# NCERT Class 8 Maths Direct and Inverse Proportions

The chapter 13 begins with an introduction to Direct and Inverse Proportions by citing situations in our day-to-day life, where variation in one quantity bringing in variation in the other to delve into the subject of interest.After this , Direct Proportion is explained by stating some real-life scenario and there are some related examples to dig deep into it.Similarly, Inverse Proportion i.e if one quantity increases, the other quantity decreases and vice versa, is also discussed in the same manner with some real-life examples and both of them have some exercise problems to work on.

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## Chapter 13 Ex.13.1 Question 1

Following are the car parking charges near a railway station up to:

$$4$$ hours          ₹$$60$$

$$8$$ hours          ₹$$100$$

$$12$$ hours        ₹$$140$$

$$24$$ hours        ₹$$180$$

Check if the parking charges are in direct proportion to the parking time.

### Solution

What is Known?

Parking charges for different hours.

What is Unknown?

Parking charges are direct proportion to the parking time or not.

Reasoning:

If two quantities are related in such a way that an increase in one lead to a corresponding proportional increase in the other, then such a variation is called direct variation.

Steps:

The parking charge for $$1$$ hour in all the four cases then the variation is direct.

We have:

\begin{align}\frac{{60}}{4}&=\frac{{15}}{1}\\&=15\\ \\ \frac{{100}}{8}&=\frac{{25}}{2}\\ &= 12.5\\ \\ \frac{{140}}{{12}} &=\frac{{35}}{3}\\&=11.67 \\ \\\frac{{180}}{{24}}&=\frac{{15}}{2}\\&= 7.50\end{align}

Since all the values are not the same, the parking charges are not in direct proportion to parking times.

## Chapter 13 Ex.13.1 Question 2

A mixture of paint is prepared by mixing $$1$$ part of red pigments with $$8$$ parts of base. In the following table find the parts of base that needed to be added.

 Parts of red pigment 1 4 7 12 20 Parts of base 8 ... ... ... ...

### Solution

What is Known?

Parts of red pigments used.

What is Unknown?

Parts of base used.

Reasoning:

Two numbers $$x$$ and $$y$$ are said to vary in direct proportion if

$\frac{x}{y} = k,x = yk$

Where, $$k$$ is a constant.

Steps:

Let the parts of red pigments used be $$x$$ and parts of base used be $$y.$$

\begin{align}\therefore \frac{{{x_1}}}{{{y_1}}} = \frac{{{x_2}}}{{{y_2}}}\end{align}

Here

\begin{align} \quad{x_1} &= 1, \qquad {x_2} = 4 \\ {y_1} &= 8,\qquad {y_2} = ?\\ \frac{{{x_1}}}{{{y_1}}} &= \frac{{{x_2}}}{{{y_2}}} \\ \frac{1}{8} &= \frac{4}{{y2}}\\ \,{y_2} &= 8 \times 4 \\ &= 32{\text{ parts }}\end{align}

$$32$$ parts of base is needed for $$4$$ parts of red pigment.

Here

\begin{align} {{x}_{1}}&=1,\quad \ {{x}_{2}}=7 \\ {{y}_{1}}&=8,\quad{{y}_{2}}=? \\ \frac{{{x}_{1}}}{{{y}_{1}}}&=\frac{{{x}_{2}}}{{{y}_{2}}} \\ \,\,\frac{1}{8}&=\frac{7}{{{y}^{2}}} \\ {{y}_{2}}&=8\times 7 \\ \,\,\,\,\,\,&=56\,\ \text{parts} \end{align}

$$56$$ parts of base is needed for $$7$$ parts of red pigment.

Here

\begin{align} \,\,\,\,\,\,\,\,{{x}_{1}}&=1,\quad{{x}_{2}}=12 \\ {{y}_{1}}&=8,\quad{{y}_{2}}=? \\ \frac{{{x}_{1}}}{{{y}_{1}}}&=\frac{{{x}_{2}}}{{{y}_{2}}} \\ \frac{1}{8}&=\frac{12}{{{y}_{2}}} \\ {{y}_{2}}&=8\times 12\quad \\ &=96 \end{align}

$$96$$ parts of base is needed for $$12$$ parts of red pigment.

Here

\begin{align} {{x}_{1}}&=1,\quad{{x}_{2}}=20 \\ {{y}_{1}}&=8,\quad{{y}_{2}}=? \\ \frac{{{x}_{1}}}{{{y}_{1}}}&=\frac{{{x}_{2}}}{{{y}_{2}}} \\ \frac{1}{8}&=\frac{20}{{{y}_{2}}} \\ {{y}_{2}}&=8\times 20 \\ &=160 \end{align}

$$160$$ parts of base are needed for $$20$$ parts of red pigment.

## Chapter 13 Ex.13.1 Question 3

In question ($$2$$), above if $$1$$ part of red pigment requires $$75\,\rm{ml}$$ of base, how much red pigment should we mix with $$1800\,\rm{ml}$$ of base?

### Solution

What is Known?

$$1$$ part of red pigment requires $$75\,\rm{ml}$$ of base.

What is Unknown?

$$1800$$ mL of base needed how much red pigment?

Reasoning:

Two numbers $$x$$ and $$y$$ are said to vary in direct proportion if,

\begin{align}\frac{x}{y} = k, \qquad x = y\,k\end{align}

Where $$k$$ is a constant.

Steps:

Let the number of parts of red pigment be $$x$$

As the number of parts of red pigment increases, amount of base also increases in the same ratio. So it is a case of direct proportion.

Here,

\begin{align} \frac{{{x}_{1}}}{{{y}_{1}}}&=\frac{{{x}_{2}}}{{{y}_{2}}} \\ {{x}_{1}}&=1,\qquad {{x}_{2}}=? \\ {{y}_{1}}&=8,\qquad{{y}_{2}}=1800 \\ \frac{1}{75}&=\frac{{{x}_{2}}}{1800} \\ {{x}_{2}}&=\frac{1\ \times \ 1800}{75} \\ {{x}_{2}}&=24 \end{align}

$$24$$ parts of red pigment should be mixed with $$1800\,\rm{ml}$$ of base.

## Chapter 13 Ex.13.1 Question 4

A machine in a soft drink factory fills $$840$$ bottles in $$6$$ hours. How many bottles it will fill in five hours?

### Solution

What is Known?

$$840$$ bottles can be filled in $$6$$ hours.

What is Unknown?

Bottles filled in $$5$$ hours.

Reasoning:

Two numbers $$x$$ and $$y$$ are said to be in direct proportion, if

\begin{align}\frac{x}{y}=k,\quad x=y\,k\end{align}

Where $$k$$ is a constant.

Steps:

 No of bottles Time in hours $${x}_{1}=840$$ $${y}_{1}=6$$ $${x}_{2}=?$$ $${y}_{2}=5$$

So, the number of bottles filled, and numbers of hours are directly proportional to each other.

\begin{align} \frac{{{x}_{1}}}{{{y}_{1}}}&=\frac{{{x}_{2}}}{{{y}_{2}}} \\ \frac{840}{6}&=\frac{{{x}_{2}}}{5} \\ 6{{x}_{2}}&=840\,\times \,5 \\ {{x}_{2}}&=\frac{840\times 5}{6} \\ {{x}_{2}}&=700 \end{align}

$$700$$ bottles will be filled in $$5$$ hours.

## Chapter 13 Ex.13.1 Question 5

A photograph of a bacteria is enlarged $$50,000$$ times attains a length of $$​​5 ​​\rm{cm}$$ as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged $$20,000$$ times only, what would be its enlarged length?

### Solution

What is Known?

Bacteria enlarged $$50,000$$ times attain a length of $$5 \,\rm{cm.}$$

What is Unknown?

Actual length of the bacteria

if $$20,000$$ times enlarged what will be the length of the bacteria?

Reasoning:

Two numbers $$x$$ and $$y$$ are said in direct proportion if,

\begin{align}\frac{x}{y}=k,\quad x=y\,k\end{align}

Where $$k$$ is a constant.

Steps:

\begin{align} \text{Actual length,}\ l&=\frac{{{y}_{1}}}{{{x}_{1}}} \\ l&=\frac{5}{50000} \\ l&=0.0001\ \rm{cm} \\ \end{align}

 Number  of  times  enlarged Length attained ${{x_1} = {\rm{50,000}}}$ ${{y_{\rm{1}}} = {\rm{5}}}$ ${{x_2} = {\rm{20,000}}}$ ${{y_{\rm{2}}} = {\rm{?}}}$

The number of times enlarged is directly proportional to the length attained.

Actual length $$= 0.0001 \,\rm{cm}$$

Enlarged length will be $$2 \,\rm{cm.}$$

## Chapter 13 Ex.13.1 Question 6

In a model of a ship, the mast is $$9 \,\rm{cm}$$ high, while the mast of the actual ship is $$12\, \rm{cm}$$ high. If the length of the actual ship is $$28\;\rm{m}$$, how long is the model ship?

### Solution

What is Known?

The mast is $$9 \,\rm{cm}$$ high while the mast of actual ships is $$12\, \rm{cm}$$ high.

What is Unknown?

If the length of the ship is $$28\;\rm{m}$$ How long is the model ship?

Reasoning:

Two numbers $$x$$ and $$y$$ are said in direct proportion if,

\begin{align}\frac{x}{y} = k,\quad x = y\,k\end{align}

Where $$k$$ is a constant.

Steps:

 Actual ship Model ship $${y}_{1}=12\;\rm{m}$$ $${y}_{2}=9\;\rm{cm}$$ $${x}_{1}=28\rm{m}$$ $${x}_{2}=?$$

More the length of the ship more would be the length of its mast. Hence, this is a direct proportion.

\begin{align}\frac{{{x_1}}}{{{y_1}}} &= \frac{{{x_2}}}{{{y_2}}}\\\frac{{28}}{{12}} &= \frac{{{x_2}}}{9}\\12 \times {x_2} &= 28 \times 9\\{x_2} &= \frac{{28 \times 9}}{{12}}\\{x_2} &= 21\;{\rm{m}}\end{align}

Length of the model ship is $$21\;\rm{ m.}$$

## Chapter 13 Ex.13.1 Question 7

Suppose $$2\,\rm{kg}$$ of sugar contains $$9 \times {10^6}$$ crystals. How many sugar crystals are there in

(1)  $$5\,\rm{kg}$$  of sugar?

(2) $$1.2\,\rm{kg}$$ of sugar?

### Solution

(i) How many crystals are there in $$5\, \rm{kg}$$ of crystals?

What is Known?

$$2\,\rm{kg}$$ of sugar contains $$9 × 10^6$$ crystals.

What is Unknown?

(i) $$5\,\rm{kg}$$ of sugar contains how many crystals?

Reasoning:

Two numbers $$x$$ and $$y$$ are said in direct proportion if,

\begin{align}\frac{x}{y} = k,\quad\;\; x = yk\,\end{align}

Where $$k$$ is a constant.

Steps (i):

 Amount of sugar No. of crystals $$2$$ $${9 \times 10^6}$$ $$5$$ ?

More the amount of sugar more will be the number of crystals. Hence this is a direct proportion.

\begin{align}\frac{{{x_1}}}{{{y_1}}} &= \frac{{{x_2}}}{{{y_2}}}\\\frac{2}{{9 \times {{10}^6}}}&= \frac{5}{{{y_2}}}\\2 \times y_2 &= 9 \times {10^6} \times 5\\ y_2 &= \frac{{9 \times {{10}^6} \times 5}}{2}\\{y_2} &= 22.5 \times {10^6}\\{y_2} &= 2.25 \times {10^7}\end{align}

Hence there are $$2.25 \times {10^7}$$ crystals.

(ii) How many crystals are there in $$1.2\,\rm{kg}$$ of crystals?

What is Known?

$$2\,\rm{kg}$$ of sugar contains $$9 × 10^6$$ crystals.

What is UnKnown?

How many crystals are there in $$1.2\,\rm{kg}$$ of crystals?

Steps (ii):

\begin{align}\frac{{{x_1}}}{{{y_1}}} &= \frac{{{x_2}}}{{{y_2}}}\\\frac{2}{{9 \times {{10}^6}}} &= \frac{{1.2}}{{{y_2}}}\\2 \times y_2 &= 9 \times {10^6} \times 1.2\\ y_2 &=\frac{{9 \times {{10}^6} \times 5}}{2}\\{y_2} &= 5.4 \times {10^6}\end{align}

Hence there are $$5.4 \times {10^6}$$ crystals.

## Chapter 13 Ex.13.1 Question 8

Rashmi has a road map with a scale of $$1 \;\rm{cm}$$ representing $$18 \,\rm{km}$$. She drives on a road for $$72 \,\rm{km}$$. What would be her distance covered in the map?

### Solution

What is Known:

The scale of representing $$1\; \rm{cm}$$ $$=$$ $$18 \,\rm{km}$$

What is Unknown:

The distance covered in map when the distance on road is $$72 \,\rm{km.}$$

Reasoning:

Two numbers $$x$$ and $$y$$ are said in direct proportion if

\begin{align}\frac{x}{y} = k,\, \qquad x = y\,k\end{align}

Where $$k$$ is a constant.

The map is a representation of very large region. The scale shows the representation of the actual length and the length represented in map.

Steps:

$$1\,\rm{cm}$$ on map represents $$18 \,\rm{km}$$ of actual distance then $$2\, \rm{cm}$$ on the map represents $$36 \,\rm{km}$$. Hence the scale is based on the concept of direct proportion.

\begin{align}1:18 &= x:72\\\frac{{1}}{18} &= \frac{x}{{72}}\\18 \times x &= 72 \times 1\\x &= \frac{{72}}{{18}}\\x &= 4\end{align}

The distance covered in the map would be $$4 \,\rm{cm}$$.

## Chapter 13 Ex.13.1 Question 9

A $$5\,\rm{m}$$ $$60\,\rm {cm}$$ high vertical pole casts a shadow $$3\;\rm{m}$$ $$20 \;\rm{cm}$$ long. Find at the same time

(i) Length of the shadow cast by another pole $$10 \;\rm{m}$$ $$50 \;\rm{cm}$$ high.

(ii) The height of the pole which casts a shadow of $$5\;\rm{m}$$ long.

### Solution

(i) Length of the shadow cast by another pole $$10\; \rm{m}$$ $$50 \;\rm{cm}$$ high.

Waht is Known?

$$5.6\, \rm{m}$$ vertical pole casts a shadow of $$3.2\; \rm{m}$$ long.

Waht is Unknown?

The length of a shadow cast by a pole $$10.5\; \rm{m}$$ high.

Reasoning:

Two numbers $$x$$ and $$y$$ are said in direct proportion if

\begin{align}\frac{x}{y} = k,\quad x= y\,k\end{align}

Where $$k$$ is a constant.

Steps:

 Height of the pole Length of the shadow $${5.6{\rm{m}}}$$ $${3.2{\rm{m}}}$$ $${10.5{\rm{m}}}$$ $$?$$

As the height of the pole increases the length of the shadow also increases. So, it is a direct proportion.

\begin{align}\frac{{{x_1}}}{{{y_1}}}&= \frac{{{x_2}}}{{{y_2}}}\\\frac{{5.6}}{{3.2}}&= \frac{{10.5}}{{{y_2}}}\\\,\,5.6\,\, \times &= 10.5 \times 3.2\\{y_2}&= \frac{{10.5 \times 3.2}}{{5.6}}\\ y_2 & = 6\end{align}

If the height of the pole is $$10.5\; \rm{m}$$, then length of the shadow is $$6\;\rm{m}$$.

(ii) The height of the pole which casts a shadow of $$5\,\rm{m}$$ long.

Waht is Known?

$$5.6 \,\rm{m}$$ vertical pole casts a shadow of $$3.2\,\rm{m}$$ long.

Waht is Unknown?

The height of the pole when the length of the shadow is $$5\,\rm{m}$$ long.

Steps:

\begin{align}\frac{{{x_1}}}{{{y_1}}}&= \frac{{{x_2}}}{{{y_2}}}\\\frac{{5.6}}{{3.2}}&= \frac{{{x_2}}}{5}\\3.2x &= 5 \times 5.6\\{x_2}&= \frac{{5 \times 5.6}}{{3.2}}\\{x_2}&= 8.75\end{align}

If the height of the pole is $$5\,\rm{m}$$, then length of the shadow is $$8.75\, \rm{m.}$$

## Chapter 13 Ex.13.1 Question 10

A loaded truck travels $$14\,\rm{km}$$ in $$25$$ minutes. If the speed remains the same, how far it travels in $$5$$ hours?

### Solution

What is Known?

Truck travels $$14\,\rm{km}$$ in $$25$$ minutes.

What is Unknown?

Distance travelled in $$5$$ hours.

Reasoning:

Two numbers $$x$$ and $$y$$ are said in direct proportion if,

\begin{align}\frac{x}{y} = k,\quad x = k\,y\end{align}

Where $$k$$ is a constant.

Steps:

In $$25$$ minutes, it travels $$14 \,\rm{km}$$. In $$5$$ hours, it will travel more distance. So, it is a case of direct proportion.

 Distance Time in  minutes $${{\rm{14}}}$$ $${{\rm{25}}}$$ $${\,{\rm{?}}}$$ $$5 \times 60$$ ($$1$$ hour $$=$$ $$60$$ minutes)

[For comparison the unit should be same]

\begin{align}\frac{{{x_1}}}{{{y_1}}} &= \frac{{{x_2}}}{{{y_2}}}\\\frac{{14}}{{25}} &= \frac{{{x_2}}}{{5 \times 60}}\\25\,{x_2} &= 5 \times 60 \times 14\\{x_2} &= \frac{{5 \times 60 \times 14}}{{25}}\\{x_2} &= 168\;{\rm{km}}\end{align}

Hence the truck can travel $$168 \,\rm{km}$$ in $$5$$ hours.

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