There was an equal number of stools and chairs in the morning. At 4 p.m., some stools were added and some chairs were removed, resulting in the number of stools being increased by 20% while the number of chairs decreased by 40%. 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 chairs were there in the end?

At 4 p.m.	Stools	Chairs
Before	1 u	1 u
Change	+ 0.2 u	- 0.4 u
After	1.2 u	0.6 u

At 5 p.m.	Stools	Chairs
Before	1.2 u	0.6 u
Change	+ 0.6 u	No change
After	1.8 u	0.6 u

Number of stools added
= 20% x 1 u
= ²⁰₁₀₀ x 1 u
= 0.2 u

Number of chairs removed
= 40% x 1 u 
= ⁴⁰₁₀₀ x 1 u
= 0.4 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 chairs in the end
= 1.8 u - 0.6 u
= 1.2 u
= 1.2 x 75
= 90

Answer(s): 90