Question

As part of a new advertising campaign, a beverage company wants to increase the dimensions of their cans by a multiple of 1. 10. If the cans are currently 12 cm tall, 6 cm in diameter, and have a volume of 339. 12 cm3, how much more will the new cans hold? Use 3. 14 for π and round your answer to the nearest hundredth. 112. 25 cm3 373. 03 cm3 451. 37 cm3 790. 49 cm3.

Answer 1

The extra amount the new cans can hold is provided by: Option A: 112.25 cm³ approx.

What is the volume of a right circular cylinder?

Suppose that the radius of considered right circular cylinder be 'r' units.

And let its height be 'h' units.

Then, its volume is given as:

$V = \pi r^2 h \: \rm unit^3$

Right circular cylinder is the cylinder in which the line joining center of top circle of the cylinder to the center of the base circle of the cylinder is perpendicular to the surface of its base, and to the top.

For the considered case, assuming the shape of cans is cylindrical, the dimension of old cans were:

• Height of 12 cm,
• and diameter of 6 cm.

Therefore, volume of an old can is:

$V_{\text{old}} = \pi r^2 h = \pi (d/2)^2 h = \pi. (3)^2.(12) = 108\pi \: \rm cm^3$

Since each dimension is increased by a multiple of 1.10, therefore, the new cans will have dimensions as:

• Height of $12 \times 1.10 = 13.2 \: \rm cm$
• and diameter of $6 \times 1.10 = 6.6 \: \rm cm$

Therefore, volume of a new can is:

$V_{\text{new}} = \pi r^2 h = \pi (d/2)^2 h = \pi. (3.3)^2.(13.2) = 143.748\pi \: \rm cm^3$

Therefore, the extra amount the each of the new cans can hold compared to old cans is the extra volume they have. It is evaluated as:

$V_{\text{extra}} = V_{\text{new}} - V_{\text{old}} = (143.748- 108)\pi \approx 35.748 \times 3.14 \approx 112.25\: \rm cm^3$

Thus, the extra amount the new cans can hold is provided by: Option A: 112.25 cm³ approx.

Learn more about volume of cylinder here:

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