Question

NO LINKS!! Fredrico's Bank will you decide how often your interest will be computed, but with certain restrictions. Read the following options. What is the best deal after one year? Justify your answer.​

Answer 1

Answer: Option B

Compounded quarterly; annual interest rate is 7.875%

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Step-by-step explanation:

Let's say we are depositing P = 100 dollars, and we leave it in the account for a timespan of t = 1 year.

This will apply for options A through E.

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If we go with option A, then we would have n = 1 and r = 0.08

Use the compound interest formula to get the following:

A = P*(1+r/n)^(n*t)

A = 100*(1+0.08/1)^(1*1)

A = 108

We'll have $108 dollars in the account at the end of 1 year if we go with option A. The amount of interest only is A-P = 108-100 = 8 dollars.

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For option B, we have: n = 4 and r = 0.07875

A = P*(1+r/n)^(n*t)

A = 100*(1+0.07875/4)^(4*1)

A = 108.110625948487

A = 108.11

So far option B is slightly better since you've earned an extra $0.11, aka 11 cents.

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For option C, we have: n = 4 and r = 0.07875

A = P*(1+r/n)^(n*t)

A = 100*(1+0.0775/12)^(12*1)

A = 108.031299777277

A = 108.03

This amount is less compared to the $108.11 with option B. Therefore, option B is still the best so far.

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For option D, we have: n = 52 and r = 0.07625

A = P*(1+r/n)^(n*t)

A = 100*(1+0.07625/52)^(52*1)

A = 107.917207523206

A = 107.92

This is worse compared to option C.

Option B is still the best so far.

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For option E, we have: n = 365 and r = 0.075

A = P*(1+r/n)^(n*t)

A = 100*(1+0.075/365)^(365*1)

A = 107.787584644

A = 107.79

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Here's a table comparing the balance after 1 year for options A through E. Those balances all have a starting point of $100 as the deposit.

`\begin{array}c \cline{1-2}\text{Option} & \text{Balance after 1 year}\\\cline{1-2}A & \$108\\\cline{1-2}\boldsymbol{B} & \boldsymbol{\$108.11}\\\cline{1-2}C & \$108.03\\\cline{1-2}D & \$107.92\\\cline{1-2}E & \$107.79\\\cline{1-2}\end{array}`

Option B is the best choice earning the most ($108.11) if you deposit $100.

There's nothing special about the $100, so feel free to change it to something else to try other starting deposits. You'll find that option B will still provide the highest return.

Answer 2

B

Step-by-step explanation:

Compound Interest Formula

`\large \text{$ \sf A=P\left(1+(r)/(n)\right)^(nt) $}`

where:

• A = final amount
• P = principal amount
• r = interest rate (in decimal form)
• n = number of times interest applied per time period
• t = number of time periods elapsed

Use the Compound Interest Formula with the given values for each option to calculate a final equation where A equals a multiple of P.

The option with the greatest coefficient of P is the account which yields the highest interest after one year, and is therefore the best deal.

Option A

Given:

• r = 8% = 0.08
• n = 1 (annually)
• t = 1 year

Substitute the given values into the formula:

`\implies \sf A=P\left(1+(0.08)/(1)\right)^1`

`\implies \sf A=1.08P`

Option B

Given:

• r = 7.875% = 0.07875
• n = 4 (quarterly)
• t = 1 year

Substitute the given values into the formula:

`\implies \sf A=P\left(1+(0.07875)/(4)\right)^(4 \cdot 1)`

`\implies \sf A=P\left(1.0196875)^4`

`\implies \sf A=1.081106259P`

Option C

Given:

• r = 7.75% = 0.0775
• n = 12 (monthly)
• t = 1 year

Substitute the given values into the formula:

`\implies \sf A=P\left(1+(0.0775)/(12)\right)^(12\cdot 1)`

`\implies \sf A=P\left(1.006458333\right)^(12)`

`\implies \sf A=1.080312998P`

Option D

Given:

• r = 7.625% = 0.07625
• n = 52 (weekly)
• t = 1 year

Substitute the given values into the formula:

`\implies \sf A=P\left(1+(0.0765)/(52)\right)^(52\cdot 1)`

`\implies \sf A=P\left(1.001471154\right)^(52)`

`\implies \sf A=1.079441506P`

Option E

Given:

• r = 7.5% = 0.075
• n = 365 (daily)
• t = 1 year

Substitute the given values into the formula:

`\implies \sf A=P\left(1+(0.075)/(365)\right)^(365\cdot 1)`

`\implies \sf A=P\left(1.000205479\right)^(365)`

`\implies \sf A=1.077875846P`

Summary

`\begin{array}l\cline{1-2} \sf Answer\;Option & \sf Coefficient\;of\;P\\\cline{1-2} A & 1.08\\\cline{1-2} B & 1.081106259\\\cline{1-2} C & 1.080312998\\\cline{1-2} D & 1.079441506\\\cline{1-2} E & 1.077875846\\\cline{1-2} \end{array}`

Therefore, the option with the greatest coefficient of P and therefore the best deal is option B.