Examples (Revised) - Chapter 13 - Direct & Inverse Proportions - Ncert Solutions class 8 - Maths

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Chapter 11 - Direct & Inverse Proportions | NCERT Solutions for Class 8 Maths

Example 1:

The cost of 5 metres of a particular quality of cloth is $₹ 210$. Tabulate the cost of $2,4,10$ and 13 metres of cloth of the same type.

Solution:

Suppose the length of cloth is $x$ metres and its cost, in ₹, is $y$.

As the length of cloth increases, cost of the cloth also increases in the same ratio. It is a case of direct proportion.
We make use of the relation of type $\frac{x_1}{y_1}=\frac{x_2}{y_2}$
(i) Here $x_1=5, y_1=210$ and $x_2=2$
Therefore, $\frac{x_1}{y_1}=\frac{x_2}{y_2}$ gives $\frac{5}{210}=\frac{2}{y_2}$ or $5 y_2=2 \times 210$ or $y_2=\frac{2 \times 210}{5}=84$
(ii) If $x_3=4$, then $\frac{5}{210}=\frac{4}{y_3}$ or $5 y_3=4 \times 210$ or $y_3=\frac{4 \times 210}{5}=168$ [Can we use $\frac{x_2}{y_2}=\frac{x_3}{y_3}$ here? Try!]
(iii) If $x_4=10$, then $\frac{5}{210}=\frac{10}{y_4}$ or $y_4=\frac{10 \times 210}{5}=420$
(iv) If $x_5=13$, then $\frac{5}{210}=\frac{13}{y_5}$ or $y_5=\frac{13 \times 210}{5}=546$
$\left[\right.$ Note that here we can also use $\frac{2}{84}$ or $\frac{4}{168}$ or $\frac{10}{420}$ in the place of $\left.\frac{5}{210}\right]$

Example 2:

An electric pole, 14 metres high, casts a shadow of 10 metres. Find the height of a tree that casts a shadow of 15 metres under similar conditions.
Solution:

Let the height of the tree be $x$ metres. We form a table as shown below:

Note that more the height of an object, the more would be the length of its shadow.
Hence, this is a case of direct proportion. That is, $\frac{x_1}{y_1}=\frac{x_2}{y_2}$
We have
$\frac{14}{10}=\frac{x}{15}$ (Why?)
or
$\frac{14}{10} \times 15=x$
or
$\frac{14 \times 3}{2}=x$

So
$
21=x
$

Thus, height of the tree is 21 metres.
Alternately, we can write $\frac{x_1}{y_1}=\frac{x_2}{y_2}$ as $\frac{x_1}{x_2}=\frac{y_1}{y_2}$

so
$
x_1: x_2=y_1: y_2
$

Therefore,
or
$14: x=10: 15$
$10 \times x=15 \times 14$
$
x=\frac{15 \times 14}{10}=21
$

Example 3:

If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same paper would weigh $2 \frac{1}{2}$ kilograms?
Solution:
Let the number of sheets which weigh $2 \frac{1}{2} \mathrm{~kg}$ be $x$. We put the above information in the form of a table as shown below:

1 kilogram $=1000$ grams
$2 \frac{1}{2}$ kilograms $=2500$ grams

More the number of sheets, the more would their weight be. So, the number of sheets and their weights are directly proportional to each other.

So,
$\frac{12}{40}=\frac{x}{2500}$
or
$
\frac{12 \times 2500}{40}=x
$
or
$
750=x
$

Thus, the required number of sheets of paper $=750$.
Alternate method:

Two quantities $x$ and $y$ which vary in direct proportion have the relation $x=k y$ or $\frac{x}{y}=k$
Here, $\quad k=\frac{\text { number of sheets }}{\text { weight of sheets in grams }}=\frac{12}{40}=\frac{3}{10}$
Now $x$ is the number of sheets of the paper which weigh $2 \frac{1}{2} \mathrm{~kg}[2500 \mathrm{~g}]$.
Using the relation $x=k y, x=\frac{3}{10} \times 2500=750$
Thus, 750 sheets of paper would weigh $2 \frac{1}{2} \mathrm{~kg}$.
Example 4:

A train is moving at a uniform speed of $75 \mathrm{~km} / \mathrm{hour}$.
(i) How far will it travel in 20 minutes?
(ii) Find the time required to cover a distance of $250 \mathrm{~km}$.

Solution:

Let the distance travelled (in $\mathrm{km}$ ) in 20 minutes be $x$ and time taken (in minutes) to cover $250 \mathrm{~km}$ be $y$.

$1 \text { hour }=60 \text { minutes }$

Since the speed is uniform, therefore, the distance covered would be directly proportional to time.
(i) We have $\frac{75}{60}=\frac{x}{20}$
or $\frac{75}{60} \times 20=x$
or $\quad x=25$
So, the train will cover a distance of $25 \mathrm{~km}$ in 20 minutes.
(ii) Also, $\frac{75}{60}=\frac{250}{y}$
or $y=\frac{250 \times 60}{75}=200$ minutes or 3 hours 20 minutes.
Therefore, 3 hours 20 minutes will be required to cover a distance of 250 kilometres.

Alternatively, when $x$ is known, then one can determine $y$ from the relation $\frac{x}{20}=\frac{250}{y}$.

You know that a map is a miniature representation of a very large region. A scale is usually given at the bottom of the map. The scale shows a relationship between actual length and the length represented on the map. The scale of the map is thus the ratio of the distance between two points on the map to the actual distance between two points on the large region.
For example, if $1 \mathrm{~cm}$ on the map represents $8 \mathrm{~km}$ of actual distance [i.e., the scale is $1 \mathrm{~cm}: 8 \mathrm{~km}$ or $1: 800,000]$ then $2 \mathrm{~cm}$ on the same map will represent $16 \mathrm{~km}$. Hence, we can say that scale of a map is based on the concept of direct proportion.

Example 5:

The scale of a map is given as 1:30000000. Two cities are $4 \mathrm{~cm}$ apart on the map. Find the actual distance between them.

Solution:

Let the map distance be $x \mathrm{~cm}$ and actual distance be $y \mathrm{~cm}$, then
$
1: 30000000=x: y
$
or
$
\frac{1}{3 \times 10^7}=\frac{x}{y}
$

Since
$
x=4 \quad \text { so, } \quad \frac{1}{3 \times 10^7}=\frac{4}{y}
$
or
$
y=4 \times 3 \times 10^7=12 \times 10^7 \mathrm{~cm}=1200 \mathrm{~km} \text {. }
$

Thus, two cities, which are $4 \mathrm{~cm}$ apart on the map, are actually $1200 \mathrm{~km}$ away from each other.

Example 7:

6 pipes are required to fill a tank in 1 hour 20 minutes. How long will it take if only 5 pipes of the same type are used?

Solution:
Let the desired time to fill the tank be $x$ minutes. Thus, we have the following table.

Lesser the number of pipes, more will be the time required by it to fill the tank. So, this is a case of inverse proportion.

Hence, $\quad 80 \times 6=x \times 5 \quad\left[x_1 y_1=x_2 y_2\right]$
or $\quad \frac{80 \times 6}{5}=x$
or $\quad x=96$
Thus, time taken to fill the tank by 5 pipes is 96 minutes or 1 hour 36 minutes.
Example 8:

There are 100 students in a hostel. Food provision for them is for 20 days. How long will these provisions last, if 25 more students join the group?
Solution:

Suppose the provisions last for $y$ days when the number of students is 125 . We have the following table.

Note that more the number of students, the sooner would the provisions exhaust. Therefore, this is a case of inverse proportion.
So, $\quad 100 \times 20=125 \times y$
or $\quad \frac{100 \times 20}{125}=y \quad$ or $\quad 16=y$
Thus, the provisions will last for 16 days, if 25 more students join the hostel.
Alternately, we can write $x_1 y_1=x_2 y_2$ as $\frac{x_1}{x_2}=\frac{y_2}{y_1}$.
That is,

$
x_1: x_2=y_2: y_1
$
or
$
100: 125=y: 20
$
or
$
y=\frac{100 \times 20}{125}=16
$

Example 9:

If 15 workers can build a wall in 48 hours, how many workers will be required to do the same work in 30 hours?
Solution:
Let the number of workers employed to build the wall in 30 hours be $y$.

We have the following table. 

Obviously more the number of workers, faster will they build the wall. So, the number of hours and number of workers vary in inverse proportion.
So $48 \times 15=30 \times y$

Therefore, $\quad \frac{48 \times 15}{30}=y \quad$ or $\quad y=24$ i.e., to finish the work in 30 hours, 24 workers are required.