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## NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3

**NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3**

Ex 8.3 Class 8 Maths Question 1.

Calculate the amount and compound interest on

(a) ₹ 10,800 for 3 years at 12(frac < 1 >< 2 >) % per annum compounded annually.

(b) ₹ 18,000 for 2(frac < 1 >< 2 >) years at 10% per annum compounded annually.

(c) ₹ 62,500 for 1(frac < 1 >< 2 >) years at 8% per annum compounded half yearly.

(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify).

(e) ₹ 10,000 for 1 year at 8% per annum compounded half yearly.

Solution:

(a) Given:

P = ₹ 10,800, n = 3 years,

CI = A – P = ₹ 15,377.35 – ₹ 10,800 = ₹ 4,577.35

Hence amount = ₹ 15,377.34 and CI = ₹ 4,577.34

(b) Given: P = ₹ 18,000, n = 2(frac < 1 >< 2 >) years = (frac < 5 >< 2 >) years

R = 10% p.a.

The amount for 2(frac < 1 >< 2 >) years, i.e., 2 years and 6 months can be calculated by first calculating the amount to 2 years using CI formula and then calculating the simple interest by using SI formula.

The amount for 2 years has to be calculated

Total CI = ₹ 3780 + ₹ 1089 = ₹ 4,869

Amount = P + I = ₹ 21,780 + ₹ 1,089 = ₹ 22,869

Hence, the amount = ₹ 22,869

and CI = ₹ 4,869

(c) Given: P = ₹ 62,500, n = 1(frac < 1 >< 2 >) years = (frac < 3 >< 2 >) years per annum compounded half yearly

= (frac < 3 >< 2 >) × 2 years = 3 half years

R = 8% = (frac < 8 >< 2 >) % = 4% half yearly

CI = A – P = ₹ 70,304 – ₹ 62,500 = ₹ 7,804

Hence, amount = ₹ 70304 and CI = ₹ 7804

(d) Given: P = ₹ 8,000, n = 1 years R = 9% per annum compounded half yearly

Since, the interest is compounded half yearly n = 1 × 2 = 2 half years

CI = A – P = ₹ 8,736.20 – ₹ 8,000 = ₹ 736.20

Hence, the amount = ₹ 8736.20 and CI = ₹ 736.20

(e) Given: P = ₹ 10,000, n = 1 year and R = 8% pa compounded half yearly

Since the interest is compounded half yearly n = 1 × 2 = 2 half years

CI = A – P = ₹ 10,816 – ₹ 10,000 = ₹ 816

Hence the amount = ₹ 10,816 and Cl = ₹ 816

Ex 8.3 Class 8 Maths Question 2.

Kamala borrowed ₹ 26,400 from a 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 clear the loan? (Hint: Find amount for 2 years with interest is compounded yearly and then find SI on the 2nd

year amount for (frac < 4 >< 12 >) years).

Solution:

Given:

P = ₹ 26,400

R = 15% p.a. compounded yearly

n = 2 years and 4 months

Amount after 2 years and 4 months = ₹ 34,914 + ₹ 1745.70 = ₹ 36,659.70

Hence, the amount to be paid by Kamla = ₹ 36,659.70

Ex 8.3 Class 8 Maths Question 3.

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?

Solution:

For Fabina: P = ₹ 12,500, R = 12% p.a. and n = 3 years

Difference between the two interests = ₹ 4500 – ₹ 4137.50 = ₹ 362.50

Hence, Fabina pays more interest by ₹ 362.50.

Ex 8.3 Class 8 Maths Question 4.

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?

Solution:

Given: P = ₹ 12,000, R = 6% p.a., n = 2 years

Difference between two interests = ₹ 1483.20 – ₹ 1440 = ₹ 43.20

Hence, the extra amount to be paid = ₹ 43.20

Ex 8.3 Class 8 Maths Question 5.

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?

Solution:

(i) Given: P = ₹ 60,000, R = 12% p.a. compounded half yearly

Hence, the required amount = ₹ 67416

Ex 8.3 Class 8 Maths Question 6.

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.

Solution:

(i) Given: P = ₹ 80,000

R = 10% p.a.

n = 1(frac < 1 >< 2 >) years

Since the interest is compounded annually

Difference between the amounts = ₹ 92,610 – ₹ 92,400 = ₹ 210

Ex 8.3 Class 8 Maths Question 7.

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 third year.

Solution:

(i) Given: P = ₹ 8,000, R = 5% p.a.

and n = 2 years

Hence, interest for the third year = ₹ 441

Ex 8.3 Class 8 Maths Question 8.

Find the amount and the compound interest on ₹ 10,000 for 1(frac < 1 >< 2 >) years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Solution:

Given: P = ₹ 10,000, n = 1(frac < 1 >< 2 >) years

R = 10% per annum

Since the interest is compounded half yearly

Total interest = ₹ 1,000 + ₹ 550 = ₹ 1,550

Difference between the two interests = ₹ 1,576.25 – ₹ 1,550 = ₹ 26.25

Hence, the interest will be ₹ 26.25 more when compounded half yearly than the interest when compounded annually.

Ex 8.3 Class 8 Maths Question 9.

Find the amount which Ram will get on ₹ 4,096, if he gave it for 18 months at 12(frac < 1 >< 2 >) per annum, interest being compounded half yearly.

Solution:

Given: P = ₹ 4,096, R = 12(frac < 1 >< 2 >) % pa, n = 18 months

Hence, the required amount = ₹ 4913

Ex 8.3 Class 8 Maths Question 10.

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?

Solution:

(i) Given: Population in 2003 = 54,000

Rate = 5% pa

Time = 2003 – 2001 = 2 years

Population in 2003 = Population in 2001

Ex 8.3 Class 8 Maths Question 11.

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.

Solution:

Given: Initial count of bacteria = 5,06,000

Rate = 2.5% per hour

n = 2 hours

Number of bacteria at the end of 2 hours = Number of count of bacteria initially

Thus, the number of bacteria after two hours = 5,31,616 (approx).

Ex 8.3 Class 8 Maths Question 12.

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

Solution:

Given: Cost price of the scooter = ₹ 42,000

Rate of depreciation = 8% p.a.

Time = 1 year

Final value of the scooter

Hence, the value of scooter after 1 year = ₹ 38,640.

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## NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.1

NCERT Solutions for Class 6 Maths Chapter 8 Exercise 8.1 (Ex. 8.1) Decimals in Hindi Medium and English Medium updated for new academic session 2021-2022. All the contents are free to use without any login or other formalities.

Videos related to solutions and explanation are given separately. Use PDF solutions and take help from videos, if required, to understand properly. The method, we have used, is very helpful for all the students.

## Class 6 Maths Chapter 8 Exercise 8.1 Solution

### CBSE NCERT Class 6 Maths Chapter 8 Exercise 8.1 Solution in Hindi and English Medium

Download App for Class 6 all Subjects

### Class 6 Maths Chapter 8 Exercise 8.1 Solution in Videos

#### Decimals

The word decimal comes from Latin word ‘Decem’ meaning 10. We have learnt in the earlier classes that in a place value chart, each place has a value ten times the value of the next place on its right. For example, the value of hundreds place is ten times that of the value of tens place, the value of thousands place is ten times that of the value of hundreds place and so on. On the other hand, the value of a place is one-tenth of the value of the next place on its left. For example, the value of hundreds place is one-tenth that of the value of thousands place, the value of tens place is one-tenth that of the value of hundreds place etc.

##### Let us consider the place value of 7 in the following numbers:

(i) 7083 – The place value of 7 is 7000

(ii) 5769 – The place value of 7 is 700

(iii) 2573 The place value of 7 is 70 and

(iv) 4567 The place value of 7 is 7.

From the above discussion, we observe that when the digit 7 moves one place to the right, its value becomes one-tenth (1/10) of its previous value, when it moves two places to the right, its value becomes one-hundredth (1/100) of its previous value and so on.

If we wish to continue moving towards right, we shall have to extend the place-value chart beyond ones place by opening the places for tenths, hundredths, thousandths and so on.

###### Thus, the place value table takes the following shape:

If we have to represent the number 157 + 3/10 + 8/100 in the place value chart, then 1 goes to hundreds place, 5 goes to tens place, 7 goes to ones place, 3 goes to tenths place and 8 goes to hundredths place.

The number shown in the chart is written as 157.38 and is called a decimal fraction or simply decimal. It is read as one hundred fifty-seven point three eight. Here the point (.) called the decimal point. separates whole number and fractional parts.

Becoming an algebra expert will lead you to open new doors to help you build a better career. You need to be good at maths solving to understand the concepts of computer science, and even in the medicine stream, this Rs Aggarwal maths class 8 exercise 8a solution is going to help you. In Rs Aggarwal class 8 maths ch 8 ex 8a, you will be learning to solve one of the most used mathematical equations. Which you will be using over and over again in incoming classes.

The linear equation is something which students need to put their time and effort into understanding completely. With maths, Rs Aggarwal solutions class 8 exercise 8a, students get to have experience solving different types of problems, including linear equations and algebra. Not only with this article, but you will also be able to have Rs Aggarwal class 8 maths chapter 8 exercise 8a solutions. But you will be learning innovative ways and shortcuts to check if your answer is correct or not?

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## NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

**NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.1**

Ex 9.1 Class 8 Maths Question 1.

Identify the terms, their coefficients for each of the following expressions.

(i) 5xyz 2 – 3zy

(ii) 1 + x + x 2

(iii) 4x 2 y 2 – 4x 2 y 2 z 2 + z 2

(iv) 3 – pq + qr – rp

(v) (frac < x >< 2 >) + (frac < y >< 2 >) – xy

(vi) 0.3a – 0.6ab + 0.5b

Solution:

Ex 9.1 Class 8 Maths Question 2.

Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

x + y, 1000, x + x 2 + x 3 + x 4 , 7 + y + 5x, 2y – 3y 2 , 2y – 3y 2 + 4y 3 , 5x – 4y + 3xy, 4z – 15z 2 , ab + bc + cd + da, pqr, p 2 q + pq 2 , 2p + 2q

Solution:

Ex 9.1 Class 8 Maths Question 3.

Add the following:

(i) ab – bc, bc – ca, ca – ab

(ii) a – b + ab, b – c + bc, c – a + ac

(iii) 2p 2 q 2 – 3pq + 4, 5 + 7pq – 3p 2 q 2

(iv) l 2 + m 2 , m 2 + n 2 , n 2 + l 2 , 2lm + 2mn + 2nl

Solution:

(i) Given: ab – bc, bc – ca, ca – ab

We have

(ab – bc) + (bc – ca) + (ca – ab) (Adding all the terms)

= ab – bc + bc – ca + ca – ab

= (ab – ab) + (bc – bc) + (ca – ca) (Collecting the like terms together)

= 0 + 0 + 0

= 0

(ii) Given:

a – b + ab, b – c + bc, c – a + ac

We have (a – b + ab) + (b – c + bc) + (c – a + ac) (Adding all the terms)

= a – b + ab + b – c + bc + c – a + ac

= (a – a) + (b – b) + (c – c) + ab + bc + ac (Collecting all the like terms together)

= 0 + 0 + 0 + ab + bc + ac

= ab + bc + ac

(iii) Given:

2p 2 q 2 – 3pq + 4, 5 + 7pq – 3p 2 q 2

By arranging the like terms in the same column, we have

(Adding columnwise)

(iv) Given: l 2 + m 2 , m 2 + n 2 , n 2 + l 2 , 2lm + 2mn + nl

By arranging the like terms in the same column, we have

Thus, the sum of the given expressions is 2(l 2 + m 2 + n 2 + lm + mn + nl)

Ex 9.1 Class 8 Maths Question 4.

(a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

(6) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz

(c) Subtract 4p 2 q – 3pq + 5pq 2 – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq 2 + 5p 2 q

Solution:

(a) Arranging the like terms column-wise, we have

[Change the signs of all the terms of lower expressions and then add]

(b) Arranging the like terms column-wise, we have

[Change the signs of all the terms of lower expressions and then add]

(c) Arranging the like terms column-wise, we have

[Change the signs of all the terms of lower expressions and then add]

The terms are p 2 q – 7pq 2 + 8pq – 18q + 5p + 20

## NCERT Solutions for Class 8 Science

The links below provide the detailed solutions for **NCERT science class 8 textbook**.

### NCERT Solutions for Class 8 Science

NCERT textbooks are prescribed by CBSE as the best books for preparation of the school as well as board examinations. The textbooks are deemed as more than enough, without any aid from other refreshers. The solutions are designed keeping in mind the lucid language and the simplicity of the explanations that are given in the NCERT textbooks. Not just the board and school examinations, NCERT textbooks are known to play a very important role in JEE and NEET.

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We prepared **Class 8 science NCERT Solutions** as per CBSE remodeled assessment structure. We have discussed extensively how are different food crops produced. Which living organisms do we see under a microscope in a drop of water? Are some of our clothes synthetic? How do we conserve biodiversity? What is the internal structure of a plant? and many more.

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### Class 8 Science Chapter 1 Crop Production and Management

This chapter explains rural practices related to the creation of yield and their administration.It includes the presentation of various harvest, their climatic and nourishing necessities and the preparation of soil for comprehensive improvement of the crops

### Class 8 Science Chapter 2 Microorganisms: Friends and Foe

This chapter manages destructive and valuable kinds of microbes and their significance in different areas. The places where microorganisms live, microorganisms and us, harmful microorganisms, food preservation, nitrogen fixation, and nitrogen cycle are a portion of the primary topics covered in this chapter.

### Class 8 Science Chapter 3 Synthetic Fibres and Plastic

This chapter manages artificial texture and plastic that is utilized for an assortment of purpose. Kind of synthetic fibers, characteristics of synthetic fibers, plastics, plastic as materials of choice and plastic and the environment are a portion of the fundamentals points of this chapter.

### Class 8 Science Chapter 4 Materials: Metals and Non-metals

This chapter explains the world of metals, non-metals, their properties, and employment. This section will assist the students with classifying the components into metals and non-metals based on their properties. Physical properties of metals and non-metals, chemical properties of metals and non-metals, use of metals and non-metals are a portion of the vital topics talked about in this chapter.

### Class 8 Science Chapter 5 Coal and Petroleum

Coal and petroleum manage the two most imperative fuels ever found by humankind. The chapter additionally explains natural gas and how the characteristics assets are getting depleted because of huge and uncontrolled usage.

### Class 8 Science Chapter 6 Combustion and Flame

In this chapter, you will learn about different sorts of fuels are utilized for a different purpose at home, in industry and for running autos. Powers like cow dung, wood, coal, charcoal, petroleum, diesel, packed gaseous petrol (CNG) and so on are known to the students. In combustion and flame, the chemical procedure of burning and the sorts of flames created amid this procedure are considered in detail.

### Class 8 Science Chapter 7 Conservation of Plants and Animals

Conservation of plants and animals is a critical subject of science. This part will push the students to not just to understand the human exercise that negatively affects the nature’s abundance yet in addition to manners by which they can ensure the fauna and flora of the ecosystem.

### Class 8 Science Chapter 8 Cell: Structure and Functions

It explains the inside and out information about the major unit of life -cell. It enlightens the students regarding the discovery of the cell, its inclination, properties, and composition

### Class 8 Science Chapter 9 Reproduction in Animals

It explains the modes and procedure of reproduction in creatures. It is fundamental for the continuation of a species. In this chapter, u will figure out how reproduction happens in animals is talked about in detail.

### Class 8 Science Chapter 10 Reaching the Age of Adolescence

In this chapter, the student will find out about changed that happen in the human body after which an individual ends up fitting for reproduction.in this chapter we will learn about the human conceptive organs and the role of hormones in the development of an individual.

### Class 8 Science Chapter 11 Fore and Pressure

This chapter will take the students through the brilliant world of force and pressure. It will push them to build u uphold of the point as well as ace the ideas that will assist them with performing better.

### Class 8 Science Chapter 12 Friction

It is an essential section of physics that manages the backing off or opposition of movement. Its causes, impacts, applications and other key properties are talked about in his chapter.

### Class 8 Science Chapter 13 Sound

Sound talks about the distinction between music and noise. Noise pollution, its causes, impacts, and control measures are explained towards the finish off the chapter. This chapter will strengthen your base for future complex topics.

### Class 8 Science Chapter 14 Chemical Effects of Electric Current

In this chapter, we will discover that metals, for example, copper and aluminum conducts electricity while materials, for example, rubber, plastic, and wood don’t conduct electricity.

### Class 8 Science Chapter 15 Some Natural Phenomenon

This chapter throws light on the wonderful phenomenon of nature, for example, lightning and earthquake. Causes, impacts, magnitude and properties of lightning are discussed in this chapter.

### Class 8 Science Chapter 16 Light

Light talks about the key properties, impacts, and utilization of light and its subordinates. This chapter also gives information about regular reflection, diffused reflection, laws of reflection, etc.

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This chapter explains about the celestial objects. Position of those objects, their structure and centrality are portrayed in this chapter. This part also discusses the moon, periods of the moon its situation in the planetary system and surface of the moon.

### Class 8 Science Chapter 18 Pollution of Air and Water

This chapter addresses the solution for the unsafe wonder of pollution and how might one add to its decrease in the individual dimension. Air pollution, how does air get polluted are some of the significant topics discussed in this chapter.

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## NCERT Class 8 Maths Solutions – Download Class VIII Maths Solved Exercises

**NCERT Solutions for Class 8 Maths**: Mathematics is one of the most important subjects for **CBSE Class 8**. However, this subject requires continuous efforts and practice as well as a good understanding of mathematical concepts. That’s where Class 8 Maths **NCERT Solutions** help students. Prepared by a team of academic experts, these solutions will guarantee that you don’t miss a single question in the final exam. Maths NCERT Solutions Class 8 are prepared in accordance with the latest CBSE curriculum and NCERT syllabus.

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With topics like percentage, practical geometry, profit, loss, and value-added tax, the 8th grade is one of the most important foundational years for mathematics students. It sets up a practical knowledge that students will employ for the rest of their lives, allowing them to gain perspective on many of life's most important tasks. This is what makes it extremely important for CBSE students in class 8 to comprehensively test their knowledge by using the RD Sharma class 8 solutions, will provide easy access to the solutions for all the problems the books will challenge them with.

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Students of Class 8th can download RS Aggarwal Class 8 Maths Exercise 18A solution PDF here and clear all of their doubts if any. The PDF is designed to help the students in undeRStanding the concepts of the area of trapezium and polygons. Class 8 is a foundational class, and the mathematical concepts you learn here will help you in your higher studies.

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