XAT 2020 Question 68
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XAT
Given, a+b+c=50 and a+b+c+x+y+z=190 => x+y+z=140.
Also, let E1=k => E2=2k and E3=3k
E1+E2+E3= 6k=190+50=240 => k=40.
Option A: If the number of students choosing only E2, the number of students choosing both E2 and E3, are given then the number of students who choose E2 and E1, E1 and E3 can be found. From this only E1, only E3 can be calculated.
Option B: knowing the number of students choosing both E1 and E2, the number of students choosing both E2 and E3, and a number of students choosing both E3 and E1 is insufficient. This information is not enough to calculate the number of students who choose only
E1, only E2, and only E3.
Option C: If x and c are known, we can't find y and z.