Question: 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?
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.
Now 3d + e = 50
Also, d = 3 or 10
But it is given that e is a multiple of 5, so
e = 20
Now 20 + c 20 + a = 2 1 , 20 + c 20 + b = 4 1 20 + a 20 + b = 2 1
Solving the above expression we get the following equations:
c – 2a = 20 … (I)
c – 4b = 60 … (II)
a – 2b = 20 … (III)
Adding (I), (II) and (III) we get
–5b + 3c = 250 … (VI)
Solving (II) and (VI) we get
b = 10 and c = 100
∴ a =150 – 10 – 100 = 40
Now the number of candidates who scored 90 percentile overall and above and who score 80 percentile and above in P and M is a + e = 40 + 20 = 60
Answer : 60