# The length of one parallelepiped is 6 times less than the length of the other, the width is 9 times

**The length of one parallelepiped is 6 times less than the length of the other, the width is 9 times the width of the other, the height is 3 times less than the height of the other.**

Let’s say that the length of the first parallelepiped is x.

Since it is 6 times smaller than the width of another figure, its length will be equal to: 6 * x.

Write down the width of the second parallelepiped as y.

Since the width of the first is 9 times larger, it will be equal to: 9 * y.

Let’s designate the height of the first box as z.

In this case, the height of the second box will be: 3 * z.

The volume of the first figure will be equal to:

x * 9 * y * z = 9 * x * y * z.

The volume of the second figure will be:

6 * x * y * 3 * z = 18 * x * y * z.

It follows from this that the volume of the second parallelepiped is larger in:

18 * x * y * z / 9 * x * y * z = 2 times.

Answer: 2 times more.