There was an equal number of stools and benches in the morning. At 3 p.m., some stools were added and some benches were removed, resulting in the number of stools being increased by 20% while the number of benches decreased by 20%. At 5 p.m. the number of stools increased by 50%. If a total of 60 stools were added during the day, how many more stools than benches were there in the end?
At 3 p.m. |
Stools |
Benches |
Before |
1 u |
1 u |
Change |
+ 0.2 u |
- 0.2 u |
After |
1.2 u |
0.8 u |
At 5 p.m. |
Stools |
Benches |
Before |
1.2 u |
0.8 u |
Change |
+ 0.6 u |
No change |
After |
1.8 u |
0.8 u |
Number of stools added
= 20% x 1 u
=
20100 x 1 u
= 0.2 u
Number of benches removed
= 20% x 1 u
=
20100 x 1 u
= 0.2 u
Number of stools increased by stools at 5 p.m.
= 50% x 1.2 u
=
50100 x 1.2 u
= 0.6 u
Number of stools added during the day
= 1.8 u - 1 u
= 0.8 u
0.8 u = 60
1 u = 60 ÷ 0.8 = 75
Number of more stools than benches in the end
= 1.8 u - 0.8 u
= 1 u
= 1 x 75
= 75
Answer(s): 75