# Get Answers to all your Questions

#### 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 CM). 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 ATE. AET is used by ATE for final selection. For the 200 candidates who are at or above the 9oth 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 V Number of candidates below 8oth 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 ATE, 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 STE for being considered for final selection. Altogether, there are 400 candidates this year who are at or above the 80th percentile in P. Question :  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? Option: 1 3 or 10Option: 2 10Option: 3 5Option: 4 7 or 10

200 candidates scored above the 90th percentile in CET.

The Venn diagram represents the number of candidates who scored above the 80th percentile in each of the three sections.

From the given information:

From (1), n = 0.

From (2), d + e + f = 150.

From (3), a = b = c.

Total number of candidates is 200.

a + b + c + g = 50.

Using the above equations:

3a + g = 50, which implies a < 17.

From (4), (b + f + c) : (a + d + b) : (a + e + c) = 4 : 2 : 1, or 6a + (d + e + f) = 7x.

6a + 150 = 7x.

Possible values for a are 3 or 10, and x can be 24 or 30.

Using these values, we can calculate possible values for d, e, f, and g.

For the selection process:

Candidates represented by d, e, f, and g are selected for AET (candidates at or above 90th percentile and at or above 80th percentile in at least two sections).

BIE will consider candidates represented by d, e, and g for the separate test.

Given that there are 400 candidates at or above the 80th percentile in P:

The candidates at or above the 80th percentile in P and overall 90th percentile are either 104 or 80.

Therefore, there must be 296 or 320 candidates at or above the 80th percentile in P who scored less than the overall 90th percentile.

Based on the possibilities for a:

Option 1 is the correct answer.