seen that the highest number of failures(3) was in class 8 in the session of 2011-12.
    The correct option is C.

  • Question 17/20
    3 / -1

    Directions For Questions

    The two tables show the number of students studying in a class and the number of new students joining a particular class in each of the academic years. Admissions in the school start from class 6 and students who fail in a particular class in a particular academic year are dropped for the next academic year, except for class 6, where they are retained. Answer the questions based on the given data.

    ...view full instructions

    What is the pass percentage of the school in the year 2013-2014?

    Solutions

    In 2009-10, the total number of students in class 6 was 58. These students were supposed to advance to class 7 in 2010-11.
    It is given that the total number of new joinees in 2010-11 in class 7 is 3, and the total strength of class 7 is 59.
    If all 58 students had passed the exam in 2009-10, the total strength of class 7 in 2010-11 would have been 58 + 3 = 61.

    Thus, the number of students who failed in class 6 in 2009-10 = 61 - 59 = 2.

    Similarly, the number of students who failed in other classes in all the years can be found.

    The table consisting of all the failed students is as follows:

    To determine the number of failed students in 2013-14, we needed to have the total number of students in 2014-15.

    Since we do not have that data, the answer to this question cannot be determined.

    The correct option is D.

  • Question 18/20
    3 / -1

    Directions For Questions

    The two tables show the number of students studying in a class and the number of new students joining a particular class in each of the academic years. Admissions in the school start from class 6 and students who fail in a particular class in a particular academic year are dropped for the next academic year, except for class 6, where they are retained. Answer the questions based on the given data.

    ...view full instructions

    What is the total number of students who failed in class 7 from 2009-10 to 2012-13?

    Solutions

    In 2009-10, the total number of students in class 6 was 58. These students were supposed to advance to class 7 in 2010-11.
    It is given that the total number of new joinees in 2010-11 in class 7 is 3, and the total strength of class 7 is 59.
    If all 58 students had passed the exam in 2009-10, the total strength of class 7 in 2010-11 would have been 58 + 3 = 61.

    Thus, the number of students who failed in class 6 in 2009-10 = 61 - 59 = 2.

    Similarly, the number of students who failed in other classes in all the years can be found.

    The table consisting of all the failed students is as follows:

    The total number of students failed in class 7 = 2 + 4 + 2 + 5 = 13

    The correct option is B.

  • Question 19/20
    3 / -1

    Directions For Questions

    The two tables show the number of students studying in a class and the number of new students joining a particular class in each of the academic years. Admissions in the school start from class 6 and students who fail in a particular class in a particular academic year are dropped for the next academic year, except for class 6, where they are retained. Answer the questions based on the given data.

    ...view full instructions

    What is the minimum number of students who started in class 6 in the year 2009-10 and entered class 10 in the year 2013-14?

    Solutions

    In 2009-10, the total number of students in class 6 was 58. These students were supposed to advance to class 7 in 2010-11.
    It is given that the total number of new joinees in 2010-11 in class 7 is 3, and the total strength of class 7 is 59.
    If all 58 students had passed the exam in 2009-10, the total strength of class 7 in 2010-11 would have been 58 + 3 = 61.

    Thus, the number of students who failed in class 6 in 2009-10 = 61 - 59 = 2.

    Similarly, the number of students who failed in other classes in all the years can be found.

    The table consisting of all the failed students is as follows:

    Since we have to find the minimum number of students who started in class 6 and entered class 10, no student who joined in that batch in subsequent years failed in any of the years.

    The minimum number of students is therefore, 58 - 2 - 4 - 3 - 2 = 58 - 11 = 47.

    The correct option is C.

  • Question 20/20
    3 / -1

    Directions For Questions

    The two tables show the number of students studying in a class and the number of new students joining a particular class in each of the academic years. Admissions in the school start from class 6 and students who fail in a particular class in a particular academic year are dropped for the next academic year, except for class 6, where they are retained. Answer the questions based on the given data.

    ...view full instructions

    What is the ratio of the pass percentage of class 7 in 2010-2011 to the fail percentage of class 8 in 2012-13.

    Solutions

    In 2009-10, the total number of students in class 6 was 58. These students were supposed to advance to class 7 in 2010-11.
    It is given that the total number of new joinees in 2010-11 in class 7 is 3, and the total strength of class 7 is 59.
    If all 58 students had passed the exam in 2009-10, the total strength of class 7 in 2010-11 would have been 58 + 3 = 61.

    Thus, the number of students who failed in class 6 in 2009-10 = 61 - 59 = 2.

    Similarly, the number of students who failed in other classes in all the years can be found.

    The table consisting of all the failed students is as follows:

    The total number of students in class 7 in 2010-2011 = 59.

    The pass percentage of class 7 in 2010-11 = 59−459× 100= 93.22%\frac{59-4}{59}\times\ 100=\ 93.22\%......(1)

    The total number of students in class 8 in 2012-2013 = 64.

    The fail percentage of students in class 8 in 2012-2013 = 264× 100= 3.125%\frac{2}{64}\times\ 100=\ 3.125\%......(2)

    The ratio of the Pass percentage of class 7 in 2010-2011 to the fail percentage of class 8 in 2012-13 = 93.223.125=29.83\frac{93.22}{3.125}=29.83

    The answer is option (B)

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