# Exercise E13.5 Surface Areas and Volumes NCERT Solutions Class 9

Go back to  'Surface Areas and Volumes'

## Question 1

A matchbox measures $$\rm 4 cm × 2.5 cm × 1.5 cm.$$ What will be the volume of a packet containing 12 such boxes?

### Solution

Reasoning:

Volume of cuboid is base area $$\times$$height

Therefore,

Volume $$=$$ length $$\times$$ breadth $$\times$$ height

Volume is the space occupied by the solid.

What is known?

The measurements $$l, b, h$$ of a cuboid.

What is unknown?

Volume of the container containing $$12$$ such boxes.

Steps:

length $$(l) = 4\,\,\rm cm$$

breadth $$(b) = 2.5\,\,\rm cm$$

height $$(h) = 1.5\,\,\rm cm$$

Volume of a cuboid

\begin{align}=l \times b \times h \end{align}

Volume of $$1$$ matchbox

\begin{align} = 4 \times 2.5 \times 1.5 \end{align}

Volume of $$12$$ such boxes

\begin{align} = 4 \times 2.5 \times 1.5 \times 12 = 180\,\,c{m^3}\end{align}

Volume of a packet containing $$12$$ such boxes

\begin{align} = 180\,\,\rm c{m^3} \end{align}

## Question 2

A cuboidal water tank is $$6 \rm m$$ long, $$5\rm m$$ wide and $$4.5\rm m$$ deep. How many liters of water can it hold? $$(1 \text{ m}^3= 1000 \,l)$$

### Solution

Reasoning:

Capacity of the tank is nothing but the volume of the cuboid.

What is known?

The length, breadth and height the tank.

What is unknown?

Capacity of the tank in litres.

Steps:

Capacity of the tank $$=$$ Volume of the cuboidal tank.

Length\begin{align}(l) = 6\,\,\rm m \end{align}

Breadth \begin{align}(b) = 5\,\,\rm m \end{align}

Height \begin{align}(h) = 4.5\,\,\rm m \end{align}

Volume of a cuboid

\begin{align} = 6 \times 5 \times 4.5 = 135\,\,\rm{m^3} \end{align}

\begin{align} 1\,\,\rm {m^3} = 1000\,\,l \end{align}

\begin{align}\therefore 135\,\,\rm{m^3} &= 135 \times 1000\\& = 135000\,\,l\end{align}

Litres of the water the cuboidal water tank can hold \begin{align} = 135000\,\,l \end{align}.

## Question 3

A cuboidal vessel is $$10\rm m$$ long and $$8 \;\rm{ m}$$ wide. How high must it be made to hold $$380\rm \,\,cubic\, metres$$ of a liquid?

### Solution

Reasoning:

Capacity of the cuboid vessel is the cuboid.

Volume of the cuboid$$=$$length$$\times$$breadth$$\times$$height

What is known?

Length and breadth of the cuboidal vessel, and the capacity of the vessel.

What is unknown?

The height of the vessel having \begin{align}380\,\rm{m^3} \end{align} capacity.

Steps:

Capacity of the cuboidal vessel$$=$$volume of the cuboid.

Volume of the cuboid \begin{align} = l \times b \times h \end{align}

Length $$(l) = 10\,\,\rm m$$

Breadth $$(b) = 8\,\,\rm m$$

Volume $$(V) = 380\,\,\rm c{m^3}$$

Height$$(h) = ?$$

\begin{align} 380 &= 10 \times 8 \times h\\ h& = \frac{{380}}{{10 \times 8}} \\&= \frac{{19}}{4} \\ &=4.75 \rm m\end{align}

The cuboidal vessel must be made $$4.75 \rm m$$ height.

## Question 4

Find the cost of digging a cuboidal pit $$8 \rm m$$ long, $$6\,\rm m$$ broad and $$3\,\rm m$$ deep at the rate of \begin{align}\rm Rs \,30 \,\rm per\,\rm {m^3}. \end{align}

### Solution

Reasoning:

Volume of the pit will be equal to volume of a cuboid and the cost to dug the pit will be equal to product of the volume of the pit and cost to dig the pit per meter cube.

What is known?

The length, breadth and height the cuboid. Cost of digging $$\rm Rs\, 30\, per$$\begin{align}\rm {m^3} \end{align}.

What is unknown?

Cost of digging the cuboidal pit.

Steps:

length $$(l) = 8\,\,\rm m$$

breadth $$(b) = 6\,\,\rm m$$

height $$(h) = 3\,\,\rm m$$

Volume of the cuboid

\begin{align} = l \times b \times h \end{align} \begin{align} = 8 \times 6 \times 3\,\,\rm {m^3} \end{align}

Cost of the digging the pit is \begin{align}\rm Rs \,30 \,\rm per\,\rm {m^3} \end{align}

$$\therefore$$ Cost of digging for \begin{align}8\times 6\times 3\,\rm {{m}^{3}} \end{align}

\begin{align} & =30\times 8\times 6\times 3 \\ & = {\rm{Rs}}\,4320 \end{align}

Cost of digging the cuboidal pit is $$\rm= Rs \,4320.$$

## Question 5

The capacity of a cuboidal tank is $$50000$$ litres of water. Find the breadth of the tank, if its length and depth are respectively $$2.5\rm m$$ and $$10\rm m.$$

### Solution

Reasoning:

Capacity of the cuboidal tank $$=$$ Volume of the cuboidal tank.

Volume of the cuboid \begin{align} = l \times b \times h \end{align}

What is the known?

Capacity of the tank length and height of the tank.

What is the unknown?

Steps:

Capacity of the tank $$= 50000\rm \,\,\ litres.$$

\begin{align}1000\,\,l& = 1\,\,\rm {m^3} \\50000\,l &= \frac{{50000}}{{1000}} = 50\,\,\rm{m^3} \end{align}

Volume has to be changed in \begin{align}\rm {m^3} \end{align} because all the measurements are in meter.

Volume of the cuboid\begin{align}\, = l \times b \times h \end{align}

length$$\,(l)\, = 2.5\,\rm cm$$

breadth$$\,(b)\,= \,?$$

height $$(h) = 10\,\,\rm cm$$

\begin{align}2.5 \times 10 \times b &= 50\\b &=\frac{{50}}{{2.5 \times 10}}\\ &= 2\,\,\rm m \end{align}

The breadth of the cuboidal tank is $$2 \rm \,m.$$

## Question 6

A village, having a population of $$4000,$$ requires $$150$$ litres of water per head per day. It has a tank measuring $$\rm 20 \,m × 15 \,m × 6 \,m.$$ For how many days will the water of this tank last?

### Solution

Reasoning:

Capacity of the tank is the volume of cuboid.

Volume of the tank = length $$\times$$ breadth $$\times$$ height

What is the known?

Requirement of water per head per day measurements of the tank.

What is the unknown?

Number of days the water of the tank will last.

Steps:

Requirement of water per head per day $$= 150\rm\, litres.$$

Requirement of water for $$4000$$ population\begin{align}\, = 150 \times 4000 = 600000\,l \end{align}

\begin{align}1000\,l &= 1\,\,\rm{m^3} \\\therefore 600000l &= \frac{{600000}}{{1000}} = 600\,\, \rm {m^3} \end{align}

length of the tank$$(l) = 20\,\,\rm m$$

breadth of the tank$$(b) = 15\,\,\rm m$$

height of the tank$$(h) = 6\,\,\rm m$$

Capacity of the tank $$=$$ Volume of the cuboid

\begin{align}& = l \times b \times h \\ &= 20 \times 15 \times 6 \\ &= 1800\,\, \rm {m^3} \end{align}

Number of days for which the water of the tank will last

\begin{align} & = \frac{\text{Capacity of the tank}}{\left[\begin{array}{l} {\text{Requirement of water }} \\ {\text{for the total population}} \\ \end{array} \right]} \\ & = \frac{{1800}}{{600}} \\ & = 3{\text{ days}}\end{align}

The water of tin tank will last for $$3$$ days.

## Question 7

A godown measures $$\rm 40 m \!×\! 25 m \!×\! 15 m.$$ Find the maximum number of wooden crates each measuring $$\rm 1.5 m \!×\! 1.25 m \!×\! 0.5 m$$ that can be stored in the godown.

### Solution

Reasoning:

The maximum number of wooden crates that can be stored in the godown will be the ratio of volume of the godown to the volume of wooden crate.

What is known?

Measurements of the godown measurements of the wooden crates.

What is unknown?

Number of crates that can be stored in the godown.

Steps:

Volume of the godown $$=$$ Capacity of godown

length $$(l) = 40\,\,\rm m$$

breadth $$(b) = 25\,\,\rm m$$

height $$(h) = 15\,\,\rm m$$

Capacity of the godown

\begin{align} = 40 \times 25 \times 15 \end{align} \begin{align} = 1500\,\,{m^3} \end{align}

Volume of the wooden crate:

length $$(l) = 1.5\,\,\rm m$$

breadth $$(b) = 1.25\,\,\rm m$$

height $$(h) = 0.5\,\,\rm m$$

Volume of the wooden crate

\begin{align} &= 1.5 \times 1.25 \times 0.5 \\& = 0.9375\,\,{m^3} \end{align}

\left[ \begin{align} {\text{Number}} \\ {\text{of crates}} \\ \end{align} \right] \!=\! \left[ {\frac{{{\text{Number of crates}}}}{{{\text{Volume of the crate}}}}} \right]

The maximum number of wooden crates that can be stored in the godown is $$1600.$$

## Question 8

A solid cube of side $$12 \rm \,cm$$ is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

### Solution

Reasoning:

Volume of the cube is \begin{align}\,{a^3} \end{align} when the side measurement is a. Surface area is nothing but the sum of the area of the $$6$$ faces.

What is  known?

Side length of the cube. Number of cubes cut from the bigger cube.

What is  unknown?

Volume of the small cube and ratio between the surface areas of the bigger and the smaller cube.

Steps:

Side of the solid cube $$= 12\rm cm$$

Volume of the solid cube

\begin{align}={a^3} = {(12)^3} = 1728\,\,\rm {m^3} \end{align}

It is cut in to $$8$$ equal cubes of same volume.

Therefore,

Volume of one small cube

\begin{align} & = \frac{{1728}}{8} \\& = 216 \rm \,{m^3} \end{align}

Let $$x$$ be side of the small cube.

\begin{align}\therefore \end{align}Volume of one small cube\begin{align}\,{x^3}\rm c{m^3} \end{align}

\begin{align}{x^3} &= 216\\x &= \sqrt[3]{{216}}\\\,\,\,\, &= {(216)^{\frac{1}{3}}}\\\,\,\,\, &= {(6 \times 6 \times 6)^{\frac{1}{3}}}\\\,\,\,\, &= 6\,\,cm \end{align}

For the ratio of this surface area we have to find the surface area of the two cubes having side 12cm and 6 cm respectively.

Surface area of the cube \begin{align}\, = 6 \times {a^2} \end{align}

Surface area of the cube having $$12 \rm cm$$ side

\begin{align} = 6 \times {12^2} \end{align}

Surface area of the cube having $$6\rm cm$$ side \begin{align}= 6 \times {6^2} \end{align}

\begin{align} \therefore \rm Ratio& = \frac{{6 \times 12 \times 12}}{{6 \times 6 \times 6}} \\ &= \frac{4}{1} = 4:1 \end{align}

The side of the new cube is $$6\rm m.$$

Ratio between the surface area $$= 4:1.$$

## Question 9

A river $$3\rm m$$ deep and $$40 \rm m$$ wide is flowing at the rate of $$2 \rm km$$ per hour. How much water will fall into the sea in a minute?

### Solution

Reasoning:

Water that falls in to the sea in the volume of the cuboid.

What is known?

Depth, wide and rate of water following per hour.

What is unknown?

Amount of water will fall in to the sea in a minute.

Steps:

length $$(l) \!=\! 2\,\rm km \!=\! 2 \!\times\! 1000 \!=\! 2000\,\rm m$$

breadth $$(b) = 40\,\,\rm m$$

height $$(h) = 3\,\,\rm m$$

Water fall in the sea $$=$$ Volume of cuboid $$= l\times b\times h$$

\begin{align} = 2000 \times 40 \times 3 \end{align}

This is for $$1$$ hour

Therefore, water fell in to the sea in a minute.

\begin{align} &= \frac{{2000 \times 40 \times 3}}{{60}}\\ &= 4000\,\,\rm{m^3} \end{align}

\begin{align}4000\,\,\rm{m^3} \end{align} of water will fall into the sea in a minute.