Class 8 maths ch 8 ex 8.3 solutions

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3, are part of NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 Question 1. Calculate the amount and compound interest on Solution. (a) By using year by year calculation (b) By using year by year calculation (c) By using half year by half year calculation (d) By using half-year by half-year calculation Question 2. Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan? (Hint: Find A for 2 years if interest is compounded yearly and then find SI on the 2nd year amount for \(\frac % \) per annum, interest being compounded half yearly. Solution. 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 he its population in 2005? Solution. (i) Let the population in 2001 be P. R = 5% p.a. n = 2 years (ii) initial population in 2003 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. Initial count of bacteria Question 12. A scooter was bought at ₹ 42,000. It’s value depreciated at the rate of 8% per annum. Find its value after one year. Solution. Initial value of the scooter We hope the NCERT Solutions for Class 8 Maths Chapter 8 Comp...

Important Important Deleted for CBSE Board 2024 Exams Important Deleted for CBSE Board 2024 Exams Important Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams Important Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams Important Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams Important Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams Deleted for CBSE Board 2024 Exams You are here Deleted for CBSE Board 2024 Exams Transcript Question 8 Find the amount and the compound interest on Rs 10,000 for 1 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? Given, Principal (P) = 10000 Rate (R) = 10 % P.A Compounded half Yearly = 10/2 % Per Half year = 5 % per half year Time = 1 1/2 Years = 3/2 years = 3/2 × 2 half years = 3 Half Years Amount(A) = P (1+𝑅/100)^𝑛 = 10000 × (1+5/100)^3 = 10000 × (1+1/20)^3 = 10000 × ((20 + 1)/20)^3 = 10000 × (21/20)^3 = 10,000 × (21 × 21 × 21)/(20 × 20 × 20) = 10 × (441 × 21)/(2 × 2 × 2) = 10 × 9261/8 = 92610/8 = 11576.25 Now, Amount = Principal + Interest 11576.25 = 10,000 + Interest 11576.25 − 10000 = Interest 1576.25 = Interest ∴ Interest = 1576.25 Interest when Compounded Annually P = 10000 R = 10% P.A. N = 1 1/2 years Since n is in fraction, We use the formula Compound Interest for 1 1/2 years = Compound Interest for 1 Year + Simple Interest for next 1/2 ...

RD Sharma Class 8 Solutions (2020-2021 Edition) Get Latest Edition of RD Sharma Class 8 Solutions Pdf Download on LearnInsta.com. It provides step by step solutions RD Sharma Class 8 Solutions Pdf Download. You can download the RD Sharma Class 8 Solutions with Free PDF download option, which contains chapter wise solutions. In RD Sharma Solutions for Class 8 all questions are solved and explained by expert Mathematic teachers as per CBSE board guidelines. By studying these RD Sharma Class 8 Solutions you can easily get good marks in CBSE Class 8 Examinations. RD Sharma Solutions Class 8 is given here to help students of class 8 while solving the exercise problems. Also RD Sharma Solutions Class 8 is useful for various entrance/ competitive examination preparation. Here you will get all RD Sharma Class 8 Solutions PDF for free. RD Sharma Class 8 Solutions 2020 Edition for 2021 Examinations RD Sharma Solutions for Class 8 are given below for all chapter wise. Select chapter number to view solutions exercise wise. • Chapter 1 Rational Numbers • Chapter 2 Powers • Chapter 3 Squares and Square Roots • Chapter 4 Cubes and Cube Roots • Chapter 5 Playing With Numbers • Chapter 6 Algebraic Expressions and Identities • Chapter 7 Factorization • Chapter 8 Division of Algebraic Expressions • Chapter 9 Linear Equations in One Variable • Chapter 10 Direct and Inverse variations • Chapter 11 Time and Work • Chapter 12 Percentage • Chapter 13 Profits, Loss, Discount and Value Added Tax (V...

NCERT Solutions Class 8 Maths Chapter 8 Comparing Quantities The NCERT solutions for Class 8 maths Chapter 8 Comparing Quantities is a part of already learned mathematical principles about simple mathematics to calculate the value of money. At the very beginning of Now instead of using the number zero as the absolute, concepts like • • • NCERT Solutions for Class 8 Maths Chapter 8 PDF The concept of comparing quantities forms the basis for many higher math topics ahead, as well as it is a practically important skill to have in order to solve real-life problems. Hence, sufficient practice on the same is required to build up quick mental math skills, which will help students in the examinations, enabling them save enough time on calculations. A brief of the exercises in the ☛ Download Class 8 Maths NCERT Solutions Chapter 8 NCERT Class 8 Maths Chapter 8 Download PDF NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Comparing quantities holds an important place in everyday lives, it is one of the core reasons mathematics was devised. Hence, in mathematics, they devised a way to compare quantities even when numerically they are not comparable. The comparing quantities chapter provides the description of the theory and its application in order to understand the concepts better and faster. An exercise-wise detailed analysis of the class 8 maths NCERT solutions chapter 8 can be found in the pdf below : • Class 8 Maths Chapter 8 Ex 8.1 - 6 Questions • Class 8 Maths ...

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 are part of • • Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 8 Chapter Name Comparing Quantities Exercise Ex 8.3 Number of Questions Solved 12 Category NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 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 \(\frac \) 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: Ex 8.3 Class 8 Maths Question 2. Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan? [Hint : Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for \(\frac \) % per annum, interest being compounded half yearly. Solution: Here, Principal = ₹ 4096, Time = 18 months = 3 half years 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: 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 bac...

Class VIII Mathematics Exercise 8.1, Exercise 8.2, Exercise 8.3 and Exercise 8.4 in English and Hindi Medium updated for new academic session. Download Prashnavali 8.1, Prashnavali 8.2, Prashnavali 8.3 and Prashnavali 8.4 in Hindi Medium to study online or in PDF format to free download for offline use. NCERT (https://ncert.nic.in/) Solutions 2023-24 are updated for the new academic session based on updated In Chapter 8 Algebraic Expression and Identities, we have to learn about the fundamental terms like factors, coefficients, monomial, binomial, trinomial and all other polynomials, like and unlike terms in an algebraic expression, additions and subtractions of expressions taking like terms aside. Multiplications of two or three monomial and the multiplication of a monomial or binomial with a polynomial. Simplification of different 1. (a – b)² = a² – 2ab + b² 2. (a + b)² = a² + 2ab + b² 3. (a + b)(a – b) = a² – b²