Question

What best can be concluded about the number of candidates sitting for the separate test for BIE who were at or above the 90th percentile overall in CET?

• A. 3 or 10
• B. 10
• C. 5
• D. 7 or 10

Answer 1

The correct option is A

3 or 10

Refer to the figure above,
Given, P + C + M + PC + CM + PM + PCM = 200
PC + CM + PM = 150
P = C = M
Therefore, 3P + PCM = 50 ....... (1)
Also given,
(C + M + CM) : (P + M + PM) : (P + C + PC) = 4 : 2 : 1
Putting P = C = M in the above ratio, we get
(2P + CM) : (2P + PM) : (2P + PC) = 4 : 2 : 1
Adding the above, we see that ( 6P + CM + PM + PC) i.e. ( 6P + 150) must be a multiple of 7 i.e. divisible by 7.
or (6P + 3) must be divisible by 7.
Also we see that for P = 3, 10, 17…… (6P + 3) is divisible by 7.
Now, from (1), we know that P < 17
Hence, P = C = M = 3 or 10
Correspondingly,
PCM = 41 or 20
CM = 90 or 100
PM = 42 or 40
PC = 18 or 10
Therefore, the number of candidates sitting for the separate test for BIE who were at or above 90 percentile in CET overall = 3 or 10.