# NCERT Solutions Class 8 Maths Chapter 8

## NCERT Solutions for Class 8 Mathematics Chapter 8: Comparing Quantities

Mathematics is a subject that requires a deep knowledge of the concepts and rapid calculation abilities. Furthermore, it contains complex formulas and theorems. Therefore, students in Class 8 need to study with a greater focus to excel against the other students in the class. For those interested in this subject, it offers a variety of opportunities in science and engineering. It is a significant learning experience for them as it helps them prepare for new opportunities.

Class 8 Mathematics Chapter 8: Comparing quantities covers the more significant part of the mathematical principles. It talks about simple mathematics to calculate the value of money. At the beginning of the chapter, students will learn to compare different quantities. In comparing quantities in Class 8, students will understand how to calculate discounts on the marked price and sale price. Further, they will learn the compound interest and properties of the simple interest. Finally, all the essential formulas and concepts are elaborated in our NCERT Solutions for Class 8 Mathematics Chapter 8.

The Extramarks NCERT Solutions for Class 8 Mathematics Chapters 8 are available to students. It covers essential subjects and assists in revising the chapter correctly. It consists of 30 questions, including 21 that are easy, four of them are moderately easy, and five are long answer-type questions. Our NCERT Class 8 Mathematics Chapter 8 solution revolves around certain fundamental concepts in entrance tests.

In addition, thousands of students believe in us since we are India’s top online education platform. Our NCERT-related solutions and study materials are favoured by students pursuing secondary and primary schooling. All questions are designed using the new format of CBSE to ensure they have an in-depth understanding of the exam.

To get the latest updates and news related to exams, students can check out the Extramarks website. Additionally, they can look up other solutions for class, like NCERT Solutions Class 10, NCERT Solutions Class 11 and NCERT Solutions Class 12.

## Key Topics Covered in NCERT Solutions for Class 8 Mathematics Chapter 8:

Highly experienced subject matter experts have designed NCERT Solutions for Class 8 Mathematics Chapter 8. They have created simple solutions using stepwise summaries of answers that make them easy to understand. It can help students increase their confidence and aid them in managing their time in the crucial exam. Extramarks offer NCERT Solutions that help students overcome their doubts and problems in solving issues.

Some of the key topics featured in NCERT Solutions for Class 8 Mathematics Chapter 8 are:

### Recalling Ratios and Percentages

The ratio to percent conversion in Mathematics is beneficial for comparing each part and the whole. A ratio is a comparison of any two parts of a total. The ratio of sugar to salt in a solution, for example, is 3:1. This means that the sugar in the solution is three times as high as salt. Percentages can be used to describe a specific type of ratio. Percentage compares one part of the total against another, rather than comparing the parts individually. For example, the solution contains 75% sugar and 25% salt.

To understand the key fundamentals of ratios and percentages, students can refer to NCERT Solutions for Class 8 Mathematics Chapter 8.

### Finding the increase or decrease percent

Before we get into the conversion of ratio percent, let’s briefly explain the meaning of percentages and ratios. A ratio is a relationship or comparison between quantities of the same type and the same unit. So, first, identify the ratio to convert it to percent. To get the percentage value, we need first to identify the ratio.

### Finding discounts

Students can extract the discount rate formula by either subtracting the product’s price from the marked price or multiplying the discount rate and the product’s price. The formula for discount can be represented in mathematical terms as follows,

Discount = Marked Price – Selling Price

OR

Discount Percentage Formula = Marked Price × Discount Rate

Other basis Discount formulas are as below: –

Discount = List Price – Selling Price

Therefore

Selling Price = List Price – Discount

List Price = Selling Price + Discount

A discount is a pricing system where the price of a commodity (goods or services) is lower than its listed price. Simply put, a discount is a percentage of the cost. A discount is a price reduction in products. It is usually noticed in consumer transactions where buyers propose a percentage of rebates for various products to increase sales. A discount is a rebate that the seller offers to the buyer.

Students can refer to our NCERT Solutions for Class 8 Mathematics Chapter 8 to understand the increase and decrease of percentage in market price.

### Prices related to buying and selling (profit and loss)

Profit and loss can be used to determine if a deal is profitable. These terms are used frequently in our daily lives. The difference between the selling and cost prices is called profit. The selling price must be greater than the Cost to make it profitable.

Cost price: This is the price at which an item is bought. Example: Neil buys an umbrella for $8. This is the umbrella’s cost price. It is also abbreviated CP. Selling price: This is the price an article is sold for. If Neil sells the same umbrella for$10, then \$10 would be the selling price. It is also abbreviated as SP.

Profit: A transaction in which the selling price exceeds the cost price results in a gain.

Marked price: The marked price is the price the seller has set on the article’s label. This is the price at which the seller offers discounts. It is then sold at a lower price known as the selling cost after the Marked Price has been reduced. However, students can refer to NCERT Solutions for Class 8 Mathematics Chapter 8 to get a clear idea of the discount and selling price.

Sales tax, value-added tax, goods and services tax.

Tax is the name given to the government’s fees to collect revenue for public facilities and infrastructure, such as electricity, road maintenance, education, etc. The government uses this income to pay for various economic activities.

### Sales Tax

Sales tax is a tax that is levied on products purchased. This tax is levied primarily by the government on any product sold. This indirect tax is a percentage of the product’s selling price after being purchased from a shop.

The sale tax can be calculated by:

Sales tax = percentage of tax of Bill Amount

Total Bill = Sales Tax + Cost cost of the product

Another tax is the Value Added Tax or VAT. This is a tax taken by the government to pay for public welfare expenses. It can also be considered a consumption tax that is added to every product at each stage of production and distribution. It is mainly paid in the product costs.

### Goods and Services Tax

This indirect tax is used to supply goods and services for any product. This tax is considered a destination-based tax in multiple stages and comprehensives, as it can be used in all indirect taxes, except a few state taxes.

### Compound Interest

Compound interest is the interest imposed on a loan or deposit amount. It is the most common concept we use in our everyday lives. The amount of compound interest depends on the principal and the interest earned over time. This is the significant difference between compound interest and simple interest.

### Rate compounded annually

The compound annual growth rate (CAGR) is growing over several years, with each year’s growth adding to the original value. The compound annual growth rate (CAGR), also known as compound interest, is the average yearly rate for growth when you reinvest your returns over several years.

Calculating the compounded rate can be difficult. Therefore, students can refer to our NCERT Solutions for Class 8 Mathematics Chapter 8. We have elaborated with illustrative examples for a better understanding of the students.

## NCERT Solutions for Class 8 Mathematics Chapter 8: Exercise &  Solutions

Mathematical subjects are a favourite for many students, yet it is equally problematic because it involves complicated multiplication. As a result, many students struggle with understanding the theories and concepts. This is why it’s helpful to them to work on various questions with more difficulty. In addition, it assists students with remembering the formulas and how to use them in different stages of the problem.

The NCERT exemplar covers every topic and concept, which are explained in comprehensible language. It provides a solid foundation for the various ideas. The majority of questions that are asked in the annual exam are drawn from the NCERT textbooks. Additionally, examples offer multiple-choice, descriptive-type questions, and objective-type questions.

Some of the problems taken from the NCERT examples are included within NCERT Solutions for Class 8 Mathematics Chapter 8. In addition, students can find answers or solutions for every question in each chapter, intext, and other miscellaneous questions.

The NCERT Solutions for Class 8 Mathematics Chapter 8 explains various methods used to determine compound and simple interest. In addition, students can sign up on our website to obtain an account trial and clear their doubts.

Students can click the links below for specific questions and solutions:

Along with this, students can also refer to other solutions for primary and secondary classes:

## NCERT Exemplar for Class 8 Mathematics:

Mathematics is an intriguing subject. However, some concepts are difficult to grasp. Students have difficulty remembering these concepts and do not know how to solve issues involving compound interest. Therefore, it is beneficial to answer different questions with more tests. In addition, it is helpful to remember the formulas and how to use them for different stages of problems.

The NCERT exemplar includes the entire range of topics and concepts that are explained in easily comprehensible language. Class 8 Mathematics exemplar aids in establishing a solid foundation for all concepts. Most questions asked during the annual exam come entirely based on the NCERT textbooks. The NCERT books are excellent for questions in yearly exams. It assists students in solving questions on a more challenging level and also helps students to pass competitions in maths.

Specific questions come from NCERT Solutions for Class 8 Mathematics Chapter 8. This helps students understand how to calculate the discount on marked price and selling price.

### Key Features of NCERT Solutions for Class 8 Mathematics Chapter 8:

The class 8 Mathematics class discusses the characteristics of compound interest and simple interest. The most important subjects include selling price, cost price, compound interest, and simple interest. The Extramarks Solutions from NCERT Solutions for Class 8 Mathematics Chapter 8 covers the most important topics and concepts.

The fundamental concepts of NCERT Solutions for Class 8 Mathematics Chapter 8 are explained in this are:

• The NCERT Solutions for Class 8 Mathematics Chapter 8 includes excellently curated examples and questions covering all the subject’s fundamental ideas.
• Students should be able to comprehend and solve problems to develop a solid understanding.
• To access NCERT Solutions for Class 8 Mathematics Chapter 8 from Extramarks students can register themselves on its official website.
• Students can tackle problems of various difficulty levels, to enhance their confidence levels.
• The review questions provide an additional challenge, and students can try them again to increase their knowledge of the subject.
• Additional exercises are provided to encourage active learning. Students can test what they’ve learned on a more advanced degree.
• Students can understand the language and concepts better and comprehend the formulas provided. This helps them be more successful in exams.

Q.1 Find the ratio of the following.

(a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.

(b) 5 m to 10 km

(c) 50 paise to ₹ 5

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{Ratio}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{speed}\text{}\mathrm{of}\text{}\mathrm{cycle}\text{}\mathrm{to}\text{}\mathrm{the}\text{}\mathrm{speed}\text{}\mathrm{of}\text{}\mathrm{scooter}=\frac{15}{30}=1:2\\ \left(\mathrm{b}\right)\text{}\mathrm{Since}\text{}1\text{}\mathrm{km}\text{}=\text{}1000\text{}\mathrm{m},\\ \therefore \mathrm{Required}\text{}\mathrm{ratio}=\frac{5\text{​}\mathrm{m}}{10\text{}\mathrm{km}}=\frac{5\text{​}\mathrm{m}}{10×1000\text{}\mathrm{m}}=1:2000\\ \\ \left(\mathrm{c}\right)\text{}\mathrm{Since}\text{₹}1\text{}=\text{}100\text{}\mathrm{paise},\\ \therefore \mathrm{Required}\text{}\mathrm{ratio}=\frac{50\text{}\mathrm{paise}}{\text{₹\hspace{0.17em}}5}=\frac{50\text{\hspace{0.17em}}\mathrm{paise}}{500\text{\hspace{0.17em}}\mathrm{paise}}=1:10\end{array}$

Q.2 Convert the following ratios to percentages.

(a) 3 : 4

(b) 2 : 3

Ans

$\begin{array}{l}\left(\text{a}\right)3:4=\frac{3}{4}=\frac{3}{4}×\frac{100}{100}=\frac{3}{4}×100%=75%\\ \left(\text{b}\right)2:3=\frac{2}{3}=\frac{2}{3}×\frac{100}{100}=\frac{2}{3}×100%=\frac{200}{3}%=66\frac{2}{3}%\end{array}$

Q.3 72% of 25 students are good in mathematics. How many are not good in mathematics?

Ans

$\begin{array}{l}\text{Given that 72}\mathrm{%}\text{of 25 students are good in mathematics}.\\ \\ \text{Therefore},\text{Percentage of students who are not good}\\ \text{in mathematics}=\left(\text{1}00\text{}-\text{72}\right)\mathrm{%}=\text{28}\mathrm{%}.\\ \\ \mathrm{Thus},\mathrm{the}\text{\hspace{0.17em}}\mathrm{n}\text{umber of students who are not good in}\\ \text{mathematics}=\frac{28}{100}×25=\text{7}\\ \\ \text{Hence},\text{7 students are not good in mathematics}.\end{array}$

Q.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?

Ans

$\begin{array}{l}\text{Let the total number of matches played by the team be}\mathrm{x}.\\ \\ \text{It is given that the team won 10 matches and the winning}\\ \text{percentage of the team was 40}\mathrm{%}\text{.}\\ \\ \text{Therefore,}\\ \frac{40}{100}×\mathrm{x}=10\\ ⇒\mathrm{x}=10×\frac{100}{40}\\ ⇒\mathrm{x}=25\\ \\ \text{Hence, the team played 25 matches in all.}\end{array}$

Q.5 If Chameli had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?

Ans

$\begin{array}{l}\text{Let the amount of money which Chameli had in the}\\ \text{beginning be}\mathrm{x}.\\ \text{It is given that after spending}75\mathrm{%}\text{of ₹}\mathrm{x},\text{she}\\ \text{was left with ₹}600.\\ \text{Therefore},\left(100\text{}-\text{}75\right)\mathrm{%}\text{of}\mathrm{x}=\text{Rs}600\end{array}$ $\begin{array}{l}⇒\text{}25\mathrm{%}\text{of}\mathrm{x}=\text{₹}600\\ ⇒\frac{25}{100}×\mathrm{x}=\text{₹}600\\ ⇒\mathrm{x}=₹\left(600×\frac{100}{25}\right)=₹\text{}2400\\ \\ \text{Hence, she had ₹ 2400 in the beginning.}\end{array}$

Q.6 If 60% people in a city like cricket, 30% like football and the remaining like other games, then what per cent of the people like other games? If the total number of people are 50 lakh, find the exact number who like each type of game.

Ans

$\begin{array}{l}\mathrm{Percentage}\text{}\mathrm{of}\text{}\mathrm{people}\text{}\mathrm{who}\text{}\mathrm{like}\text{}\mathrm{other}\text{}\mathrm{games}=\left(100-60-30\right)\mathrm{%}\\ \text{}=\left(100-90\right)\mathrm{%}\\ \text{}=10\mathrm{%}\\ \text{It is given that the total num}\mathrm{ber}\text{}\mathrm{of}\text{}\mathrm{people}=50\text{}\mathrm{lakh}\\ \\ \therefore \mathrm{Number}\text{}\mathrm{of}\text{}\mathrm{people}\text{}\mathrm{who}\text{}\mathrm{like}\text{}\mathrm{cricket}=\left(\frac{60}{100}×50\right)\mathrm{lakh}=30\text{​}\mathrm{lakh}\\ \mathrm{Number}\text{}\mathrm{of}\text{}\mathrm{people}\text{}\mathrm{who}\text{}\mathrm{like}\text{}\mathrm{football}=\left(\frac{30}{100}×50\right)\mathrm{lakh}=15\text{​}\mathrm{lakh}\\ \mathrm{Number}\text{}\mathrm{of}\text{}\mathrm{people}\text{}\mathrm{who}\text{}\mathrm{like}\text{}\mathrm{other}\text{}\mathrm{games}=\left(\frac{10}{100}×50\right)\mathrm{lakh}=5\text{​}\mathrm{lakh}\end{array}$

Q.7 A man got a 10% increase in his salary. If his new salary is ₹ 1,54,000, find his original salary.

Ans

$\begin{array}{l}\text{Let the original salary be}\mathrm{x}.\text{}\\ \text{It is given that the new salary is ₹}1,54,000.\\ \\ \text{Therefore, Original salary + Increment = New salary}\\ \\ \text{But it is given that the increment is}10\mathrm{%}\text{of the original salary.}\\ \\ \text{Therefore,we have}\\ \mathrm{x}\text{+}\frac{10}{100}×\mathrm{x}=1,54,000\\ ⇒\frac{110\mathrm{x}}{100}=1,54,000\\ ⇒\mathrm{x}=1,54,000×\frac{100}{110}\\ ⇒\mathrm{x}=1,40,000\\ \\ \text{Thus, the original salary was ₹}1,40,000.\end{array}$

Q.8 On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the per cent decrease in the people visiting the Zoo on Monday?

Ans

$\begin{array}{l}\text{It is given that, 845 people went to the zoo on Sunday and}\\ \text{169 people went on Monday}.\\ \\ \text{Thus,decrease in the number of people}=\text{845}-\text{169}=\text{676}\\ \\ \text{Percentage decrease}=\left(\frac{\text{Decrease in the number of people}}{\begin{array}{l}\text{Number of people who went to}\\ \text{the zoo on sunday}\end{array}}×100\right)%\\ \\ \text{}=\left(\frac{676}{845}×100\right)%\\ \\ \text{}=80%\end{array}$

Q.9 A shopkeeper buys 80 articles for ₹ 2,400 and sells them for a profit of 16%. Find the selling price of one article.

Ans

$\begin{array}{l}\mathrm{It}\text{}\mathrm{is}\text{}\mathrm{given}\text{}\mathrm{that}\text{}\mathrm{the}\text{}\mathrm{shopkeeper}\text{}\mathrm{buys}\text{}80\text{}\mathrm{articles}\text{}\mathrm{for}\text{}₹\text{}2,400.\\ \mathrm{Cost}\text{}\mathrm{Price}\text{}\mathrm{of}\text{}\mathrm{one}\text{}\mathrm{article}=₹\frac{2,400}{80}=\text{₹}30\\ \mathrm{Profit}\text{}\mathrm{percent}=16\mathrm{%}\\ \mathrm{Profit}\text{}\mathrm{percent}=\frac{\mathrm{Profit}}{\mathrm{Cost}\text{}\mathrm{Price}}×100\\ 16=\frac{\mathrm{Profit}}{\text{₹}30}×100\\ \mathrm{Profit}=₹\frac{16×30}{100}=₹\text{}4.80\text{}\\ \\ \mathrm{Selling}\text{}\mathrm{price}\text{}\mathrm{of}\text{}\mathrm{one}\text{}\mathrm{article}=\mathrm{C}.\mathrm{P}.\text{}+\text{}\mathrm{Profit}\\ =\text{₹}\left(30+4.80\right)\\ =₹\text{}34.80\end{array}$

Q.10 The cost of an article was ₹ 15,500. ₹ 450 were spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.

Ans

The cost of an article was ₹ 15,500
The amount spent on its repairs ₹ 450

$\begin{array}{l}\therefore \text{Total cost of an article}=\text{Cost}+\text{Overhead expenses}\\ \\ =\text{₹ 155}00+\text{Rs 45}0\\ \\ =\text{₹ 1595}0\\ \mathrm{Profit}\mathrm{%}=\frac{\mathrm{Profit}}{\mathrm{C}.\mathrm{P}.}×100\\ ⇒\text{}15=\frac{\text{Profit}}{\text{₹}15,950}×100\\ \therefore \text{Profit}=₹\text{}\left(\frac{15,950×15}{100}\right)\\ \text{}=₹\text{2392.50}\\ \\ \text{We know, S.P. of an article}=\text{C}.\text{P}.+\text{Profit}\\ \\ =\text{₹}\left(\text{1595}0+\text{2392}.\text{5}0\right)\\ \\ =\text{₹ 18,342}.\text{5}0\end{array}$

Q.11 A VCR and TV were bought for ₹ 8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss percent on the whole transaction.

Ans

$\begin{array}{l}\mathrm{C}.\text{P}.\text{of a VCR}=\text{₹ 8}000\\ \text{Loss% = 4}%\text{}\\ \end{array}$ $\begin{array}{l}\text{Let us suppose that C}.\text{P}.\text{is ₹ 1}00\\ \text{Since loss is}4\mathrm{%}\text{,so S}.\text{P}.\text{is ₹ 96}.\\ \\ \therefore \text{When C}.\text{P}.\text{is Rs 8}000,\text{S}.\text{P}.=₹\left(\frac{96}{100}×8000\right)=₹\text{}7680\\ \\ \text{Now, C}.\text{P}.\text{of a TV}=\text{₹ 8}000\\ \text{The shopkeeper made a profit of 8}\mathrm{%}\text{on TV}.\\ \\ \text{Let us suppose that C}.\text{P}.\text{of TV is ₹ 1}00\\ \text{Then S}.\text{P}.\text{is ₹ 1}0\text{8}.\\ \text{When C}.\text{P}.\text{is Rs 8}000,\text{S}.\text{P}.=₹\left(\frac{108}{100}×8000\right)=₹\text{}8640\\ \\ \text{To find the profit or loss on whole transaction,we have}\\ \text{to find total C.P.and S.P.}\\ \\ \text{Total S}.\text{P}.=₹\text{768}0+₹\text{864}0=₹\text{1632}0\\ \text{Total C}.\text{P}.=₹\text{8}000+₹\text{8}000=₹\text{16}000\\ \\ \text{Since total S}.\text{P}.\text{is greater than total C}.\text{P}.,\mathrm{therefore},\mathrm{we}\text{}\mathrm{have}\text{a profit}.\end{array}$ $\begin{array}{l}\text{Profit}=\text{Total S}.\text{P}-\text{Total C}.\text{P}.\\ \text{}=₹\text{1632}0-₹\text{16}000\\ \text{}=₹\text{32}0\\ \\ \text{Now,Profit}\mathrm{%}=\frac{\text{Profit}}{\mathrm{C}.\mathrm{P}.}×100\\ =\frac{\text{320}}{16000}×100\\ =2\mathrm{%}\\ \\ \text{Hence},\text{the shopkeeper had a gain of 2}\mathrm{%}\text{on the whole transaction}.\end{array}$

Q.12 During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at ₹ 1450 and two shirts marked at ₹ 850 each?

Ans

$\begin{array}{l}\text{Total marked price}=₹\text{}\left(\text{1},\text{45}0+\text{2}×\text{85}0\right)\\ =₹\text{}\left(\text{1},\text{45}0+\text{1},\text{7}00\right)\\ =₹\text{3},\text{15}0\\ \text{Given that},\text{discount}%=\text{1}0%\\ \therefore \text{Discount}=₹\text{}\left(\text{3},\text{15}0×\frac{10}{100}\right)=₹\text{315}\\ \text{Also},\text{Discount}=\text{Marked price}-\text{Sale price}\\ ⇒\text{₹ 315}=₹\text{315}0-\text{Sale price}\\ ⇒\text{Sale price}=₹\text{}\left(\text{315}0-\text{315}\right)\\ \text{}=₹\text{2835}\\ \\ \text{Thus},\text{the customer will have to pay ₹ 2},\text{835}.\end{array}$

Q.13 A milkman sold two of his buffaloes for ₹ 20,000 each.On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss. (Hint: Find CP of each)

Ans

$\begin{array}{l}\mathrm{Given},\mathrm{S}.\mathrm{P}.\text{}\mathrm{of}\text{}\mathrm{each}\text{}\mathrm{buffalo}=\text{₹}20000\\ \mathrm{Gain}\mathrm{%}=5\mathrm{%}\text{}\\ \\ \mathrm{This}\text{}\mathrm{means}\text{}\mathrm{if}\text{}\mathrm{C}.\mathrm{P}.\text{}\mathrm{is}\text{₹}100,\text{}\mathrm{then}\text{}\mathrm{S}.\mathrm{P}.\text{}\mathrm{is}\text{₹}105.\\ \\ \therefore \mathrm{C}.\mathrm{P}.\text{}\mathrm{of}\text{}\mathrm{one}\text{}\mathrm{buffalo}=₹\text{}20000×\frac{100}{105}=₹\text{}19047.62\\ \\ \mathrm{Also},\mathrm{the}\text{}\mathrm{second}\text{}\mathrm{buffalo}\text{}\mathrm{was}\text{}\mathrm{sold}\text{}\mathrm{at}\text{}\mathrm{a}\text{}\mathrm{loss}\text{}\mathrm{of}\text{}10\mathrm{%}.\\ \\ \mathrm{This}\text{}\mathrm{means}\text{}\mathrm{if}\text{}\mathrm{C}.\mathrm{P}.\text{}\mathrm{is}\text{₹}100,\text{}\mathrm{then}\text{}\mathrm{S}.\mathrm{P}.\text{}\mathrm{is}\text{₹}90.\\ \\ \therefore \mathrm{C}.\mathrm{P}.\text{}\mathrm{of}\text{}\mathrm{other}\text{}\mathrm{buffalo}=₹\text{}20000×\frac{100}{90}=₹\text{}22222.22\\ \\ \therefore \mathrm{Total}\text{}\mathrm{C}.\mathrm{P}.=₹\text{}19047.62+₹\text{}22222.22=₹\text{}41269.84\\ \mathrm{Total}\text{}\mathrm{S}.\mathrm{P}.=\text{₹}20000\text{}+\text{₹}20000=\text{₹}40000\\ \\ \mathrm{Loss}=₹\text{}41269.84-\text{₹}40000=₹\text{}1269.84\\ \\ \mathrm{Thus},\text{}\mathrm{the}\text{}\mathrm{overall}\text{loss}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{m}\text{ilkman}\mathrm{was}\text{₹}1,269.84.\end{array}$

Q.14 The price of a TV is ₹ 13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.

Ans

$\begin{array}{l}\text{The price of the TV = ₹ 13,000}\\ \text{Sales tax =12%}\\ \text{i}\text{.e}\text{.= ₹}\left(\frac{12}{100}\text{×13,000}\right)\\ \text{}=₹\text{}1,560\\ \therefore \text{Required amount}=\text{Cost}+\text{Sales Tax}\\ =\text{₹ 13}000+\text{₹ 156}0\\ =₹\text{1456}0\\ \\ \text{Therefore},\text{Vinod have to pay ₹ 14},\text{56}0\text{for the T}.\text{V}.\end{array}$

Q.15 Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is ₹ 1,600, find the marked price.

Ans

$\begin{array}{l}\text{Let the marked price be}\mathrm{x}.\\ \text{Discount}\mathrm{%}=\frac{\text{Discount}}{\text{Marked price}}×100\\ 20=\frac{\text{Discount}}{\mathrm{x}}×100\\ \text{Discount}=\frac{20\mathrm{x}}{100}=\frac{1}{5}\mathrm{x}\\ \text{Also,Discount = Marked price}-\text{Sale price}\\ \frac{1}{5}\mathrm{x}=\mathrm{x}-₹\text{}1600\\ \mathrm{x}-\frac{1}{5}\mathrm{x}=₹\text{}1600\\ \frac{4}{5}\mathrm{x}=₹\text{}1600\\ \mathrm{x}=\text{₹}1600×\frac{5}{4}=₹\text{}2000\\ \\ \therefore \text{The marked price of the scates was ₹ 2000.}\end{array}$

Q.16 I purchased a hair-dryer for ₹ 5,400 including 8% VAT. Find the price before VAT was added.

Ans

$\begin{array}{l}\text{It is given that the price of the hair drier includes VAT}.\\ \\ \text{Let us suppose the price without VAT be ₹ 1}00,\text{then}\\ \text{the price including VAT will be ₹ 1}0\text{8}.\\ \\ \text{When price including VAT is ₹ 1}0\text{8},\text{original price}=₹\text{1}00\\ \text{When price including VAT is ₹ 5400,original price}\\ =₹\left(\frac{100}{108}×5400\right)\\ =₹\text{}5000\\ \\ \text{Thus},\text{the price of the hair}-\text{dryer before the}\\ \text{addition of VAT was ₹ 5},000.\end{array}$

Q.17

$\begin{array}{l}\text{Calculate the amount and compound interest on}\\ \left(\mathrm{a}\right)\text{ ₹}10,800\mathrm{for}3\text{ }\mathrm{years}\text{ }\mathrm{at}\text{ }12\frac{1}{2}\mathrm{%}\mathrm{per}\mathrm{annum}\\ \mathrm{compounded}\text{ }\mathrm{annually}.\\ \left(\mathrm{b}\right)\text{ ₹}18,000\mathrm{for}2\frac{1}{2}\text{ }\mathrm{years}\text{ }\mathrm{at}\text{ }10\mathrm{%}\text{ }\mathrm{per}\mathrm{annum}\\ \mathrm{compounded}\text{ }\mathrm{annually}.\\ \left(\mathrm{c}\right)\text{ ₹}62,500\mathrm{for}1\frac{1}{2}\text{ }\mathrm{years}\text{ }\mathrm{at}\text{ }8\mathrm{%}\text{ }\mathrm{per}\mathrm{annum}\\ \mathrm{compounded}\text{ }\mathrm{half}\text{ }\mathrm{yearly}.\\ \left(\mathrm{d}\right)\text{ ₹ }8,000\mathrm{for}1\text{ }\mathrm{year}\text{ }\mathrm{at}\text{ }9\mathrm{%}\text{ }\mathrm{per}\mathrm{annum}\\ \mathrm{compounded}\text{ }\mathrm{half}\text{ }\mathrm{yearly}.\\ \mathrm{}\left(\begin{array}{l}\mathrm{You}\text{ }\mathrm{could}\text{ }\mathrm{use}\text{ }\mathrm{they}\text{ }\mathrm{ear}\text{ }\mathrm{by}\text{ }\mathrm{year}\text{ }\mathrm{calculation}\\ \mathrm{using}\text{ }\mathrm{SI}\text{ }\mathrm{for}\text{ }\mathrm{mulatoverify}\end{array}\right).\\ \left(\mathrm{e}\right)\text{ ₹}10,000\mathrm{for}1\text{ }\mathrm{yearat}\text{ }8\mathrm{%}\text{ }\mathrm{per}\mathrm{annum}\\ \mathrm{compounded}\text{ }\mathrm{half}\text{ }\mathrm{yearly}.\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\text{Principal}=₹\text{}10,800\\ \text{}\mathrm{R}=12\frac{1}{2}\mathrm{%}=\frac{25}{2}\mathrm{%}\\ \text{}\mathrm{n}=3\\ \therefore \mathrm{Amount}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}\\ =10800{\left(1+\frac{25}{2×100}\right)}^{3}\\ =10800{\left(1+\frac{25}{200}\right)}^{3}\\ =10800{\left(\frac{225}{200}\right)}^{3}\\ =10800×\frac{225}{200}×\frac{225}{200}×\frac{225}{200}\\ =15377.34\\ \\ \therefore \mathrm{C}.\mathrm{I}=\mathrm{Amount}-\mathrm{Principal}\\ =15377.34-10800\\ =4,577.34\end{array}$ $\begin{array}{l}\left(\mathrm{b}\right)\text{Principal}=₹\text{}18,000\\ \text{}\mathrm{R}=10\mathrm{%}\\ \text{}\mathrm{n}=2\frac{1}{2}\\ \therefore \mathrm{Amount}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}\\ \text{}=18000{\left(1+\frac{10}{100}\right)}^{2\frac{1}{2}}\\ \\ \text{Since,the years are in fractional form,so we will calculate}\\ \text{the Amount for the whole part first and then we will calculate}\\ \text{the simple interest for the remaining fractional part.}\\ \therefore \mathrm{Amount}=18000{\left(1+\frac{10}{100}\right)}^{2}\\ =18000{\left(\frac{11}{10}\right)}^{2}\\ =18000×\frac{11}{10}×\frac{11}{10}\\ =₹\text{}21,780\end{array}$ $\begin{array}{l}\text{Now,we will take ₹ 21,780 as principal and calculate the simple}\\ \text{interest for the next}\frac{1}{2}\mathrm{year}.\\ \therefore \mathrm{S}.\mathrm{I}=\frac{\mathrm{PRT}}{100}\\ =\frac{21780×10×1}{2×100}\\ =₹\text{1089}\\ \\ \therefore \text{Interest for the first 2 years}=₹\left(\text{2178}0-\text{18}000\right)=₹\text{378}0\\ \text{And interest for the next}\frac{1}{2}\text{year}=₹\text{1}0\text{89}\\ \therefore \text{Total C}.\text{I}.=₹\text{378}0+₹\text{1}0\text{89}=₹\text{4},\text{869, and}\\ \\ \text{A}=\text{P}+\text{C}.\text{I}.=₹\text{18}000+₹\text{4869}=₹\text{22},\text{869}\end{array}$ $\begin{array}{l}\left(\mathrm{c}\right)\text{Principal}=₹\text{}62,500\\ \text{}\mathrm{R}=8\mathrm{%}\text{}\mathrm{yearly}=4\mathrm{%}\text{}\mathrm{half}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=1\frac{1}{2}=3\text{}\mathrm{half}\text{}\mathrm{years}\\ \therefore \mathrm{Amount}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}\\ \text{}=62500{\left(1+\frac{4}{100}\right)}^{3}\\ \text{}=62500{\left(\frac{26}{25}\right)}^{3}\\ \text{}=62500×\frac{26}{25}×\frac{26}{25}×\frac{26}{25}\\ \text{}=₹\text{}70,304\\ \\ \therefore \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}=₹\text{}70,304-₹\text{}62,500\\ \text{}=₹\text{}7,804\\ \left(\mathrm{d}\right)\text{Principal}=₹\text{}8000\\ \text{}\mathrm{R}=9\mathrm{%}\text{}\mathrm{yearly}=\frac{9}{2}\mathrm{%}\text{}\mathrm{half}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=1=2\text{}\mathrm{half}\text{}\mathrm{years}\\ \therefore \mathrm{Amount}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}\\ \text{}=8000{\left(1+\frac{9}{200}\right)}^{2}\\ \text{}=8000{\left(\frac{209}{200}\right)}^{2}\\ \text{}=₹\text{}8,736.20\\ \\ \therefore \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}=₹\text{}8,736.20-\mathrm{Rs}\text{}8000\\ \text{}=₹\text{}736.20\\ \left(\mathrm{e}\right)\text{}\mathrm{Principal}=\mathrm{Rs}\text{}10,000\\ \text{}\mathrm{R}=8\mathrm{%}\text{}\mathrm{yearly}=4\mathrm{%}\text{}\mathrm{half}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=1=2\text{}\mathrm{half}\text{}\mathrm{years}\\ \therefore \mathrm{Amount}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}\\ \text{}=10,000{\left(1+\frac{4}{100}\right)}^{2}\\ \text{}=10,000{\left(1+\frac{1}{25}\right)}^{2}\\ \text{}=10,000{\left(\frac{26}{25}\right)}^{2}\\ \text{}=10,000×\frac{26}{25}×\frac{26}{25}\\ \text{}=₹\text{}10,816\\ \\ \therefore \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}=\mathrm{Rs}\text{}10,816-\mathrm{Rs}\text{}10,000\\ =₹\text{}816\end{array}$

Q.18 Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for 4/12 years).

Ans

$\begin{array}{l}\mathrm{Principal}=₹\text{}26,400\\ \text{}\mathrm{R}=15\mathrm{%}\text{}\mathrm{per}\text{}\mathrm{annum}\\ \text{}\mathrm{n}=2\frac{4}{12}\mathrm{years}\\ \mathrm{Since},\mathrm{the}\text{}\mathrm{years}\text{}\mathrm{are}\text{}\mathrm{in}\text{}\mathrm{fractional}\text{}\mathrm{form},\mathrm{so}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{calculate}\\ \mathrm{the}\text{}\mathrm{Amount}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{whole}\text{}\mathrm{part}\text{}\mathrm{first}\text{}\mathrm{and}\text{}\mathrm{then}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{calculate}\\ \mathrm{the}\text{}\mathrm{simple}\text{}\mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{remaining}\text{}\mathrm{fractional}\text{}\mathrm{part}.\\ \\ \therefore \mathrm{Amount}=26,400{\left(1+\frac{15}{100}\right)}^{2}\\ =26,400{\left(\frac{23}{20}\right)}^{2}\end{array}$ $\begin{array}{l}=26,400×\frac{23}{20}×\frac{23}{20}\\ =₹\text{}34,914\\ \mathrm{Now},\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{take}\text{}\mathrm{Rs}\text{}34,914\text{}\mathrm{as}\text{}\mathrm{principal}\text{}\mathrm{and}\text{}\mathrm{calculate}\text{}\mathrm{the}\text{}\mathrm{simple}\text{}\\ \mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{next}\text{}\frac{1}{3}\mathrm{years}.\\ \therefore \mathrm{S}.\mathrm{I}=\frac{\mathrm{PRT}}{100}\\ =\frac{34,914×15×1}{3×100}\\ =₹\text{}1745.70\\ \\ \therefore \text{}\mathrm{Interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{first}\text{}2\text{}\mathrm{years}=₹\left(34,914-26,400\right)=₹\text{}8,514\\ \mathrm{And}\text{}\mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{next}\text{}\frac{1}{3}\mathrm{year}=₹\text{}1745.70\\ \therefore \text{}\mathrm{Total}\text{}\mathrm{C}.\mathrm{I}.=₹\text{}8,514+₹\text{}1745.70=₹\text{}10,259.70,\text{}\mathrm{and}\\ \\ \mathrm{A}\text{}=\text{}\mathrm{P}\text{}+\text{}\mathrm{C}.\mathrm{I}.=₹\text{}26,400+₹\text{}10,259.70=₹\text{}36,659.70\end{array}$

Q.19 Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

Ans

$\begin{array}{l}\mathrm{Interest}\text{}\mathrm{paid}\text{}\mathrm{by}\text{}\mathrm{Fabina}=\frac{\mathrm{PRT}}{100}\\ =₹\left(\frac{12,500×12×3}{100}\right)\\ =₹\text{}4,500\\ \\ \mathrm{Amount}\text{}\mathrm{paid}\text{}\mathrm{by}\text{}\mathrm{Radha}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}\\ =₹\text{}12,500{\left(1+\frac{10}{100}\right)}^{3}\\ \\ =₹\text{}12,500×{\left(\frac{110}{100}\right)}^{3}\\ =₹\text{}16,637.50\\ \\ \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}=₹\text{}16,637.50\text{}-₹\text{}12,500\\ =4,137.50\\ \\ \mathrm{The}\text{}\mathrm{interest}\text{}\mathrm{paid}\text{}\mathrm{by}\text{}\mathrm{Fabina}\text{}\mathrm{is}\text{₹}4,500\text{}\mathrm{and}\text{}\mathrm{by}\text{}\mathrm{Radha}\text{}\mathrm{is}\text{₹}4,137.50.\\ \mathrm{Thus},\text{}\mathrm{Fabina}\text{}\mathrm{pays}\text{}\mathrm{more}\text{}\mathrm{interest}.\\ \mathrm{i}.\mathrm{e}.\text{₹}4500-₹\text{}4137.50=₹\text{}362.50\\ \\ \mathrm{Hence},\text{}\mathrm{Fabina}\text{}\mathrm{will}\text{}\mathrm{have}\text{}\mathrm{to}\text{}\mathrm{pay}\text{₹}362.50\text{}\mathrm{more}.\end{array}$

Q.20 I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?

Ans

$\begin{array}{l}\text{}\mathrm{Principal}=₹\text{}12,000\\ \text{}\mathrm{R}=6\mathrm{%}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=2\text{}\mathrm{years}\\ \mathrm{Simple}\text{}\mathrm{Interest}\text{}=\frac{\mathrm{PRT}}{100}\\ \text{}=₹\left(\frac{12,000×6×2}{100}\right)\\ \text{}=₹\text{}1,440\\ \therefore \mathrm{Amount}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}\\ \text{}=12,000{\left(1+\frac{6}{100}\right)}^{2}\\ \text{}=12,000{\left(1+\frac{3}{50}\right)}^{2}\\ \text{}=12,000{\left(\frac{53}{50}\right)}^{2}\\ \text{}=12,000×\frac{53}{50}×\frac{53}{50}\\ \text{}=₹\text{}13,483.20\\ \\ \therefore \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}=₹\text{}13,483.20-₹\text{}12,000=₹\text{}1,483.20\\ \text{}\mathrm{C}.\mathrm{I}-\mathrm{S}.\mathrm{I}=₹\text{}1,483.20-₹\text{}1,440=₹\text{}43.20\\ \\ \mathrm{Therefore},\text{}\mathrm{the}\text{}\mathrm{extra}\text{}\mathrm{amount}\text{}\mathrm{to}\text{}\mathrm{be}\text{}\mathrm{paid}\text{}\mathrm{is}\text{₹}43.20.\end{array}$

Q.21 Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get

(i) after 6 months?

(ii) after 1 year?

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{}\mathrm{Principal}=₹\text{}60,000\\ \text{}\mathrm{R}=12\mathrm{%}\text{}\mathrm{yearly}=6\mathrm{%}\text{}\mathrm{half}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=6\text{}\mathrm{months}=1\text{}\mathrm{half}\text{}\mathrm{year}\\ \\ \therefore \mathrm{Amount}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}\\ \text{}=60,000{\left(1+\frac{6}{100}\right)}^{1}\\ \text{}=60,000\left(\frac{106}{100}\right)\\ \text{}=\text{}₹\text{}63,600\\ \left(\mathrm{ii}\right)\text{}\mathrm{Principal}=₹\text{}60,000\\ \text{}\mathrm{R}=12\mathrm{%}\text{}\mathrm{yearly}=6\mathrm{%}\text{}\mathrm{half}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=1\text{}\mathrm{year}=2\text{}\mathrm{half}\text{}\mathrm{years}\\ \\ \therefore \mathrm{Amount}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}\\ \text{}=60,000{\left(1+\frac{6}{100}\right)}^{2}\\ \text{}=60,000{\left(\frac{106}{100}\right)}^{2}\\ \text{}=\text{}60,000×\frac{106}{100}×\frac{106}{100}\\ \text{}=\text{}₹\text{}67,416\end{array}$

Q.22 Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after

$1\frac{1}{2}$

years if the interest is

(i) compounded annually.

(ii) compounded half yearly.

Ans

$\begin{array}{l}\text{(i)}\mathrm{Principal}=₹\text{80},000\\ \mathrm{R}=10%\text{}\mathrm{per}\text{}\mathrm{annum}\\ \mathrm{n}=1\frac{1}{2}\mathrm{years}\\ \mathrm{Since},\mathrm{the}\text{}\mathrm{years}\text{}\mathrm{are}\text{}\mathrm{in}\text{}\mathrm{fractional}\text{}\mathrm{form},\mathrm{so}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{calculate}\\ \mathrm{the}\text{}\mathrm{Amount}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{whole}\text{}\mathrm{part}\text{}\mathrm{first}\text{}\mathrm{and}\text{}\mathrm{then}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{calculate}\\ \mathrm{the}\text{}\mathrm{simple}\text{}\mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{remaining}\text{}\mathrm{fractional}\text{}\mathrm{part}.\\ \therefore \mathrm{Amount}=80,000{\left(1+\frac{10}{100}\right)}^{1}\\ =80,000\left(1+\frac{1}{10}\right)\\ =80,000{\left(\frac{11}{10}\right)}^{2}\\ =₹\text{}88,000\\ \\ \mathrm{Now},\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{take}\text{}\mathrm{Rs}\text{}88,000\text{}\mathrm{as}\text{}\mathrm{principal}\text{}\mathrm{and}\text{}\mathrm{calculate}\text{}\mathrm{the}\text{}\mathrm{simple}\text{}\\ \mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{next}\text{}\frac{1}{2}\mathrm{year}.\\ \therefore \mathrm{S}.\mathrm{I}=\frac{\mathrm{PRT}}{100}\\ =\frac{88,000×10×1}{2×100}\\ =₹\text{}4,400\\ \text{Interest for the first year}=\text{₹ 88}000\text{}-\text{₹ 8}0000\text{}=\text{₹ 8},000\\ \text{And interest for the next}\frac{1}{2}\text{year}=\text{₹ 4},\text{4}00\\ \text{Total C}.\text{I}.\text{}=\text{₹ 8}000\text{}+\text{₹ 4},\text{4}00=\text{₹ 1},\text{24}00\\ \\ \therefore \text{A}=\text{P}+\text{C}.\text{I}.\text{}=\text{₹}\left(\text{8}0000\text{}+\text{124}00\right)\text{}=\text{₹ 92},\text{4}00\end{array}$ $\begin{array}{l}\left(\text{ii}\right)\text{Now,the interest is compounded half yearly}.\\ \text{Rate}=\text{1}0%\text{per annum}=\text{5}%\text{per half year}\\ \text{n=}1\frac{1}{2}\mathrm{years}=3\text{half years}\\ \therefore \mathrm{Amount}=80,000{\left(1+\frac{5}{100}\right)}^{3}\\ =80,000\left(1+\frac{1}{20}\right)3\\ =80,000{\left(\frac{21}{20}\right)}^{2}\\ =₹\text{92},610\end{array}$

Therefore, the difference between the amounts = ₹ 92,610 − ₹ 92,400 = ₹ 210

Q.23 Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find

(i) The amount credited against her name at the end of the second year.

(ii) The interest for the 3rd year.

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{}\mathrm{Principal}=₹\text{8},000\\ \text{}\mathrm{R}=5\mathrm{%}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=2\text{}\mathrm{years}\\ \\ \therefore \mathrm{Amount}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}\\ \text{}=8,000{\left(1+\frac{5}{100}\right)}^{2}\\ \text{}=8,000{\left(\frac{21}{20}\right)}^{2}\\ \text{}=\text{8},000×\frac{21}{20}×\frac{21}{20}\\ \text{}=\text{}₹\text{}8,820\end{array}$ $\begin{array}{l}\text{(ii)}\mathrm{Now},\text{we have to calculate the interest for the third year.}\\ \text{So we}\mathrm{will}\text{}\mathrm{take}\text{}\mathrm{Rs}\text{8,820}\mathrm{as}\text{}\mathrm{principal}\text{}\mathrm{and}\text{}\mathrm{calculate}\text{}\mathrm{the}\text{}\mathrm{simple}\text{}\\ \mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{next}\text{}\mathrm{year}.\\ \therefore \mathrm{S}.\mathrm{I}=\frac{\mathrm{PRT}}{100}\\ =\frac{8820×5×1}{100}\\ =₹\text{}441\end{array}$

Q.24 Find the amount and the compound interest on ₹10,000 for years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Ans

$\begin{array}{l}\mathrm{Principal}=₹\text{}10,000\\ \text{}\mathrm{R}=10\mathrm{%}\text{}\mathrm{yearly}=5\mathrm{%}\text{}\mathrm{half}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=1\frac{1}{2}\text{}\mathrm{years}=3\text{}\mathrm{half}\text{}\mathrm{years}\\ \\ \therefore \mathrm{Amount}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}\\ \text{}=10,000{\left(1+\frac{5}{100}\right)}^{3}\\ \text{}=10,000{\left(1+\frac{1}{20}\right)}^{3}\\ \text{}=10,000{\left(\frac{21}{20}\right)}^{3}\end{array}$ $\begin{array}{l}\text{}=\text{}60,000×\frac{21}{20}×\frac{21}{20}×\frac{21}{20}\\ \text{}=\text{}₹\text{}11,576.25\\ \\ \mathrm{C}.\mathrm{I}=\mathrm{A}-\mathrm{P}\\ \text{}=₹\text{}11,576.25-\mathrm{Rs}\text{}10,000\\ \text{}=₹\text{}1,576.25\\ \\ \mathrm{Now}\text{we have to calculate the interest compounded annually.}\\ \mathrm{Since},\mathrm{the}\text{}\mathrm{years}\text{}\mathrm{are}\text{}\mathrm{in}\text{}\mathrm{fractional}\text{}\mathrm{form},\mathrm{so}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{calculate}\\ \mathrm{the}\text{}\mathrm{Amount}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{whole}\text{}\mathrm{part}\text{}\mathrm{first}\text{}\mathrm{and}\text{}\mathrm{then}\text{}\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{calculate}\\ \mathrm{the}\text{}\mathrm{simple}\text{}\mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{remaining}\text{}\mathrm{fractional}\text{}\mathrm{part}.\\ \\ \therefore \mathrm{Amount}=₹\text{}10,000{\left(1+\frac{10}{100}\right)}^{1}\\ =₹\text{}10,000\left(\frac{11}{10}\right)\\ =11,000\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\mathrm{Now},\mathrm{we}\text{}\mathrm{will}\text{}\mathrm{take}\text{}₹11,000\text{}\mathrm{as}\text{}\mathrm{principal}\text{}\mathrm{and}\text{}\mathrm{calculate}\text{}\mathrm{the}\text{}\mathrm{simple}\text{}\\ \mathrm{interest}\text{}\mathrm{for}\text{}\mathrm{the}\text{}\mathrm{next}\text{}\frac{1}{2}\mathrm{years}.\\ \therefore \mathrm{S}.\mathrm{I}=\frac{\mathrm{PRT}}{100}\\ =\frac{11,000×10×1}{2×100}\\ =₹\text{}550\\ \\ \therefore \text{Interest for the first year}=₹\text{11}000-₹\text{1}0000=₹\text{1},000\\ \therefore \text{Total compound interest}=₹\text{1}000+₹\text{55}0=₹\text{1},\text{55}0\\ \\ \text{Yes,the interest would be more when compounded half yearly}\\ \text{than the interest when compounded annually}.\end{array}$

Q.25 Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at

$12\frac{1}{2}$

% 2 per annum, interest being compounded half yearly.

Ans

$\begin{array}{l}\mathrm{Principal}=₹\text{}4,096\\ \text{}\mathrm{R}=12\frac{1}{2}\mathrm{%}\text{}\mathrm{yearly}=\frac{25}{4}\mathrm{%}\text{}\mathrm{half}\text{}\mathrm{yearly}\\ \text{}\mathrm{n}=18\text{}\mathrm{months}=3\text{}\mathrm{half}\text{}\mathrm{years}\\ \\ \therefore \mathrm{Amount}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{n}}\\ \text{}=₹\text{}4096{\left(1+\frac{25}{400}\right)}^{3}\\ \text{}=₹\text{}4096{\left(1+\frac{1}{16}\right)}^{3}\\ \text{}=₹\text{}4096{\left(\frac{17}{16}\right)}^{3}\\ \text{}=\text{₹}4096×\frac{17}{16}×\frac{17}{16}×\frac{17}{16}\\ \text{}=\text{₹}4,913\\ \\ \text{Therefore},\text{the required amount is ₹ 4},\text{913}.\end{array}$

Q.26 The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

(i) find the population in 2001.

(ii) what would be its population in 2005?

Ans

$\begin{array}{l}\text{(i) The population in the year 2003= 54},000\\ \text{}\\ \therefore 54,000=\left(\text{Population in 2001}\right){\left(1+\frac{5}{100}\right)}^{2}\\ \text{}54,000=\left(\text{Population in 2001}\right){\left(\frac{21}{20}\right)}^{2}\\ \text{Population in 2001}=54,000×{\left(\frac{20}{21}\right)}^{2}\\ \text{}=48979.59\\ \\ \text{Therefore},\text{the population in the year 2}00\text{1 was approximately 48},\text{98}0.\\ \text{(ii) Population in 2005}=54,000×{\left(1+\frac{5}{100}\right)}^{2}\\ =54,000×{\left(1+\frac{1}{20}\right)}^{2}\\ =54,000×{\left(\frac{21}{20}\right)}^{2}\\ =59,535\\ \\ \text{Therefore},\text{the population in the year 2}00\text{5 would be 59},\text{535}.\end{array}$

Q.27 In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5, 06,000.

Ans

$\begin{array}{l}\text{The initial count of bacteria is given as}5,06,000.\\ \text{Bacteria at the end of 2 hours}=5,06,000{\left(1+\frac{1}{40}\right)}^{2}\\ =5,06,000{\left(\frac{41}{40}\right)}^{2}\text{}\\ =5,06,000×\frac{41}{40}×\frac{41}{40}\\ =\text{}531616.25\\ \\ \text{Thus, the count of bacteria at the end of 2 hours will be}5,31,616\text{}\left(\text{approx}.\right).\end{array}$

Q.28 A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Ans

$\begin{array}{l}\mathrm{Principal}=\mathrm{Cost}\text{}\mathrm{price}\text{}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{scooter}=₹\text{}42,000\\ \mathrm{Depreciation}=8\mathrm{%}\text{​}\mathrm{of}\text{₹}42,000\text{}\\ \text{}=\text{}42000×\frac{8}{100}=₹\text{}3,360\\ \\ \therefore \mathrm{Value}\text{}\mathrm{after}\text{}1\text{}\mathrm{year}=₹\text{}42000-₹\text{}3360=₹\text{}38,640.\end{array}$