6983manish wrote:What percent of the MIS students enrolled at Wisconsin University are female?
1. 5% of female students at Wisconsin University are studying MIS.
2. 12% of male students at Wisconsin University are studying MIS.
Let the no. of females at MIS enrolled at Wisconsin University = F, and
Let the no. of males at MIS enrolled at Wisconsin University = M
and let total no. of females enrolled at Wisconsin University = x and total no. of males enrolled at Wisconsin University = y.
So, we have to determine the value of F/(F + M).
(1) 5% of x = F or 0.05x = F. It's evident that we cannot find the value of F/(F + M) from this information.
So, (1) is NOT SUFFICIENT.
(2) 12% of y = M or 0.12y = M. Again we cannot find the value of F/(F + M) from this information.
So, (2) is NOT SUFFICIENT.
Next combining (1) and (2), we get 0.05x/{0.05x + 0.12y}= F/(F + M). Again it can be seen that we cannot find the required value from the above equation.
The correct answer is
E.