Percentage of IT graduates in state A,
\(=\frac{31}{46+31+23+22}\times100=\frac{3100}{122}=25.4\%\)
Percentage of IT graduates in state B,
\(=\frac{38}{41+38+28+28}\times100=\frac{38}{135}=28.14\%\)
Percentage of IT graduates in state C,
\(=\frac{22}{62+34+26+23}\times100=\frac{22}{145}=15.17\%\)
Percentage of IT graduates in state D,
\(=\frac{18}{59+45+31+18}\times100=\frac{18}{153}=11.76\%\)
Percentage of IT graduates in state E,
\(=\frac{27}{38+28+24+12}\times100=\frac{27}{102}=26.47\%\)
Percentage of IT graduates in state F,
\(=\frac{16}{54+25+16+11}\times100=\frac{16}{106}=15.09\%\)
\(∴\) Percentage of IT graduates in C and F is mostly same.
Hence, option A is the correct answer.
The correct option is (A): C and F