# Exercise 13.1 Direct and Inverse Proportions- NCERT Solutions Class 8

Go back to  'Direct and Inverse Proportions'

## 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.

## 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.

## 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.

## 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.

## 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 $$2 \,\rm{cm.}$$

## 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 $$21\;\rm{ m.}$$

## 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.

## 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 $$4 \,\rm{cm}$$.

## 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 $$8.75\, \rm{m.}$$

## 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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