Question
A builder borrows Rs.2550 to be paid back with compound interest at the rate of 4% per annum by the end of 2 years in two equal yearly instalments. How much will each instalment be ?
Answer: Option B
Answer: (b)A = Rs.2550R = 4% per annumn = 2 yearsLet each of the two equal instalments be xPresent worth = $\text"Instalment"/(1 + r/100)^n$$P_1 = x/(1 + 4/100)^1 = x/{1 + 1/25} = x/{26/25}$or $P_1 = 25/26x$Similarly,$P_2 = (25/26)^2x = 625/676x$$P_1 + P_2$ = A$25/26x + 625/676x$ = 2550${(650 + 625)x}/676 = 2550$$1275/676x = 2550$x = 2550 $× 676/1275$ ⇒ x = Rs.1352Using Rule 9,If a sum 'P' is borrowed at r% annual compound interest which is to be paid in 'n' equal annual installments including interest, then(i) For n = 2, Each annual installment= $p/{(100/{100 + r}) + (100/{100 + r})^2$(ii) For n = 3, Each annual installment= $p/{(100/{100 + r}) + (100/{100 + r})^2 + (100/{100 + r})^3$
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Answer: (b)A = Rs.2550R = 4% per annumn = 2 yearsLet each of the two equal instalments be xPresent worth = $\text"Instalment"/(1 + r/100)^n$$P_1 = x/(1 + 4/100)^1 = x/{1 + 1/25} = x/{26/25}$or $P_1 = 25/26x$Similarly,$P_2 = (25/26)^2x = 625/676x$$P_1 + P_2$ = A$25/26x + 625/676x$ = 2550${(650 + 625)x}/676 = 2550$$1275/676x = 2550$x = 2550 $× 676/1275$ ⇒ x = Rs.1352Using Rule 9,If a sum 'P' is borrowed at r% annual compound interest which is to be paid in 'n' equal annual installments including interest, then(i) For n = 2, Each annual installment= $p/{(100/{100 + r}) + (100/{100 + r})^2$(ii) For n = 3, Each annual installment= $p/{(100/{100 + r}) + (100/{100 + r})^2 + (100/{100 + r})^3$
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