Question
A man borrows Rs.21000 at 10% compound interest. How much he has to pay annually at the end of each year, to settle his loan in two years ?
Answer: Option D
Answer: (d)If each instalment be x, then Present worth of first instalment= $x/{1 + 10/100} = {10x}/11$= Present worth of second instalment= $x/(1 + 10/100)^2 = 100/121x$$10/11x + 100/121x$ = 21000${110x + 100x}/121 = 21000$210x = 21000 × 121$x = {21000 × 121}/210$ = Rs.12100 Using Rule 9,If a sum 'P' is borrowed at r% annual compound interest which is to be paid in 'n' equal annual installments including interest, then(i) For n = 2, Each annual installment= $p/{(100/{100 + r}) + (100/{100 + r})^2$(ii) For n = 3, Each annual installment= $p/{(100/{100 + r}) + (100/{100 + r})^2 + (100/{100 + r})^3$
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Answer: (d)If each instalment be x, then Present worth of first instalment= $x/{1 + 10/100} = {10x}/11$= Present worth of second instalment= $x/(1 + 10/100)^2 = 100/121x$$10/11x + 100/121x$ = 21000${110x + 100x}/121 = 21000$210x = 21000 × 121$x = {21000 × 121}/210$ = Rs.12100 Using 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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