There was an equal number of desks and stools in the morning. At 1 p.m., some desks were added and some stools were removed, resulting in the number of desks being increased by 20% while the number of stools decreased by 40%. At 7 p.m. the number of desks increased by 50%. If a total of 44 desks were added during the day, how many more desks than stools were there in the end?

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

At 7 p.m.	Desks	Stools
Before	1.2 u	0.6 u
Change	+ 0.6 u	No change
After	1.8 u	0.6 u

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

Number of stools removed
= 40% x 1 u 
= ⁴⁰₁₀₀ x 1 u
= 0.4 u

Number of desks increased by desks at 7 p.m.
= 50% x 1.2 u
= ⁵⁰₁₀₀ x 1.2 u
= 0.6 u

Number of desks added during the day
= 1.8 u - 1 u
= 0.8 u

0.8 u = 44

1 u = 44 ÷ 0.8 = 55

Number of more desks than stools in the end
= 1.8 u - 0.6 u
= 1.2 u
= 1.2 x 55
= 66

Answer(s): 66