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A shipping box in the shape of a cube has an edge length of 8 inches. It can hold 8 mugs with no extra space between them. If the dimensions of the shipping box are all doubled, what is the maximum number of mugs that the larger box can hold? 16 24 48 64

2 Answers

3 votes

Answer:

The maximum number of mugs that the larger box can hold is:

64

Explanation:

The edge length of box is: 8 inches.

This means that the volume of box is: 8×8×8=512 cubic inches.

Also, the number of mugs a box of edge length 8 in. can hold is: 8

Hence, volume of 1 mug= Volume of box/Number of mugs

Volume of 1 mug= 512/8=64 cubic inches

Now, if the dimension of the box is doubled.

This means that the Volume of new box= 8×Volume of old box

Volume of new box=8×512=4096 cubic inches

Let n be the number of mugs that the new box can hold.

Volume of new box=n×Volume of 1 mug

⇒ 4096=n×64

⇒ n=4096/64

⇒ n=64

Hence, the answer is:

64

User Gitbox
by
8.1k points
5 votes

Answer:

64

Explanation:

Volume of a cube is:

V = s³

When s = 8 in:

V = (8 in)³ = 512 in³

When s is doubled (s = 16 in):

V = (16 in)³ = 4096 in³

Now we can write a proportion:

8 mugs / 512 in³ = x mugs / 4096 in³

x = 64

User Zane XY
by
8.9k points