it seems valid --> 80% study BM and 40% study French, BM&French is 20%
%% 80-20=60 BM
%% 40-20=20 French
%% 20 BM&French
%% 20+60 + 20=100 Class

vardhans83 wrote:In a club, people study either business management(BM) or French. If 80% study BM, 40% study french. How many people study both business management and French?

If there are 100 people in the club,

80 BM + 40 F - Both = 100

120-both=100

Both =20

I don't see the other 10%

yes you are correct it makes 60 alone BM, 20 alone French and 20 both BM and French

jk2010 wrote:If there are 100 people in the club,

80 BM + 40 F - Both = 100

120-both=100

Both =20

I don't see the other 10%

vardhans83 wrote:In a club, people study either business management(BM) or French. If 80% study BM, 40% study french. How many people study both business management and French?

This question is lame, unless it is supposed to be a data sufficiency problem. The question only gives percentages, and then it suddenly wants a integer. Only way that this can be a PS is if the end of the stem asked for a percentage. With the available info, we can only figure out that 20% take both though.

320 is common multiple of 80 and 40

320*.80=256

320*.40=128

256+128=384

384-320=64

64/320=.20