Examples (Revised) - Chapter 8 - Comparing Quantities - Ncert Solutions class 8 - Maths

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NCERT Solutions for Class 8 Maths Chapter 7 - Comparing Quantities

Example 1: A picnic is being planned in a school for Class VII. Girls are $60 \%$ of the total number of students and are 18 in number.
The picnic site is $55 \mathrm{~km}$ from the school and the transport company is charging at the rate of ₹ 12 per $\mathrm{km}$. The total cost of refreshments will be ₹ 4280 .

Can you tell.
1. The ratio of the number of girls to the number of boys in the class?
2. The cost per head if two teachers are also going with the class?
3. If their first stop is at a place $22 \mathrm{~km}$ from the school, what per cent of the total distance of $55 \mathrm{~km}$ is this? What per cent of the distance is left to be covered?

Solution:
1. To find the ratio of girls to boys.
Ashima and John came up with the following answers.
They needed to know the number of boys and also the total number of students.

Ashima did this
Let the total number of students be $x .60 \%$ of $x$ is girls.
Therefore, $60 \%$ of $x=18$
$
\begin{aligned}
& \frac{60}{100} \times x=18 \\
& \text { or, } x=\frac{18 \times 100}{60}=30
\end{aligned}
$
or, $x=\frac{18 \times 100}{60}=30$
Number of students $=30$.

or

John used the unitary method
There are 60 girls out of 100 students.
There is one girl out of $\frac{100}{60}$ students.
So, 18 girls are out of how many students?
$
\begin{aligned}
\text { Number of students } & =\frac{100}{60} \times 18 \\
& =30
\end{aligned}
$

So, the number of boys $=30-18=12$.
Hence, ratio of the number of girls to the number of boys is $18: 12$ or $\frac{18}{12}=\frac{3}{2}$. $\frac{3}{2}$ is written as $3: 2$ and read as 3 is to 2 .
2. To find the cost per person.
Transportation charge $=$ Distance both ways $\times$ Rate
$$
\begin{aligned}
& =₹(55 \times 2) \times 12 \\
& =₹ 110 \times 12=₹ 1320
\end{aligned}
$
$
\begin{aligned}
\text { Total expenses }= & \text { Refreshment charge } \\
& + \text { Transportation charge } \\
= & ₹ 4280+₹ 1320 \\
= & ₹ 5600
\end{aligned}
$
$
\begin{aligned}
\text { Total number of persons } & =18 \text { girls }+12 \text { boys }+2 \text { teachers } \\
& =32 \text { persons }
\end{aligned}
$

Ashima and John then used unitary method to find the cost per head.
For 32 persons, amount spent would be ₹ 5600 .
The amount spent for 1 person $=₹ \frac{5600}{32}=₹ 175$.
3. The distance of the place where first stop was made $=22 \mathrm{~km}$.

To find the percentage of distance:
Ashima used this method:
$
\frac{22}{55}=\frac{22}{55} \times \frac{100}{100}=40 \%
$

She is multiplying
the ratio by $\frac{100}{100}=1$ and converting to percentage.

or

John used the unitary method:
Out of $55 \mathrm{~km}, 22 \mathrm{~km}$ are travelled.
Out of $1 \mathrm{~km}, \frac{22}{55} \mathrm{~km}$ are travelled.
Out of $100 \mathrm{~km}, \frac{22}{55} \times 100 \mathrm{~km}$ are travelled.
That is $40 \%$ of the total distance is travelled.

Both came out with the same answer that the distance from their school of the place where they stopped at was $40 \%$ of the total distance they had to travel. Therefore, the percent distance left to be travelled $=100 \%-40 \%=60 \%$.

Example 2:

An item marked at ₹ 840 is sold for ₹ 714 . What is the discount and liscount $\%$ ?

Solution:
$
\begin{aligned}
\text { Discount } & =\text { Marked Price }- \text { Sale Price } \\
& =₹ 840-₹ 714 \\
& =₹ 126
\end{aligned}
$

Since discount is on marked price, we will have to use marked price as the base.
On marked price of ₹ 840 , the discount is ₹ 126 .
On MP of ₹ 100 , how much will the discount be?
$
\text { Discount }=\frac{126}{840} \times 100 \%=15 \%
$

You can also find discount when discount \% is given.

Example 3:

The list price of a frock is ₹ 220 . A discount of $20 \%$ is announced on sales. What is the amount of discount on it and its sale price.

Solution:

Marked price is same as the list price. $20 \%$ discount means that on ₹ 100 (MP), the discount is ₹ 20 . By unitary method, on ₹1 the discount will be $₹ \frac{20}{100}$. On $₹ 220$, discount $=₹ \frac{20}{100} \times 220=₹ 44$
The sale price $=(₹ 220-₹ 44)$ or ₹ 176
Rehana found the sale price like this -
A discount of $20 \%$ means for a MP of $₹ 100$, discount is ₹ 20 . Hence the sale price is $₹ 80$. Using unitary method, when MP is ₹ 100 , sale price is ₹ 80 ;

When MP is ₹ 1 , sale price is $₹ \frac{80}{100}$. Hence when MP is $₹ 220$, sale price $=₹ \frac{80}{100} \times 220=₹ 176$.

Example 4:

(Finding Sales Tax) The cost of a pair of roller skates at a shop was ₹ 450 . The sales tax charged was $5 \%$. Find the bill amount.
Solution:

On ₹ 100 , the tax paid was ₹ 5 .

$
\begin{aligned}
\text { On ₹ } 450, \text { the tax paid would be } & =₹ \frac{5}{100} \times 450 \\
& =₹ 22.50
\end{aligned}
$
$
\text { Bill amount }=\text { Cost of item }+ \text { Sales tax }=₹ 450+₹ 22.50=₹ 472.50 \text {. }
$

Example 5:

(Value Added Tax (VAT)) Waheeda bought an air cooler for ₹ 3300 including a tax of $10 \%$. Find the price of the air cooler before VAT was added.
Solution:

The price includes the VAT, i.e., the value added tax. Thus, a $10 \%$ VAT means if the price without VAT is ₹ 100 then price including VAT is ₹ 110 .
Now, when price including VAT is $₹ 110$, original price is $₹ 100$.
Hence when price including tax is $₹ 3300$, the original price $=₹ \frac{100}{110} \times 3300=₹ 3000$.
Example 6: Salim bought an article for ₹ 784 which included GST of $12 \%$. What is the price of the article before GST was added?
Solution: Let original price of the article be $₹ 100$. GST $=12 \%$.
Price after GST is included $=₹(100+12)=₹ 112$
When the selling price is $₹ 112$ then original price $=₹ 100$.
When the selling price is $₹ 784$, then original price $=₹ \frac{100}{12} \times 784=₹ 700$

Example 7: A sum of $₹ 10,000$ is borrowed at a rate of interest $15 \%$ per annum for 2 years. Find the simple interest on this sum and the amount to be paid at the end of 2 years.
Solution: On ₹ 100 , interest charged for 1 year is ₹ 15 .
$$
\begin{aligned}
& \text { So, on } ₹ 10,000 \text {, interest charged }=\frac{15}{100} \times 10000=₹ 1500 \\
& \text { Interest for } 2 \text { years }=₹ 1500 \times 2=₹ 3000 \\
& \text { Amount to be paid at the end of } 2 \text { years }=\text { Principal }+ \text { Interest } \\
& =₹ 10000+₹ 3000=₹ 13000 \\
&
\end{aligned}
$$

Example 8: Find CI on ₹ 12600 for 2 years at $10 \%$ per annum compounded annually. Solution: We have, $\mathrm{A}=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^n$, where Principal $(\mathrm{P})=₹ 12600$, Rate $(\mathrm{R})=10$,
$$
\begin{gathered}
\text { Number of years }(n)=2 \\
=₹ 12600\left(1+\frac{10}{100}\right)^2=₹ 12600\left(\frac{11}{10}\right)^2
\end{gathered}
$$

$\begin{aligned}
& =₹ 12600 \times \frac{11}{10} \times \frac{11}{10}=₹ 15246 \\
\mathrm{CI}=\mathrm{A}-\mathrm{P} & =₹ 15246-₹ 12600=₹ 2646
\end{aligned}$

Example 9: The population of a city was 20,000 in the year 1997. It increased at the rate of $5 \%$ p.a. Find the population at the end of the year 2000 .

Solution: There is $5 \%$ increase in population every year, so every new year has new population. Thus, we can say it is increasing in compounded form.
Population in the beginning of $1998=20000$ (we treat this as the principal for the 1st year)

Increase at $5 \%=\frac{5}{100} \times 20000=1000$
Population in $1999=20000+1000=21000$
Treat as the Principal for the 2nd year.

Increase at $5 \%=\frac{5}{100} \times 21000=1050$
2nd year.

Population in $2000=21000+1050$ $\qquad$ Treat as the Principal  for the 3rd year.
Increase at $5 \%=\frac{5}{100} \times 22050$

At the end of 2000 the population $=22050+1102.5=23152.5$
or, Population at the end of $2000=20000\left(1+\frac{5}{100}\right)^3$
$
\begin{aligned}
& =20000 \times \frac{21}{20} \times \frac{21}{20} \times \frac{21}{20} \\
& =23152.5
\end{aligned}
$

So, the estimated population $=23153$.

Aruna asked what is to be done if there is a decrease. The teacher then considered the following example.
Example 10:

ATV was bought at a price of ₹ 21,000 . After one year the value of the TV was depreciated by $5 \%$ (Depreciation means reduction of value due to use and age of the item). Find the value of the TV after one year.
Solution:
$
\begin{aligned}
\text { Principal } & =₹ 21,000 \\
\text { Reduction } & =5 \% \text { of } ₹ 21000 \text { per year } \\
& =₹ \frac{21000 \times 5 \times 1}{100}=₹ 1050
\end{aligned}
$
value at the end of 1 year $=₹ 21000-₹ 1050=₹ 19,950$
Alternately, We may directly get this as follows:

$
\begin{aligned}
\text { value at the end of } 1 \text { year } & =₹ 21000\left(1-\frac{5}{100}\right) \\
& =₹ 21000 \times \frac{19}{20}=₹ 19,950
\end{aligned}
$