Examples with solutions for Volume of a Orthohedron: Worded problems

Exercise #1

If we increase the side of a cube by 6, how many times will the volume of the cube increase by?

Video Solution

Step-by-Step Solution

Let's denote the initial cube's edge length as x,

The formula for the volume of a cube with edge length b is:

V=b3 V=b^3

Therefore the volume of the initial cube (meaning before increasing its edge) is:

V1=x3 V_1=x^3

Proceed to increase the cube's edge by a factor of 6, meaning the edge length is now: 6x . Therefore the volume of the new cube is:

V2=(6x)3=63x3 V_2=(6x)^3=6^3x^3

In the second step we simplified the expression for the new cube's volume by using the power rule for multiplication in parentheses:

(zy)n=znyn (z\cdot y)^n=z^n\cdot y^n

We applied the power to each term inside of the parentheses multiplication.

Next we'll answer the question that was asked - "By what factor did the cube's volume increase", meaning - by what factor do we multiply the old cube's volume (before increasing its edge) to obtain the new cube's volume?

Therefore to answer this question we simply divide the new cube's volume by the old cube's volume:

V2V1=63x3x3=63 \frac{V_2}{V_1}=\frac{6^3x^3}{x^3}=6^3

In the first step we substituted the expressions for the volumes of the old and new cubes that we obtained above. In the second step we reduced the common factor between the numerator and denominator,

Therefore we understood that the cube's volume increased by a factor of -63 6^3 when we increased its edge by a factor of 6,

The correct answer is b.

Answer

63 6^3

Exercise #2

A building is 21 meters high, 15 meters long, and 14+30X meters wide.

Express its volume in terms of X.

(14+30X)(14+30X)(14+30X)212121151515

Step-by-Step Solution

We use a formula to calculate the volume: height times width times length.

We rewrite the exercise using the existing data:

21×(14+30x)×15= 21\times(14+30x)\times15=

We use the distributive property to simplify the parentheses.

We multiply 21 by each of the terms in parentheses:

(21×14+21×30x)×15= (21\times14+21\times30x)\times15=

We solve the multiplication exercise in parentheses:

(294+630x)×15= (294+630x)\times15=

We use the distributive property again.

We multiply 15 by each of the terms in parentheses:

294×15+630x×15= 294\times15+630x\times15=

We solve each of the exercises in parentheses to find the volume:

4,410+9,450x 4,410+9,450x

Answer

4410+9450x 4410+9450x

Exercise #3

After cleaning the public quadrilateral pool,
To be completed again
we fill it with buckets,
The volume of each bucket is 8 liters.
The quadrilateral pool with a depth of 3 meters and a width of 10 meters,
How many buckets are needed to refill the pool?

Video Solution

Answer

37500

Exercise #4

A cargo ship anchored in port, its volume is 360 cm³.
The vessel is composed of 6 small vessels.
They disassembled 3 small tanks whose total volume is half of the cargo tanks.
Suggest sizes for the edges of each small tank if its height is known to be 3 meters.

Video Solution

Answer

2X10, 4X5