Problem 1
Part (a)
V1 = volume of sink = (4000/3)*pi = (4000pi)/3
V2 = volume of cup A
V2 = volume of cylinder with diameter 4 in, height 8 in
V2 = pi*r^2*h
V2 = pi*2^2*8 ... radius r = 2 is half the diameter d = 4
V2 = 32pi
x = number of cups of water needed to empty sink
x*V2 = V1
x = V1/V2
x = V1 divided by V2
x = (4000pi/3) divided by (32pi)
x = 41.667 approximately
Round up to the nearest whole number to get 42. It will take 42 scoops to empty the sink if you use cup A. The last scoop (the 42nd scoop) isn't a full cup, but it is needed or else the sink will still be full. Put another way, 41 scoops is not enough.
<h3>Answer: 42 cups</h3>
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Part (b)
V3 = volume of cup B
V3 = volume of cylinder with diameter 8 in, height 8 in
V3 = pi*r^2*h
V3 = pi*4^2*8 ... radius r = 4 is half the diameter d = 8
V3 = 128pi
y = number of cups of water needed to empty sink if you use cup B
y*V3 = V1
y = V1/V3
y = V1 divided by V3
y = (4000pi/3) divided by (128pi)
y = 10.4167 approximately
Round up to the nearest whole number to get 11
You need 11 scoops of cup B to empty the sink. Ten cups won't be enough.
<h3>Answer: 11 cups</h3>
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Problem 2
Part (a)
V1 = volume of sink = (4000/3)*pi = (4000pi)/3
V2 = volume of cup A
V2 = volume of cone with diameter 4, height 8
V2 = (1/3)*pi*r^2*h
V2 = (1/3)*pi*4^2*8
V2 = (32/3)*pi ..... note how we multiply problem 1's V2 value by 1/3
V2 = 32pi/3
x = number of cups of water needed to empty sink
x*V2 = V1
x = V1/V2
x = V1 divided by V2
x = (4000pi/3) divided by (32pi/3)
x = 125
<h3>Answer: 125 cups</h3>
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Part (b)
V3 = volume of cup B
V3 = volume of cone with diameter 8 in, height 8 in
V3 = (1/3)*pi*r^2*h
V3 = (1/3)*pi*4^2*8 ... radius r = 4 is half the diameter d = 8
V3 = (128/3)pi ..... this is 1/3 of the V3 value in problem 1 above
V3 = 128pi/3
y = number of cups of water needed to empty sink if you use cup B
y*V3 = V1
y = V1/V3
y = V1 divided by V3
y = (4000pi/3) divided by (128pi/3)
y = 31.25
Round up to the nearest whole number to get 32. Using 31 cups won't be enough.
<h3>Answer: 32 cups</h3>
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Problem 3
A = volume of cone at the bottom
A = (1/3)*pi*r^2*h
A = (1/3)*pi*8^2*12
A = (1/3)*pi*768
A = (1/3)*768*pi
A = 256pi
B = volume of hemisphere on top
B = half of the volume of a sphere
B = (1/2)*(volume of sphere)
B = (1/2)*( (4/3)*pi*r^3 )
B = (2/3)*pi*r^3
B = (2/3)*pi*8^3
B = (2/3)*pi*512
B = (2/3)*512*pi
B = (1024/3)*pi
B = 1024pi/3
C = volume of entire 3D solid
C = A+B
C = 256pi + 1024pi/3
C = 256pi*(3/3) + 1024pi/3
C = 768pi/3 + 1024pi/3
C = (768pi+1024pi)/3
C = 1792pi/3 ... is the exact total volume
C = 1792*3.14/3
C = 1875.62666666667 ... is the approximate total volume
C = 1875.6
<h3>Answer: 1875.6 cubic inches</h3>
