At the school, Level Y and Level Z had an equal number of desks at first. If 7 desks were removed from Level Z and 5 times as many desks were removed from Level Y as Level Z, the number of desks in Level Z would be 5 times as many as in Level Y. Find the number of desks on each level.
| |
Level Y |
Level Z |
| Before |
1 u + 35 |
5 u + 7 |
| Change |
- 35 |
- 7 |
| After |
1 u |
5 u |
Number of desks removed from Level Y
= 5 x 7
= 35
Number of desks on Level Y and Level Z at first is the same.
5 u + 7 = 1 u + 35
5 u - 1 u = 35 - 7
4 u = 28
1 u = 28 ÷ 4 = 7
Number of desks on each level
= 1 u + 35
= 7 + 35
= 42
Answer(s): 42
