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 80th percentile in both P and M in CET?___

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Solution

As explained above, we have PCM = 41 or 20. Given PCM is a multiple of 5.
Hence PCM = 20
Correspondingly, PM = 40
Therefore, the number of candidates at or above 90 percentile overall and at or above 80 percentile in both P and M = PM + PCM = 40 + 20 = 60.

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As explained above, we have PCM = 41 or 20. Given PCM is a multiple of 5. Hence PCM = 20 Correspondingly, PM = 40 Therefore, the number of candidates at or above 90 percentile overall and at or above 80 percentile in both P and M = PM + PCM = 40 + 20 = 60.

As explained above, we have PCM = 41 or 20. Given PCM is a multiple of 5.
Hence PCM = 20
Correspondingly, PM = 40
Therefore, the number of candidates at or above 90 percentile overall and at or above 80 percentile in both P and M = PM + PCM = 40 + 20 = 60.