If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Volume word problems | Lesson

What are volume word problems, and how frequently do they appear on the test?

Volume word problems focus on using the volume formulas of three-dimensional solids: rectangular prisms, cylinders, etc.
In this lesson, we'll learn to:
1. Calculate the volumes and dimensions of three-dimensional solids
2. Determine how dimension changes affect volume
On your official SAT, you'll likely see 1 question that is a volume word problem.
You can learn anything. Let's do this!

How do I calculate the volumes and dimensions of shapes?

Volume word problem: gold ring

Volume word problem: gold ringSee video transcript

Volume of a cone

Volume of a coneSee video transcript

The volumes of three-dimensional solids

Good news: You do not need to remember any volume formulas for the SAT! At the beginning of each SAT math section, the following volume formulas are provided as reference.
ShapeFormula
Right rectangular prismV, equals, ell, w, h
Right circular cylinderV, equals, pi, r, squared, h
SphereV, equals, start fraction, 4, divided by, 3, end fraction, pi, r, cubed
Right circular coneV, equals, start fraction, 1, divided by, 3, end fraction, pi, r, squared, h
Rectangular pyramidV, equals, start fraction, 1, divided by, 3, end fraction, ell, w, h
If the test asks for the volume of a different shape, the volume formula will be provided alongside the question.
To calculate the volume of a solid:
1. Find the volume formula for the solid.
2. Plug the dimensions into the formula.
3. Evaluate the volume.

Example: Fei Fei has a model of the Moon in the shape of a sphere. If the model has a radius of 10 centimeters, what is the volume of the model in cubic centimeters?

Some questions will provide the volume of the solid and ask us to find a linear dimension such as length or radius.
To find an unknown dimension when the volume of a solid is given:
1. Find the volume formula for the solid.
2. Plug the volume and any known dimensions into the formula.
3. Isolate the unknown dimension.

Example: A puzzle box is shaped like a rectangular prism and has a volume of 240 cubic inches. If the puzzle box has a length of 10 inches and a width of 8 inches, what is the height of the puzzle box in inches?

Try it!

try: find the volume of a pyramid
A pyramid has a square base with a side length of 8 centimeters. The height of the pyramid is start fraction, 3, divided by, 4, end fraction as long as the side length of its base.
What is the height of the pyramid in centimeters?
What is the volume of the pyramid in cubic centimeters?

How do changing dimensions affect volume?

How volume changes when dimensions change

How volume changes from changing dimensionsSee video transcript

Impact of increasing the radiusSee video transcript

The effect of changing dimensions on volume

When a linear dimension to the first power, e.g., the length of a rectangular prism or the height of a cylinder, changes by a factor, the volume of the solid changes by the same factor.
However, when a linear dimension to the second power, e.g., the radius of a cylinder or cone, changes by a factor, the volume of the solid changes by the square of the factor.

Try it!

try: compare the volumes of two cylinders
Right circular cylinder A has a volume of 64, pi cubic feet. Which of the following right circular cylinders have the same volume as cylinder A ?

practice: calculate a volume
What is the volume, in cubic meters, of a right rectangular prism that has a length of 2 meters, a width of 0, point, 4 meter, and a height of 5 meters?

practice: calculate a linear dimension
A medicine bottle is in the shape of a right circular cylinder. If the volume of the bottle is 144, pi cubic centimeters, what is the diameter of the base of the bottle, in centimeters?

practice: determine the effect of scaling on volume
The volume of a right circular cone A is 225 cubic inches. What is the volume, in cubic inches, of a right circular cone with twice the radius and twice the height of cone A?

Want to join the conversation?

• The volume of a right circular cone A is 225 cubic inches. What is the volume, in cubic inches, of a right circular cone with twice the radius and twice the height of cone A?

Since r is squared in the formula, the area of the cone's base is 2^2=42
2
=42, squared, equals 4 times the area of the base of cone A.

• The SAT asks these kinds of problems a lot. "If x does this, then what does 2x do?" Basically, it wants to see how well you understand the area and volume formulas.
The formula for volume of a cone is 1/3 * pi * r^2 * h.
We want to know the volume of a cone with twice the radius and twice the height. Since radius is squared, twice the radius would mean 4 times the volume, and twice the height would multiply the volume by 2 again leading to 8 times the volume.
If you ever feel unsure, you can always plug "2r" and "2h" into the formula and see what multiple of V you get:
V = 1/3 pi*(2r)^2*2h
V = 1/3 pi * 4r^2 * 2h
V = (4*2) 1/3 pi * r^2 * h
V = 8 * 1/3 pi * r^2 * h
This equation tells us that the volume of the new cone will be 8x the volume of the old cone.
• Does ayone have the time to explain me the last "try it" question because I really didn't catch it.
I would be very happy!
Thank you1
• Sure! I hope you meant the question that starts with "Right circular cylinder A has a volume of 64pi cubic feet."
For me, the easiest way to solve this question and others like it is to set up an equation where you can track how the final volume changes. The normal volume formula for a cylinder is V = pi*r^2*h. Let's have r and h be the radius and height of cylinder A, and then we can test the volumes of all the cylinders in the options to see which ones fit.
A)
V_a = pi * (2r)^2 * 0.5h
V_a = 4/2 * (pi * r^2 * h)
A) doesn't give us the volume we want, instead being 2x the volume of the starting cylinder.

B)
V_b = pi * (2r)^2 * .25h
V_b = 4/4 * (pi * r^2 * h)
Choice B) works, as we get the same volume formula back, with no coefficient like in A).

C)
V_c = pi * (.5r)^2 * 8h
V_c = 8/4 * (pi * r^2 * h)
C) again gives us a cylinder 2x bigger.

D)
V_d = pi * (1/3 r)^2 * 9h
V_d = 9/9 (pi * r^2 * h)
D) again works.

If you're really familiar with the formula, you can solve problems like these a lot faster. You know that the exponent on r is twice as much as that on h, so the factor of h would have to be the inverse of the square of the factor of r in order to cancel the two coefficients that result. The only answer choices that have a pair of r and h values like this are B) and D).