CAT 2017 Slot 1 — DILR Question 10
Answer the following question based on the information given below.
Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Enginneering (BIE) have to appear for a Common Entrance Test (CET). The test has three sections: Physics (P), Chemistry (C), and Maths (M). Among those appearing for CET, those at or above the 80th percentile in at least two sections, and at or above the 90th percentile overall, are selected for Advanced Entrance Test (AET) conducted by AIE. AET is used by AIE for final selection.
For the 200 candidates who are at or above the 90th percentile overall based on CET, the following are known about their performance in CET:
- No one is below the 80th percentile in all 3 sections.
- 150 are at or above the 80th percentile in exactly two sections.
- The number of candidates at or above the 80th percentile only in P is the same as the number of candidates at or above the 80th percentile only in C. The same is the number of candidates at or above the 80th percentile only in M.
- Number of candidates below 80th percentile in P : Number of candidates below 80th percentile in C : Number of candidates below 80th percentile in M = 4 : 2 : 1.
BIE uses a different process for selection. If any candidates is appearing in the AET by AIE, BIE consider their AET score for final selection provided the candidates is at or above the 80th percentile in P. Any other candidate at or above the 80th percentile in P in CET, but who is not eligible for AET, is required to appear in a separate test to be conducted by BIE for being considered for final selection. Altogether, there are 400 candidates this year who are at or above the 80th percentile in P.
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?
Answer & solution
Answer: 60
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 and
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