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  1. Chapter 8 Class 8 Comparing Quantities
  2. Serial order wise
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Ex 8.3, 2 Kamala borrowed Rs 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 4/12 years). Given, Principal (P) = 26400 Rate (R) = 15% p.a Time (n) = 2 Years 4 Months = 2 4/12 years = 21/3 years Since n is in fraction We use the formula Compound interest for 2 1/3 years = Compound Interest for 2 years + Simple interest for the next 1/3 years Compound interest for 2 years Principal = 26,400 Rate = 15% Time (n) = 2 Amount = P (1+๐‘…/100)^๐‘› = 26400 (1+15/100)^2 = 26400 ((100 + 15)/100)^2 = 26400 (115/100)^2 = 26400 ร— ((115 ร— 115)/(100 ร—100))^2 = 264 ร— 13225/100 = (264 ร— 13225)/100 = 3491400/100 = 34914 โˆด Amount = Rs 34914 Now, Amount = Principal + Interest 34914 = 26400 + Interest 34914 โˆ’ 26400 = Interest 8514 = Interest โˆด Interest = 8514 โˆด Interest for 2 years = Rs 8514 & Amount after 2 years = Rs 34914 Simple interest for next ๐Ÿ/๐Ÿ‘ year Principal will be the amount after 1 year P = Rs 34914 R = 15% p.a T = 1/3 years SI = ๐‘ƒ๐‘…๐‘‡/100 = (34914 ร— 15 ร— 1/3 )/100 = (34914 ร—15)/(100 ร— 3) = (34914 ร— 5)/100 = 34914/20 = 17457/10 = 1745.7 โˆด Simple interest for 1/3 years = Rs 1745.7 Now, Compound Interest for 2 1/3 years = Compound Interest for 2 years + Simple interest for 1/3 year = 8514 + 1745.7 = 10259.7 Now, Amount = Principal + interest = 26400 + 10259.7 = 36659.70 โˆด Kamala will Pay Rs 36659.70 to the Bank.

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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.