Question
Kamal took Rs.6800 as a loan which along with interest is to be repaid in two equal annual instalments. If the rate of interest is 12$1/2$%, compounded annually, then the value of each instalment is
Answer: Option C
Answer: (c)Let the annual instalment be xA = P$(1 + R/T)^T$$x = P_1(1 + 25/200)$$x = P_1 × 9/8$$P_1 = 8/9x$Similarly, $P_2 = 64/81x$$P_1 + P_2$ = 6800$8/9x + 64/81x$ = 6800${72x + 64x}/81 = 6800$${136x}/81 = 6800$$x = {6800 × 81}/136$ = Rs.4050Using Rule 9(i),Here, P = Rs.6800, R = $25/2$%, n = 2Each instalment= $p/{(100/{100 + r}) + (100/{100 + r})^2$= $6800/{(100/{100 +{25/2}}) + (100/{100 + {25/2}})^2$= $6800/{200/225 + (200/225)^2}$= $6800/{200/225(1 + {200/225})}$= ${6800 × 225 × 225}/{200 × 425}$ = Rs.4050
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Answer: (c)Let the annual instalment be xA = P$(1 + R/T)^T$$x = P_1(1 + 25/200)$$x = P_1 × 9/8$$P_1 = 8/9x$Similarly, $P_2 = 64/81x$$P_1 + P_2$ = 6800$8/9x + 64/81x$ = 6800${72x + 64x}/81 = 6800$${136x}/81 = 6800$$x = {6800 × 81}/136$ = Rs.4050Using Rule 9(i),Here, P = Rs.6800, R = $25/2$%, n = 2Each instalment= $p/{(100/{100 + r}) + (100/{100 + r})^2$= $6800/{(100/{100 +{25/2}}) + (100/{100 + {25/2}})^2$= $6800/{200/225 + (200/225)^2}$= $6800/{200/225(1 + {200/225})}$= ${6800 × 225 × 225}/{200 × 425}$ = Rs.4050
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