# NCERT Solutions Class 7 Maths Chapter 8

## NCERT Solutions for Class 7 Mathematics Chapter 8

NCERT Solutions for Class 7th Mathematics Chapter 8 Comparing Quantities are available on Extramarks to encourage and motivate students with their preparation for this chapter. In NCERT Solutions Class 7 Mathematics Chapter 8, students will find detailed step-by-step solutions to all exercises given in Chapter 8 of their Class 7 NCERT Mathematics textbook.

The important topics covered in NCERT Class 7 Mathematics Chapter 8 are as follows:

•  Equivalent Ratios
• Percentage and its uses
• Converting Fraction Numbers to Percentage
• Converting Decimals to Percentage
• Converting Percentages to Fractions or Decimals
• Converting Ratios to Percent
• Increase or Decrease as Percent
• Profit or Loss
• Simple Interest
• Interest in Multiple Years

### NCERT Solutions for Class 7 Mathematics Chapter 8 - Comparing Quantities

(Include NCERT Solutions for Class 7 Mathematics Chapter 8)

Access NCERT Solutions for Class 7 Mathematics Chapter 8 - Comparing Quantities

## NCERT Solutions for Class 7 Mathematics Chapter 8 - Comparing Quantities

The quantitative relation between two quantities which reflects the relative size of both the quantities is what is covered in the chapter: comparing quantities. It is simply the means to compare any two given quantities.

In daily life, there are many occasions when we compare two quantities.

While comparing quantities, the ratio for two different comparisons must be the same. But there's a rule in comparing quantities that states if the unit of the two quantities is not the same, it cannot be compared.

Some important facts covered in this chapter are:

• Various quantities that are of the same kind are compared using their ratios.
• If two fractions are equal, their ratios are also equal.
• When two ratios are equal, then the four quantities are in proportion.
• Percentages are numerators of fractions with the denominator 100 and they are also a way of comparing quantities.
• For conversion of a percentage into a decimal, you need to drop the sign of percentage and then shift the decimal point two places to the left.
• For conversion of a fraction into a percentage, you need to multiply the fraction by 100 and write the "%" sign on the right of the number.
• Profit = SP – CP (when SP > CP).
• Loss = CP – SP (when CP > SP).
• Profit or Loss per cent is always calculated on CP.
• Money borrowed is known as the principal.
• Simple interest = (P x R x T)/100
• Amount = Principal + Interest

Ratio:

Comparison through ratio means to know "how many times one quantity is of the other", or to know "what part of one quantity is the other".

The ratio of two numbers ‘b’ and ‘c’ (c ≠ 0) is b/c and it is denoted by b:c. A ratio in the simplest form is also known as the ratio in the lowest terms.

Note: To compare two quantities or to find the ratio of two quantities, the general rule remains the same i.e. their units must be the same.

Different ratios can also be compared by writing them as ‘like-fractions’.

To make your concepts more clear, you can  refer to the exercises in NCERT Solutions Class 7 Mathematics Chapter 8 from Extramarks.

Equivalent Ratios:

As mentioned earlier, we can compare various ratios by converting them to like fractions. If the like fractions obtained are equal, then the given ratios are said to be equivalent.

Percentage:

Percentages are numerators of fractions having denominators as 100 which are used in comparing results. As mentioned earlier, the percentage is represented by "%" and it is expressed as parts of hundredths.

For example, 8 % means 8 out of 100. It is written as 8% = 8/100 = 0.08.

If the total is not a hundred, then we need to convert the fraction into an equivalent fraction with denominator 100.

Ratios To Percent:

Sometimes, the parts of a whole quantity are given in the form of ratios that we can convert into percentages.

Note:  Here, by converting we mean to convert the increase or decrease in a quantity as a percentage of the initial amount as per the given question.

Profit Or Loss As A Percentage:

We know that the price of any item at which it is bought is called its cost price (CP) and the price at which it is sold is called its selling price (SP).

Note:

If CP < SP, then there is a profit in the monetary transaction, and Profit = SP – CP.

If CP > SP, then there is a loss in the monetary transaction and Loss = CP – SP.

The percentage of profit or loss is always calculated on the CP.

Simple Interest:

The money borrowed is known as the Principal amount. For using a bank's money, the borrower pays some extra money to the bank which is called Interest. The total money paid back with interest is known as the Amount.

Thus, Amount = Principal + Interest

Interest is generally given in per cent for a period for which the loan is taken. Generally, the rate of interest is expressed as a percentage per year or annum.

So, Principal is denoted by P, Rate of interest by R, and Time by T.

Now,  simple interest, or Interest = (Principal x Rate x Time)/100, or I = PRT/100

### NCERT Solutions for Class 7 Mathematics–

You can access the NCERT Solutions for Class 7 Mathematics  from Extramarks and practise the questions from the link given below.

### NCERT Solutions for Class 7 Mathematics

In case you found the solutions of Class 7 Mathematics Chapter 8 useful and want to continue studying Mathematics through our website, you can click on the link below to find solutions for the other chapters.

## NCERT Solutions for Class 7

Extramarks  offers  multiple  benefits beyond what was presented to you through the conventional teaching methods. The solutions offered by Extramarks can  help you in the following ways:

1. Engaging study material - The solutions of Mathematics Class 7 made available on our website are simple and easy to comprehend. As a result, you don't get confused, in fact, they are motivated to practise more sample test papers, and questions to improve their score.
2. Concepts made easier- Solving Mathematics questions through a study guide makes your mind actively involved in learning at a deeper level to understand better.
3. Important concepts emphasised and explained- As you will practice Mathematics with Extramarks' solutions, you’ll know key topics and highlight the main concepts of the chapter which will help you gain confidence and score good marks in your exams.

Q.1

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{ratio}\mathrm{of}:\\ \left(\mathrm{a}\right)\mathrm{Rs}5\mathrm{to}50\mathrm{paise}\left(\mathrm{b}\right)15\mathrm{kg}\mathrm{to}210\mathrm{g}\\ \left(\mathrm{c}\right)9\mathrm{m}\mathrm{to}27\mathrm{cm}\left(\mathrm{d}\right)30\mathrm{days}\mathrm{to}36\mathrm{hour}\end{array}$

Ans

$\begin{array}{l}\left(\text{a}\right)\text{Rs 5 to 5}0\text{paise}\\ \text{}1\text{ruppe}=\text{100 paise}\\ \text{So,}\text{\hspace{0.17em}}\text{5 rupee}=\text{500 paise}\\ \text{So,}\frac{\text{Rs}5}{50\text{paise}}=\frac{500}{50}=\frac{10}{1}\\ \text{Thus, the required ratio is}\overline{)\text{10:1}}.\\ \left(\text{b}\right)\text{15 kg to 21}0\text{g}\\ \text{1 kg}=\text{1000 g}\\ \text{So, 15 kg}=\text{15000 g}\\ \text{So,}\frac{15\text{kg}}{210\text{g}}=\frac{15000}{210}=\frac{500}{7}\\ \text{Thus, the required ratio is}\overline{)\text{500:7}}.\\ \left(\text{c}\right)\text{9 m to 27 cm}\\ \text{1 m}=\text{100 cm}\\ \text{So, 9 m}=\text{900 cm}\\ \text{So,}\frac{9\text{m}}{27\text{cm}}=\frac{900}{27}=\frac{100}{3}\\ \text{Thus, the required ratio is}\overline{)\text{100:3}}.\\ \left(\text{d}\right)\text{3}0\text{days to 36 hour}\\ \text{1 day}=\text{24 hours}\\ \text{So, 30 days}=\text{720 hours}\\ \text{So,}\frac{30\text{days}}{36\text{hour}}=\frac{720}{36}=\frac{20}{1}\\ \text{Thus, the required ratio is}\overline{)\text{20:1}}.\end{array}$

Q.2

$\begin{array}{l}\mathrm{In}\mathrm{a}\mathrm{computer}\mathrm{lab},\mathrm{there}\mathrm{are}3\mathrm{computers}\mathrm{for}\mathrm{every}6\\ \mathrm{students}.\mathrm{How}\mathrm{many}\mathrm{computers}\mathrm{will}\mathrm{be}\mathrm{needed}\mathrm{for}24\\ \mathrm{students}?\end{array}$

Ans

$\begin{array}{l}\text{For 6 students, number of computers required}=\text{3}\\ \text{So, for 1 student, number of computers required}=\frac{3}{6}=\frac{1}{2}\\ \text{Thus, for 24 students, number of computers required}=\frac{1}{2}×24\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\overline{)12}\\ \text{Therefore, 12 computers are needed for 24 students}\text{.}\end{array}$

Q.3

$\begin{array}{l}\mathrm{Population}\mathrm{of}\mathrm{Rajasthan}=570\mathrm{lakhs}\mathrm{and}\\ \mathrm{population}\mathrm{of}\mathrm{UP}=1660\mathrm{lakh}\\ \mathrm{Area}\mathrm{of}\mathrm{Rajasthan}= 3\mathrm{lakh}{\mathrm{km}}^{\mathrm{2}}\mathrm{and}\\ \mathrm{area}\mathrm{of}\mathrm{UP}= 2\mathrm{lakh}{\mathrm{km}}^{\mathrm{2}}\mathrm{.}\\ \left(\mathrm{i}\right)\mathrm{How}\mathrm{many}\mathrm{people}\mathrm{are}\mathrm{there}\mathrm{per}{\mathrm{km}}^{\mathrm{2}}\mathrm{in}\mathrm{both}\mathrm{these}\mathrm{States}?\\ \left(\mathrm{ii}\right)\mathrm{Which}\mathrm{State}\mathrm{is}\mathrm{less}\mathrm{populated}?\end{array}$

Ans

$\begin{array}{l}{\text{(i) Population of Rajasthan in 3 lakh km}}^{\text{2}}=\text{570 lakhs}\\ {\text{Population of Rajasthan in 1 lakh km}}^{\text{2}}\text{}=\frac{\text{570}}{3}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}=\overline{)190\text{}}\\ {\text{Population of UP in 2 lakh km}}^{\text{2}}\text{}\text{}\text{}=\text{1660 lakhs}\\ {\text{Population of UP in 1 lakh km}}^{\text{2}}\text{}\text{}\text{}=\frac{1660}{2}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}=\overline{)830\text{\hspace{0.17em}}}\\ \text{(ii)}\\ \text{From above data, clearly Rajasthan is less populated}\text{.}\end{array}$

Q.4

$\begin{array}{l}\mathrm{Convert}\mathrm{the}\mathrm{given}\mathrm{fractional}\mathrm{numbers}\mathrm{to}\mathrm{per}\mathrm{cents}.\\ \left(\mathrm{a}\right)\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{b}\right)\frac{\mathrm{5}}{\mathrm{4}}\left(\mathrm{c}\right)\frac{\mathrm{3}}{\mathrm{40}}\left(\mathrm{d}\right)\frac{\mathrm{2}}{\mathrm{7}}\end{array}$

Ans

$\begin{array}{l}\text{(a}\right)\frac{1}{8}\\ =\frac{1}{8}×\frac{100}{100}=\frac{1}{8}×100%=12.5%\\ \left(\text{b}\right)\frac{5}{4}\\ =\frac{5}{4}×\frac{100}{100}=\frac{5}{4}×100%=125%\\ \left(\text{c}\right)\frac{3}{40}\\ =\frac{3}{40}×\frac{100}{100}=\frac{3}{40}×100%=7.5%\\ \left(\text{d}\right)\frac{2}{7}\\ =\frac{2}{7}×\frac{100}{100}=\frac{2}{7}×100%=28\frac{4}{7}%\end{array}$

Q.5 Convert the given decimal fractions to percents.
a. 0.65 b. 2.1 c. 0.02 d. 12.35

Ans

$\begin{array}{l}\left(\text{a}\right)\text{}0.\text{65}\\ =0.65×100%=\frac{65×100}{100}%=65%\\ \left(\text{b}\right)\text{2}.\text{1}\\ =2.1×100%=\frac{21×100}{10}%=210%\\ \left(\text{c}\right)\text{}0.0\text{2}\\ =0.02×100%=\frac{2×100}{100}%=2%\\ \left(\text{d}\right)\text{12}.\text{35}\\ =12.35×100%=\frac{1235×100}{100}%=1235%\end{array}$

Q.6 Estimate what part of the figures is coloured and hence find the percent which is coloured.

Ans

$\begin{array}{l}\text{(i)}\\ \text{From the figure, we observe that 1 part out of 4 is shaded,}\\ \text{so its}\frac{1}{4}.\\ \frac{1}{4}×\frac{100}{100}=\frac{1}{4}×100%=25%\\ \text{(ii)}\\ \text{From the figure, we observe that 3 part out of 5 is shaded,}\\ \text{so its}\frac{3}{5}.\\ \frac{3}{5}×\frac{100}{100}=\frac{3}{5}×100%=60%\\ \text{(iii)}\text{\hspace{0.17em}}\\ \text{From the figure, we observe that 3 part out of 8 is shaded,}\\ \text{so its}\frac{3}{8}.\\ \frac{3}{8}×\frac{100}{100}=\frac{3}{8}×100%=37.5%\end{array}$

Q.7

$\begin{array}{l}\mathrm{Find}:\\ \left(\mathrm{a}\right)15%\mathrm{of}250 \left(\mathrm{b}\right)1%\mathrm{of}1\mathrm{hour}\\ \left(\mathrm{c}\right)20%\mathrm{of}\mathrm{Rs}2500 \left(\mathrm{d}\right)75%\mathrm{of}1\mathrm{k}\end{array}$

Ans

$\begin{array}{l}\left(\text{a}\right)\text{15}%\text{of 25}0\\ =\frac{15}{100}×250=\frac{75}{2}=37.5\\ \left(\text{b}\right)\text{1}%\text{of 1 hour}\\ \text{1 hour}=\text{60 minutes}\text{.}\\ \text{So, 1}%\text{of 1 hour}\\ =\frac{1}{100}×60=\frac{6}{10}=\frac{3}{5}\\ \left(\text{c}\right)\text{2}0%\text{of Rs 25}00\\ =\frac{20}{100}×2500=20×25=500\\ \left(\text{d}\right)\text{75}%\text{of 1 kg}\\ \text{1 kg}=\text{1000 g}\text{.}\\ \text{So, 75}%\text{of 1 kg}\\ =\frac{75}{100}×1000=75×10=750\end{array}$

Q.8

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{whole}\mathrm{quantity}\mathrm{if}\\ \left(\mathrm{a}\right)5%\mathrm{of}\mathrm{it}\mathrm{is}600.\left(\mathrm{b}\right)12%\mathrm{of}\mathrm{it}\mathrm{is}₹1080.\\ \left(\mathrm{c}\right)40%\mathrm{of}\mathrm{it}\mathrm{is}500\mathrm{km}.\left(\mathrm{d}\right)70%\mathrm{of}\mathrm{it}\mathrm{is}14\mathrm{minutes}\\ \left(\mathrm{e}\right)8%\mathrm{of}\mathrm{it}\mathrm{is}40\mathrm{litre}\end{array}$

Ans

$\begin{array}{l}\left(\text{a}\right)\text{5}%\text{of it is 6}00\text{is same as 5}%\text{of x is 6}00.\\ \frac{5}{100}×x=600\\ x=\frac{600×100}{5}=12000\\ \left(\text{b}\right)\text{12}%\text{of it is}₹\text{1}0\text{8}0\text{is same as 12}%\text{of x is}₹\text{\hspace{0.17em}}1080.\\ \frac{12}{100}×x=1080\\ x=\frac{1080×100}{12}=\text{\hspace{0.17em}}₹\text{\hspace{0.17em}}9000\\ \left(\text{c}\right)\text{4}0%\text{of it is 5}00\text{km is same as 40}%\text{of x is}500\text{\hspace{0.17em}}\text{km}.\\ \frac{40}{100}×x=500\\ x=\frac{500×100}{40}=1250\\ \left(\text{d}\right)\text{7}0%\text{of it is 14 minutes is same as 70}%\text{of x is}14\text{\hspace{0.17em}}\text{minutes}\\ \frac{70}{100}×x=14\\ x=\frac{14×100}{70}=20\\ \left(\text{e}\right)\text{8}%\text{of it is 4}0\text{litres is same as 8}%\text{of x is}40\text{\hspace{0.17em}}\text{litres}\\ \frac{8}{100}×x=40\\ x=\frac{40×100}{8}=500\end{array}$

Q.9

$\begin{array}{l}\mathrm{Convert}\mathrm{given}\mathrm{per}\mathrm{cents}\mathrm{to}\mathrm{decimal}\mathrm{fractions}\mathrm{and}\mathrm{also}\mathrm{to}\\ \mathrm{fractions}\mathrm{in}\mathrm{simplest}\mathrm{forms}:\\ \left(\mathrm{a}\right)\mathrm{25%}\left(\mathrm{b}\right)150%\left(\mathrm{c}\right)\mathrm{20%}\left(\mathrm{d}\right)\mathrm{5%}\end{array}$

Ans

$\begin{array}{l}\left(\text{a}\right)\text{25}%\\ =\frac{25}{100}=\frac{1}{4}=0.25\\ \left(\text{b}\right)\text{15}0%\text{}\\ =\frac{150}{100}=\frac{15}{10}=\frac{3}{2}=1.5\\ \left(\text{c}\right)\text{2}0%\\ =\frac{20}{100}=\frac{2}{10}=\frac{1}{5}=0.2\\ \left(\text{d}\right)\text{5}%\\ =\frac{5}{100}=\frac{1}{20}=0.05\end{array}$

Q.10

$\begin{array}{l}\mathrm{In}\mathrm{a}\mathrm{city}, 30%\mathrm{are}\mathrm{females}, 40%\mathrm{are}\mathrm{males}\mathrm{and}\mathrm{remaining}\\ \mathrm{are}\mathrm{children}.\mathrm{What}\mathrm{percent}\mathrm{are}\mathrm{children}?\end{array}$

Ans

$\begin{array}{l}\text{Children}=\left(100-\left(30+40\right)\right)%\\ =\left(100-70\right)%\\ =30%\end{array}$

Q.11

$\begin{array}{l}\mathrm{Out}\mathrm{of}15,000\mathrm{voters}\mathrm{in}\mathrm{a}\mathrm{constituency}, 60%\mathrm{voted}.\\ \mathrm{Find}\mathrm{the}\mathrm{percentage}\mathrm{of}\mathrm{voters}\mathrm{who}\mathrm{did}\mathrm{not}\mathrm{vote}.\mathrm{Can}\mathrm{you}\\ \mathrm{now}\mathrm{find}\mathrm{how}\mathrm{many}\mathrm{actually}\mathrm{did}\mathrm{not}\mathrm{vote}?\end{array}$

Ans

$\begin{array}{l}\text{Percentage of voters who voted}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=60%\text{}\\ \text{Percentage of voters who did not vote}\text{\hspace{0.17em}}=\text{100%-60%}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{40%}\\ \text{Number of voters who did not vote}=\text{40% of 15,000}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{40}{100}×15000\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=6000\end{array}$

Q.12

$\begin{array}{l}\mathrm{Meeta}\mathrm{saves}₹400\mathrm{from}\mathrm{her}\mathrm{salary}.\mathrm{If}\mathrm{this}\mathrm{is}10%\mathrm{of}\mathrm{her}\\ \mathrm{salary}.\mathrm{What}\mathrm{is}\mathrm{her}\mathrm{salay}?\end{array}$

Ans

$\begin{array}{l}\text{Let her salary be x}\text{.}\\ \text{Then according to the question, we have}\\ \text{\hspace{0.17em}}\text{10% of x}=\text{400}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{10}{100}×x=400\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\frac{400×100}{10}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4000\\ \text{Therefore Meeta’s salary is ₹ 4000}\end{array}$

Q.13

$\begin{array}{l}\mathrm{A}\mathrm{local}\mathrm{cricket}\mathrm{team}\mathrm{played}20\mathrm{matches}\mathrm{in}\mathrm{one}\mathrm{season}.\\ \mathrm{It}\mathrm{won}25%\mathrm{of}\mathrm{them}.\mathrm{How}\mathrm{many}\mathrm{matches}\mathrm{did}\mathrm{they}\mathrm{win}?\end{array}$

Ans

$\begin{array}{l}\text{Since team played 20 matches,}\\ \text{So, number of matches won}=\text{25% of 20}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{25}{100}×20\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{500}{100}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=5\\ \text{So, number they won 5 matchces}\end{array}$

Q.14

$\begin{array}{l}\mathrm{Tell}\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{profit}\mathrm{or}\mathrm{loss}\mathrm{in}\mathrm{the}\mathrm{following}\mathrm{transactions}.\\ \mathrm{Also}\mathrm{find}\mathrm{profit}\mathrm{per}\mathrm{cent}\mathrm{or}\mathrm{loss}\mathrm{per}\mathrm{cent}\mathrm{in}\mathrm{each}\mathrm{case}.\\ \left(\mathrm{a}\right)\mathrm{Gardening}\mathrm{shears}\mathrm{bought}\mathrm{for}₹250\mathrm{and}\mathrm{sold}\mathrm{for}₹325.\\ \left(\mathrm{b}\right)\mathrm{A}\mathrm{refrigerater}\mathrm{bought}\mathrm{for}₹12,000\mathrm{and}\mathrm{sold}\mathrm{at}₹13,500.\\ \left(\mathrm{c}\right)\mathrm{A}\mathrm{cupboard}\mathrm{bought}\mathrm{for}₹2,500\mathrm{and}\mathrm{sold}\mathrm{at}₹3,000.\\ \left(\mathrm{d}\right)\mathrm{A}\mathrm{skirt}\mathrm{bought}\mathrm{for}₹250\mathrm{and}\mathrm{sold}\mathrm{at}₹150.\end{array}$

Ans

$\begin{array}{l}\text{(a) Cost price}=\text{₹ 250}\\ \text{Selling price}=\text{₹ 325}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Profit}=\text{₹}\text{\hspace{0.17em}}\text{325}-\text{₹}\text{\hspace{0.17em}}\text{250}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}=\text{₹}\text{\hspace{0.17em}}\text{75}\\ \text{}\text{Profit%}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\text{Profit}}{\text{Cost Price}}×100\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{75}{250}×100\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=30%\\ \text{(b) Cost price}=\text{₹ 12000}\\ \text{Selling price}=\text{₹ 13,500}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Profit}=\text{₹}\text{\hspace{0.17em}}\text{13500}-\text{₹}\text{\hspace{0.17em}}\text{1200}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{₹}\text{\hspace{0.17em}}\text{1500}\\ \text{}\text{Profit%}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\text{Profit}}{\text{Cost Price}}×100\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1500}{12000}×100\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=12.5%\\ \text{(c) Cost price}=\text{₹ 2500}\\ \text{Selling price}=\text{₹ 3,000}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Profit}=\text{₹}\text{\hspace{0.17em}}\text{3000}-\text{₹}\text{\hspace{0.17em}}\text{2500}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{₹}\text{\hspace{0.17em}}\text{500}\\ \text{Profit%}\text{}\text{}=\frac{\text{Profit}}{\text{Cost Price}}×100\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{500}{2500}×100\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=20%\\ \text{(d) Cost price}=\text{₹ 250}\\ \text{Selling price}=\text{₹ 150}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Loss}=\text{₹}\text{\hspace{0.17em}}\text{250}-\text{₹}\text{\hspace{0.17em}}1\text{50}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{₹}\text{\hspace{0.17em}}\text{1000}\\ \text{}\text{Loss%}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\text{Loss}}{\text{Cost Price}}×100\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{100}{250}×100\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=40%\end{array}$

Q.15

$\begin{array}{l}\mathrm{Convert}\mathrm{each}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{ratio}\mathrm{to}\mathrm{percentage}:\\ \left(\mathrm{a}\right)3:1 \left(\mathrm{b}\right)2:3:5 \left(\mathrm{c}\right)1:4 \left(\mathrm{d}\right)\mathrm{1: 2:5.}\end{array}$

Ans

$\begin{array}{l}\left(\text{a}\right)\text{3}:\text{1}\\ \text{Here total parts}=3+1=4\\ {1}^{\text{st}}\text{part}=\frac{3}{4}=\frac{3}{4}×100%=75%\\ {2}^{\text{nd}}\text{part}=\frac{1}{4}=\frac{1}{4}×100%=25%\\ \left(\text{b}\right)\text{2}:\text{3}:\text{5}\\ \text{Here total parts}=2+3+5=10\\ {1}^{\text{st}}\text{part}=\frac{2}{10}=\frac{2}{10}×100%=20%\\ {2}^{\text{nd}}\text{part}=\frac{3}{10}=\frac{3}{10}×100%=30%\\ {3}^{\text{rd}}\text{part}=\frac{5}{10}=\frac{5}{10}×100%=50%\\ \left(\text{c}\right)\text{1}:\text{4}\\ \text{Here total parts}=1+4=5\\ {1}^{\text{st}}\text{part}=\frac{1}{5}=\frac{1}{5}×100%=20%\\ {2}^{\text{nd}}\text{part}=\frac{4}{5}=\frac{4}{5}×100%=80%\\ \left(\text{d}\right)\text{1}:\text{2}:\text{5}\\ \text{Here total parts}=1+2+5=8\\ {1}^{\text{st}}\text{part}=\frac{1}{8}=\frac{1}{8}×100%=12.5%\\ {2}^{\text{nd}}\text{part}=\frac{2}{8}=\frac{2}{8}×100%=25%\\ {3}^{\text{rd}}\text{part}=\frac{5}{8}=\frac{5}{8}×100%=62.5%\end{array}$

Q.16

$\begin{array}{l}\mathrm{The}\mathrm{population}\mathrm{of}\mathrm{a}\mathrm{city}\mathrm{decreased}\mathrm{from}25,000\mathrm{to}24,500.\\ \mathrm{Find}\mathrm{the}\mathrm{percentage}\mathrm{decrease}.\end{array}$

Ans

$\begin{array}{l}\text{Initial population}=2500\text{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Final populaton}=24500\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Decrease}=25000-2\text{4500}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=500\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Decrease}%=\frac{500}{25000}×100\\ =2%\end{array}$

Q.17

$\begin{array}{l}\mathrm{Arun}\mathrm{bought}\mathrm{a}\mathrm{car}\mathrm{for}₹3,50,000.\mathrm{The}\mathrm{next}\mathrm{year},\mathrm{the}\mathrm{price}\\ \mathrm{went}\mathrm{up}\mathrm{to}₹3,70,000.\mathrm{What}\mathrm{was}\mathrm{the}\mathrm{Percentage}\mathrm{of}\mathrm{price}\\ \mathrm{increase}?\end{array}$

Ans

$\begin{array}{l}\text{Initial Price}=\text{₹ 3,50,000}\\ \text{Final Price}=\text{₹ 3,70,000}\\ \text{Increase}=\text{₹}\text{\hspace{0.17em}}\text{3,70,000}-\text{₹}\text{\hspace{0.17em}}\text{3,50,000}\\ \text{}\text{}\text{}=\text{₹}\text{\hspace{0.17em}}\text{20,000}\\ \text{Increase%}=\frac{20,000}{3,50,000}×100\\ \text{}=5\frac{5}{7}%\end{array}$

Q.18

$\begin{array}{l}\mathrm{I}\mathrm{buy}\mathrm{a}\mathrm{T}.\mathrm{V}.\mathrm{for}₹10,000\mathrm{and}\mathrm{sell}\mathrm{it}\mathrm{at}\mathrm{a}\mathrm{profit}\mathrm{of}20%.\mathrm{How}\\ \mathrm{much}\mathrm{money}\mathrm{do}\mathrm{I}\mathrm{get}\mathrm{for}\mathrm{it}?\end{array}$

Ans

$\begin{array}{l}\text{We know that}\\ \text{Profit%}=\frac{\text{Profit}}{\text{Cost price}}×100\\ \text{So,}\\ \text{}20=\frac{\text{Profit}}{10,000}×100\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{\text{Profit}}{100×\overline{)100}}×\overline{)100}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Profit}=\text{20}×\text{100}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{₹}\text{\hspace{0.17em}}\text{2000}\\ \text{Now,}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Profit}=\text{Selling price}-\text{Cost price}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{₹}\text{\hspace{0.17em}}\text{2000}=\text{Selling price}-\text{₹}\text{\hspace{0.17em}}\text{10000}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Selling price}=\text{₹}\text{\hspace{0.17em}}\text{10000}+\text{₹}\text{\hspace{0.17em}}\text{2000}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{₹}\text{\hspace{0.17em}}\text{12000}\end{array}$

Q.19

$\begin{array}{l}\mathrm{Juhi}\mathrm{sells}\mathrm{a}\mathrm{washing}\mathrm{machine}\mathrm{for}₹13,500.\mathrm{She}\mathrm{loses}20%\mathrm{in}\\ \mathrm{the}\mathrm{bargain}.\mathrm{What}\mathrm{was}\mathrm{the}\mathrm{price}\mathrm{at}\mathrm{which}\mathrm{she}\mathrm{bought}\mathrm{it}?\end{array}$

Ans

$\begin{array}{l}\text{Selling price}=\text{₹ 135}00\\ \text{Loss}%\text{}=\text{2}0%\\ \text{Let the cost price be}x.\\ \therefore \text{Loss}=\text{2}0%\text{of}x\\ \text{Cost price}-\text{Loss}=\text{Selling price}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}-20%\text{of x}=\text{₹}\text{\hspace{0.17em}}\text{13500}\\ \text{}\text{}\text{x}-\frac{20}{100}×x=13500\\ \text{}\text{}\frac{100x-20x}{100}=13500\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{80x}{100}=13500\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}x=\frac{13500×100}{80}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{₹}\text{\hspace{0.17em}}16875\end{array}$

Q.20

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Chalk}\mathrm{contains}\mathrm{calcium},\mathrm{carbon}\mathrm{and}\mathrm{oxygen}\mathrm{in}\mathrm{the}\mathrm{ratio}\\ 10:3:12.\mathrm{Find}\mathrm{the}\mathrm{percentage}\mathrm{of}\mathrm{carbon}\mathrm{in}\mathrm{chalk}.\\ \left(\mathrm{ii}\right)\mathrm{If}\mathrm{in}\mathrm{a}\mathrm{stick}\mathrm{of}\mathrm{chalk},\mathrm{carbon}\mathrm{is}3\mathrm{g},\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{weight}\\ \mathrm{of}\mathrm{the}\mathrm{chalk}\mathrm{stick}.\end{array}$

Ans

$\begin{array}{l}\text{(i) Given ratio}=\text{10:3:12}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Total}=\text{10}+\text{3}+\text{12}\\ =\text{25}\\ \text{Percentage of Carbon}=\frac{3}{25}×100\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=12%\\ \text{(ii) Let the weight of the stick be x g}\\ \text{So, 12% of x}=\text{3}\\ \text{}\frac{12}{100}×x=3\\ \text{}\text{}x=\frac{300}{12}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=25\text{\hspace{0.17em}}\text{g}\end{array}$

Q.21 Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for?

Ans

$\begin{array}{l}\text{Cost price of book}=\text{₹ 275}\\ \text{Loss%}=\text{15}\\ \text{We have,}\\ \text{Loss%}=\frac{\text{Loss}}{\text{Cost price}}×100\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}15=\frac{\text{Loss}}{275}×100\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Loss}=\frac{15×275}{100}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=41.25\\ \text{Selling price}=\text{Cost price}-\text{Loss}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{₹}\text{\hspace{0.17em}}\text{275}-\text{₹}\text{\hspace{0.17em}}\text{41}\text{.25}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{₹ 233}\text{.75}\end{array}$

Q.22 Find the amount to be paid at the end of 3 years in each case:
(a) Principal = ₹1,200 at 12% p.a.
(b) Principal = ₹7,500 at 5% p.a.

Ans

$\begin{array}{l}\text{(a)}\\ \text{\hspace{0.17em}}\text{Principal (P)}=\text{₹ 1200}\\ \text{}\text{Rate (R)}=\text{12%}\\ \text{}\text{Time (T)}=3\text{years}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{S}\text{.I}\text{.}=\frac{\text{P}×\text{R}×\text{T}}{100}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}=\frac{1200×12×3}{100}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}=\text{₹}432\\ \text{}\text{Amount}=\text{P}+\text{S}\text{.I}\text{.}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}=\text{₹}\text{\hspace{0.17em}}\text{1200}+\text{₹}\text{\hspace{0.17em}}\text{432}=\text{₹ 1632}\\ \text{(b)}\\ \text{Principal (P)}=\text{₹ 7500}\\ \text{}\text{Rate (R)}=\text{5%}\\ \text{}\text{Time (T)}=\text{3 years}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{S}\text{.I}\text{.}=\frac{\text{P}×\text{R}×\text{T}}{100}\\ \text{}\text{}\text{}=\frac{7500×5×3}{100}\\ \text{}\text{}\text{}=\text{₹}1125\\ \text{}\text{Amount}=\text{P}+\text{S}\text{.I}\text{.}\\ \text{}\text{}\text{}=\text{₹}\text{\hspace{0.17em}}\text{7500}+\text{₹}\text{\hspace{0.17em}}\text{1125}\\ \text{}\text{}\text{}=\text{₹ 8625}\end{array}$

Q.23 What rate gives ₹280 as interest on a sum of ₹56000 in 2 years?

Ans

$\begin{array}{l}\mathrm{We}\mathrm{know}\mathrm{that}\\ \mathrm{S}.\mathrm{I}=\frac{\mathrm{P}×\mathrm{R}×\mathrm{T}}{100}\\ 280=\frac{\mathrm{56000}×\mathrm{R}×2}{100}\\ \mathrm{R}=\frac{280×100}{56000×2}\\ =0.25\\ \mathrm{Therefore},\mathrm{the}\mathrm{rate}\mathrm{is}0.25%.\end{array}$

Q.24

$\begin{array}{l}\mathrm{If}\mathrm{Meena}\mathrm{gives}\mathrm{an}\mathrm{interest}\mathrm{of}₹45\mathrm{for}\mathrm{one}\mathrm{year}\mathrm{at}9%\\ \mathrm{rate}\mathrm{p}.\mathrm{a}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{sum}\mathrm{she}\mathrm{has}\mathrm{borrowed}?\end{array}$

Ans

$\begin{array}{l}\text{We know that}\\ \text{S}\text{.I}=\frac{\text{P}×\text{R}×\text{T}}{100}\\ \text{So, we get}\\ \text{45=}\frac{\text{P×9×1}}{\text{100}}\\ \text{P=}\frac{\text{45×100}}{\text{9}}\\ \text{=₹}\text{\hspace{0.17em}}\text{500}\\ \text{Therefore, she borrowed ₹ 500}\text{.}\end{array}$

1. Give a summary of the topics covered in NCERT Solutions for Class 7 Mathematics Chapter 8?

Chapter 8 of NCERT Solutions for Class 7 Mathematics has 3 exercises. The topics covered in NCERT Solutions for Class 7 Mathematics Chapter 8 are as follows:

8.1 – Introduction

8.2 – Equivalent Ratios

8.3 – Percentage – Another Way of Comparing Quantities

8.4 – Use of Percentages

8.5 – Prices Related to an Item or Buying or Selling

8.6 – Charge Given on Borrowed Money or Simple Interest

2. Define comparing quantities.

It is the quantitative relation between two quantities that reflects the relative size of both quantities. It is simply a means to compare the given two quantities.

3. What does PA stand for in comparing quantities?

The full form of PA in comparing quantities is Per Annum. It is a term used in a financial context to refer to a financial year.

4. What is a ratio?

A ratio is a comparison of two or more quantities having the same units. A ratio compares two quantities by division, with the number being divided termed as the antecedent and the divisor or number dividing termed as the consequent. The mathematical symbol used to denote ratio is “:” (which is read as “is to”).

5. Define equivalent ratios.

As mentioned before, for comparing two quantities, their units must be the same. In the case of ratios, two ratios can be compared by converting them into like fractions. If the two like fractions obtained are equal, we say that the two given ratios are equivalent ratios (which are obtained by dividing or multiplying both the antecedent and consequent of the given ratio by the same number).

6. How does Extramarks teach Comparing Quantities?

When teaching the comparison of two or more quantities, a student needs to understand the order of numbers and this is explained through the use of a number line. The useful tricks shared in   Extramarks Solutions will ultimately make the concepts easier for you.