20.01.2022 ANSWER KEY

Answer Key

Abhimanyu Kumar

ᴛᴇᴀᴄʜᴇʀ

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8 - Maths - T3

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Pdf Description

Page 1 :
STD 8 MATHEMATICS – CHAPTER 4 – LIFE MATHEMATICS, HOMEWORK 20.01.2022 (ANSWER), Sum 1 : Dinesh has deposited ₹10000 in bank at the rate of, 15% per annum. Calculate the simple interest after 5 years, and also find the total amount after 5 years., Solution:, 15% of 10000 =, , 15, , 100, , × 10000 = 15 × 100 = 1500, , Interest per year = ₹1500, Interest after 5 year = ₹1500 × 5 = ₹7500, Total amount after 5 year = Principal Amount + Interest Amount =, 10000 + 7500 = ₹17500, Simple Interest =, Simple Interest =, , 𝐏𝐧𝐫, , 𝟏𝟎𝟎, , 10000 × 5 × 15, , 100, , = 100 × 5 × 15 = 100 × 75 =, , 7500, Simple Interest after 5 years = ₹7500, Total amount after 5 year = Principal Amount + Interest Amount =, 10000 + 7500 = ₹17500, Sum 2 : Mukesh has deposited ₹15000 in bank at the rate of, 10% per annum. Calculate the simple interest after 5 years, and also find the total amount after 5 years., Solution:

Page 2 :
Simple Interest =, Simple Interest =, , 𝐏𝐧𝐫, , 𝟏𝟎𝟎, , 15000 × 5 × 10, , = 150 × 5 × 10 = 150 × 50 =, , 100, , 7500, Simple Interest after 5 years = ₹7500, Total amount after 5 year = Principal Amount + Interest Amount =, 10000 + 7500 = ₹17500, ANSWER = ₹17500, , Sum 3 : Dinesh has deposited ₹10000 in bank at the rate of, 15% per annum. Calculate the compound interest after 5, years and also find the total amount after 5 years., Solution:, First year , P = ₹10000, r = 15%, 15% of 10000 =, , 15, , 100, , × 10000 = 15 × 100 = ₹1500, , T = P + I = ₹10000 + ₹1500 = ₹11500, Second year , P = ₹11500, r = 15%, 15% of 11500 =, , 15, , 100, , × 11500 = 15 × 115 = ₹1725, , T = P + I = 11500 + 1725 = ₹13225, Third year , P = ₹13225, r = 15%, 15% of 13225 =, ₹1983.75, , 15, , 100, , × 13225 =, , 3, , 20, , × 13225 =, , T = P + I = ₹13225 + ₹1983.75 = ₹15208.75, Fourth year , P = ₹15208.75, r = 15%, , 3, , 4, , × 2645 =, , 7635, , 4, , =

Page 3 :
15% of 15208.75 =, =, , 3, , 4, , × 3041.75 =, , 15, , × 15208.75 =, , 100, , 9125.25, , 3, , 20, , × 15208.75, , = ₹2281.3125, , 4, , T = P + I = ₹15208.75 + ₹2281.3125 = ₹17490.0625, Fifth year , P = ₹17490.0625, r = 15%, 15% of 17490.0625 =, =, , 3, , 4, , × 3498.0125 =, , 15, , × 17490.0625 =, , 100, , 10494.0375, , 3, , 20, , × 17490.0625, , = ₹2623.509375, , 4, , T = P + I = ₹17490.0625 + ₹2623.509375 = 20113.571875, Total amount After 5 years = ₹20113.57, Compound Interest for 5 year = ₹20113.57 – ₹10000 =, ₹10113.57, Sum 4 : Raghul has deposited ₹15000 in bank at the rate of, 10% per annum. Calculate the compound interest after 5, years and also find the total amount after 5 years., 𝒓, , Amount = P(𝟏 + 𝟏𝟎𝟎) 𝒏 = P(11 +, Amount = 15000(, , 𝒓, , )𝒏, 𝟏𝟎𝟎, , 100 + 10 5, 100, , 110 5, , ) = 15000(, , 100, , 𝒓, , = P(100 + 𝟏𝟎𝟎) 𝒏 = P (, , 𝟏𝟎𝟎+ 𝒓 𝒏, ), 𝟏𝟎𝟎, , 100 + 10 5, 100, , ), , 11 5, , = 15000(100) = 15000(10), 11, , 11, , 11, , 11, , 11, , =15000 (10) (10) (10) (10) (10) =, =, =, , 80×104×104×104×104×104, 100×100×100×100, 8×104×104×104×10504, 100×100×100×10, , =, , =, , 15000×11×11×11×11×11, , 10×10×10×10×10, 8×104×104×104×104×104, , 100×100×100×10, 8×104×10504×10504 832×10504×10504, 100×100×1000, , =, , 100×100000

Page 4 :
=, , 832×110334016, 10000000, , =, , 91797901312, 10000000, , = 9179.7901312 = 9179.79, , Amount = ₹9179.79, Compound Interest = A – P = 9179.79 – 8000 = ₹1179.79, ANSWER = ₹1179.79

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