Question

1. nathalie invests a capital of $5000 at a simple interest rate of 4% for 18 months. at the end of the investment, nathalie invests the capital and the interest received at a simple interest rate of 6% for 6 months. what amount will nathalie receive at the end of the 2nd investment?
2. eric invests an amount, for 2 years, at a simple interest rate of 3%. at the end of the investment, eric invests the accumulated value at a simple interest rate of 5% for 3 years. what is the initial amount of eric’s first investment if the value accumulated at the end of the second investment is $4876?

Explanation:

Question 9

Step 1: Calculate simple interest for first investment

The formula for simple interest is \( I = P \times r \times t \), where \( P \) is principal, \( r \) is rate (decimal), \( t \) is time (years).
Principal \( P = 5000 \), rate \( r = 4\% = 0.04 \), time \( t = \frac{18}{12} = 1.5 \) years.
\( I_1 = 5000 \times 0.04 \times 1.5 \)
\( I_1 = 300 \)

Step 2: Find amount after first investment

Amount \( A_1 = P + I_1 = 5000 + 300 = 5300 \)

Step 3: Calculate interest for second investment

New principal \( P_2 = 5300 \), rate \( r_2 = 6\% = 0.06 \), time \( t_2 = \frac{6}{12} = 0.5 \) years.
\( I_2 = 5300 \times 0.06 \times 0.5 \)
\( I_2 = 159 \)

Step 4: Find amount after second investment

Amount \( A_2 = P_2 + I_2 = 5300 + 159 = 5459 \)

Step 1: Let initial principal be \( P \)

First investment: time \( t_1 = 2 \) years, rate \( r_1 = 3\% = 0.03 \).
Amount after first investment: \( A_1 = P + P \times 0.03 \times 2 = P(1 + 0.06) = 1.06P \)

Step 2: Second investment

Principal \( P_2 = A_1 = 1.06P \), time \( t_2 = 3 \) years, rate \( r_2 = 5\% = 0.05 \).
Amount after second investment: \( A_2 = P_2 + P_2 \times 0.05 \times 3 = 1.06P(1 + 0.15) = 1.06P \times 1.15 \)
\( A_2 = 1.219P \)

Step 3: Solve for \( P \)

Given \( A_2 = 4876 \), so \( 1.219P = 4876 \)
\( P = \frac{4876}{1.219} = 4000 \)

Answer:

\( \$5459 \)

Question 10