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  1. Chapter 8 Class 8 Comparing Quantities
  2. Serial order wise
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Ex 8.3, 7 Maria invested Rs 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. Given, Principal (P) = Rs 8000 Rate (R) = 5% per annum Time = 2 Years We need to find the amount after 2 years So, we use the formula A = P ("1 + " ๐‘Ÿ/100)^๐‘› Putting values A = 8000 ("1 + " 5/100)^2 = 8000 ("1 + " 1/20)^2 = 8000 ((20 + 1)/20)^2 = 8000 (21/20)^2 = 8000 ร— ((21 ร— 21)/(20 ร— 20)) = 80 ร— ((21 ร— 21)/(2 ร— 2)) = 20 ร— 21 ร— 21 = 20 ร— 441 = 8820 โˆด Amount after 2 years = Rs 8,820 Ex 8.3, 7 (Method 1) Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find (ii) The interest for the 3rd year. Interest in 3rd year = Amount after 2 years ร— Rate = 8820 ร— 5/100 % = 8820 ร— 5/100 = 8820 ร— 1/20 = 882/2 = 441 โˆด Interest for the 3rd year = Rs 441 Ex 8.3, 7 (Method 2) Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find (ii) The interest for the 3rd year. Now, Interest for the 3rd year = Amount after 3 years โˆ’ Amount after 2 years Finding Amount after 3 years Amount after 3 years = P (1+๐‘…/100)^๐‘› Putting values Amount = 8000 (1+5/100)^3 = 8000 (1+1/20)^3 = 8000 ((20 + 1)/20)^3 = 8000 (21/20)^3 = 8000 ร— (21 ร— 21 ร—21)/(20 ร— 20 ร—20) = 8000 ร— (441 ร— 21)/(400 ร— 20) = 8000 ร— 9261/8000 = 9261 Therefore, Interest for the 3rd year = Amount after 3 years โˆ’ Amount after 2 years = 9261 โˆ’ 8820 = 441 โˆด Interest for the 3rd year = Rs 441

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.