# Comparing Quantities - NCERT Class 7 Maths

Exercise 8.1

## Question 1

Find the ratio of:

(a) \(\rm Rs\, 5\) to \(50\) paise

(b) \(15\,\rm kg\) to \(210\,\rm g\)

(c) \(9 \,\rm m\) to \(27 \,\rm cm\)

(d) \(30\) days to \(36\) hours

### Solution

**Video Solution**

**What is Known?**

Quantities in different units

**What is Unknown?**

Our task is to find the ratio of given quantities.

**Reasoning:**

These questions are based on the basic concept of comparing two quantities i.e. to find ratio. For finding ratio, first we should observe that both the quantities should be in same unit for e.g. we cannot compare metres with kilometres or grams with kilograms. So, we will have to convert both quantities into same units.

**Steps:**

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

Since, \(1 {\rm rupee} = 100 \,\rm paise\)

\[\begin{align}{\text{So, 5 rupees}}\,\,&= \frac{{{\rm{100}}}}{{\rm{1}}} \times {\rm{5}}\,\\&= \,{\rm{500}}\,{\rm{paisa}}\end{align}\]

Therefore the ratio

\[\begin{align}&=\frac{500}{50} \\ & =\frac{10}{1}\text{or }10\text{:}\,1 \end{align}\]

(b)\(15\,\rm kg\) to \(210\,\rm g\)

Since\(, 1 \,\rm kg = 1000\, grams\)

\[\begin{align}{\text{So,}}\,\,15\,{\rm{kg}} &= \frac{{1000}}{1} \times 15\\ &= 15000\,{\rm{grams}}\end{align}\]

There for the ratio

\[\begin{align}&= \frac{{15000}}{{210}}\\ &= \frac{{500}}{7} {\rm{or}}\,\,500:7\end{align}\]

(c) \(9 \,\rm m\) to \(27 \,\rm cm\)

\(1 \,\rm m = 100\, cm\)

\[\begin{align}{\rm{So,}}\,\,{\rm{9}}\,{\rm{m}} &= \frac{{100}}{1} \times 9\\ &= 900\,{\rm{m}}\end{align}\]

\[\begin{align}{\rm{Thus,}}\,{\rm{ratio}} &= \frac{{900}}{{27}}\\& = \frac{{100}}{3}{\rm{or}}\,\,100\,{\rm{:}}\,3\end{align}\]

(d) \(30\) days to \(36\) hours

Since, \(1\) day = \(24\) hrs

\[\begin{align}{\rm{So,}}\,{\rm{30}}\,{\rm{days}} = \frac{{24}}{1} \times 30\\{\rm{ = 720}}\,{\rm{hrs}}\end{align}\]

\[\begin{align}{\rm{So,}}\,{\rm{ratio}} &= \frac{{720}}{{36}}\\&= \frac{{20}}{1}{\rm{or}}\,\,20\,{\rm{:}}\,1\end{align}\]

## Question 2

In a computer lab, there are \(3\) computers for every \(6\) students. How many computers will be needed for \(24\) students?

### Solution

**Video Solution**

**What is Known?**

Number of computers for \(6\) students

**What is Unknown?**

Number of computers for \(24\) students

**Reasoning:**

This question can be solved using unitary method. Since number of computers required for six students are given, we can find how many computers are required for one student and from the answer, number of computers required for \(24\) students can be calculated.

**Steps:**

For \(6\) students, \(3\) computers needed

So for one student computers needed

\[\begin{align}&=\frac{3}{6} \\ &=\frac{1}{2} \end{align}\]

And for \(24\) students number of computers required

\[\begin{align}&=\,\text{24}\times\frac{\text{1}}{\text{2}}\, \\ &=\,\text{12}\,\text{computers} \end{align}\]

## Question 3

Population of Rajasthan \(= 570\) lakhs and population of \(\rm UP = 1660\) lakhs. Area of Rajasthan \(= 3 \rm \,lakh\, km^2\) and area of \({\rm UP} = 2\,\rm lakh\, km^2.\)

(i) How many people are there per \(\rm km^2\) in both these States?

(ii) Which state is less populated?

### Solution

**Video Solution**

**What is Known?**

Area and population of Rajasthan and UP.

**What is Unknown?**

Number of people \(\rm per\, km^2\) and which state is less populated.

**Reasoning:**

To solve this question, we will divide total population with total area to know how many people are there \(\rm per\, km^2\)

**Steps:**

(i)** Rajasthan. **Number of people \(\rm per\, km^2\)

\[\begin{align} &= \frac {{570,00,000}} {{3,00,000}}\\ &= 190 \end{align}\]

It means \(190\) people are there in \(\rm per\, km^2\) in Rajasthan

**UP. **Number of people \(\rm per\, km^2\)

\[\begin{align} &= \frac {{1660,00,000}} {{2,00,000}}\\ &= 830 \end{align}\]

It means \(830\) people are there in \(\rm per\, km^2\) in UP

(ii) Which State is less populated?

Rajasthan is less populated because in Rajasthan less people live in \(\rm per\, km^2\) as compared to UP.

The chapter 8 begins with an introduction to comparing quantities by citing some real life examples. Then the concept of Equivalent Ratios is introduced and under it keeping things in proportion and getting solutions is also briefly explained. Then, the concept of percentage is introduced and the meaning of percentage, converting fractional numbers to percentage, converting decimals to percentage are discussed under it. This is followed by the concept of use of percentages, interpreting and converting of percentages. The concept of prices related to an item or buying and selling and profit or loss as a percentage is explained in the subsequent section. Lastly, the charge on the borrowed money or the simple interest is the last topic of discussion.

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