For the 200 candidates who are at or above the 90th percentile overall based on CET, the following are known about their performance in CET:
1. No one is below the 80th percentile in all 3 sections.
2. 150 are at or above the 80th percentile in exactly two sections.
3. The number of candidates at or above the 80th percentile only in P is the same as the number of candidates at or above the 80th percentile only in C. The same is the number of candidates at or above the 80th percentile only in M.
4. Number of candidates below 80th percentile in P: Number of candidates below 80th percentile in C: Number of candidates below 80th percentile in M = 4:2:1.
BIE uses a different process for selection. If any candidate is appearing in the AET by AIE, BIE considers their AET score for final selection provided the candidate is at or above the 80th percentile in P. Any other candidate at or above the 80th percentile in P in CET, but who is not eligible for the AET, is required to appear in a separate test to be conducted by BIE for being considered for final selection. Altogether, there are 400 candidates this year who are at or above the 80th percentile in P.
What best can be concluded about the number of candidates sitting for the separate test for BIE who were at or above the 90th percentile overall in CET?
- 3 or 10
- 10
- 5
- 7 or 10
The Correct Option is A
Solution and Explanation
It is stated that 200 candidates scored above the 90th percentile overall in the Common Entrance Test (CET). The subsequent Venn diagram illustrates the distribution of individuals who scored above the 80th percentile in each of the three sections.
Deducing from equation 1, h equals 0.
- Deriving from equation 2, the sum of d, e, and f is 150.
- Establishing from equation 3 that a equals b equals c. Given a total of 200 candidates, The equation 3a + g equals 200 – 150, resulting in 50.
- Examining equation 4, (2a + f) : (2a + e) : (2a + d) is in the ratio of 4 : 2 : 1. Therefore, 6a + (d + e + f) is divisible by 4 + 2 + 1, which is 7. As d + e + f equals 150, 6a + 150 is divisible by 7, implying that 6a + 3 is divisible by 7. Consequently, a can be 3, 10, 17, and so on. Furthermore, since 3a + g equals 50, a must be less than 17. Therefore, only two cases are feasible for the value of a, namely, 3 or 10. The values of the other variables can be computed for these two cases. For a = 3 or 10:
- d can be 18 or 10
- e can be 42 or 40
- f can be 90 or 100
- g can be 41 or 20
Candidates at or above the 90th percentile who are also at or above the 80th percentile in at least two sections are selected for the Admission Eligibility Test (AET). Therefore, candidates represented by d, e, f, and g are chosen for AET. The Board of Intermediate Education (BIE) will consider candidates appearing for AET and scoring at or above the 80th percentile in P. Hence, BIE will evaluate candidates represented by d, e, and g, resulting in either 104 or 80 candidates. BIE will conduct a separate test for other students scoring at or above the 80th percentile in P. Given a total of 400 candidates at or above the 80th percentile in P, and considering 104 or 80 candidates at or above the 80th percentile in P who are also at or above the 90th percentile overall, there must be 296 or 320 candidates at or above the 80th percentile in P who scored less than the 90th percentile overall. The number of candidates taking the separate test for BIE who were at or above the 90th percentile in CET (a) is either 3 or 10.
If the number of candidates who are at or above the 90th percentile overall and also at or above the 80th percentile in all three sections in CET is actually a multiple of 5, what is the number of candidates who are at or above the 90th percentile overall and at or above the 80th percentile in both P and M in CET?
Correct Answer: 60
Solution and Explanation
It is stated that 200 candidates scored above the 90th percentile overall in the Common Entrance Test (CET). The subsequent Venn diagram illustrates the distribution of individuals who scored above the 80th percentile in each of the three sections.
Inference from equation 1: h equals 0.
- Deriving from equation 2: the sum of d, e, and f is 150.
- Establishing from equation 3: a equals b equals c. Given a total of 200 candidates, The equation 3a + g equals 200 – 150, resulting in 50.
- Examining equation 4: the ratio (2a + f) : (2a + e) : (2a + d) is 4 : 2 : 1. Therefore, 6a + (d + e + f) is divisible by 4 + 2 + 1, which is 7. As d + e + f equals 150, 6a + 150 is divisible by 7, implying that 6a + 3 is divisible by 7. Hence, a can be 3, 10, 17, and so on. Further, since 3a + g equals 50, a must be less than 17. Therefore, only two cases are feasible for the value of a, namely, 3 or 10. The values of the other variables can be computed for these two cases. For a = 3 or 10:
- d can be 18 or 10
- e can be 42 or 40
- f can be 90 or 100
- g can be 41 or 20
Candidates at or above the 90th percentile who are also at or above the 80th percentile in at least two sections are selected for the Admission Eligibility Test (AET). Therefore, the candidates represented by d, e, f, and g are chosen for AET. The Board of Intermediate Education (BIE) will consider candidates appearing for AET and scoring at or above the 80th percentile in P. Hence, BIE will evaluate candidates represented by d, e, and g, resulting in either 104 or 80 candidates. BIE will conduct a separate test for other students scoring at or above the 80th percentile in P. Given a total of 400 candidates at or above the 80th percentile in P, and considering 104 or 80 candidates at or above the 80th percentile in P who are also at or above the 90th percentile overall, there must be 296 or 320 candidates at or above the 80th percentile in P who scored less than the 90th percentile overall. Based on the given condition, g is a multiple of 5; hence, g equals 20. The number of candidates at or above the 90th percentile overall and at or above the 80th percentile in both P and M is e + g, resulting in 60.
If the number of candidates who are at or above the 90th percentile overall and also at or above the 80th percentile in all three sections in CET is actually a multiple of 5, then how many candidates were shortlisted for the AET for AIE?
Correct Answer: 170
Solution and Explanation
It is stated that 200 candidates scored above the 90th percentile overall in the Common Entrance Test (CET). The subsequent Venn diagram illustrates the distribution of individuals who scored above the 80th percentile in each of the three sections.
In case 1, h equals 0, and based on equation 2, the sum of d, e, and f is 150. Equation 3 establishes that a equals b equals c. With a total of 200 candidates, equation 3a + g equals 200 – 150, resulting in 50. Examining equation 4, the ratio (2a + f) : (2a + e) : (2a + d) is 4 : 2 : 1. Therefore, 6a + (d + e + f) is divisible by 4 + 2 + 1 = 7. Since d + e + f equals 150, 6a + 150 is divisible by 7, i.e., 6a + 3 is divisible by 7. Hence, a can be 3, 10, 17, and so on.
Further, since 3a + g equals 50, a must be less than 17. Therefore, only two cases are feasible for the value of a, namely, 3 or 10. The values of the other variables can be computed for these two cases.
For a = 3 or 10:
- d can be 18 or 10
- e can be 42 or 40
- f can be 90 or 100
- g can be 41 or 20
Among the candidates at or above the 90th percentile, those at or above the 80th percentile in at least two sections are selected for the Admission Eligibility Test (AET). Therefore, candidates represented by d, e, f, and g are chosen for AET.
The Board of Intermediate Education (BIE) will consider candidates appearing for AET and scoring at or above the 80th percentile in P. Hence, BIE will evaluate candidates represented by d, e, and g, resulting in either 104 or 80 candidates.
BIE will conduct a separate test for other students scoring at or above the 80th percentile in P. Given a total of 400 candidates at or above the 80th percentile in P, and considering 104 or 80 candidates at or above the 80th percentile in P and at or above the 90th percentile overall, there must be 296 or 320 candidates at or above the 80th percentile in P who scored less than the 90th percentile overall.
In this case, g equals 20, and the number of candidates shortlisted for AET is d + e + f + g, resulting in 10 + 40 + 100 + 20 = 170.
If the number of candidates who are at or above the 90th percentile overall and also are at or above the 80th percentile in P in CET, is more than 100,how many candidates had to sit for the separate test for BIE?
- 299
- 310
- 321
- 330
The Correct Option is A
Solution and Explanation
It is stated that 200 candidates scored above the 90th percentile overall in the Common Entrance Test (CET). The subsequent Venn diagram illustrates the distribution of individuals who scored above the 80th percentile in each of the three sections.
The analysis starts with the given conditions:
- From 1, h = 0.
- From 2, d + e + f = 150.
- From 3, a = b = c.
- Since there are a total of 200 candidates, 3a + g = 200 – 150 = 50.
- From 4, the ratio (2a + f) : (2a + e) : (2a + d) = 4 : 2 : 1.
- Therefore, 6a + (d + e + f) is divisible by 4 + 2 + 1 = 7.
- Since d + e + f = 150, 6a + 150 is divisible by 7, i.e., 6a + 3 is divisible by 7.
The values of variables are calculated for two possible cases:
- Case 1: a = 3, d = 18, e = 42, f = 90, g = 41.
- Case 2: a = 10, d = 10, e = 40, f = 100, g = 20.
Among candidates at or above the 90th percentile, those at or above the 80th percentile in at least two sections (d, e, f, and g) are selected for the Admission Eligibility Test (AET). For Case 1, the candidates for AET are d, e, f, and g, and for Case 2, they are d, e, and g, resulting in either 104 or 80 candidates.
The Board of Intermediate Education (BIE) will consider candidates appearing for AET and scoring at or above the 80th percentile in P. Therefore, BIE will evaluate candidates represented by d, e, and g, leading to either 104 or 80 candidates.
BIE will conduct a separate test for other students scoring at or above the 80th percentile in P. Given a total of 400 candidates at or above the 80th percentile in P, and since there are 104 or 80 candidates at or above the 80th percentile in P and at or above the 90th percentile overall, there must be 296 or 320 candidates at or above the 80th percentile in P who scored less than the 90th percentile overall.
From the given condition, the number of candidates at or above the 90th percentile overall and at or above the 80th percentile in P in CET is 104. The number of candidates who have to sit for a separate test is 296 + 3 = 299.
Top Questions on Conclusion
- Faculty members in a management school can belong to one of four departments – Finance and Accounting (F&A), Marketing and Strategy (M&S), Operations and Quants (O&Q) and Behaviour and Human Resources (B&H). The numbers of faculty members in F&A, M&S, O&Q and B&H departments are 9, 7, 5 and 3 respectively. Prof. Pakrasi, Prof. Qureshi, Prof. Ramaswamy and Prof. Samuel are four members of the school's faculty who were candidates for the post of the Dean of the school. Only one of the candidates was from O&Q. Every faculty member, including the four candidates, voted for the post. In each department, all the faculty members who were not candidates voted for the same candidate. The rules for the election are listed below.
1. There cannot be more than two candidates from a single department.
2. A candidate cannot vote for himself/herself.
3. Faculty members cannot vote for a candidate from their own department.
After the election, it was observed that Prof. Pakrasi received 3 votes, Prof. Qureshi received 14 votes, Prof. Ramaswamy received 6 votes and Prof. Samuel received 1 vote. Prof. Pakrasi voted for Prof. Ramaswamy, Prof. Qureshi for Prof. Samuel, Prof. Ramaswamy for Prof. Qureshi and Prof. Samuel for Prof. Pakrasi - There are only four neighbourhoods in a city-Levmisto, Tyhrmisto,Pesmisto and Kitmisto. During the onset of a pandemic,the number of new cases of a disease in each of these neighbourhoods was recorded over a period of five days. On each day, the number of new cases recorded in any of the neighbourhoods was either 0,1,2 or 3.
The following facts are also known:
1. There was at least one new case in every neighbourhood on Day 1.
2. On each of the five days, there were more new cases in Kitmisto than in Pesmisto.
3. The number of new cases in the city in a day kept increasing during the five-day period. The number of new cases on Day 3 was exactly one more than that on Day 2.
4. The maximum number of new cases in a day in Pesmisto was 2 and this happenedonly once during the five-day period.
5. Kitmisto is the only place to have 3 new cases on Day 2.
6. The total numbers of new cases in Levmisto,Tyhrmisto,Pesmisto and Kitmisto over the five-day period were 12,12,5 and 14 respectively. - The Humanities department of a college is planning to organize eight seminars, one for each of the eight doctoral students - A, B, C, D, E, F, G and H. Four of them are from Economics, three from Sociology and one from Anthropology department. Each student is guided by one among P, Q, R, S and T. Two students are guided by each of P, R and T, while one student is guided by each of Q and S. Each student is guided by a guide belonging to their department.
Each seminar is to be scheduled in one of four consecutive 30-minute slots starting at 9:00 am, 9:30 am, 10:00 am and 10:30 am on the same day. More than one seminars can be scheduled in a slot, provided the guide is free. Only three rooms are available and hence at the most three seminars can be scheduled in a slot. Students who are guided by the same guide must be scheduled in consecutive slots.
The following additional facts are also known.
1. Seminars by students from Economics are scheduled in each of the four slots.
2. A’s is the only seminar that is scheduled at 10:00 am. A is guided by R.
3. F is an Anthropology student whose seminar is scheduled at 10:30 am.
4. The seminar of a Sociology student is scheduled at 9:00 am.
5. B and G are both Sociology students, whose seminars are scheduled in the same slot. The seminar of an Economics student, who is guided by T, is also scheduled in the same slot.
6. P, who is guiding both B and C, has students scheduled in the first two slots.
7. A and G are scheduled in two consecutive slots. - Ten musicians (A, B, C, D, E, F, G, H, I and J) are experts in at least one of the following three percussion instruments: tabla, mridangam, and ghatam. Among them, three are experts in tabla but not in mridangam or ghatam, another three are experts in mridangam but not in tabla or ghatam, and one is an expert in ghatam but not in tabla or mridangam. Further, two are experts in tabla and mridangam but not in ghatam, and one is an expert in tabla and ghatam but not in mridangam.
The following facts are known about these ten musicians.
1. Both A and B are experts in mridangam, but only one of them is also an expert in tabla.
2. D is an expert in both tabla and ghatam.
3. Both F and G are experts in tabla, but only one of them is also an expert in mridangam.
4. Neither I nor J is an expert in tabla.
5. Neither H nor I is an expert in mridangam, but only one of them is an expert in ghatam. - A large store has only three departments, Clothing, Produce, and Electronics. The following figure shows the percentages of revenue and cost from the three departments for the years 2016, 2017 and 2018. The dotted lines depict percentage levels. So for example, in 2016, 50% of store's revenue came from its Electronics department while 40% of its costs were incurred in the Produce department.
In this setup, Profit is computed as (Revenue – Cost) and Percentage Profit as Profit/Cost × 100%.
It is known that
1. The percentage profit for the store in 2016 was 100%.
2. The store’s revenue doubled from 2016 to 2017, and its cost doubled from 2016 to 2018.
3. There was no profit from the Electronics department in 2017.
4. In 2018, the revenue from the Clothing department was the same as the cost incurred in the Produce department.
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