If the number of candidates who are at or above the 90th percentile overall and also are at or above 80th percentile in P in CET, is more than 100, how many candidates had to sit for the separate test for BIE?
299
We know that the number of candidates at or above 90 percentile overall and at or above 80 percentile in P = P + PCM + PM + PC = (3 + 41 + 42 + 18) or (10 + 20 + 40 + 10) = 104 or 80
Given (P + PCM + PM + PC) > 100
Therefore, (P + PCM + PM + PC) = 104
Also given, number of candidates at or above 80 percentile in P = 400. Hence, number of candidates to sit for the separate test for BIE = 400 - (PCM + PM + PC) = 400 - (104 -3) = 400 - 101 = 299.