The answers are:
(a) 21Ā
(b) 42
It is given:
SINK (sphere): dā = 2rā = 20 in Ā Ā Ā Ā ā rā = 20Ā Ć· 2 = 10 in
CUP (cylinder and conical): hā = 8 in, dā = 2rā = 4 in Ā Ā ā rā = 4 Ć· 2 = 2 in
Ļ = 3.14
The volume of the sphere is: V = 4/3Ā Ļ rĀ³
The volume of the half-sphereĀ is: Vā = 1/2 * 4/3Ā Ļ rāĀ³ = 2/3Ā Ļ rāĀ³
Ā Ā Ā Ā Ā Ā Ā Ā ā Vā = 2/3Ā Ā· 3.14Ā Ā· 10Ā³ = 2,093.3 inĀ³
The volume of the cylindrical cup is: Vā =Ā Ļ rāĀ² h
Ā Ā Ā Ā Ā Ā Ā Ā āĀ Vā = 3.14Ā Ā· 2Ā²Ā Ā· 8 = 100.5 inĀ³
The volume of the conical cupĀ is: Vā = 1/3Ā Ļ rāĀ² h
Ā Ā Ā Ā Ā Ā Ā Ā āĀ Vā = 1/3Ā Ā· 3.14Ā Ā· 2Ā²Ā Ā· 8 = 33.5 inĀ³
(a)Ā How many cups of water did heĀ scoop out of the sink to empty it?
The answer is ratio between the volume of the half-sphere and the volume of theĀ cylindrical cup:
[tex] \frac{V_1}{V_2} = \frac{2,093.3in^{2} }{100.5 in^{2} } = 20.8 = 21[/tex]
(b)Ā 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 betweenĀ ratio between the volume of the half-sphere and the volume of theĀ conical cup andĀ ratio between the volume of the half-sphere and the volume of theĀ cylindrical cup:
Ā [tex] \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[/tex]
