Two cylinders are given. The base radius and the height of the first are 2 and 3, respectively, and the second 8 and 3.

Two cylinders are given. The base radius and the height of the first are 2 and 3, respectively, and the second 8 and 3. How many times is the volume of the second cylinder greater than the volume of the first?

Let’s introduce the notation:
base radius and height of the first cylinder – r1; h1;
base radius and height of the second cylinder – r2; h2.
The volume of a cylinder (V) is equal to the product of its base area (S) and height (h):
V = S * h.
Since the base of the cylinder is a circle, and the area of the circle is S = n * r ^ 2, then:
V = n * r ^ 2 * h,
where n is the number “pi” (n = 3.14);
r is the radius of the cylinder base.
Volumes of the first and second cylinders:
V1 = n * r1 ^ 2 * h1 = n * 2 ^ 2 * 3 = 12n;
V2 = n * r2 ^ 2 * h2 = n * 8 ^ 2 * 3 = 192n.
Let’s find the ratio of the volumes of the cylinders:
V2 / V1 = 192p / 12p = 16.
This means that the volume of the second cylinder is 16 times the volume of the first.
Answer: 16 times.