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, then how many candidates were shortlisted for the AET for AIE?___

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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
CM = 100
PC = 10
Hence, the number of candidates shortlisted for the AET for AIE = PCM + PM + CM + PC = 20 + 40 + 100 + 10 = 170.

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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, then how many candidates were shortlisted for the AET for AIE?___

As explained above, we have PCM = 41 or 20. Given PCM is a multiple of 5. Hence PCM = 20 Correspondingly, PM = 40 CM = 100 PC = 10 Hence, the number of candidates shortlisted for the AET for AIE = PCM + PM + CM + PC = 20 + 40 + 100 + 10 = 170.

As explained above, we have PCM = 41 or 20. Given PCM is a multiple of 5.
Hence PCM = 20
Correspondingly, PM = 40
CM = 100
PC = 10
Hence, the number of candidates shortlisted for the AET for AIE = PCM + PM + CM + PC = 20 + 40 + 100 + 10 = 170.