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Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET). The test has three sections: Physics (P), Chemistry (C), and Maths (M). Among those appearing for CET, those at or above the 80th percentile in at least two sections, and at or above the 90th percentile overall, are selected for Advanced Entrance Test (AET) conducted by AIE. AET is used by AIE for final selection.
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.
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?
It is given that 200 candidates scored above 90th percentile overall in CET. Let the following Venn diagram represent the number of persons who scored above 80 percentile in CET in each of the three sections:
From (1), n = 0
From (2), d + e + f = 150
From (3), a = b = c
Since, there are a total of 200 candidates
a + b + c + g = 200 - 150 = 50
3a + g = 50 => a < 17
From (4), (b + f + c) : (a + d + b) : (a + e + c) = 4 : 2: 1
Or (2a + f) : (2a + d) : (2a + e) = 4 : 2 : 1
or, 6a + (d + e + f) = 7x
Or, 6a + 150 = 7x
So, a can be 3 or 10
x can be 24 or 30
2a + e can be 24 or 30 => e can be 18 or 10
2a + d can be 48 or 60 = > d can be 42 or 40
2a + f can be 96 or 120 => f can be 90 or 100
3a + g = 50 => g can be 41 or 20
Among the candidates who are at or above 90th percentile, the candidates who are at or above 80th percentile in at least two sections are selected for AET. Hence, the candidates represented by d, e, f and g are selected for AET.
BIE will consider the candidates who are appearing for AET and are at or above 80th percentile in P. Hence, BIE will consider the candidates represented by d, e and g, which can be 104 or 80.
BIE will conduct a separate test for the other students who are at or above 80th percentile in P. Given that there are a total of 400 candidates at or above 80th percentile in P, and since there are 104 or 80 candidates at or above 80th percentile in P and are at or above 90th percentile overall, there must be 296 or 320 candidates at or above 80th percentile in P who scored less than 90th percentile overall.
From the given condition, the number of candidates at or above 90th percentile overall and at or above 80th percentile in P in CET = (3 + 18 + 42 + 41) = 104.
The number of candidates who have to sit for separate test = (400 - 104 + 3) = 296 + 3 = 299 (we have added 3 for those who have scored more than 80th percentile only in P which is ‘a’).
Hence, option A is the correct answer.
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