At the school, Floor U and Floor V had an equal number of tables at first. If 6 tables were removed from Floor V and 4 times as many tables were removed from Floor U as Floor V, the number of tables in Floor V would be 4 times as many as in Floor U. Find the number of tables on each floor.
| |
Floor U |
Floor V |
| Before |
1 u + 24 |
4 u + 6 |
| Change |
- 24 |
- 6 |
| After |
1 u |
4 u |
Number of tables removed from Floor U
= 4 x 6
= 24
Number of tables on Floor U and Floor V at first is the same.
4 u + 6 = 1 u + 24
4 u - 1 u = 24 - 6
3 u = 18
1 u = 18 ÷ 3 = 6
Number of tables on each floor
= 1 u + 24
= 6 + 24
= 30
Answer(s): 30
