Question

There are two vendors at the fair sleeping snow cones for the same price. If the containers are completely filled and then leveled off across their tops, what is the difference in the amount of snow cone each will hold

Answer 1

Answer:

$d = 787.81cm^3$

Step-by-step explanation:

Given

See attachment for the containers

Required

The difference in the amount of snow cone they hold

The amount they hold is determined by calculating the volume of the containers.

The traditional snow cone has the following dimension

Shape: Cone

$r=4cm$ --- radius

$h = 13cm$ --- height

The volume is calculated as:

$Volume = \frac{1}{3} \pi r^2h$

So, we have:

$V_1 = \frac{1}{3} \pi * 4^2 * 13$

$V_1 = \frac{208}{3} \pi$

The snow cone in a cup has the following dimension

Shape: Cylinder

$r=8cm$ --- radius

$h = 5cm$ --- height

The volume is calculated as:

$Volume = \pi r^2h$

So, we have:

$V_2 = \pi *8^2 * 5$

$V_2 = \pi *320$

$V_2 = 320\pi$

The difference (d) in the amount they hold is:

$d = V_2 - V_1$

$d = 320\pi - \frac{208\pi}{3}$

Take LCM

$d = \frac{3*320\pi -208\pi}{3}$

$d = \frac{960\pi -208\pi}{3}$

$d = \frac{752\pi}{3}$

Take $\pi = 22/7$

$d = \frac{752}{3} * \frac{22}{7}$

$d = \frac{752 * 22}{3*7}$

$d = \frac{16544}{21}$

$d = 787.81cm^3$