Question

Applicants for Indian Institute of Technology (IT) and Birla Institute of Tech and Sciences (BITS) 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 Both percentile in at least two sections, and at or above the 90th percentile
overall, are selected for Advanced Entrance Test (AET) conducted by IIT. AET is used by IIT 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 8th percentile in
exactly two sections. 3. The number of candidates at or above the 8th percentile only in P is the same as the number of candidates at or above the Both 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. BITS uses a different process for selection. If any candidate is appearing in the AET by IT, BITS considers their AET
score for final selection provided the candidate is at or above the 8th percentile in P. Any other candidate at or above the Both percentile in Pin CET, but who is not eligible for the
AET, is required to appear in a separate test to be conducted by BITS 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 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 8th percentile in both P and M in CET?​

Answer 1

Answer:

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Answer 2

Explanation:

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