NCERT Solutions Class 8 Maths Chapter 8 Exercise 8.2
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The NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 are intended to help students fully understand the ideas and gain a thorough understanding of the many subjects covered in the CBSE Class 8 Mathematics problems. The subject matter specialists at Extramarks have created the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 based on Comparing Quantities.Students can prepare for exams by downloading and using the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2.. The students will be able to answer the questions more accurately and will be better prepared for their exams with the aid of these NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2.
The means of comparing quantities are the subject of a number of problems that must be solved by students in this chapter of Mathematics Class 8. But the PDF provided mainly focuses on NCERT Solutions for Class 8 Maths chapter 8 Exercise 8.2. Therefore, students are suggested to use the Extramarkscurated NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 for a better approach and understanding. The NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 based on Comparing Quantities are available online, but it is advised for readers to download the PDF version of the NCERT Solutions for Class 8 Maths chapter 8 Exercise 8.2 and save it on their computer or tablet. This way, they can study it at any time and anywhere.
In the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 numerous stories in the Comparing Quantities literature are based on the principles of ratios and percentages. There are instances where the original quantity’s value rises and the percentagemust be calculated, and vice versa, where the original quantity’s value decreases and the percentage or ratio must be changed.
The lengthy problems in the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 might often obscure how the ratios and percentages theory can be used. These sums have also offered tips, but it’s more enjoyable to forego them at first and attempt the problem on your own. These inquiries will aid in the development of the logical thinking abilities required to properly structure the problem’s component mathematical equations. The link on the Extramarks website and mobile application will take students to the PDF file of the questions in the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 based on Comparing Quantities.
Students must comprehend how the addition of percentages and ratios enhances the dissemination of knowledge. It’s a terrific technique to scale them down so that they can compare unlike quantities on the same scale.After practising these NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2, students can determine the quantitative value of two quantities in relation to one another using percentages and ratios that only require one number.Additionally, by transforming numbers into percentages, students can compare two numbers that would otherwise be incomparable. Before proceeding to the questions in the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 based on Comparing Quantities, it will be advisable for the students to practise the examples in the chapter by themselves.
NCERT Solutions for Class 8 Maths Chapter 8 (Ex 8.2) Exercise 8.2
In the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2, finding ratios between two quantities stated in different units is the first step in the Comparing Quantities section of the test. These questions will therefore reinforce both the interconversion of fundamental units of measurement and the ideas of ratios.
The NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 demonstrates how the conversion of fractions into percentages and vice versa in Class 8 Mathematics Exercise 8.2 will help demonstrate how simple it is to integrate all of these into ways of measuring quantity comparisons.Towards the end of the exercises, there are enjoyable realworld problems that show how beneficial these ideas can be and how they can convey a lot of information while also giving the student the opportunity to play while comparing two quantities.
There are 11 questions based on the aforementioned ideas in the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 ,on Comparing Quantities.
Extramarks’ NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 based on Comparing Quantities will appear simple to students if they are meticulous with the calculation portion.. Additionally, having strong mental math abilities can help students finish the questions quickly and save them a tonne of time.
The NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 provides examples to practise using the numbers in the percentage and ratio realms while also assisting in the development of a better grasp of the world of percentages and ratios.
Access NCERT Solutions for Class 8 Mathematics Chapter 8 – Comparing Quantities
Among the numerous concepts covered in the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 are: theConcept of Percent and Percentage, Concept of Ratio, Increase or Decrease as percent, Estimation in Percentages, Concept of Discount, Concepts of Cost Price, Selling price, Total Cost price, and Profit and Loss, Discount, Overhead Expenses and GST, Sales and Value Added Tax, Concept of Principle, Interest, Amount and Simple Interest, Concept of Compound Interest, Deducing a Formula for Compound Interest, Rate Compounded Annually or Half yearly (Semi Annually), Applications of Compound Interest Formula.
There may be a number of options available online, but the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 are guaranteed to be among the best ones. With the elf of these NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2, students have an easy route to preparingfor their final examinations.
Exercise 8.2
People encounter several circumstances, amounts, activities, etc. in daily life. It is frequently needed to measure and compare these amounts. Humans, in fact, occasionally evaluate themselves against others in terms of their size, height, grades, speed, quantity, etc. The number of items on this list grows. A person’s weight cannot be compared to another person’s height because one can only compare two exact quantities. Consequently, there needs to be a standard point of comparison at all times.
A common yardstick or method for assessing quantities is also required. Among them are proportion and ratio. The comparison between two quantities in terms of ratio and proportion can be taught to students. The NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 will teach students all there is to know about comparing various quantities.
Comparing amounts refers to the quantitative relationship between two quantities that captures their respective sizes. It is merely a method for comparing quantities.
NCERT Solutions for Class 8 Exercise 8.2 Solutions – Brief Overview
The units of the two quantities must be the same in order to compare them. When two ratios are converted into like fractions, they can be compared. It is said that the two specified ratios are equivalent if the two fractions are equal. The four quantities involved will be in proportion if two ratios are equivalent (or equal). The formulas for ratio, proportion, and Comparing Quantities are explained in this article. Additionally, it aids in the resolution of proportional and ratio issues.
A number line can be used to teach children the importance of understanding the order of numbers when teaching them how to compare two or more items. Students can practise counting objects once the sequence of numbers can be seen.
Comparing quantities, as is known, is the quantitative relationship between two quantities that reflects the relative sizes of both quantities. It is merely a tool for comparing quantities. In order to determine if one quantity is more than, smaller than, or equal to another, it is necessary to compare the differences between the two numbers, quantities, or values.
Here is an overview of the main concepts in the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2:
 The units of two quantities must match in order to compare them.
 • By transforming two ratios into similar fractions, two ratios can be compared. We say that the two specified ratios are equivalent if the two fractions are equal.
 • Four quantities are said to be in proportion if two ratios are equivalent (or equal).
 • Percentage comparisons of amounts are one method of comparison. The word “percent” is derived from the Latin “per centum,” which means “per hundred.”
 • The symbol %, which also stands for percent, denotes a hundredth.
 • Percentages can be changed into fractions and vice versa.
 Additionally, decimals can be changed into percentages and vice versa.
 • An item’s cost price is its price at the time of purchase. It is abbreviated as CP.
 The selling price, or SP for short, is the cost at which an object is sold.
 • If CP > SP, a profit is made, and profit = CP SP.
 • There is no gain or loss if CP Equals SP.
 • If CP>SP, a loss is created, and Loss = CPSP.
 Profit per cent = Profit ×100 CP
 Loss per cent = Loss ×100
 Principal (P) is short for borrowed funds.
 • “Interest” I is the additional payment made by the borrower for using borrowed funds for a specific period of time.
 “Time Period” T is the name of the duration for which the loan is made.
 ‘Rate of Interest’ to determine the amount of interest to be paid.
 The rate of interest is often expressed as a percentage per year.
 The total money paid alongwith interest or principal P is called the amount (A). Thus A = P + I.
Question 1
Question 1 of the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 discusses the following concepts in detail:
 Ratios and percentages to Recall
 Finding the Percent Increase or Decrease
Question 2
Question 2 of the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 discusses the following concepts in detail:
 Finding Savings
Question 3
Question 3 of the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 discusses the following concepts in detail:
 Prices for Purchasing and Selling (Profit and Loss)
Question 4
Question 4 of the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 discusses the following concepts in detail:
 calculating cost, selling price, and profit/loss ratios
Question 5
Question 5 of the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 discusses the following concepts in detail:
 Value Added Tax/Sales Tax
Question 6
Question 6 of the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 discusses the following concepts in detail:
 Compound Interest
Question 7
Question 7 of the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 discusses the following concepts in detail:
 Deducing a Formula for Compound Interest
Question 8
Question 8 of the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 discusses the following concepts in detail:
 Rate Compounding Every Year or Every Half Year (SemiAnnually)
Question 9
Question 9 of the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 discusses the following concepts in detail:
 Compound interest formula applications
Question 10
Question 10 of the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 discusses the following concepts in detail:
 Percentagebased estimation
Chapter 8 Comparing Quantities of Class 8:
Because it covers such topics as tax, interest, profit, loss, and much more, the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 is crucial to practice. The ideas covered in these NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 include:
 Discount refers to a price decrease that is offered. Discount is equal to Marked Price – Sale Price.
 When the discount % is provided, the discount can be calculated. Discount equals a percentage off the marked price.
 Overhead costs are additional costs incurred after purchasing an item and are included in the cost price. Buying price plus overhead costs equals CP.
 The government levies a sales tax on purchases, which is then added to the bill’s total. Tax% of the total bill equals sales tax.
 Products and services tax, or GST, is charged on the supply of either goods or services or both.
 The interest based on the amount from the prior year is known as compound interest (A = P + I)
Below is a list of reasons as to why students should prefer downloading the NCERT Solutions for Class 8 Maths Chapter 8 Exercise 8.2 provided by Extramarks. The students are able to do the following things after learning the chapter on Comparing Quantities:
 Applications of percentages, profit & loss, overhead costs, discount, and tax are used in slightly complex tasks.
 Difference between simple and compound interest, formula for compound interest derived from patterns, and application to simple situations (compounded annually up to 3 years or halfyearly up to 3 stages only).
 Simple and simple word problems with direct variation
 Inverse variant – straightforward word puzzles
 Simple and direct word puzzles involving time and work
NCERT Solutions for Class 8
Most of the fundamentals for the various disciplines are covered up through the eighth grade, and students are given a quick overview of all the ideas and subjects. By practising all the subjects thoroughly, including the questions from the exercises, students can grasp every subject quickly and be the top performer. All the NCERT Solutions for Class 8 have been carefully created by Extramarks’ knowledgeable mentors and instructors in order to assist students in passing with distinction.
A subjectwise list of all the subjects for which Extramarks provides NCERT Solutions for Class 8 has been provided below:
 NCERT Solutions Class 8 Maths
 NCERT Solutions Class 8 English
 NCERT Solutions Class 8 Hindi
 NCERT Solutions Class 8 Science
 NCERT Solutions Class 8 Social Science
Q.1 A man got a 10% increase in his salary. If his new salary is ₹ 1,54,000, find his original salary.
Ans
$\begin{array}{l}\text{Let the original salary be}\mathrm{x}.\text{}\\ \text{It is given that the new salary is \u20b9}1,54,000.\\ \\ \text{Therefore, Original salary + Increment = New salary}\\ \\ \text{But it is given that the increment is}10\mathrm{\%}\text{of the original salary.}\\ \\ \text{Therefore,we have}\\ \mathrm{x}\text{+}\frac{10}{100}\times \mathrm{x}=1,54,000\\ \Rightarrow \frac{110\mathrm{x}}{100}=1,54,000\\ \Rightarrow \mathrm{x}=1,54,000\times \frac{100}{110}\\ \Rightarrow \mathrm{x}=1,40,000\\ \\ \text{Thus, the original salary was \u20b9}1,40,000.\end{array}$
Q.2 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}}\times 100\right)\%\\ \\ \text{}=\left(\frac{676}{845}\times 100\right)\%\\ \\ \text{}=80\%\end{array}
Q.3 A shopkeeper buys 80 articles for ₹ 2,400 and sells them for a profit of 16%. Find the selling price of one article.
Ans
$\begin{array}{l}\mathrm{It}\text{}\mathrm{is}\text{}\mathrm{given}\text{}\mathrm{that}\text{}\mathrm{the}\text{}\mathrm{shopkeeper}\text{}\mathrm{buys}\text{}80\text{}\mathrm{articles}\text{}\mathrm{for}\text{}\u20b9\text{}2,400.\\ \mathrm{Cost}\text{}\mathrm{Price}\text{}\mathrm{of}\text{}\mathrm{one}\text{}\mathrm{article}=\u20b9\frac{2,400}{80}=\text{\u20b9}30\\ \mathrm{Profit}\text{}\mathrm{percent}=16\mathrm{\%}\\ \mathrm{Profit}\text{}\mathrm{percent}=\frac{\mathrm{Profit}}{\mathrm{Cost}\text{}\mathrm{Price}}\times 100\\ 16=\frac{\mathrm{Profit}}{\text{\u20b9}30}\times 100\\ \mathrm{Profit}=\u20b9\frac{16\times 30}{100}=\u20b9\text{}4.80\text{}\\ \\ \mathrm{Selling}\text{}\mathrm{price}\text{}\mathrm{of}\text{}\mathrm{one}\text{}\mathrm{article}=\mathrm{C}.\mathrm{P}.\text{}+\text{}\mathrm{Profit}\\ =\text{\u20b9}(30+4.80)\\ =\u20b9\text{}34.80\end{array}$
Q.4 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
$\begin{array}{l}\therefore \text{Total cost of an article}=\text{Cost}+\text{Overhead expenses}\\ \\ =\text{\u20b9 155}00+\text{Rs 45}0\\ \\ =\text{\u20b9 1595}0\\ \mathrm{Profit}\mathrm{\%}=\frac{\mathrm{Profit}}{\mathrm{C}.\mathrm{P}.}\times 100\\ \Rightarrow \text{}15=\frac{\text{Profit}}{\text{\u20b9}15,950}\times 100\\ \therefore \text{Profit}=\u20b9\text{}\left(\frac{15,950\times 15}{100}\right)\\ \text{}=\u20b9\text{2392.50}\\ \\ \text{We know, S.P. of an article}=\text{C}.\text{P}.+\text{Profit}\\ \\ =\text{\u20b9}(\text{1595}0+\text{2392}.\text{5}0)\\ \\ =\text{\u20b9 18,342}.\text{5}0\end{array}$
Q.5 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
$\begin{array}{l}\mathrm{C}.\text{P}.\text{of a VCR}=\text{\u20b9 8}000\\ \text{Loss\% = 4}\%\text{}\\ \end{array}$
$\begin{array}{l}\text{Let us suppose that C}.\text{P}.\text{is \u20b9 1}00\\ \text{Since loss is}4\mathrm{\%}\text{,so S}.\text{P}.\text{is \u20b9 96}.\\ \\ \therefore \text{When C}.\text{P}.\text{is Rs 8}000,\text{S}.\text{P}.=\u20b9(\frac{96}{100}\times 8000)=\u20b9\text{}7680\\ \\ \text{Now, C}.\text{P}.\text{of a TV}=\text{\u20b9 8}000\\ \text{The shopkeeper made a profit of 8}\mathrm{\%}\text{on TV}.\\ \\ \text{Let us suppose that C}.\text{P}.\text{of TV is \u20b9 1}00\\ \text{Then S}.\text{P}.\text{is \u20b9 1}0\text{8}.\\ \text{When C}.\text{P}.\text{is Rs 8}000,\text{S}.\text{P}.=\u20b9(\frac{108}{100}\times 8000)=\u20b9\text{}8640\\ \\ \text{To find the profit or loss on whole transaction,we have}\\ \text{to find total C.P.and S.P.}\\ \\ \text{Total S}.\text{P}.=\u20b9\text{768}0+\u20b9\text{864}0=\u20b9\text{1632}0\\ \text{Total C}.\text{P}.=\u20b9\text{8}000+\u20b9\text{8}000=\u20b9\text{16}000\\ \\ \text{Since total S}.\text{P}.\text{is greater than total C}.\text{P}.,\mathrm{therefore},\mathrm{we}\text{}\mathrm{have}\text{a profit}.\end{array}$
$\begin{array}{l}\text{Profit}=\text{Total S}.\text{P}\text{Total C}.\text{P}.\\ \text{}=\u20b9\text{1632}0\u20b9\text{16}000\\ \text{}=\u20b9\text{32}0\\ \\ \text{Now,Profit}\mathrm{\%}=\frac{\text{Profit}}{\mathrm{C}.\mathrm{P}.}\times 100\\ =\frac{\text{320}}{16000}\times 100\\ =2\mathrm{\%}\\ \\ \text{Hence},\text{the shopkeeper had a gain of 2}\mathrm{\%}\text{on the whole transaction}.\end{array}$
Q.6 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
$\begin{array}{l}\text{Total marked price}=\u20b9\text{}(\text{1},\text{45}0+\text{2}\times \text{85}0)\\ =\u20b9\text{}(\text{1},\text{45}0+\text{1},\text{7}00)\\ =\u20b9\text{3},\text{15}0\\ \text{Given that},\text{discount}\%=\text{1}0\%\\ \therefore \text{Discount}=\u20b9\text{}(\text{3},\text{15}0\times \frac{10}{100})=\u20b9\text{315}\\ \text{Also},\text{Discount}=\text{Marked price}\text{Sale price}\\ \Rightarrow \text{\u20b9 315}=\u20b9\text{315}0\text{Sale price}\\ \Rightarrow \text{Sale price}=\u20b9\text{}(\text{315}0\text{315})\\ \text{}=\u20b9\text{2835}\\ \\ \text{Thus},\text{the customer will have to pay \u20b9 2},\text{835}.\end{array}$
Q.7 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
$\begin{array}{l}\mathrm{Given},\mathrm{S}.\mathrm{P}.\text{}\mathrm{of}\text{}\mathrm{each}\text{}\mathrm{buffalo}=\text{\u20b9}20000\\ \mathrm{Gain}\mathrm{\%}=5\mathrm{\%}\text{}\\ \\ \mathrm{This}\text{}\mathrm{means}\text{}\mathrm{if}\text{}\mathrm{C}.\mathrm{P}.\text{}\mathrm{is}\text{\u20b9}100,\text{}\mathrm{then}\text{}\mathrm{S}.\mathrm{P}.\text{}\mathrm{is}\text{\u20b9}105.\\ \\ \therefore \mathrm{C}.\mathrm{P}.\text{}\mathrm{of}\text{}\mathrm{one}\text{}\mathrm{buffalo}=\u20b9\text{}20000\times \frac{100}{105}=\u20b9\text{}19047.62\\ \\ \mathrm{Also},\mathrm{the}\text{}\mathrm{second}\text{}\mathrm{buffalo}\text{}\mathrm{was}\text{}\mathrm{sold}\text{}\mathrm{at}\text{}\mathrm{a}\text{}\mathrm{loss}\text{}\mathrm{of}\text{}10\mathrm{\%}.\\ \\ \mathrm{This}\text{}\mathrm{means}\text{}\mathrm{if}\text{}\mathrm{C}.\mathrm{P}.\text{}\mathrm{is}\text{\u20b9}100,\text{}\mathrm{then}\text{}\mathrm{S}.\mathrm{P}.\text{}\mathrm{is}\text{\u20b9}90.\\ \\ \therefore \mathrm{C}.\mathrm{P}.\text{}\mathrm{of}\text{}\mathrm{other}\text{}\mathrm{buffalo}=\u20b9\text{}20000\times \frac{100}{90}=\u20b9\text{}22222.22\\ \\ \therefore \mathrm{Total}\text{}\mathrm{C}.\mathrm{P}.=\u20b9\text{}19047.62+\u20b9\text{}22222.22=\u20b9\text{}41269.84\\ \mathrm{Total}\text{}\mathrm{S}.\mathrm{P}.=\text{\u20b9}20000\text{}+\text{\u20b9}20000=\text{\u20b9}40000\\ \\ \mathrm{Loss}=\u20b9\text{}41269.84\text{\u20b9}40000=\u20b9\text{}1269.84\\ \\ \mathrm{Thus},\text{}\mathrm{the}\text{}\mathrm{overall}\text{loss}\mathrm{of}\text{}\mathrm{the}\text{}\mathrm{m}\text{ilkman}\mathrm{was}\text{\u20b9}1,269.84.\end{array}$
Q.8 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
$\begin{array}{l}\text{The price of the TV = \u20b9 13,000}\\ \text{Sales tax =12\%}\\ \text{i}\text{.e}\text{.= \u20b9}\left(\frac{12}{100}\text{\xd713,000}\right)\\ \text{}=\u20b9\text{}1,560\\ \therefore \text{Required amount}=\text{Cost}+\text{Sales Tax}\\ =\text{\u20b9 13}000+\text{\u20b9 156}0\\ =\u20b9\text{1456}0\\ \\ \text{Therefore},\text{Vinod have to pay \u20b9 14},\text{56}0\text{for the T}.\text{V}.\end{array}$
Q.9 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
$\begin{array}{l}\text{Let the marked price be}\mathrm{x}.\\ \text{Discount}\mathrm{\%}=\frac{\text{Discount}}{\text{Marked price}}\times 100\\ 20=\frac{\text{Discount}}{\mathrm{x}}\times 100\\ \text{Discount}=\frac{20\mathrm{x}}{100}=\frac{1}{5}\mathrm{x}\\ \text{Also,Discount = Marked price}\text{Sale price}\\ \frac{1}{5}\mathrm{x}=\mathrm{x}\u20b9\text{}1600\\ \mathrm{x}\frac{1}{5}\mathrm{x}=\u20b9\text{}1600\\ \frac{4}{5}\mathrm{x}=\u20b9\text{}1600\\ \mathrm{x}=\text{\u20b9}1600\times \frac{5}{4}=\u20b9\text{}2000\\ \\ \therefore \text{The marked price of the scates was \u20b9 2000.}\end{array}$
Q.10 I purchased a hairdryer for ₹ 5,400 including 8% VAT. Find the price before VAT was added.
Ans
$\begin{array}{l}\text{It is given that the price of the hair drier includes VAT}.\\ \\ \text{Let us suppose the price without VAT be \u20b9 1}00,\text{then}\\ \text{the price including VAT will be \u20b9 1}0\text{8}.\\ \\ \text{When price including VAT is \u20b9 1}0\text{8},\text{original price}=\u20b9\text{1}00\\ \text{When price including VAT is \u20b9 5400,original price}\\ =\u20b9(\frac{100}{108}\times 5400)\\ =\u20b9\text{}5000\\ \\ \text{Thus},\text{the price of the hair}\text{dryer before the}\\ \text{addition of VAT was \u20b9 5},000.\end{array}$
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