Question

Anthony has a sink that is shaped like a half-sphere. The sink has a diameter of 20 in. One day, his sink clogged and he had to use a cylindrical cup to scoop the water out of the sink. The cup had a diameter of 4 in. and a height of 8 in.
(a) The sink was completely full when Anthony begon scooping. How many cups of water did he scoop out of the sink to empty it? round to the nearest whole number of scoops.
(b) If the cup had been cone-shaped with the same diameter and height as the cylindrical cup, how many more cups of water would he have needed to scoop to empty the sink?

Answer 1

The answers are:
(a) 21 
(b) 42

It is given:
SINK (sphere): d₁ = 2r₁ = 20 in       ⇒ r₁ = 20 ÷ 2 = 10 in
CUP (cylinder and conical): h₂ = 8 in, d₂ = 2r₂ = 4 in    ⇒ r₂ = 4 ÷ 2 = 2 in
π = 3.14

The volume of the sphere is: V = 4/3 π r³
The volume of the half-sphere is: V₁ = 1/2 * 4/3 π r₁³ = 2/3 π r₁³
              ⇒ V₁ = 2/3 · 3.14 · 10³ = 2,093.3 in³
The volume of the cylindrical cup is: V₂ = π r₂² h
              ⇒ V₂ = 3.14 · 2² · 8 = 100.5 in³
The volume of the conical cup is: V₃ = 1/3 π r₂² h
              ⇒ V₃ = 1/3 · 3.14 · 2² · 8 = 33.5 in³


(a) How many cups of water did he scoop out of the sink to empty it?
The answer is ratio between the volume of the half-sphere and the volume of the cylindrical cup:
$\frac{V_1}{V_2} = \frac{2,093.3in^{2} }{100.5 in^{2} } = 20.8 = 21$


(b) If the cup had been cone-shaped with the same diameter and height as the cylindrical cup, how many more cups of water would he have needed to scoop to empty the sink?
The answer is the difference between ratio between the volume of the half-sphere and the volume of the conical cup and ratio between the volume of the half-sphere and the volume of the cylindrical cup:
 $\frac{V_1}{V_3}- \frac{V_1}{V_2} = \frac{2,093.3in^{2} }{33.5 in^{2} } - \frac{2,093.3in^{2} }{100.5 in^{2} }=62.5-20.8=41.7=42$