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

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310

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321

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330

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Solution

The correct option is A
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.

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	Favorable	Unfavorable	Not Sure
Candidate M	40	20	40
Candidate N	30	35	35

The table above shows the results of a survey of 100 voters who each responded "Favorable" or "Unfavorable" or "Not Sure" when asked about their impressions of Candidate M and of Candidate N. What was the number of voters who responded "Favorable" for both candidates?
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