Question: If the number of candidates who are at or above the 90th percentile overall and also are at or above the 80th percentile in P in CET, is more than 100, how many candidates had to sit for the separate test for BIE?
Explanation:
Using the information given in the question let us represent it in the Venn diagram shown below. The diagram depicts the number of candidates getting 80 percentile and above in at least one or more of the subjects amongst students getting 90 percentile overall.
The number of candidates scoring 80 percentile and above in exactly each of Physics, chemistry and Math is the same. Let this be ‘d’
Let ‘a’ – number of candidates scoring 80 percentile and above only in Physics and Math.
Let ‘b’ – number of candidates scoring 80 percentile above only in Physics and Chemistry.
Let ‘c’ – number of candidates scoring 80 percentile and above in Chemistry and Math.
Let ‘e’ – number of candidates scoring 80 percentile and above in all 3 subjects.
a +b + c = 150
Also a + b + c + 3d + e = 200
⇒3d+e=50
Given that (2d + c) : (2d + a) : (2d + b)
= 4: 2: 1
This implies 6d + a + b + c is a multiple of 7. We already know that a + b + c= 150.
So 6d + 150 is a multiple of 7. This implies that 6d + 3 will also be a multiple of 7. So d will be 3, 10, 17. But as 3d + e = 50, it implies that d < 17. So d will be either 3 or 10.
The number of candidates who are at or above 90th percentile overall and also at or above 80th percentile in P (as indicated in the Venn diagram) = a + b + d + e. As indicated in the answers to the previous question a = 3 or 10. Now we have already seen that if d = 10, a = 40, e As indicated in the answers to the previous question a = 3 or 10. Now we have So then a + b + d + e
= 42 + 18 + 3 + 41 = 104, which satisfies the condition given in the question.
Now the number of candidates who appear separately for the BIE test will be those who got 90 percentile and above overall and got 80 percentile and above only in P plus there who got 80 percentile only in P but got less than 90th percentile overall. Candidates who got 80th percentile only in P and got less than 90 in percentile overall = 400 – 104 =296
Number of candidates who get 80 percentile and above only in P and got 90 in percentile and above overall = d = 3
So, the number of candidates who sit for the separate test for BIE = 296 + 3 = 299
Hence, option (a).