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Volume Word Problems

Volume of a solid is defined as the measure of the amount of space contained in the solid. The solid could be of any shapes like cube, cuboid, cone, cylinder, prism, sphere, pyramid and many more. The solid for which we try to find out the volume need to be three dimensional in shape. The unit of measurement of volume of a solid figure is always cubic.

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Definition

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The measure of the quantity of space that a solid figure can hold is said to be the volume of the solid figure. Few of the most commonly used volume formulas are as follows:

Volume of a cube = Side $\times$ Side $\times$ Side

Volume of a cuboid = length $\times$ width $\times$ height

Volume of a prism = area of the base $\times$ height. The base of the prism could be of any shape

Volume of cylinder = $\pi\ \times$ radius2 $\times$ height

Volume of hollow cylinder = $\pi\ \times$ (Outer radius2 – Inner radius2) $\times$ height

Volume of cone = $\frac{1}{3}$ $\times\ \pi\ \times$ radius2 $\times$ height

Volume of pyramid = $\frac{1}{3}$ $\times$ area of the base $\times$ height

Volume of sphere = $\frac{4}{3}$ $\times\ \pi\ \times$ radius3

Volume of hemisphere = $\frac{3}{3}$ $\times\ \pi\ \times$ radius3

Examples

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Example 1: 

Metal spheres are taken $15 in$ number and melted to form a cuboid. The dimension of each metal sphere is radius of $7 m$ and the dimension of each cuboid has a length of $5m$ and width of $2 m$. Find out the height of the cuboid.

Solution: 

Metal Sphere: Volume of metal sphere = $\frac{4}{3}$ $\times \pi\ \times\ radius^3$

Substituting the value of radius in the formula of volume, we compute

Volume of each metal sphere = $\frac{4}{3}$ $\times\ \pi\ \times\ 7^3$ = $1437.3333\ m^3$

Volume of $15$ metal spheres = $15 \times$ Volume of each metal sphere = $15 \times 1437.333$ = $21560\ m^3$

Cuboid: Volume of a cuboid = length $\times$ width $\times$ height

Let the height of the cuboid be $h$

Volume of a cuboid = $5 \times 2 \times h$ = $10h\ m^3$

According to the problem, $15$ metal spheres has been melted to form a cuboid so their volume shall be equal.

Volume of $15$ metal spheres = Volume of a cuboid

Substituting the values of volume of $15$ metal spheres and volume of a cuboid

$21560$ = $10h$

$H$ = $2156\ m$

Thus, the height of the cuboid is $2156\ m$.
Example 2: 

A scoop of chocolate ice cream is placed on an ice cream cone. The radius of the ice cream cone is $5 cm$. On melting the ice cream fills up $\frac{2}{3}$rd of the ice cream cone. Find the volume of the ice cream cone.

Solution: 

The scoop of the ice cream is placed over the ice cream cone, so the radius of the cone and the radius of the sphere is the same. The amount of melted ice cream is the volume of the ice cream scoop. 

Volume of the ice cream scoop = Volume of the sphere = $\frac{4}{3}$ $\times\ \pi\ \times\ radius^3$

Substituting the value of the radius of the sphere in the volume of the sphere formula, we get

Volume of the ice cream scoop = $\frac{4}{3}$ $\times\ \pi\ \times\ 5^3$

According to the problem, $\frac{2}{3}$rd of the volume of the ice cream cone = Volume of the ice cream scoop

Formula of volume of cone = $\frac{1}{3}$ $\times\ \pi\ \times\ radius^2\ \times\ height$

Radius given = $5 cm$ and let the height of the cone be '$h$'. Substituting these values, we get $\frac{2}{3}$ $\times\ \pi\ \times\ 5^2\ \times\ h$ = $\frac{4}{3}$ $\times\ \pi\ \times\ 5^3$

Separating out $h$, 

'$h$ = $\frac{(\frac{4}{3}\ \times\ \pi\ \times\ 5^3)}{(\frac{2}{3}\ \times\ \pi\ \times\ 5^2)}$

'$h$ = $2 \times 5$

'$h$ = $10 cm$

$V$ = $\frac{1}{3}$ $\times\ \pi\ \times\ radius^2\ \times\ height$

$V$ = $\frac{1}{3}$ $\times\ \pi\ \times\ 5^2\ \times\ 10$

$V$ = $261.9 cm^3$

Thus, the height of the ice cream cone is $10 cm$ and the volume of the ice cream cone is $261.9 cm^3$.
Example 3: 

$24$ cans of sweets are packed in a rectangular box in $4$ rows of $6$ each. Each can is cylinder in shape having radius of $4 cm$ and height of $15 cm$. The packing is done in such a manner that no space is left at the top, side or bottom. Find out the volume of empty space unoccupied by the cans.

Solution: 

Number of rows = $4$ and number of columns = $6$. The height of each row is equal to the height of the cylinder. The height of each row is 15 cm so the height of $4$ rows is $4 \times 15$ = $60 cm$. Similarly, radius of each cylinder = $4 cm$ and number of radius which occupies the total width of the rectangular box = $2 \times 6$ = $12$. So, the width of the rectangular box = $12 \times 4$ = $48 cm$. The base of the rectangular box has $4$ cans length wise each having diameter of $2 \times 4$ = $8 cm$. Length of the rectangular box would be $4 \times 8$ = $32 cm$. The rectangular box is cuboid in shape. So, the volume of the rectangular box = length $\times$ width $\times$ height. Volume of the rectangular box = $32 \times 48 \times 60$ = $92160 cc$. 

Volume of each cylinder = $\pi\ \times\ radius^2\ \times\ height$

Substituting the values of radius = $4 cm$, height = $15 cm$, we get volume of each cylinder = $\pi\ \times\ 4^2\ \times\ 15$ = $753.98 cc$. So, volume of $24$ cans = $24\ \times$ Volume of each can = $24\ \times\ 753.98$ = $18095.57 cc$.

Volume of empty space unoccupied by the cans = Volume of the rectangular box – Volume of $24$ cylindrical cans 

Thus, the volume of empty space unoccupied by the cans = $92160 cc\ -\ 18095.57 cc$ = $74064.43 cc$
Example 4: 

Mark buys a perfume bottle which is in the shape of triangular prism. He applies the perfume fast and is left with only half of it. Find out the volume of the perfume left in the bottle. Given, length of the base is $6 cm$, height of the base is $4 cm$ and the height of the bottle is $18 cm$.

Solution: 

Triangular prism mean the base of the perfume bottle is triangular in shape. So, in order to find out the volume we need to first find out the surface area of the triangular base. 

Area of the base = Area of the triangle

Area of the base = $\frac{1}{2}$ $\times$ base $\times$ height

Area of the base = $\frac{1}{2}$ $\times\ 6\ \times 4$

Area of the base = $12 cm^2$

Height of the perfume contained in the bottle = $\frac{1}{2}$ of the height of the perfume bottle.

Height of the perfume contained in the bottle = $\frac{1}{2}$ $\times\ 18$

Height of the perfume contained in the bottle = $9 cm$

Volume of the filled up perfume in the bottle = Area of the triangular base $\times$ height of the perfume contained in the bottle

Volume of the filled up perfume in the bottle = $12 \times 9$ = $108 cc$

Thus, the volume of the perfume left in the bottle = $108 cc$
Example 5: 

A swimming pool has dimensions: length = $8 m$, width = $6 m$, height = $1.5 m$. The inside walls of the swimming pool is painted in order to make it water resistant. The cost of painting is $ \$6$ per square meter. Find out:

a) The cost of painting the interior walls of the swimming pool

b) The storage capacity of the swimming pool when filled up to the brim

Solution: 

i) In order to find out the cost of painting the interior walls of the swimming pool we need to find out the surface area of the swimming pool. The surface area of the swimming pool = Area of the base + area of the four walls

Surface area of the swimming pool = length $\times$ width + $2\ \times$ length $\times$ height + $2\ \times$ width $\times$ height

Surface area of the swimming pool = $8\ \times\ 6\ +\ 2\ \times\ 8\ \times\ 1.5\ +\ 2\ \times\ 6\ \times\ 1.5$

Surface area of the swimming pool = $90 m^2$

Cost of painting per sq m of swimming pool = $ \$6$

The cost of painting the interior walls of the entire swimming pool = Cost of painting per square meter $\times$ Surface area of the swimming pool

Thus, cost of painting the interior walls of the swimming pool = $90 \times 6$ = $ \$540$

ii) The storage capacity of the swimming pool when filled up to the brim is the volume of the rectangular swimming pool.

Volume of the swimming pool = Volume of cuboid

Volume of swimming pool = length $\times$ width $\times$ height

Volume of swimming pool = $8\ \times\ 6\ \times\ 1.5$ = $72 m^3$

Thus, the storage capacity of the swimming pool when filled up to the brim is $72 m^3$.
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