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RS Aggarwal Solutions for Class 8 Maths Chapter 11 Exercise 11.2 Compound Interest

Here, we have provided RS Aggarwal Solutions for Class 8 Maths Chapter 11 Exercise 11.2. Students can view these RS Aggarwal Solutions for Class 8 Maths Chapter 11 Exercise 11.2 before exams for better understanding.
authorImageAnanya Gupta6 Aug, 2024
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RS Aggarwal Solutions for Class 8 Maths Chapter 11 Exercise 11.2

RS Aggarwal Solutions for Class 8 Maths Chapter 11 Exercise 11.2: RS Aggarwal Solutions for Class 8 Maths Chapter 11 Exercise 11.2 provides detailed explanations and step-by-step solutions for problems related to compound interest. This exercise focuses on applying the compound interest formula to various scenarios, helping students understand how to calculate compound interest for different time periods and compounding frequencies.

By working through the problems students will gain a deeper understanding of how compound interest accumulates over time, including the effects of different compounding intervals on the final amount. The detailed solutions guide students through the process of solving compound interest problems, enhancing their problem-solving skills and preparing them effectively for exams.

RS Aggarwal Solutions for Class 8 Maths Chapter 11 Exercise 11.2 Compound Interest Overview

RS Aggarwal Solutions for Class 8 Maths Chapter 11 Exercise 11.2 provide an in-depth overview of problems related to compound interest. This exercise is designed to help students master the concept of compound interest through practical application. It includes a variety of problems that require students to use the compound interest formula to determine the amount accumulated over time, considering different interest rates and compounding periods. By working through these exercises students can better understand how compound interest grows exponentially compared to simple interest. The solutions provided explain each step clearly, ensuring that students grasp the principles of compound interest and can apply them confidently in different scenarios.

RS Aggarwal Solutions for Class 8 Maths Chapter 11 Exercise 11.2 PDF

RS Aggarwal Solutions for Class 8 Maths Chapter 11 Exercise 11.2 PDF provides a detailed guide to solving compound interest problems. By referring to this PDF, students can gain clarity on how to calculate compound interest, manage different interest rates, and handle various compounding periods. Access the PDF through the link below to enhance your understanding and practice of compound interest calculations.

RS Aggarwal Solutions for Class 8 Maths Chapter 11 Exercise 11.2 PDF

RS Aggarwal Solutions for Class 8 Maths Chapter 11 Exercise 11.2 (Ex 11B)

RS Aggarwal Solutions for Class 8 Maths Chapter 11 Exercise 11.2 are available below. This resource provide detailed solutions and explanations for problems related to operations on compound interest.

By using the formula, find the amount and compound interest on:

(Question1) Rs 6000 for 2 years at 9% per annum compounded annually.

Solution:
Principal (P) =6000 Rate of the interest (R) = 9% Time of the interest (N) = 2 years Amount = P( 1+R/100)*n Amount = 6000 (1+9/100)*2 Amount = 6000 (1.09)*2 Amount = 6000 (1.09×1.09) Amount = 6000× 1.1881 Amount = Rupees 7128.60 Compounded Interest = Amount - Principal C.I = 7128.6- 6000.0 C.I = 1128.6​

(Question 2) Rs 10000 for 2 years at 11% per annum compounded annually.

Solution:

Given,
Principal = Rs. 10000
Rate = 11 %
Time = 2 years
Compound = P ( 1 + r 100 ) 2
= 10000 ( 1 + 11 100 ) 2
= 10000 × ( 111 ) 2 10000
= 12321
Amount = 12321 10000 = 2321 .

(Question 3) Rs 31250 for 3 years at 8% per annum compounded annually.

Solution:

Given,
Principal = Rs. 31250
Rate r = 8 %
Time = 3 years
Amount = p ( 1 + r 100 ) t
= 31250 ( 1 + 8 100 ) 3
= 31250 × ( 108 100 ) 3
= 39 , 366 R s
compound = amount principal
= 39 , 366 31250
= Rs. 8116 .
(Question 4) Rs 10240 for 3 years at 12(1/2)% per annum compounded annually.
Solution:
Given p = Rs. 10240
r = 12 1 2 % 25 2 %
Time = 3 years
Amount = p ( 1 + r 100 ) t = 10240 ( 1 + 25 / 2 100 ) 3
= 10240 ( 1 + 1 8 ) 3
= 10240 × ( 9 8 ) 3
= 14 , 580
Amount = 14 , 580

(Question 5) Rs 62500 for 2 years 6 months at 12% per annum compounded annually.

Solution:
Principal (P) = R s . 62500 Rate (R) = 12 % p . a . Period (n) = 2 years 6 months = 2 1 2 years Here for first 2 years, we apply the compound interest formula and next 6 months, we apply simple interest formula. The amount for 2 years is Amount (A) = P ( 1 + R 100 ) n = R s . 62500 ( 1 + 12 100 ) 2 = R s . 62500 × ( 28 25 ) 2 × 53 50 = R s . 62500 × 28 25 × 28 25 100 × 784 = R s . 78400 The amount for next 6 months: 78400 × ( 1 + 12 × 1 100 × 2 ) = 78400 × 53 50 = 83104 Therefore, amount after 2 1 2 years is R s . 83104 C.I = A- P = R s . 83104 R s . 62500 = R s . 20604

(Question 6) Rs 9000 for 2 years 4 months at 10% per annum compounded annually.

Solution:
Given: Principal ( P ) = R s . 9000 Rate ( R ) = 10 % per annum Time ( n ) = 2 years 4 months = 2 1 3 years [ 1 year = 12 months] Amount for 2 years Amount ( A ) = P ( 1 + R 100 ) n = R s . 9000 ( 1 + 10 100 ) 2 = R s . 9000 × 11 10 × 11 10 = R s . 10890 For the next 1 3 years, Principal = Amount for 2 years = R s . 10890. Interest for 1 3 years = P × R × n 100 = ( 10890 × 10 × 1 3 100 ) = R s . 363 So, the total Amount for 2 1 3 years = R s . 10890 + 363 = R s . 11253 C . I . = A P = R s . 11253 R s . 9000 = R s . 2253

(Question 7) Find the amount of Rs 8000 for 2 years compounded annually and the rates being 9% per annum during the first year and 10% per annum during the second year.

Solution:
Principal (P) = Rs. 8000 Period (n) = 2 years Rate ( R 1 ) = 9 % for the first year R 2 = 10 % the second year Amount (A) = P ( 1 + R 1 100 ) 1 ( 1 + R 2 100 ) 1 = 8000 ( 1 + 9 100 ) ( 1 + 10 100 ) = R s 8000 × 109 100 × 110 100 = R s . 9592

(Question 8) Anand obtained a loan of Rs 125000 from the Allahabad Bank for buying computers. The bank charges compound interest at 8% per annum, compounded annually. What amount will he have to pay after 3 years to clear the debt?

Solution:
Principal (p) = 1,25,000 Rate of interest (r) = 8 % p . a . Period (n) = 3 years Amount (A) = P ( 1 + r 100 ) n = R s . 125000 × ( 1 + 8 100 ) 3 = R s . 125000 × ( 27 25 ) 3 = R s . 125000 × 27 25 × 27 25 × 27 25 = R s . 157464 A n s .

(Question 9) Three years ago, Beeru purchased a buffalo from Surjeet for Rs 11000. What payment will discharge his debt now, the rate of interest being 10% per annum, compounded annually?

Solution:
Price of a buffalo (P) = Rs. 11000 Rate of interest (R) = 10 % p . a . Period (n) = 3 years Price of buffalo at present ( A ) = ( 1 + R 100 ) n = R s . 11000 ( 1 + 10 100 ) 3 = R s . 11000 × ( 11 10 ) 3 = R s . 11000 × 11 10 × 11 10 × 11 10 = R s . 14641

(Question 10) Shubhalaxmi took a loan of Rs 18000 from Surya Finance to purchase a TV set. If the company charges interest at 12% per annum during the first year and 12(1/2)% per annum during the second year, how much will she have to pay after 2 years?

Solution:
Amount of loan taken (P) = Rs. 18000 Rate ( R 1 ) = 12 % p.a. during first year R 1 = 12 1 2 % = 25 2 % p.a. during second year Period (n) = 2 years Total amount (A) = P ( 1 + R 1 100 ) 1 ( 1 + R 2 100 ) 1 = R s . 18000 ( 1 + 12 100 ) ( 1 + 25 2 × 100 ) = R s . 18000 × 28 25 × 9 8 = R s . 22680

Benefits of RS Aggarwal Solutions for Class 8 Maths Chapter 11 Exercise 11.2

  • Clear Explanations: Each solution is detailed and easy to understand, which helps students grasp the concept of compound interest and its applications more effectively.
  • Step-by-Step Guidance: The solutions break down complex problems into manageable steps, making it easier for students to follow and learn the correct methods for calculating compound interest.
  • Practice Opportunities: By working through these solutions, students can practice a variety of problems, reinforcing their understanding and improving their problem-solving skills.
  • Conceptual Clarity: The solutions help clarify common doubts and misconceptions about compound interest, ensuring a solid grasp of the topic.
  • Enhanced Exam Preparation: Regular use of these solutions can enhance students’ preparation for exams by familiarizing them with different types of compound interest questions and solutions.

RS Aggarwal Solutions for Class 8 Maths Chapter 11 Exercise 11.2 FAQs

What is the focus of Exercise 11.2 in Chapter 11 of RS Aggarwal Solutions?

Exercise 11.2 focuses on solving problems related to compound interest, specifically dealing with different cases and scenarios of compound interest calculations.

What type of problems can I expect in Exercise 11.2?

You can expect problems that involve calculating compound interest for various time periods, interest rates, and principal amounts. The problems may include finding the amount, interest earned, or principal based on given conditions.

How do the solutions in Exercise 11.2 help in understanding compound interest?

The solutions provide step-by-step explanations for each problem, helping students understand the process of calculating compound interest. They clarify the use of the compound interest formula and its application in different scenarios.

How can practicing Exercise 11.2 benefit my exam preparation?

Practicing this exercise helps reinforce your understanding of compound interest concepts and improves your problem-solving skills. It prepares you for similar questions that may appear in exams, enhancing your overall performance in mathematics.
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