题目内容
甲、乙两班相差15人,把甲班人数的
调入乙班后,两班人数相等,原来乙班有
| 1 | 6 |
30
30
人.分析:设乙班原来有x人,那么甲班就有x+15人,依据题意可列方程:(x+15)×(1-
)=x+(x+15)×
,依据等式的性质即可解答.
| 1 |
| 6 |
| 1 |
| 6 |
解答:解:设乙班原来有x人,
(x+15)×(1-
)=x+(x+15)×
,
x+
=x+
x+
,
x+
-
x=
x+
-
x,
-
=
x+
-
,
10=
x,
10÷
=
x÷
,
x=30.
答:原来乙班有30人,
故答案为:30.
(x+15)×(1-
| 1 |
| 6 |
| 1 |
| 6 |
| 5 |
| 6 |
| 25 |
| 2 |
| 1 |
| 6 |
| 5 |
| 2 |
| 5 |
| 6 |
| 25 |
| 2 |
| 5 |
| 6 |
| 7 |
| 6 |
| 5 |
| 2 |
| 5 |
| 6 |
| 25 |
| 2 |
| 5 |
| 2 |
| 1 |
| 3 |
| 5 |
| 2 |
| 5 |
| 2 |
10=
| 1 |
| 3 |
10÷
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
x=30.
答:原来乙班有30人,
故答案为:30.
点评:解答本题的关键是:依据题意列出方程,解方程时注意对齐等号.
练习册系列答案
相关题目