Class 8 maths ex 8.3

  1. ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 8 Simple and Compound Interest Ex 8.3 – Learn Cram
  2. ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 8 Simple and Compound Interest Ex 8.3
  3. NCERT Solutions for Class 6 Maths Exercise 8.3 Chapter 8 Decimals


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ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 8 Simple and Compound Interest Ex 8.3 – Learn Cram

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 8 Simple and Compound Interest Ex 8.3 Question 1. Calculate the amount and compound interest on (i) ₹15000 for 2 years at 10% per annum compounded annually. (ii) ₹156250 for \(1 \frac\) years, compound interest payable half-yearly, find the rate of interest per annum. Solution: Question 12. In what time will ₹15625 amount to ₹ 17576 at 4% per annum compound interest? Solution: Question 13. ₹ 16000 invested at 10% p.a. compounded semi-annually, amounts to ₹18522. Find the time period of investment. Solution: Categories Post navigation

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 8 Simple and Compound Interest Ex 8.3

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 8 Simple and Compound Interest Ex 8.3 Question 1. Calculate the amount and compound interest on (i) ₹15000 for 2 years at 10% per annum compounded annually. (ii) ₹156250 for \(1 \frac\) years at 8% per annum compounded half-yearly. (iii) ₹ 100000 for 9 months at 4% per annum compounded quarterly. Solution: Question 2. Find the difference between the simple interest and compound interest on ₹4800 for 2 years at 5% per annum, compound interest being reckoned annually. Solution: Question 3. Find the compound interest on ₹3125 for 3 years if the rates of interest for the first, second and third year are respectively 4%, 5% and 6% per annum. Solution: Question 4. Kamla borrowed ₹26400 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? Solution: Question 5. Anil borrowed ₹18000 from Rakesh at 8% per annum simple interest for 2 years. If Anil had borrowed this sum at 8% per annum compound interest, what extra amount would he has to pay? Solution: Question 6. Mukesh borrowed 775000 from a bank. If the rate of interest is 12% per annum, find the amount he would be paying after \(1 \frac\) years, compound interest payable half-yearly, find the rate of interest per annum. Solution: Question 12. In what time will ₹15625 amount to ₹ 17576 at 4% per annum compound interest? Solution: Question 13. ₹ 16000 invested at 10% p.a. compounded se...

NCERT Solutions for Class 6 Maths Exercise 8.3 Chapter 8 Decimals

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Exercise 8.3 NCERT Solutions for Class 6 Maths Chapter 8 Decimals Exercise 8.3 covers the concepts, such as the comparison of decimals with respect to which is greater and smaller. The primary aim of creating exercise-wise Maths is a subject which contains a lot of concepts to be understood. Scoring good marks in the exam has become easy with the help of the Previous Next  Access NCERT Solutions for Class 6 Maths Chapter 8: Decimals Exercise 8.3 1. Which is greater? (a) 0.3 or 0.4 (b) 0.07 or 0.02 (c) 3 or 0.8 (d) 0.5 or 0.05 (e) 1.23 or 1.20 (f) 0.099 or 0.19 (g) 1.5 or 1.50 (h) 1.431 or 1.490 (i) 3.3 or 3.300 (j) 5.64 or 5.603 Solutions: (a) 0.3 or 0.4 The whole parts for both numbers are the same. We know that the tenth part of 0.4 is greater than that of 0.3. ∴ 0.4 > 0.3. (b) 0.07 or 0.02 Both the numbers have the same parts up to the tenth place, but the hundredth part of 0.07 is greater than that of 0.02. ∴ 0.07 > 0.02. (c) 3 or 0.8 The whole part of 3 is greater than that of 0.8. ∴ 3 > 0.8. (d) 0.5 or 0.05 The whole parts for both numbers are the same. But, the tenth part of 0.5 is greater than that of 0.05. ∴ 0.5 > 0.05. (e) 1.23 or 1.20 Here, both the numbers have the same parts up to the tenth place. The hundredth part of 1.23 is greater than that of 1.20. ∴ 1.23 > 1.20. (f) 0.099 or 0.19 The whole parts for both numbers are the same. But, the tenth part of 0.19 is greater than that of 0.099. ∴ 0.0...

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