HELLONOISE

A room (8 m X 6 m X 4 m (height)) with hard sound reflecting surfaces is to be decorated for meeting purpose. The reverberation time is measured and found to be 3 seconds in a particular octave band. Calculate how much extra sound absorption must be provided to reduce the reverberation time to 0.8 s, that is suitable for speech and discussion. Assume that a carpet is added on the entire floor with the value of absorption coefficient = 0.2 in the same band and sound absorbing panels are added to cover areas of the wall and ceiling with the value of absorption coefficient= 0.4. What is the area for these panels required? You may assume that the absorption coefficient of the hard sound reflective surfaces is negligible and other settings (e.g., equipment and furniture) inside the room is the same. My work: for the first part: Calculate how much extra sound absorption must be provided to reduce the reverberation time to 0.8 s, since by the following equation for T=3, aisi=10.24 m2 sabins and for 0.8s , aisi=38.4 , so the extra sound absorption is 38.4-10.24 =28.16 m2 sabins (Is that correct??????any missing data I didnt use???) for the area for these panels required? (my work is as follows) (0.4*(8*4*2)+(6*4*2))+(0.2*6*8)+(0.2*8*6)=73.6 m2 The answer of this question is 59.7 m2 , can someone point out my mistake????