Xavier Aptitude Test 2020 Paper

Given, $\mathit{a}+\mathit{b}+\mathit{c}=50$ and $\mathit{a}+\mathit{b}+\mathit{c}+\mathit{x}+\mathit{y}+\mathit{z}=190\Rightarrow \mathit{x}+\mathit{y}+\mathit{z}=140$
Also, let ${\mathrm{E}}_1=\mathit{k}\Rightarrow {\mathrm{E}}_2=2\mathit{k}$ and ${\mathrm{E}}_3=3\mathit{k}$
${\mathrm{E}}_1+{\mathrm{E}}_2+{\mathrm{E}}_3=6\mathit{k}=190+50=240\Rightarrow \mathit{k}=40$
Option A:
If the number of students choosing only ${\mathrm{E}}_2$, the number of students choosing both ${\mathrm{E}}_2$ and ${\mathrm{E}}_3$, are given then the number of students who choose ${\mathrm{E}}_2$ and ${\mathrm{E}}_1$, ${\mathrm{E}}_1$ and ${\mathrm{E}}_3$ can be found. From this only ${\mathrm{E}}_1$, only ${\mathrm{E}}_3$ can be calculated.
Option B: knowing the number of students choosing both ${\mathrm{E}}_1$ and ${\mathrm{E}}_2$, the number of students choosing both ${\mathrm{E}}_2$ and ${\mathrm{E}}_3$, and a number of students choosing both ${\mathrm{E}}_3$ and ${\mathrm{E}}_1$ is insufficient. This information is not enough to calculate the number of students who choose only${\mathrm{E}}_1$, only ${\mathrm{E}}_2$, and only ${\mathrm{E}}_3$.
Option C: If $\mathit{x}$ and $\mathit{c}$ are known, we can't find $\mathit{y}$ and $\mathit{z}$.
Option E is not sufficient