Xavier Aptitude Test 2020 Paper

Given, $a+b+c=50$ and $a+b+c+x+y+z=190 \Rightarrow x+y+z=140$ Also, let $E_1=k \Rightarrow E_2=2k$ and $E_3=3k$ $E_1+E_2+E_3 = 6k = 190+50 = 240 \Rightarrow k = 40$ Option A: If the number of students choosing only $E_2$, the number of students choosing both $E_2$ and $E_3$, are given then the number of students who choose $E_2$ and $E_1$, $E_1$ and $E_3$ can be found. From this only $E_1$, only $E_3$ can be calculated. Option B: knowing the number of students choosing both $E_1$ and $E_2$, the number of students choosing both $E_2$ and $E_3$, and a number of students choosing both $E_3$ and $E_1$ is insufficient. This information is not enough to calculate the number of students who choose only$E_1$, only $E_2$, and only $E_3$. Option C: If $x$ and $c$ are known, we can't find $y$ and $z$. Option E is not sufficient