22. Raymond plans to join a savings scheme in a bank to prepare for his study in Australia 3 years later. He will deposit \( \$ P \) in the bank at the beginning of each of the three consecutive years at an interest rate of \( 5 \% \) per annum compounded continuously. He will take out all the money from his account at the beginning of the 4th year to pay for his tuition fee. (a) It is given that the required tuition fee is \( \$ 450000 \), find the least value of \( P \). (b) At least how much is he going to deposit in total? (c) Raymond is going to postpone the plan for studying abroad by 1 year. Therefore, he decides to extend the savings scheme by 1 year, and thus he will deposit \( \$ Q \) in the bank at the beginning of each of the four consecutive years. If the way in calculating the interest does not change, at least how much is he going to deposit in total? (Give your answers correct to 3 significant figures.)
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To solve these financial questions, let's break them down step-by-step. (a) For continuous compounding, the formula we use is \( A = Pe^{rt} \). Since Raymond will make three deposits at the beginning of each consecutive year, we need to find the accumulated value of each deposit at the end of the three years. 1. The first deposit \( P \) will accumulate for 3 years: \[ A_1 = Pe^{0.05 \times 3} = Pe^{0.15} \] 2. The second deposit \( P \) will accumulate for 2 years: \[ A_2 = Pe^{0.05 \times 2} = Pe^{0.10} \] 3. The third deposit \( P \) will accumulate for 1 year: \[ A_3 = Pe^{0.05 \times 1} = Pe^{0.05} \] Thus, the total amount Raymond will have at the beginning of the 4th year is: \[ A = A_1 + A_2 + A_3 = P(e^{0.15} + e^{0.10} + e^{0.05}) \] We set this equal to the tuition fee: \[ P(e^{0.15} + e^{0.10} + e^{0.05}) = 450000 \] Calculating the exponentials: - \( e^{0.15} \approx 1.161834 \) - \( e^{0.10} \approx 1.105171 \) - \( e^{0.05} \approx 1.051271 \) Now, add these together: \[ e^{0.15} + e^{0.10} + e^{0.05} \approx 1.161834 + 1.105171 + 1.051271 \approx 3.318276 \] Now solve for \( P \): \[ P \cdot 3.318276 = 450000 \] \[ P = \frac{450000}{3.318276} \approx 135,000.78 \] Rounded to three significant figures, the least value of \( P \) is: \[ P \approx 135,000 \] (b) To find out how much Raymond will deposit in total: Since he will deposit \( P \) at the beginning of the three years: Total amount deposited = \( 3P \) Calculating: \[ 3P = 3 \times 135000 = 405000 \] Thus, the total amount he is going to deposit is: \[ 405,000 \] (c) Now, if Raymond postpones his plan and extends the savings for one more year, he will deposit \( Q \) at the beginning of each of the four years. The new total amount with four deposits can be calculated similarly: For each deposit: 1. First deposit \( Q \) will accumulate for 4 years: \[ A_1 = Qe^{0.05 \times 4} = Qe^{0.20} \] 2. Second deposit \( Q \) will accumulate for 3 years: \[ A_2 = Qe^{0.15} \] 3. Third deposit \( Q \) will accumulate for 2 years: \[ A_3 = Qe^{0.10} \] 4. Fourth deposit \( Q \) will accumulate for 1 year: \[ A_4 = Qe^{0.05} \] The total will now be: \[ A = A_1 + A_2 + A_3 + A_4 = Q(e^{0.20} + e^{0.15} + e^{0.10} + e^{0.05}) \] Setting this equal to \( 450000 \): \[ Q(e^{0.20} + e^{0.15} + e^{0.10} + e^{0.05}) = 450000 \] Calculating \( e^{0.20} \approx 1.221402 \): Adding gives: \[ e^{0.20} + e^{0.15} + e^{0.10} + e^{0.05} \approx 1.221402 + 1.161834 + 1.105171 + 1.051271 \approx 4.539678 \] Now, solve for \( Q \): \[ Q \cdot 4.539678 = 450000 \] \[ Q = \frac{450000}{4.539678} \approx 99,020.46 \] Total deposited will be \( 4Q \): \[ 4Q = 4 \times 99,020
