Question
An amount of money appreciates to Rs.7,000 after 4 years and to Rs.10,000 after 8 years at a certain compound interest compounded annually. The initial amount of money was
Answer: Option D
Answer: (d)A = P$(1 + R/100)^T$7000 = P$(1 + R/100)^4$....(i)10000 = P$(1 + R/100)^8$.......(ii)Dividing equation (ii) by (i)$10000/7000 = (1 + R/100)^4$$10/7 = (1 + R/100)^4$From equation (i),7000 = P × $10/7$ ⇒ P = Rs.4900Using Rule 7(iii),Here, b - a = 8 - 4 = 4B = Rs.10,000, A = Rs.7000R% = $((B/A)^{1/n} - 1)$ × 100%R% = $[(10000/7000)^{1/4} - 1]$= $[(10/7)^{1/4} - 1]$$1 + R/100 = (10/7)^{1/4}$$(1 + R/100)^4 = 10/7$7000 = $P × 10/7$Since, Amount = P$(1 + R/100)^4$P = Rs.4900
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Answer: (d)A = P$(1 + R/100)^T$7000 = P$(1 + R/100)^4$....(i)10000 = P$(1 + R/100)^8$.......(ii)Dividing equation (ii) by (i)$10000/7000 = (1 + R/100)^4$$10/7 = (1 + R/100)^4$From equation (i),7000 = P × $10/7$ ⇒ P = Rs.4900Using Rule 7(iii),Here, b - a = 8 - 4 = 4B = Rs.10,000, A = Rs.7000R% = $((B/A)^{1/n} - 1)$ × 100%R% = $[(10000/7000)^{1/4} - 1]$= $[(10/7)^{1/4} - 1]$$1 + R/100 = (10/7)^{1/4}$$(1 + R/100)^4 = 10/7$7000 = $P × 10/7$Since, Amount = P$(1 + R/100)^4$P = Rs.4900
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