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標題:
F5 Maths
發問:
Danny wnats to buy a car which is worth $250 000 at present. He makes a deposit of $20 000 and decides to deposit an additional of $20 000 every6 months. The interest is compounded half-yearly at a rate of 8% per annum. If the value of the car depreciates by 4% each year, will Danny buy the car within 5 years... 顯示更多 Danny wnats to buy a car which is worth $250 000 at present. He makes a deposit of $20 000 and decides to deposit an additional of $20 000 every 6 months. The interest is compounded half-yearly at a rate of 8% per annum. If the value of the car depreciates by 4% each year, will Danny buy the car within 5 years from now? Explain your answer.
The value of first deposit for 5 years =20000(1.04)^10 The value of second deposit for 4.5 years=20000(1.04)^9 The value of third deposit for 4 years =20000(1.04)^8 . . The value of tenth deposit for 0.5 years=20000(1.04)^1 The value of last deposit =20000 The total value of all deposit at 5 years later =20000(1.04)^10+20000(1.04)^9+...+20000(1.04)^1+20000 =20000(1.04^10+1.04^9+...+1.04+1) =20000[(1)(1.04^10-1)/(1.04-1)] (summation of geometric sequence) =$240122 The value of the car at 5 years later =250000x(1-0.04)^5 =$203843 Therefore at 5 years later, value of deposit>value of car, Danny can buy the car within 5 years
其他解答:
F5 Maths
發問:
Danny wnats to buy a car which is worth $250 000 at present. He makes a deposit of $20 000 and decides to deposit an additional of $20 000 every6 months. The interest is compounded half-yearly at a rate of 8% per annum. If the value of the car depreciates by 4% each year, will Danny buy the car within 5 years... 顯示更多 Danny wnats to buy a car which is worth $250 000 at present. He makes a deposit of $20 000 and decides to deposit an additional of $20 000 every 6 months. The interest is compounded half-yearly at a rate of 8% per annum. If the value of the car depreciates by 4% each year, will Danny buy the car within 5 years from now? Explain your answer.
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最佳解答:The value of first deposit for 5 years =20000(1.04)^10 The value of second deposit for 4.5 years=20000(1.04)^9 The value of third deposit for 4 years =20000(1.04)^8 . . The value of tenth deposit for 0.5 years=20000(1.04)^1 The value of last deposit =20000 The total value of all deposit at 5 years later =20000(1.04)^10+20000(1.04)^9+...+20000(1.04)^1+20000 =20000(1.04^10+1.04^9+...+1.04+1) =20000[(1)(1.04^10-1)/(1.04-1)] (summation of geometric sequence) =$240122 The value of the car at 5 years later =250000x(1-0.04)^5 =$203843 Therefore at 5 years later, value of deposit>value of car, Danny can buy the car within 5 years
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