Compound Interest, Growth and Decay

Interest for the first year is added to the original investment.

1. The annual multiplier is 1.04.
2. After one year, the balance is 500\times1.04= £520.00.

Apply the growth multiplier once for each year.

1. The yearly multiplier is 1.05.
2. After two years the model gives 1200\times1.05^2=1323.

Each year the machine retains the same percentage of its current value.

1. The yearly multiplier is 0.9.
2. The value after two years is 800\times0.9^2= £648.00.

Calculate the final balance, then subtract the initial investment.

1. The balance after two years is 2000\times1.03^2= £2121.80.
2. The interest is the increase in the balance: 2121.80-2000= £121.80.

Compound interest repeatedly multiplies the current balance by the same growth factor.

1. Each year the balance is multiplied by 1+\frac{2}{100}=1.02.
2. After n years this has happened n times, so the balance is 750\times1.02^n=750(1.02)^n.
3. 750+2n adds only £2 a year, which is not 2\%; 750(0.02)^n uses just the interest part, which would shrink the balance; 750(1.2)^n is a 20\% rate.

Use the annual multiplier raised to the number of years.

1. The annual multiplier is 1.05.
2. The final balance is 1600\times1.05^3= £1852.20.

Convert the total time into the number of growth periods.

1. Two hours contains 120\div30=4 growth periods.
2. The multiplier for each period is 1.5.
3. The predicted number is 400\times1.5^4=2025.

Use the percentage remaining each hour as the decay multiplier.

1. Each hour, 75\% remains, so the multiplier is 0.75.
2. After three hours the mass is 96\times0.75^3=40.5\text{ g}.

Calculate each final balance using the correct model, then compare them.

1. The compound balance is 1250\times1.04^3= £1406.08.
2. The simple-interest balance is 1250(1+3\times0.04)= £1400.00.
3. Both started with the same amount, so the difference in interest is 1406.08-1400= £6.08.

Reverse all three years of depreciation using the full compound multiplier.

1. Each year the value is multiplied by 0.8, so over three years it is multiplied by 0.8^3=0.512.
2. So original value \times0.512=1536.
3. Divide to undo this: 1536\div0.512=3000.
4. The original value was £3000.

The factor inside the power is the fraction of the population remaining after one year.

1. The annual multiplier 0.92 means 92\% remains each year.
2. The decrease is 100\%-92\%=8\%.

Find the two-year multiplier, then take its positive square root.

1. The two-year multiplier is \frac{3025}{2500}=1.21.
2. The same yearly multiplier m is used twice, so m^2=1.21.
3. Take the positive square root: m=\sqrt{1.21}=1.1.
4. A multiplier of 1.1 is an annual increase of 10\%.

The three-year multiplier is the cube of the yearly multiplier.

1. The three-year multiplier is \frac{6400}{12\,500}=0.512.
2. The same yearly multiplier m is used three times, so m^3=0.512.
3. Take the cube root: m=\sqrt[3]{0.512}=0.8.
4. The machine keeps 80\% of its value each year, so the annual decrease is 20\%.

Calculate successive whole-year balances and check the year before your proposed answer.

1. After n years the balance is 5000\times1.04^n.
2. Try n=4: 5000\times1.04^4= £5849.29, which is less than £6000.
3. Try n=5: 5000\times1.04^5= £6083.26, which is more than £6000.
4. So the balance first exceeds £6000 after 5 years.

Use one multiplier for each of the three different annual rates.

1. The successive multipliers are 1.02, 1.04 and 1.05.
2. The final balance is 600\times1.02\times1.04\times1.05= £668.304.
3. To the nearest penny the balance is £668.30.

Complete the first two years before making the withdrawal, then apply the final year multiplier.

1. After the first two years, the balance is 2000\times1.03^2= £2121.80.
2. The withdrawal leaves 2121.80-500= £1621.80.
3. After the third year, the balance is 1621.80\times1.04= £1686.672.
4. The final balance is £1686.67 to the nearest penny.

Compare the two compounded balances, using the rate pattern stated for each account.

1. Account A finishes with 5000\times1.03^4= £5627.54405.
2. Account B finishes with 5000\times1.04^2\times1.02^2= £5626.4832.
3. The initial deposits are the same, so the difference in interest is the difference in balances.
4. The difference is £1.06085, which is £1.06 to the nearest penny.

Remove the first-year increase before finding the common multiplier for the two years of decay.

1. After the first year the population is 5000\times1.12=5600.
2. The two years of decrease multiply 5600 by the same multiplier m twice, so 5600m^2=4046.
3. m^2=\frac{4046}{5600}=0.7225, so m=\sqrt{0.7225}=0.85 (the positive root, as m is between 0 and 1).
4. Each year 85\% remains, so the decrease is 15\% and p=15.

From year 1 to year 3 the number is multiplied by k twice.

1. F_{n+1}=kF_n means each year the number of fish is multiplied by k.
2. From year 1 to year 3 it is multiplied by k twice, so 2500k^2=1600.
3. k^2=\frac{1600}{2500}=0.64. As k is positive, k=\sqrt{0.64}=0.8.
4. F_4=kF_3=0.8\times1600=1280 fish.

Use compound depreciation for the sale price, but include the service in the cost used as the percentage base.

1. The sale price is 1200\times0.85^2= £867.
2. The total spent is 1200+80= £1280.
3. The loss is 1280-867= £413.
4. The percentage loss is \frac{413}{1280}\times100\%=32.265625\%, or 32.3\% to one decimal place.