Question

(please note that this is a lengthy problem)I have a prep guide problem that I need explained and answered. I’m having trouble on it. Below is the questions for this problem (it includes 5 questions)1. What is the balance of Albert’s $2000 after 10 years??2. What is the balance of Marie’s $2000 after 10 years??3. What is the balance of Han’s $2000 after 10 years??4. What is the balance of Max’s $2000 after 10 years??AND LASTLY,5. Who is 10,000 richer at the end of the competition??5.

Answer 1

To solve this problem, we will use the following formulas:

(a) Formula for the annual interest

`P_N=P_0\cdot(1+(r)/(k))^(N\cdot k)\text{.}`

Where:

• P_N is the balance in the account after N years,

,

• P_0 is the starting balance of the account (also called an initial deposit, or principal),

,

• r is the annual interest rate in decimal form,

,

• k is the number of compounding periods in one year

(b) Formula for the annual interest compounded continuously

`P_N=P_0\cdot e^(N\cdot r).`

Where:

• P_N is the balance in the account after N years,

,

• P_0 is the starting balance of the account (also called an initial deposit, or principal),

,

• r is the annual interest rate in decimal form.

-----------------------------------------------------------------------

We will compute the balanced of each investment, and then we will sum the results to get the total balance of each son.

1) Balance of Albert’s $2000 after 10 years

1-i) $1000 earned 1.2% annual interest compounded monthly

We use the (a) formula with the values: N = 10, P_0 = 1000, r = 1.2/100 = 0.012, k = 12.

`I_(Albert,1)=1000\cdot(1+(0.012)/(12))^(10\cdot12)_{}\cong1127.43`

1-ii) $500 lost 2% over the course of the 10 years

`I_{\text{Albert,}2}=500-500\cdot(2)/(100)=500-10=490`

1-iii) $500 grew compounded continuously at a rate 0.8% annually

We use the (b) formula with the values: N = 10, P_0 = 500, r = 0.8/100 = 0.008.

`I_{\text{Albert,}3}=500\cdot e^(10\cdot0.008)=541.64`

Summing all Albert's investments, we get:

`I_{\text{Albert}}=I_{\text{Albert,}1}+I_{\text{Albert,}2}+I_{\text{Albert,}3}=1127.43+490+541.64=2159.07`

After 10 years, Albert's balance is $2159.07.

2) Balance of Marie’s $2000 after 10 years

2-i) $1500 earned 1.4% annual interest compounded quarterly

We use the (a) formula with the values: N = 10, P_0 = 1500, r = 1.4/100 = 0.014, k = 4.

`I_{\text{Marie,}1}=1500\cdot(1+(0.014)/(4))^(10\cdot4)\cong1724.99`

2-ii) $500 gained 4% over the course of 10 years

`I_{\text{Marie,}2}=500+500\cdot(4)/(100)=500+20=520`

Summing all Marie's investments, we get:

`I_{\text{Marie}}=I_{\text{Marie,1}}+I_{\text{Marie,2}}=1724.99+520=2244.99`

After 10 years, Marie's balance is $2244.99.

3) Balance of Hans’s $2000 after 10 years

$2000 grew compounded continuously at a rate of 0.9% annually

We use the (b) formula with the values: N = 10, P_0 = 2000, r = 0.9/100 = 0.009.

`I_{\text{Hans}}=2000\cdot e^(10\cdot0.009)=2188.35`

After 10 years, Hans's balance is $2188.35.

4) Balance of Max’s $2000 after 10 years

4-i) $1000 decreased in value exponentially at a rate of 0.5% annually

We use the (b) formula with the values: N = 10, P_0 = 1000, r = -0.5/100 = -0.005.

`I_{\text{Max,}1}=1000\cdot e^(10\cdot(-0.005))=951.23`

4-ii) $1000 earned 1.8% annual interest compounded biannually (twice a year)

We use the (a) formula with the values: N = 10, P_0 = 1000, r = 1.8/100 = 0.018, k = 2.

`I_{\text{Max,}2}=1000\cdot(1+(0.018)/(2))^(10\cdot2)_{}=1196.25`

Summing all Max's investments, we get:

`I_{\text{Max}}=I_{\text{Max,}1}+I_{\text{Max,}2}=951.23+1196.25=2147.48`

After 10 years, Max's balance is $2147.48.

5) From the results above, we see that Marie's balance is the highest. So Marie is 10,000 richer at the end of the competition.

Answers

0. After 10 years, Albert's balance is $2159.07.

,

1. After 10 years, Marie's balance is $2244.99.

,

2. After 10 years, Hans's balance is $2188.35.

,

3. After 10 years, Max's balance is $2147.48.

,

4. Marie is 10,000 richer at the end of the competition.