Question

Michael has a sink that is shaped like a half-sphere. The sink has a volume of 3000/3 π in3. One day, his sink is clogged. He has to use one of two cylindrical cups to scoop the water out of the sink. The sink is completely full when Michael begins scooping. (a) One cup has a diameter of 6in. and a height of 8 in. How many cups of water must Michael scoop out of the sink with this cup to empty it? Round the number of scoops to the nearest whole number. (b) One cup has a diameter of 10in. and a height of 8in.How many cups of water must he scoop out of the sink with this cup to empty it? Round the number of scoops to the nearest whole number

Answer 1

Answer:

A. 14 times

B. 5 times.

Step-by-step explanation:

We know the volume of the sink, now we need to find the volume of the cup(s) he's using to empty it out.. to find out how many scoops he'll have to do.  

We'll assume he can actually empty it all, even if it's unlikely he can scoop the last 5%-10% of it due to the shapes/sizes of the sink and the cup. We'll also assume he completely fills the cup (which is impossible due to angles).

The volume of the sink is 3000/3 π cubic inches,  let's simplify that as 1000  π cubic inches.

a - Cylindrical cup, diameter 6 inches, height: 8 inches.

We remember the volume of a cylinder is found using this formula:

V = π r²h

So, we have:

V = π * 3² * 8 = 72π

So, how many times does Michael has to scoop (S):

$S = \frac{1000 \pi }{72 \pi } = 13.88$

Of course, he can't scoop 0.83 times... so we'll round it up.

So, Michael will have to scoop 14 times with this first cup.

b - Cylindrical cup, diameter 10 inches, height: 8 inches.

We remember the volume of a cylinder is found using this formula:

V = π r² h

So, we have:

V = π * 5² * 8 = 200π

So, how many times does Michael has to scoop (S):

$S = \frac{1000 \pi }{200 \pi } = 5$

So, Michael will have to scoop 5 times with this second cup, which is not surprising since the volume of the second cup is nearly the triple of the first one.