设原一班有x人,原二班有y人.
新一班:x/3+y/4
新二班:x/4+y/3
新三班:x-(x/3+y/4)-(x/4+y/3)
=5x/12+5y/12=30
x+y=72 (1)
x/3+y/4=(x/4+y/3)×1.1
两边同时乘以12
4x+3y=(3x+4y)×1.1
(4x+3y)/(3x+4y)=11/10
交叉相成40x+30y=33x+44y
7x=14y x=2y
2y+y=72
y=24
x=2y=2×24=48(人)
答:原一班有48人.
设原一班有x人,原二班有y人.
新一班:x/3+y/4
新二班:x/4+y/3
新三班:x-(x/3+y/4)-(x/4+y/3)
=5x/12+5y/12=30
x+y=72 (1)
x/3+y/4=(x/4+y/3)×1.1
两边同时乘以12
4x+3y=(3x+4y)×1.1
(4x+3y)/(3x+4y)=11/10
交叉相成40x+30y=33x+44y
7x=14y x=2y
2y+y=72
y=24
x=2y=2×24=48(人)
答:原一班有48人.