At a grocery store, she sees a can of tomato sauce that has a radius and height that are twice that of the small can. The volume of this large can is equal to the volume of how many of the small cans?

Question

Madison Shannon

Mathematics

30 November, 17:09

At a grocery store, she sees a can of tomato sauce that has a radius and height that are twice that of the small can. The volume of this large can is equal to the volume of how many of the small cans?

+5

Answers (2)

Know the Answer?

Jayla Navarro · Answer 1

8 cans

Step-by-step explanation:

Here 2 kinds of cans of tomato sauce is available is a grocery store (small cans and big cans).

The radius and the height of the big cans are twice that of the small cans and the question says how many cans (in volume) is going to cover for the volume of that of the big one

Remember that the can of tomato is in the form of a cylinder

Volume of a cylinder is pi*r²*h

Let's now assume that the radius of the small can is 7 ans the height is 3

Then the volume of the small can will be

22/7 * 7²*3 = 462cm

Now for the volume of the big can,

The radius will be 14 and height is 6 (since the radius and the height is twice that of the small can)

22/7*14²*6 = 3696

Now divide the volume of the big one by the small one and you have the number of cans to cover for the big one: 3696/462 = 8 cans

Selena Dudley · Answer 2

The volume of the large can is equal to the volume of 8 small cans.

Step-by-step explanation:

The tomato can has a cylinder format, and the volume of a cylinder is calculated multiplying the base area by the height.

The base area is the area of a circle, which is pi*r^2, where r is the radius.

If the radius increases by 2 times, the base area increases by 4 times, as the base area has r^2 in its formula.

If the height increases by 2 times, the volume of the cylinder also increases by 2 times.

So, in total, the volume of the large can is 8 times the volume of the small can, that is, the volume of the large can is equal to the volume of 8 small cans.

We can also solve this problem using the formulas:

V = A*h, where V is the small can volume, A is the base area of the small can, and h is the height of the small can.

A = pi*r^2, where r is the radius of the small can.

h' = 2*h (h' is the height of the large can)

r' = 2*r (r' is the radius of the large can)

A' = pi * (r') ^2 = pi * (2r) ^2 = pi*4r^2 = 4A (A' is the base area of the large can)

V' = A'*h' = 4A * 2h = 8Ah = 8V (V' is the volume of the large can)