1) If there are x boys in a school, there are girls (100-x). 60% x+40% (100-x) =100 * 49.6% x = 48100-x = 52a: 48 boys and 52 girls. (2) If there are y boys in school B, there are (100-y) girls. Excellent performance of school B = [57% y+37% (100-y)]/100 = (0.2y+37)/100 (1) Excellent performance of school A = [60%. 0 0.2(y-x)>3y-x & gt； 15y & gt； X+ 1 means that when there are more boys in school B than in school A 15, the excellent rate in school B is higher than that in school A. For example, there are 68 boys and 32 girls in school B.