# Ch. - 8 Comparing Quantities

# Ch. - 8 Comparing Quantities

Comparing Quantities

Exercise 8.1

## Chapter 8 Ex.8.1 Question 1

Find the ratio of the following.

(i) Speed of a cycle \(15\,\rm{ km}\) per hour to the speed of scooter \(30\,\rm{ km}\) per hour.

(ii) \(5\,\rm{ m}\) to \(10\,\rm{ km}\)

(iii) \(50\) paise to \(\rm{Rs}\, 5\)

**Solution**

**Video Solution**

**What is known?**

Value of two quantities, which needs to be compared.

**What is unknown?**

Ratio

**Reasoning: **

A relationship between two quantities is normally expressed as the quantity of one divided by the other.

**Steps:**

**(i):**

Speed of a cycle \(= 15 \;\rm{km/hr}\)

Speed of a scooter \(= 30\,\rm{ km/hr}\)

Speed of cycle: Speed of scooter \(\begin{align}=\frac{{{15}}}{{{30}}}=\frac{{1}}{{2}} \end{align}\)

The answer is \(1:2\)

**(ii)**

Given data: \(5\,\rm{ m}\) to \(10 \,\rm{km}\)

Quantities can be compared only when the units are same.

\(1 \rm{km} = 1000 \rm{m}\)

Therefore, \(10\,\rm{ km} = 10 \times 1000 = 10000\,\rm{ m}\)

\(5\, \rm{m}\) to \(10\,\rm{ km}\) \(= 5\,\rm{m}\) to \(10000 \,\rm{m}\) \(\begin{align}=\frac{{5}}{{{10000}}}{ = }\frac{{1}}{{{2000}}}\end{align}\)

The answer is \(1:2000 \)

**(iii)**

Given data: \(50\) paise to \(\rm{Rs}\, 5\)

Quantities can be compared only when the units are same.

\[\begin{align}\rm{Rs}\, 1 &= 100\,\text{ paise}\\\rm{Rs} \,5 &= 5 \times 100\,\text{paise}\\ &= 500\, \text{paise}\end{align}\]

\(50\) paise to \(\rm{Rs}\, 5 = 50\) paise to \(500\) paise

\[\begin{align}\frac{{50}}{{500}} = \frac{1}{{10}} \end{align}\]

The answer is \(1:10\)

## Chapter 8 Ex.8.1 Question 2

Convert the following ratios to percentages

(i) \(3:4\)

(ii) \(2:3\)

**Solution**

**Video Solution**

**What is known?**

Ratios

**What is unknown?**

Percentages of given ratios

**Reasoning: **

A ratio is a comparison of any two quantities by division. A percent is a special ratio that compares any quantity to \(100,\) with \(100\) representing one whole.

**Steps:**

(i) \(3:4\)

Given data: \(3:4\)

\[\begin{align}\frac{3}{{4}}&{ \times 100}\\&= 3 \times 25\\&= 75\%\end{align}\]

The answer is \(75\% \)

(ii) \(2:3\)

Given data: \(2:3\)

**Type 1: Decimal Form**

\[\begin{align}\frac{2}{3} &\times 100\\&= 2 \times 33.33\\&= 66.67\% \end{align}\]

The answer is \(66.67\% \) (Decimal form)

**Type 2: Fractions Form**

\[\begin{align}\frac{{{2 \times 100}}}{{3}}&= \frac{{{200}}}{{3}}\\& = 66\frac{{2}}{{3}}{\% }\end{align}\]

The answer is \(\begin{align}{66}\frac{{2}}{{3}}\% \end{align}\) (Mixed fraction form)

## Chapter 8 Ex.8.1 Question 3

\(72\%\) of \(25\) students are interested in mathematics. How many are not interested in mathematics?

**Solution**

**Video Solution**

**What is known?**

Total number of students \(= 25\)

Percentage of students who are interested in Mathematics \(= 72\%\)

**What is unknown?**

Number of students who are not interested in Mathematics

**Reasoning: **

Percentage is a special ratio that compares any quantity to \(100,\) with \(100\) representing one whole.

**Steps:**

Percentage of students who are not interested in Mathematics

\[\begin{align}& = \left( {100-72} \right)\%\\&= 28\% \end{align}\]

Therefore, number of students who are not good in Mathematics

\(\begin{align}&= 28\%\,\text{of the total number of students}\\&= 28\% \,\text{of} \,25 \\&= \frac{{{28}}}{{{100}}}{ \times 25}\\&=\frac{{{28}}}{{4}}\\& = 7\end{align}\)

The total number of students who are not interested in Mathematics are \(7.\)

## Chapter 8 Ex.8.1 Question 4

A football team won \(10\) matches out of the total number of matches they played. If their win percentage was \(40\), then how many matches did they play in all?

**Solution**

**Video Solution**

**What is known?**

Total number of matches \(= 10\)

Win percentage \(= 40\%\)

**What is unknown?**

Total number of matches played.

**Reasoning: **

Assuming the total number of matches played as \(x,\) equating \(40\%\) of *\(x\)* is to \(10,\) the value of *\(x\)* can be found.

**Steps:**

Let the total number of matches played\(=x\)

\[\begin{align}40\% \;{\rm{ of}}\;x &= 10\\\frac{{40}}{{100}} \times x &= 10\\x &= \frac{{10 \times 100}}{{40}}\\&= 25\end{align}\]

The total number of matches played by football team is \(25.\)

## Chapter 8 Ex.8.1 Question 5

If Chameli had \(\rm{Rs}\, 600\) left after spending \(75\%\) of her money, how much did she have in the beginning?

**Solution**

**Video Solution**

**What is known?**

Percentage of amount Chameli spent \(= 75\%\)

Amount left with her \(= \rm{Rs} \,600\)

**What is unknown?**

Amount Chameli had in the beginning

**Reasoning: **

Since the whole is considered as \(100\%,\) Percentage of amount left with Chameli is

\[\left( {100 - 75} \right)\%= 25\% \]

Assuming the total amount in the beginning as \(x,\) and equating \(25\%\) of *\(x\)* to \(600,\) the value of *\(x\)* can be found.

**Steps:**

Let the total amount Chameli had with her in the beginning \(= x\)

Percentage of amount left with Chameli \(= 100-75 = 25%\)

\[\begin{align}25\% {\rm\;{of}}\;x &= 600\\\frac{{{25}}}{{{100}}}{ \times }x&= 600\\

x &= \frac{{{600 \times 100}}}{{{25}}}\\&= 2,400\end{align}\]

The amount that Chameli had in the beginning is \(\rm{Rs} \,2,400\)

## Chapter 8 Ex.8.1 Question 6

If \(60\%\) people in a city like cricket, \(30\%\) like football and the remaining like other games, then what percent of the people like other games? If the total number of people is \(50\) lakhs, find the exact number who like each type of game.

**Solution**

**Video Solution**

**What is known?**

Percentage of people who like cricket \(= 60\%\)

Percentage of people who like football \(= 30\%\)

Total number of people: \(50\) lakhs

**What is unknown?**

Percentage of people like other games

Exact number of people who like the game

**Reasoning: **

Since the whole is considered as \(100\%,\) percentage of people who like other games is \(100\% – (60+30) \% = 10\%\)

Number of people who like each game can be found using percentage and total number of people.

**Steps:**

Percentage of people who like other games\( = 100\% – (60+30)\% = 10\%\)

Number of people who like cricket

\[\begin{align}&= 60\% \;{\rm{ of }}\;50\,{\rm{ lakhs}}\\&= \frac{{{60}}}{{{100}}}{ \times 50,00,000}\\&= 30,00,000\\&= 30\,{\rm{ lakhs}}\end{align}\]

Number of people who like football

\[\begin{align}&= {\rm{ }}30\% \,{\rm{ of }}\,50\,{\rm{ lakhs}}\\&= \frac{{{30}}}{{{100}}}{ \times 50,00,000}\\&= 15,00,000\\&= 15\,\rm{ Lakhs}\end{align}\]

Number of people who like other games

\[\begin{align}&= 10\% \,{\rm{ of}}\;50 \;{\rm{lakhs}}\\&= \frac{{{10}}}{{{100}}}{ \times 50,00,000}\\&= 5,00,000\\&= 5\;{\rm{lakhs}}\end{align}\]

Hence the Answer is:

Percentage of people who like other games \(= 10\%\)

Number of people who like cricket \(= 30\) lakhs

Number of people who like football \(= 15\) lakhs

Number of people who like other games \(= 5\) lakhs

The chapter 8 begins with the recalling of concepts related to Ratios and Percentages.Then the procedure to find the percentage increase or decrease and discounts is explained in the next section. The next section of the chapter deals with Profit and Loss and finding cost price/selling price, profit percentage and loss percentage is dealt subsequently.Then Sales Tax, Value Added Tax, Goods and Services Tax is discussed. The formula for Compound Interest , Compound Interest compounded annually and half-yearly is explained.Lastly, the applications of Compound Interest Formula is discussed in detail.This chapter is very important to carry out day-to-day general transactions.