case study class 8 maths comparing quantities

• Comparing Quantities Class 8 Case Study Questions Maths Chapter 7

Last Updated on August 16, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 8 maths. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 8 maths. In this article, you will find case study questions for CBSE Class 8 Maths Chapter 7 Comparing Quantities. It is a part of Case Study Questions for CBSE Class 8 Maths Series.

Table of Contents

Case Study Questions on Comparing Quantities

Passage 1: A survey was conducted among primary school students, and were asked about how much time they spend on tuition and how much time on self study. It was found that 90 students take tuition for 1 h to 1.5 h. The distribution of students according to time they take tuition is 30% take tuition for 1.5 h to 2 h, 20% take tuitions for 1 h to 1.5 h, 50% did not take tuition at all.

On basis of this information given in passage answer following questions.

Q. 1. How many students do self study? (a) 300 (b) 225 (c) 375 (d) 275

Difficulty Level: Easy

Ans. Option (b) is correct. Explanation: Total students = 450 Students who study themselves = 50% of 450 = 50 x 450 / 100 = 225

Q. 2. How many students take tuitions for more than 1.5 h? (a) 135 (b) 150 (c) 110 (d) 105

Ans. Option (a) is correct. Explanation: Students who take tuitions for more than 1.5 h = 30% Total number of students = 450 Number of students = 30 x 450 / 100 = 135

Q. 3. For how much time does 90 students take tutions? (a) 1 hr (b) 1.5 hr (c) 2 hr (d) 1 hr to 1.5 hr

Difficulty Level: Medium

Ans. Option (d) is correct. Explanation: 90 students take tuition for 1 hr to 1.5 hr

Q. 4. How many students were surveyed?

Sol. Here, 90 students take tuition for 1 h to 1.5 h Percentage given = 20% Let number of students who take tuition = x Then, 20% of x = 90 20x/100 = 90 x = 90 x 100/20 = 450 Hence, 450 students were surveyed.

Q. 5. In all how many percent students take tuitions?

Sol. Number of students surveyed = 450 Number of students who did not tuitions = 225 Number of students who take tuitions = 450 – 225 = 225 Percentage of students take tuitions = 225 x 100/450 = 50%

• Cube and Cube Roots Class 8 Case Study Questions Maths Chapter 6
• Square and Square Roots Class 8 Case Study Questions Maths Chapter 5
• Data Handling Class 8 Case Study Questions Maths Chapter 4
• Understanding Quadrilaterals Class 8 Case Study Questions Maths Chapter 3

Linear Equations in One Variable Class 8 Case Study Questions Maths Chapter 2

Rational numbers class 8 case study questions maths chapter 1, download ebooks for cbse class 8 maths.

• Rational Numbers Topicwise Worksheet for CBSE Class 8 Maths
• Linear Equations in One Variable Worksheet for CBSE Class 8 Maths
• Understanding Quadrilaterals Worksheet for CBSE Class 8 Maths
• Data Handling Worksheet for CBSE Class 8 Maths
• Squares and Square Roots Worksheet for CBSE Class 8 Maths

Topics from which case study questions may be asked

• Recalling Ratios and Percentage
• Finding Discounts
• Estimating in Percentage
• Sales Tax/Value Added Tax/ Goods and Services Tax
• Compound Interest/Simple Interest
• Deducing a Formula for Compound Interest
• Application of Compound Interest Formula

Discount is on Marked Price, so marked price is used as the base .

Case study questions from the above given topic may be asked.

Frequently Asked Questions (FAQs) on Comparing Quantities Case Study

Q1: what does ‘comparing quantities’ mean in mathematics.

A1: Comparing quantities in mathematics refers to the process of analyzing two or more numbers or values to understand their relationship. This comparison can involve ratios, percentages, fractions, and differences to see how one quantity relates to another.

Q2: How is a ratio different from a fraction?

A2: A ratio compares two quantities by division and shows how many times one value is contained within another, while a fraction represents a part of a whole. For example, the ratio 3:4 indicates that for every 3 units of one quantity, there are 4 units of another, whereas the fraction 3/4 represents 3 parts out of a total of 4.

Q3: What is the percentage, and how is it calculated?

A3: A percentage is a way of expressing a number as a fraction of 100. It is calculated by dividing the part by the whole and then multiplying by 100.

Q4: How do you compare quantities using percentages?

A4: To compare quantities using percentages, convert the given values into percentages and then compare them.

Q5: What is profit and loss in comparing quantities?

A5: Profit and loss are concepts used in comparing quantities related to financial transactions. Profit is the amount gained when the selling price is higher than the cost price, while loss occurs when the selling price is lower than the cost price. Profit and loss are often expressed as percentages of the cost price.

Q6: What is the difference between Simple Interest and Compound Interest?

A6: Simple Interest is calculated only on the original principal amount throughout the investment period, while Compound Interest is calculated on the principal amount plus any interest that has been added from previous periods. Compound Interest usually results in a higher total amount compared to Simple Interest.

Q7: Why is it important to understand comparing quantities?

A7: Understanding comparing quantities is crucial in everyday life as it helps in making informed decisions related to finances, shopping, budgeting, and investments. It also lays the foundation for more advanced mathematical concepts and real-life applications.

Q8: Are there any online resources or tools available for practicing comparing quantities case study questions?

A8: We provide case study questions for CBSE Class 8 Maths on our  website . Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit  Physics Gurukul  website. they are having a large collection of case study questions for all classes.

Related Posts

• CBSE Maths Important Questions
• Class 8 Maths
• Chapter 8: Comparing Quantities

Important Question Class 8 Maths Chapter 8: Comparing Quantities

Important class 8 maths questions for chapter 8 comparing quantities provided here with solutions. The problems are given by our subject experts based on CBSE syllabus (2022-2023) and NCERT book . Students can prepare for class-test and final exams using these materials to score good marks. These questions from comparing quantities will help the students to get completely acquainted with all the concepts from this chapter.

To practice questions for all chapters, students can click on:  Important Questions Class 8 Maths . They can also do the revision before exams by solving the extra questions provided for each chapter.

Also Check:

• Important 2 Marks Questions for CBSE 8th Maths
• Important 3 Marks Questions for CBSE 8th Maths
• Important 4 Marks Questions for CBSE 8th Maths

Important Questions with Solutions For Class 8 Maths Chapter 8 (Comparing Quantities)

Several types of questions from chapter 8 (NCERT) are provided here. There are both short answer type questions and long answer type questions including HOTS type questions to help students get thorough with all the topics and solve questions more effectively in the CBSE class 8 exam.

1. Express 25% and 12% as decimals.

25% = 25/100 = 0.25

12% = 12/100 = 0.12

2. Express 45% and 78% as a fraction.

45% = 45/100 = 9/20

78% = 78/100 = 39/50

3. Calculate the ratio of 5 m to 20 km.

First, make both the units the same.

So, we need to convert 20 km to the equivalent meter, i.e. “m”.

=> 20km = (20 × 1000)m

Thus, the ratio of 5 m to 20 km = 5/20000 = 1:4000

4. A student got 150 marks out of 200 in maths and got 120 marks out of 180 in science. In which subject did the student perform better?

Express the marks in the form of ratios.

For maths, ratio = 150/200 = ¾

For science, ratio = 120/180= ⅔

Here, the ratio ¾ shows 75% (¾ × 100 = 75) and the ratio ⅔ shows 66.6% (⅔ × 100 = 66.6).

Thus, the student performed better in the maths exam.

5. If 72% of 25 students like maths, find out the number of students who do not like mathematics?

Given, 72% of 25 students like maths.

Hence, 72% of 25 = (72/100) × 25 = 18 students

Now, from 25 students, 18 students like maths

Thus, the number of students who do not like maths = 25 – 18 = 7 students

6. A person goes shopping and spends 75% of his money. If he is now left with Rs. 600, find out how much he had in the beginning.

Let “x” be the initial amount that he had in the beginning.

As per the given question, the person spent 75% of Rs.x and is left with Rs. 600.

So, the amount he spent = x – 600

=> 75% of x = x – 600

=> (75/100) × x = x – 600

=> 75x = 100x – 60,000

=>25x = 60,000

Or, x = 2400.

Thus, the person had Rs. 2400 initially.

7. A shopkeeper bought two phones for Rs. 8,000 each. After selling the phones, there was a loss of 4% on the 1st phone while a profit of 8% on the 2nd phone. Calculate the overall gain or loss per cent on the whole transaction.

As the shopkeeper bought both phones at Rs. 8000 each.

Total cost price = Rs. 16,000

Assume that the cost price of the 1st phone is Rs. 100

Consider the deal of phone 1,

As it is given, there is 4% loss, the selling price will be = Rs. 96

For CP = 100, SP = 96

So, for CP = 1, SP = 96/100

Now, given, CP = 8000

Hence, SP = 96/100 × 8000 = 7680

Thus, the selling price of 1st phone = Rs. 7680

Consider the deal of phone 2, there is an 8% profit.

Hence, the selling price will be = Rs. 108

For CP = 100, SP = 108

So, for CP = 1, SP = 108/100

Now, given CP = 8000

hence, SP = 108/100 × 8000 = 8640

Thus, the selling price of 2nd phone = Rs. 8640

Here, the total selling price will be = Rs. 7680 + Rs. 8640 = 16320

Now, it can be seen that,

Total selling price > total cost price i.e. Rs. 16320 > Rs. 16000

So, there is a profit of Rs. (16320 – 16000) = Rs. 320

Now, the overall profit percentage will be-

Profit% = (Profit/Total Cost Price) × 100 = (320/16000) × 100 = 2

Therefore, there is a total of 2% profit for the whole transaction.

8. In a certain school, there are 456 girls. Calculate the total number of students if 24% of the total students are boys.

Assume the total number of students to be 100.

It is given that 24% are boys.

So, total number of boys = 24% of 100 = 24

Thus, the total number of girls will be = (100-24)% i.e. 76%.

So, the total number of girls = 76% of 100 = 76

But, it is given that there are 456 girls.

Now, for 76 girls, total students are 100

For, 1 girl, total students will be 100/76

Therefore, for 456 girls, the total number of students will be = 100/76 × 456 = 45600/76 = 600

Thus, the total number of students = 600.

Class 8 Maths Chapter 8 Extra Questions

1. What is the ratio of 10 m to 1 km? (a) 10 : 1 (b) 1 : 10 (c) 100 : 1 (d) 1 : 100

2. What is the ratio 1 : 4 converted to percentage? (a) 50% (b) 25% (c) 75% (d) 4%

3. Apala has Rs 200 with her, and she spent 80% from the amount she had. Calculate the money left with her. (a)Rs 20 (b) Rs 10 (c) Rs 40 (d) Rs 30

4. Meenu bought a fridge for Rs 10000 and then sold it for Rs 8000. Calculate her loss. (a) Rs 12000 (b) Rs 8000 (c) Rs 2000 (d) Rs 10000

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Important Questions for CBSE Class 8 Maths Chapter 8 Comparing Quantities

Home » CBSE » Important Questions for CBSE Class 8 Maths Chapter 8 Comparing Quantities

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Chapter 8 of Class Maths introduces students to Comparing Quantities. The concept of comparing quantities sets up the basis for many higher Maths topics and is a practically essential skill to have to solve real-life problems. Hence, enough practice is required to build quick mental Maths skills, which will help students in the examinations, enabling them to save enough time on calculations.

Quick Links

Extramarks is the best study adviser for students and helps them with comprehensive online study solutions from Class 1 to Class 12. Our  Maths experts have prepared various NCERT solutions to help students in their studies and exam preparation. Students can refer to our Important Questions Class 8 Maths Chapter 8 to practise exam-oriented questions. We have collated questions from various sources such as NCERT textbooks and exemplars, CBSE sample papers , CBSE past year question papers , etc. Students can prepare well for their exams and tests by solving various chapter questions from our comparing quantities class 8 extra questions .

To get maximum benefit during exam preparations, students can register on the Extramarks website and get full access to Important Questions Class 8 Maths Chapter 8 and other study materials, including NCERT solutions, CBSE revision notes, etc.

Get Access to CBSE Class 8 Maths Extra Questions 2022-23 with Chapter-Wise Solutions

You can also find CBSE Class 8 Maths Chapter-by-Chapter Important Questions here:

Comparing Quantities Class 8 Extra Questions with Solutions

Our in-house Maths faculty experts have collected an entire list of Important Questions Class 8 Maths Chapter 8 by referring to various sources. For each question, the team experts have  prepared a step-by-step explanation that will help students understand the concepts used in each question. Also, the questions are chosen in a way that would cover full chapter topics. So by practising from our question bank, students will be able to revise the chapter and understand their strong and weak points. And improvise by further focusing on weaker sections of the chapter.

Given below are a few of the questions and answers from our question bank of Maths Class 8 Chapter 8 Important Questions:

Question 1: Find the rate of discount given on a shirt whose selling price is ₹1092 after deducting a discount of ₹208 on its marked price.

Answer 1: we know that the SP = ₹1092

The discount = ₹208

By using the formula market price = SP + Discount

= 1092 + 208 = ₹1300

∴ The discount% = (discount/market price) × 100

= (208/1300) × 100 = 16%

Question 2: A scooter was bought at the cost of ₹ 42,000. Its value drops down to the rate of 8% per annum. Find its value after one year.

Answer 2: Principal = Cost price = ₹ 42,000

Depreciation rate = 8% of ₹ 42,000 per year

                                = ( Principal x Rate x Time period)/100

                                = (42000 x 8 x 1)/100

                                = ₹ 3360

Therefore , the value after 1 year = ₹ 42000 − ₹ 3360 = ₹ 38,640.

Question 3: A particular football team won 10 matches out of all the total number of matches they played. If their winning percentage was 40 %, how many matches did they play?

Answer 3: Let the total number of matches played be x.

The team won around 10 matches, and the team’s winning percentage was 40%.

40/100 × x = 10

40x = 10 × 100

x = 1000/40

Hence, the team played 25 matches.

Question 4:In a particular school, there are 456 girls. Compute the total number of students if 24% of the total students are boys.

Answer 4: The total number of students is 100.

It is given that 24% are boys.

So, total number of boys = 24% of 100 = 24

Therefore, the total number of girls will be = (100-24)%, i.e. 76%.

So, the total number of girls = 76% of 100 = 76

But, it is given that there are 456 girls.

Now, for 76 girls, the total number of students is 100

For 1 girl, the total number of students will be 100/76

Therefore, for 456 girls, the total number of students will be = 100/76 × 456 = 45600/76 = 600

Therefore, the total number of students = 600.

Question 5: I borrowed an amount ₹ 12000 from Jamshed at a rate of 6% per annum simple interest for 2 years. Had I borrowed this sum at a rate of 6% per annum compound interest, what is the extra amount I would have to pay?

Answer 5: Principle = ₹ 12000

Rate = 6% per annum

Time period = 2 years

Simple Interest = (P x R x T)/100

= (12000 x 6 x 2)/100

To find the compound interest,

the amount (A) has to be calculated

A = P(1 + R/100)n

     = 12000(1 + 6/100)2

     = 12000(106/100)2

     = 12000(53/50)2

     = ₹ 13483.20

∴ Compound Interest = A − P

           = ₹ 13483.20 − ₹ 12000

           = ₹ 1,483.20

Compound Interest − Simple Interest = ₹ 1,483.20 − ₹ 1,440

                  = ₹ 43.20

Hence, the extra amount to be paid is ₹ 43.20.

Question 6: Find the cost price when:

SP = ₹34.40 and Gain = 7 ½ %

Answer 6: Cost price = (100 / (100+ Gain %)) × SP

= (100/ (100+ (15/2))) × 34.40

= (100/ (215/2)) × 34.40

= (200/215) × 34.40

∴ the cost price is ₹ 32

Question 7:₹ 62500 for 1½ years at 8% per annum compounded half yearly.

Answer 7: Principal (P) = ₹ 62,500

Rate = 8% per annum or 4% per half-year

Number of years = 1½

There will be 3 and half years in 1½ years

Amount, A = Principle (1 + Rate/100)time period

= 62500(1 + 4/100)3

= 62500(104/100)3

= 62500(26/25)3

Compound Interest = A – P = ₹ 70304 – ₹ 62500 = ₹ 7,804

Question 8:In a certain laboratory, the bacteria count in a particular experiment increased at 2.5% per hour. Find the bacteria at the end of exactly 2 hours if the count was initially 5,06,000.

Answer 8: The initial count of bacteria = 5,06,000

Bacteria at the end of exactly 2 hours = 506000(1 + 2.5/100)2

= 506000(1 + 1/40)2

= 506000(41/40)2

= 531616.25

Therefore, the bacteria count at the end of 2 hours will be 5,31,616 (approx.).

Question 9:A person shops and spends 75% of his money. If he is now left with Rs. 600, find out how much he had in the beginning.

Answer 9: Let “x” be the initial amount he had initially.

As per the given question, the person spent 75% of Rs.x and is left with Rs. 600.

So, the amount he spent = x – 600

=> 75% of x = x – 600

=> (75/100) × x = x – 600

=> 75x = 100x – 60,000

=>25x = 60,000

Or, x = 2400.

Thus, the person had Rs. 2400 initially.

Question 10:A student scored 150 out of 200 in the subject maths and got 120 marks out of 180 in the subject science. In which subject did the student perform better?

Answer 10: Express the following marks in the form of ratios.

For the subject maths,the ratio = 150/200 = 3/4

For the subject science, the ratio = 120/180= 2/3

Here, the ratio 3/4 shows 75% (3/4× 100 = 75) and the ratio 2/3 shows 66.6% (2/3× 100 = 66.6).

Therefore, the student performed better in the maths exam than in his science exam.

Question 11: The population of a particular place increased to 54000 in 2003 at 5% per annum

(i) find the population of the place in the year 2001

(ii) what would be the population of the place in the year 2005?

Answer 11: (i) Population in the year 2003 = 54,000

54,000 = (Population in the year 2001) (1 + 5/100)2

54,000 = (Population in the year 2001) (105/100)2

Population in the year 2001 = 54000 x (100/105)2

Thus, the population of a place in the year 2001 was approximately 48,980

(ii) Population in the year 2005 = 54000(1 + 5/100)2

= 54000(105/100)2

= 54000(21/20)2

Hence, the population of a place in the year 2005 would be 59,535.

Question 12:Compute the amount and compound interest on the principal amount ₹ 10,800 for 3 years at 12½ % per annum compounded annually.

Answer 12 :Principal (P) = ₹ 10,800

Rate (R) = 12½ % = 25/2 % (annual)

Number of years (n) = 3

Amount (A) = P(1 + R/100)n

= 10800(1 + 25/200)3

= 10800(225/200)3

= 15377.34375

= ₹ 15377.34 (approximately)

Compound interest = A – P 

                                = ₹ (15377.34 – 10800) = ₹ 4,577.34

Question 13:Evaluate the ratio of 5 m to 20 km.

Answer 13: First, make both the units the same.

So, we must convert 20 km to the equivalent metre, i.e. “m”.

=> 20km = (20 × 1000)m

Therefore, the ratio of 5 m to 20 km = 5/20000 = 1:4000

Question 14: The marked price of a particular water cooler is ₹4650. The shopkeeper proposes an off-season discount of 18% on it. Find its selling price.

Answer 14: we know that the marked price = ₹4650

The discount = 18%

And the Discount in amount = 18% of the market price

= (18/100) × 4650

By using the formula, SP = marked price – Discount

= 4650 – 837 = ₹ 3,813

Question 15: Arun bought a new pair of skates at a sale where the Discount is given 20%. If the amount Arun pays is ₹ 1,600, find out the marked price.

Answer 15: Let the marked price be x

Discount percent = Discount/Marked Price x 100

20 = Discount/x × 100

Discount = 20/100 × x

Discount = Marked price – Sale price

x/5 = x – ₹ 1600

x – x/5 = 1600

4x/5 = 1600

x = 1600 x 5/4

Therefore, the marked price was ₹ 2000.

Question 16: How much did she have in the beginning if Chameli had ₹600 left after spending 75% of her money?

Answer 16: Let the amount of money which Chameli had, in the beginning, be x

After spending amount 75% of ₹x, she was left with ₹600

So, (100 – 75)% of x = ₹600

Or, 25% of x = ₹600

25/100 × x = ₹600

x = ₹600 × 4

Hence, Chameli had ₹2,400 in the beginning.

Question 17:Express 25% and 12% as decimals.

Answer 17: 25% = 25/100 = 0.25

                      12% = 12/100 = 0.12

Question 18: A man got a 10% increase in his salary. If his new salary amounted to ₹1,54,000, what was his original salary?

Answer 18: Let the original salary be x

The new salary is ₹1,54,000

Original salary + Increment = New Salary

The increment is of 10% of the original salary

So, (x + 10/100 × x) = 154000

x + x/10 = 154000

11x/10 = 154000

x = 154000 × 10/11

Thus, the original salary was ₹1,40,000.

Question 19: A particular cell phone was marked at 40% above the cost price, and a discount of 30% was given on its marked price. Find out the gain or loss percent made by the shopkeeper.

Answer 19: Let us consider the CP of goods as x

The market price of the goods when goods marked above the 40% CP is

Market price = x + (40x/100) = 140x/100 = 1.4x

So the Discount = 30%

Discount amount = 30% of 1.40x = 0.42x

∴ The SP = Market price – Discount

= 1.4x – 0.42x= 0.98x

Since SP is less than CP, it’s a loss

We know that Loss = CP – SP

= x – 0.98x = 0.02x

∴ Loss % = (Loss × 100)/ CP

= (0.02x × 100)/ x

Question 20: Express 45% and 78% as a fraction.

Answer 20: 45% = 45/100 = 9/20

                      78% = 78/100 = 39/50

Question 21: Obtain the amount Ram will get on ₹ 4,096, and he gave it for 18 months at 12½ per annum, interest being compounded half-yearly.

Answer 21: Principal = ₹ 4,096

Rate = 12½ per annum = 25/2 per annum = 25/4 per half-year

Time period = 18 months

There will be exactly 18 months in 3 half years 

Thus, amount A = P(1 + R/100)n

= 4096(1 + 25/(4 x 100))3

= 4096 x (1 + 1/16)3

= 4096 x (17/16)3

Hence, the required amount is ₹ 4,913.

Question 22: A bookseller sells a book for ₹100, gaining ₹20. His gain percentage is

• d) None of these

Answer 22 : we know that the SP = ₹100

By using the formula CP = SP – Gain

∴ Gain% = (Gain × 100) /CP

= (20 × 100) / 80

Question 23: Find out the amount and the compound interest on ₹ 10,000 for 1½ years at the rate of 10% per annum, compounded half yearly. Could this interest be more than the interest he would get if it were compounded annually?

Answer 23: P = ₹ 10,000

Rate = 10% per annum = 5% per half-year

Time period = 1½ years

There will 3 half years in 1½ years

Amount, A = P(1 + R/100)n

= 10000(1 + 5/100)3

= 10000(105/100)3

= ₹ 11576.25

Compound interest = Amount − Principal 

= ₹ 11576.25 − ₹ 10000

= ₹ 1,576.25

The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.

= 10000(1 + 10/100)1

= 10000(110/100)

By taking ₹ 11,000 as the principal, the S.I. for the next ½ year will be calculated as

= (11000 x 10 x ½)/100

So, the interest for particularly the first year = ₹ 11000 − ₹ 10000 = ₹ 1,000

Therefore, Total compound interest = ₹ 1000 + ₹ 550 = ₹ 1,550

So the actual difference between two interests = 1576.25 – 1550 = 26.25

Thus, the interest would be 26.25 more when compounded half yearly than the interest when compounded annually.

Question 24: Maria invested ₹ 8,000 in a particular business. She would eventually be paid interest at 5% per annum compounded annually. Find out

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

(ii) The interest for the third year

Answer 24: (i) P = ₹ 8,000

R = 5% per annum

n = 2 years

= 8000(1 + 5/100)2

= 8000(105/100)2

(ii) The interest for the next year, i.e. the third year, has to be calculated. By taking ₹

8,820 as principal, the Simple Interest for the next year will be calculated.

= (8820 x 5 x 1)/100

Question 31: Vasudevan invested ₹ 60000 at an interest rate of 12% per annum compounded half yearly. What amount would he get?

(i) after 6 months?

(ii) after 1

Answer 31:(i) P = ₹ 60,000

Rate = 12% per annum = 6% per half-year

n = 6 months = 1 half-year

Amount, A = Principle(1 + Rate/100)time period

= 60000(1 + 6/100)1

= 60000(106/100)

= 60000(53/50)

(ii) There are almost 2 and half years in 1 year

So, time period = 2

= 60000(1 + 6/100)2

= 60000(106/100)2

= 60000(53/50)2

Question 25:Fabina borrows ₹ 12,500 at the rate of 12% per annum for 3 years at simple interest, and Radha borrows the same amount for the same period at 10% per annum, compounded annually. Who pays more interest, Fabina and Radha and by how much?

Answer 25: Interest paid by Fabina = (Principle x Rate x Time period)/100

= (12500 x 12 x 3)/100

Amount paid by Radha at the end of 3 years = A = Principle(1 + Rate/100)time period

Amount = 12500(1 + 10/100)3

= 12500(110/100)3

= ₹ 16637.50

Compound Interest = Amount – Principal  = ₹ 16637.50 – ₹ 12500 

                                                                      = ₹ 4,137.50

The interest paid by Fabina is ₹ 4,500 and by Radha is ₹ 4,137.50

Therefore, Fabina pays more interest

₹ 4500 − ₹ 4137.50 = ₹ 362.50

Therefore, Fabina will have to pay ₹ 362.50 more than Radha.

Question 26:Kamala borrowed ₹ 26400 from a particular Bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to pay off the loan?

(Hint: Find A for 2 years with interest compounded yearly and then find S.I. on the 2nd year amount for 4/12 years.)

Answer 26: Principal (P) = ₹ 26,400

Rate (R) = 15% per annum

Number of years (n) = 2 4/12

The amount for 2 years and 4 months can be calculated by first calculating the amount for 2

years using the compound interest formula, and then calculating the simple interest for 4 months on the amount obtained at the end of 2 years

First, the amount for 2 years has to be calculated

Amount, A = Principal (1 + Rate/100)time period

= 26400(1 + 15/100)2

= 26400(1 + 3/20)2

= 26400(23/20)2

By taking ₹ 34,914 as the principal amount, the S.I. for the next 1/3 years can be calculated.

Simple Interest = (34914 × 1/3 x 15)/100 

                            = ₹ 1745.70

Interest for the first two years

                            = ₹ (34914 – 26400) 

                            = ₹ 8,514

And interest for the next 1/3 year = ₹ 1,745.70

Total Compound Interest = ₹ (8514 + ₹ 1745.70) = ₹ 10,259.70

Amount = P + C.I. = ₹ 26400 + ₹ 10259.70 = ₹ 36,659.70

Question 27: Oranges are purchased at 5 for ₹10 and sold at 6 for ₹15. His gain percentage is

Answer 27: we know that the CP of 1 orange = ₹10/5 = ₹2

SP of 1 orange = ₹15/6 = ₹2.5

Since SP is more than CP, it’s a Gain

Gain = SP – CP

=2.5 – 2

= ((0.5) × 100) / 2

Benefits Of Solving Important Questions Class 8 Maths Chapter 8

Practice is the key to scoring 100%  in Maths. The foundation of fundamental apprehension is the Maths taught in  Classes 8, 9, and 10 and it’s important to step up their learning experience and eliminate “maths phobia” among students. . We recommend students access  Extramarks to obtain important questions in Class 8 Maths Chapter 8. By systematically solving questions and going through the required solutions, students will get  the confidence to solve any tough  questions in the given chapter ”Comparing Quantities”.

Below are a few benefits of frequently solving comparing quantities class 8 important questions:

•  Our  Maths  subject experts have carefully assembled the most important questions in  Class 8 Maths Chapter 8 by inspecting many past years’  question papers.
• The questions and solutions provided are based on the latest CBSE syllabus and as per CBSE guidelines. So the students can completely count on it.
• The questions covered in our set of important questions in Class 8 Maths Chapter 8 are entirely based on several topics covered in the chapter ”Comparing Quantities”. It is suggested that students revise and clear all their doubts before solving all these important questions. 
• By practicing class 8 maths chapter 8 extra questions students will get a general idea of how the paper will be prepared. Practising questions similar to the exam questions would help the students gain confidence, score 100%  marks in their examinations, and set their own benchmark. .

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Q.1 A washing machine was sold for 5760 after giving successive discounts of 15% and 10% respectively. What was the marked price

Marks: 4 Ans

Selling price of washing machine = 5760 Two successive discounts are 15% and 10%. Let marked price of washing machine = x S . P . of washing machine after first discount = x 100 ˆ’ 15 100 = 85 x 100 = 17 x 20 S . P . of washing machine after second discount = 17 x 20 100 ˆ’ 10 100 = 17 x 20 — 90 100 = 153 x 200 Then , according to condition 153 x 200 = Rs . 5760 x = 5760 — 200 153 x = 7529 .40 approx Thus , the marked price of washing machine is 7529 .40.

Q.2 Find the compound interest on 1,60,000 for 2 years at 10% per annum when compounded semi-annually.

Marks: 2 Ans

Principal = 160000, rate = 10% per annum = 5% per half year Time = 2 years = 4 half years. Amount = 160000 — 1 + 5 100 4 = 160000 — 21 20 — 21 20 — 21 20 — 21 20 = 1 , 94 , 481 Compound Interest = 1 , 94 , 481 ˆ’ 1 , 60 , 000 = 34,481

Q.3 Find the amount on a principal of 2000 for 2 years at 10% per annum compounded annually. Also find the compound interest.

Marks: 1 Ans

A = P 1 + r 100 n = 2000 1 + 10 100 2 = 2000 — 110 100 — 110 100 = 2420 Amount = 2420 Compound Interest = Amount – Principal = 2420 – 2000 = 420

Q.4 When 5% sale tax is added on the purchase of a bedsheet of 300, find the buying price or the cost price of the bedsheet.

5 % of 300 is 5 100 — 300 = 15 buying price of Bedsheet is 300 + 15 = 315

Q.5 Rakesh bought a watch for 800 and sold it for 1000. Mukesh bought a car for 4,00,000 and sold it for 4,20,000. Who made a better sale, Rakesh or Mukesh

C.P. of watch for Rakesh = 800 S.P. of watch for Rakesh = 1000 Profit on watch to Rakesh = 1000 ˆ’ 800 = 200 Rate of Profit = 200 1000 — 100 = 20 % C.P. of car for Mukesh = 4,00,000 S.P. of car for Mukesh = 4,20,000 Profit on car for Mukesh = 4,20,000 ˆ’ 4,00,000 Profit on car for Mukesh = 20 , 000 Rate of Profit = 20 , 000 4,00,000 — 100 = 5 % So , Rakesh made a better sale.

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Important questions for class 8 maths, chapter 1 - rational numbers.

Chapter 2 - Linear Equation in One Variable

Chapter 3 - understanding quadrilaterals, chapter 4 - practical geometry, chapter 5 - data handling, chapter 6 - squares and square roots, chapter 7 - cubes and cube roots, chapter 9 - algebraic expressions and identities, chapter 10 - visualising solid shapes, chapter 11 - mensuration, chapter 12 - exponents and powers, chapter 13 - direct and inverse proportions, chapter 14 - factorisation, chapter 15 - introduction to graphs, chapter 16 - playing with numbers, faqs (frequently asked questions), 1. what are the formulas in ”comparing quantities” class 8.

Following are the Important formulas for comparing Quantities Class 8 Maths:

• Cost Price= Selling Price – Profit
• Cost Price= Selling Price + Loss
• Selling Price=Cost Price + Profit
• Selling Price=Cost Price – Loss
• Profit= Selling Price – Cost Price
• Loss= Cost Price – Selling Price
• Profit %= Profit/ cost Price x 100
• Loss %= Loss/ Cost Price x 100

2. How can I score well in Class 8 Maths examinations?

Maths is a subject which requires a lot of practice. To score well in Maths, one must have a strong conceptual understanding of the chapter, be good in calculations, practice questions regularly, give mock tests from time to time, get feedback and avoid silly mistakes. Regular practice with discipline, working diligently and conscientiously towards your goal, will definitely ensure a 100% in your exams.  

3. What can I get from the Extramarks website?

Extramarks is one of the best educational platforms and it has its own archive of educational resources, which assists students in acing their exams. You can get all the NCERT-related material like NCERT solutions, solved exemplar solutions, NCERT-based mock tests, CBSE revision notes, and  Class 8 Maths Chapter 8 important questions on the Extramarks website. Apart from this, you can get comprehensive guidance from our subject experts and doubt-clearing sessions once you sign up on our official website for any study resources.     

4. How many total chapters will students study in Class 8 Maths?

There are 16 chapters in Class 8 Maths. The list of chapters is given below:

• Chapter 1- Rational Numbers
• Chapter 2 – Linear Equations in One Variable
• Chapter 3 – Understanding Quadrilaterals
• Chapter 4 – Practical Geometry
• Chapter 5 – Data Handling
• Chapter 6 – Square and Square Roots
• Chapter 7 – Cube and Cube Roots
• Chapter 8 – Comparing Quantities
• Chapter 9 – Algebraic Expressions and Identities
• Chapter 10 – Visualising Solid Shapes
• Chapter 11- Mensuration
• Chapter 12 – Exponents and Powers
• Chapter 13 – Direct and Inverse Proportions
• Chapter 14 – Factorisation
• Chapter 15 – Introduction to Graphs
• Chapter 16 – Playing with Numbers

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NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities

Complete Resource of NCERT Solutions Maths Chapter 7 Class 8 - Free PDF Download

Class 8 Maths NCERT Solutions for Chapter 7 Comparing Quantities is an important topic that introduces students to various concepts such as ratios, percentages, and their applications in real-life situations. This chapter helps students understand how to compare different quantities using mathematical expressions and formulas.

Students should focus on key areas like calculating simple and compound interest, understanding profit and loss, and determining discounts and taxes. The chapter is designed to build a strong foundation in these concepts, which are important for solving practical problems.  Access the latest CBSE Class 8 Maths Syllabus here.

Glance on Maths Chapter 7 Class 8 - Comparing Quantities

Formulas included in this chapter are calculating percentages, profit and loss percentages, compound interest, and simple interest.

The chapter contains solved examples, practice problems, and exercises involving direct calculation, real-life application, and conceptual questions.

This offers a varied range of questions to enhance understanding and application of concepts.

These chapter exercises are designed to reinforce mathematical concepts related to everyday calculations, financial literacy, and effective problem-solving skills in real-world contexts.

This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 7 - Comparing Quantities, which you can download as PDFs.

There are three exercises (14 fully solved questions) in Chapter 7 Comparing Quantities Class 8 Maths.

Access Exercise wise NCERT Solutions for Chapter 7 Maths Class 8

Exercises Under NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities

Exercise 7.1:.

 Introduction to Ratios

Definition and concept of ratios.

Solving problems involving the comparison of quantities using ratios.

Introduction to Percentages:

Definition and conversion between fractions, decimals, and percentages.

Solving problems related to finding percentages of given quantities.

Increase and Decrease Percentages:

Understanding how to calculate the increase or decrease in percentage.

Solving problems involving percentage increase and decrease.

Exercise 7.2:

Finding the Cost Price and Selling Price:

Concept of cost price (CP) and selling price (SP).

Solving problems to find CP and SP using given information.

Profit and Loss Calculations:

Understanding profit and loss.

Calculating profit and loss percentages based on CP and SP.

Definition and calculation of discount.

Solving problems involving discount percentages and final prices.

Exercise 7.3:

Simple Interest:

Definition and formula for calculating simple interest.

Solving problems to find simple interest, principal, rate, and time.

Compound Interest:

Basic understanding of compound interest.

Difference between simple and compound interest.

Word Problems on Comparing Quantities:

Solving various word problems that involve the application of ratios, percentages, profit and loss, and simple interest.

Access NCERT Solutions for Class 8 Maths Chapter 7 – Comparing Quantities

Exercise: 7.1.

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.

Ans: Speed of a cycle $ = $ $15$ km

Speed of scooter $ = $$30$ km

Ratio of the speed of a cycle to the speed of scooter 

$ = \dfrac{{15}}{{30}}$

$ = \dfrac{1}{2}$  

The required ratio is $1:2$ .

(b) $5$m to $10$km.

Ans:  $5$ m to $10$ km.

Since 1km $ = $$1000$ m

$ \Rightarrow $$\dfrac{{5\;{\text{m}}}}{{10\;{\text{km}}}}\; = \,\dfrac{5}{{10}} \times \dfrac{1}{{1000}}$

$ = \dfrac{1}{{2000}}$

$ \Rightarrow 1:2000$

The required ratio is $1:2000$ .

(c) $50$ paise to Rs $5$.

Ans:   $50$ paise to Rs $5$

Since Rs 1 $ = $ $100$ paise      

$ \Rightarrow \dfrac{{50paise}}{{Rs5}}$                      

$ = \dfrac{{50}}{5} \times \dfrac{1}{{100}}$                

$ = \dfrac{1}{{10}}$                     

$ \Rightarrow 1:10$  

The required ratio is $1:10$ .

2. Convert the following ratios to percentages.

Ans: $3:4$ $ = $ $\dfrac{3}{4}$

$ \Rightarrow $$\dfrac{3}{4} \times \dfrac{{100}}{{100}}$                         

$ = $$0.75 \times 100\% $  

The required ratio to percentage is $75\% $

Ans: $2:3$ $ = $ $\dfrac{2}{3}$

$ = \dfrac{2}{3} \times \dfrac{{100}}{{100}}$

$ = \dfrac{{200}}{3}\% $  

$ = 66\dfrac{2}{3}\% $

The required ratio to percentage is $66\dfrac{2}{3}\% $

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

Ans: Total number of students $ = $ 25 .

Percentage of students are good in mathematics  $ = $$72\% $

Percentage of students who are not good in mathematics  $ = $ $(100 - 72)\% $

$ \Rightarrow 28\% $

$\therefore $ Number of students who are not good in mathematics $ = $$28\%  \times 25$

$ \Rightarrow \dfrac{{28}}{{100}} \times 25$

$ \Rightarrow \dfrac{{28}}{4}$

$ \Rightarrow 7$

Students are not good in mathematics $ = 7$

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: The total number of matches won by the football team $ = 10$ .

Percentage of team $ = 40\% $

The total number of matches played by the team $ = ?$

The total number of matches played by the team 

$ \Rightarrow 40\%  \times x = 10$

$ \Rightarrow \dfrac{{40}}{{100}} \times x = 10$

$ \Rightarrow x = 10 \times \dfrac{{100}}{{40}}$

$ \Rightarrow x = \dfrac{{100}}{4}$

$ \Rightarrow x = 25$

The total number of matches played by the team $ = 25$ .

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

Ans: Chameli’s money after spend $ = 600$

Percentage of money after spend $ = 75\% $

Beginning amount of chameli $ = ?$

Percentage of beginning amount 

$ = \left( {100{\text{ }} - {\text{ }}75} \right)\% \qquad $

Beginning amount of chameli

$ \Rightarrow 25\%  \times x = 600$

$ \Rightarrow \left( {\dfrac{{25}}{{100}}} \right) \times x = 600$

$ \Rightarrow \left( {\dfrac{1}{4}} \right) \times x = 600$

$ \Rightarrow x = 600 \times 4$

$ \Rightarrow x = 2400$

Beginning amount of chameli $ = 2400$ .

6.  If $60\% $ people in 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: Total number of people in city $ = 50$ lakh

Percentage of people like cricket $ = 60\% $

Percentage of people like football $ = 30\% $

Percentage of people like other games $ = ?$

Number of people like each type of game $ = ?$

Percentage of people like other games 

$ = \left( {100 - 60 - 30} \right)\% $  

$ = \left( {100 - 90} \right)\% $

Percentage of people like other games $ = 10\% $

Number of people like cricket 

$ = \left( {60\%  \times 50} \right)$

$ = \left( {\dfrac{{60}}{{100}} \times 50} \right)$

$ = \dfrac{{60}}{2}$

$ = 30$ lakh

Number of people like cricket $ = 30$ lakh

Number of people like football

$ = \left( {30\%  \times 50} \right)$

$ = \left( {\dfrac{{30}}{{100}} \times 50} \right)$

$ = \dfrac{{30}}{2}$

$ = 15$ lakh

Number of people like football $ = 15$ lakh

Number of people like other games

$ = \left( {10\%  \times 50} \right)$

$ = \left( {\dfrac{{10}}{{100}} \times 50} \right)$

$ = \dfrac{{10}}{2}$

$ = 5$ lakh.

Number of people like other games $ = 5$ lakh.

Exercise 7.2

1. 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 Rs $1450$ and two shirts marked at Rs $850$ each?

Ans: Discount percentage $ = 10\% $

Price of pair jeans $ = $ Rs. $1450$

Price of shirt $ = $ Rs. $850$

Discount $ = $ Marked price $ - $ Sale price.

One shirt price $ = $ Rs. $850$

Two shirt price = $2 \times $ Rs. $850$

$ \Rightarrow $ Rs. $1,700$

Total marked price = Rs. $\left( {1,450 + 1,700} \right)$

= Rs. $3,150$

Discount = Rs. $\left( {10\%  \times 3150} \right)$

= Rs. $\left( {\dfrac{{10}}{{100}} \times 3150} \right)$

Discount = Rs. $315$ .

Discount $ = $ Total Marked price $ - $ Sale price

$ \Rightarrow $ Rs $315$ = Rs $3150$ − Sale price

$ \Rightarrow $ Sale price $ = $ Rs $\left( {3150 - 315} \right)$

$ \Rightarrow $ Sale price $ = $ Rs $2835$

Customer paid $ = $ Rs $2835$ .

2. The price of a TV is Rs $13,000$. 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: Price of TV $ = $ Rs. $13,000$ .

Sales tax percentage $ = 12\% $

Vinod have to pay $ = ?$

if  Rs. $100$ , then Tax to be pay is Rs. $12$ .

when  Rs. $13,000$

Tax to be pay $ = \left( {\dfrac{{12}}{{100}} \times 13,000} \right)$

Tax to be pay $ = 12 \times 130$

Tax to be pay $ = $ Rs. $1,560$ .

Vinod have to pay $ = $ price of TV $ + $ Tax to be pay

$ = $ Rs. $13,000$$ + $ Rs. $1560$

$ = $ Rs. $14560$

Vinod have to pay $ = $ Rs. $14560$ .

3. Arun bought a pair of skates at a sale where the discount given was $20\% $. If the amount he pays is Rs $1,600$ . find the marked price.

Ans: Discount in skates $ = 20\% $

Total amount $ = 1,600$

Marked price $ = x$

Discount percent \[\]$ = \left( {\dfrac{{Discount}}{{Marked{\text{ }}price}}} \right) \times 100$  

Discount percent $ = \left( {\dfrac{{Discount}}{{Marked{\text{ }}price}}} \right) \times 100$

$ \Rightarrow 20 = \left( {\dfrac{{Discount}}{x}} \right) \times 100$

Discount $ = \dfrac{{20 \times x}}{{100}}$

Discount $ = \dfrac{{1 \times x}}{5}$

Discount $ = $ Marked price $ - $ Total amount

$ \Rightarrow \dfrac{{1 \times x}}{5}$$ = x - 1600$

$ \Rightarrow 1600 = x - \dfrac{1}{5}x$

$ \Rightarrow 1600 = \dfrac{{5x - x}}{5}$

$ \Rightarrow 1600 = \dfrac{{4x}}{5}$

$ \Rightarrow \dfrac{{1600 \times 5}}{4} = x$

$ \Rightarrow x = 400 \times 5$

$ \Rightarrow x = 2000$

Marked price $ = 2000$ .

4. I purchased a hair-dryer for Rs $5,400$ including $8\% $ VAT. Find the price before VAT was added.

Ans: Hair-dryer rate include VAT $ = 5,400$

Tax percentage $ = 8\% $

Rate before VAT $ = ?$

VAT $ = 8\% $

If VAT without Rs. $100$ , then price is Rs. $108$

when  Rs. $5400$

Rate before VAT $ = \left( {\dfrac{{100}}{{108}} \times 5400} \right)$

Rate before VAT $ = 100 \times 50$

Rate before VAT $ = $ Rs. $5000$ .

5. An article was purchased for ₹1239 including GST of 18%. Find the price of the article before GST was added?

Ans: Let’s assume the price of the article before GST as ₹x.

Given that the article was purchased for 1239 rupees, including GST of 18%, so we can write the equation as

x + GST = 1239    ….(1)

We know that GST is 18% of the price before GST, so:

GST = 0.18x

Substituting this into (1)

x + 0.18x = 1239

1.18x = 1239

x = $\frac{{1239}}{{1.18}}$

So, the price of the article before GST was added is approximately 1050 rupees.

Exercise 7.3 

1. The population of a place increased to $54000$ in $2003$ at a rate of $5\% $ per annum

(i) find the population in $2001$

(ii) what would be its population in $2005$?

Ans:  Population in the year $2003$$ = 54,000$

(i) $54000$ $ = $ population in 2001 $ \times $ ${\left( {1 + \dfrac{5}{{100}}} \right)^2}$

$54000$$ = $ population in 2001 $ \times $${\left( {\dfrac{{100 + 5}}{{100}}} \right)^2}$

$54000$$ = $ population in 2001 $ \times $${\left( {\dfrac{{105}}{{100}}} \right)^2}$

$54000$$ = $ population in 2001 $ \times $${\left( {1.05} \right)^2}$

$54000$$ = $ population in 2001 $ \times $$1.1025$

$\dfrac{{54000}}{{1.1025}}$$ = $ population in 2001 

population in 2001 $ = 48979.591$

(ii) population in 2005

population in 2001 $ = $$54000{\left( {1 + \dfrac{5}{{100}}} \right)^2}$

population in 2001 $ = 54000$${\left( {\dfrac{{100 + 5}}{{100}}} \right)^2}$

population in 2001  $ = 54000{\left( {\dfrac{{105}}{{100}}} \right)^2}$

population in 2001 $ = 5400{\left( {1.05} \right)^2}$

population in 2001 $ = 54000 \times 1.1025$

population in 2001 $ = 59.535$

2. 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:  Initial count of bacteria $ = 5,06,000$

Bacteria at the end of $2$ hours

$ = 506000{\left( {1 + \dfrac{{2.5}}{{100}}} \right)^2}$

$ = 506000{\left( {1 + 0.025} \right)^2}$

$ = 506000{\left( {1.025} \right)^2}$

$ = 506000\left( {1.050625} \right)$

$ = 531616.25$ (approximately)

The count of bacteria at the end of 2 hours $ = 531616.25$ .

3. A scooter was bought at Rs $42,000$. Its value depreciated at the rate of $8\% $ per annum. Find its value after one year.

Ans:  Principal (P) $ = $ Rs $42,000$

Rate(R)             $ = $ $8\% $ per annum .

Number(n)      $ = $ $1$ year.

S.I $ = \dfrac{{P \times R \times T}}{{100}}$

S.I \[ = \] Rs \[\left( {\dfrac{{42000 \times 8 \times 1}}{{100}}} \right)\]

S.I \[ = \] Rs \[8 \times 420\]

S.I \[ = \] Rs \[3360\] .

Value after $1$ year \[ = \] Rs. \[\left( {42000 - 3360} \right)\]

\[ = \] Rs. \[38,640\] .

Overview of Deleted Syllabus for CBSE Class 8 Comparing Quantities Maths Chapter 7

Class 8 Maths Chapter 7: Exercises Breakdown

The NCERT Class 8th Maths Chapter 7, "Comparing Quantities," teaches fundamental mathematical concepts like ratios, percentages, and profit and loss. It emphasizes practical skills like discount calculation, sales price estimation, and understanding simple and compound interest. Students should master percentage calculations and conversion between percentages, fractions, and decimals for exams and real-life financial literacy. The chapter introduces concepts like compound interest through clear examples and practice problems, ensuring students understand compounded growth.

Other Study Material for CBSE Class 8 Maths Chapter 7

Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths . Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 8 Maths Chapter 7 Comparing Quantities

1. Define ratio.

A ratio means comparing two quantities of similar units. A ratio of two quantities of the same unit is derived by dividing one quantity by the other.

2. Define interest.

Interest is the extra money paid by the post offices or banks on the money deposited with them. Interest is also paid by the people when they borrow money.

3. What is known as the conversion period?

The time period after which the interest is added each time to form a new principal is known as the conversion period.

4. What is known as overhead expenses?

Overhead expenses are the additional expenses made after buying an article and are included in the cost price of the article.

5. What is the full form of GST?

GST stands for Goods and Services Tax and is imposed on the supply of goods or services or both.

6. Where can I find NCERT Solutions for Chapter 8 “Comparing Quantities” of Class 8 Maths?

Students can easily find NCERT Solutions for Chapter 8 “Comparing Quantities” of Class 8 Maths on the internet. NCERT Solutions for Class 8 maths are available at free of cost on the  Vedantu  website and on the Vedantu app. Students can download NCERT Solutions free of cost. All Solutions are explained properly. They can click on this NCERT Solutions for Chapter 8 of Class 8 Maths to download solutions. NCERT Solutions can help students to prepare well for their exams. They can understand the concepts given in Chapter 8 easily. 

7. What is the principle of comparing quantities in Class 8 Maths Chapter 7 PDF?

In Class 8 Maths Chapter 7 PDF, the principle of comparing quantities involves analysing the relationship between different items or amounts to understand their differences or similarities. This often includes looking at percentages, ratios, proportions, or growth rates to determine how one quantity relates to another.

8. What are the rules for comparing quantities in Class 8 Maths Chapter 7 PDF?

Rules for comparing quantities typically involve using a consistent basis for comparison, such as keeping the units the same or converting them to a common format. Other rules include comparing percentages to understand relative differences, using ratios and proportions to find equivalences, and applying algebraic methods to solve problems involving quantities.

9. How many exercises are there in comparing quantities?

The number of exercises in Class 8 Maths Chapter 7 can vary slightly depending on the edition of the NCERT textbook. Typically, there are around 3 to 4 exercises, each designed to help students practice and master different aspects of comparing quantities, including percentage, profit and loss, and compound interest.

10. What is the full form of CI in comparing quantities?

The word "CI" in the context of comparing quantities stands for Compound Interest. This refers to the calculation of interest where the earned interest is added to the principal sum, so that from that moment on, the interest that has been added also earns interest. This method leads to a significantly increased total return over time compared to simple interest, where interest is calculated only on the principal amount.

11. Which chart is used to compare quantities in Class 8 Maths Comparing Quantities?

Various charts can be used to compare quantities, but one common type is the bar chart. Bar charts are particularly useful for comparing quantities across different categories or groups visually, allowing for quick comparison of sizes, amounts, or changes over time.

12. What does comparing quantities mean in Class 8 Maths Comparing Quantities?

In Maths, comparing quantities means evaluating and determining the relationship between two or more quantities. This might involve calculations to find out which quantity is greater, how many times one quantity is of another, or what percentage one quantity represents of another.

13. What are the topics covered in Class 8 Maths Comparing Quantities?

Class 8 Maths Comparing Quantities covers several key concepts:

Understanding and calculating percentages.

Applying percentages to find increases or decreases, like profit and loss.

Working with simple and compound interest calculations.

Understanding ratios and proportions in different contexts.

Using conversions for comparing different units.

Solving real-world problems using the concepts of percentages, ratios, and interest.

NCERT Solutions for Class 8 Maths

Ncert solutions for class 8.