Problem : Data Sufficiency - 2

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Problem : Data Sufficiency - 2

by 6983manish » Mon Mar 28, 2011 1:17 am
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.
Source: — Data Sufficiency |

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by Anurag@Gurome » Mon Mar 28, 2011 2:55 am
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.
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