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2 years ago

Noah invested $890 in an account paying an interest rate of 4 5/8% compounded

Mathematics

1 answer:

sergiy2304 [10]2 years ago

5 0

Answer:

3.37 years

Step-by-step explanation:

\text{Rate 1: }4\tfrac{5}{8}\%=4+5/8=

Rate 1: 4

%=4+5/8=

\,\,4.625\%\rightarrow 0.04625

4.625%→0.04625

\text{Rate 2: }5\tfrac{1}{4}\%=5+1/4=

Rate 2: 5

%=5+1/4=

\,\,5.25\%\rightarrow 0.0525

5.25%→0.0525

890\text{ tripled is }2670

890 tripled is 2670

\text{Calculate Tripling Time for Noah}

Calculate Tripling Time for Noah

\overline{\phantom{\text{Calculate Tripling Time for Noah}}}

Calculate Tripling Time for Noah

\text{Compounded Annually:}

Compounded Annually:

A=P(1+r)^t

A=P(1+r)

A=2670\hspace{35px}P=890\hspace{35px}r=0.04625

A=2670P=890r=0.04625

Given values

2670=

\,\,890(1+0.04625)^{t}

890(1+0.04625)

Plug in values

2670=

\,\,890(1.04625)^{t}

890(1.04625)

Add

\frac{2670}{890}=

890

2670

\,\,\frac{890(1.04625)^{t}}{890}

890

890(1.04625)

Divide by 890

\,\,1.04625^t

1.04625

\log\left(3\right)=

log(3)=

\,\,\log\left(1.04625^t\right)

log(1.04625

)

Take the log of both sides

\log\left(3\right)=

log(3)=

\,\,t\log\left(1.04625\right)

tlog(1.04625)

Bring exponent to the front

\frac{\log\left(3\right)}{\log\left(1.04625\right)}=

log(1.04625)

log(3)

\,\,\frac{t\log\left(1.04625\right)}{\log\left(1.04625\right)}

log(1.04625)

tlog(1.04625)

Divide both sides by log(1.04625)

24.2989=

Calculate Tripling Time for Riley

Compounded Continuously:

A=Pe^{rt}

A=Pe

A=2670\hspace{35px}P=890\hspace{35px}r=0.0525

A=2670P=890r=0.0525

Given values

2670=

\,\,890e^{0.0525t}

890e

0.0525t

Plug in

\frac{2670}{890}=

890

2670

\,\,\frac{890e^{0.0525t}}{890}

890

890e

0.0525t

Divide by 890

\,\,e^{0.0525t}

0.0525t

\ln\left(3\right)=

ln(3)=

\,\,\ln\left(e^{0.0525t}\right)

ln(e

0.0525t

)

Take the natural log of both sides

\ln\left(3\right)=

ln(3)=

\,\,0.0525t

0.0525t

ln cancels the e

\frac{\ln\left(3\right)}{0.0525}=

0.0525

ln(3)

\,\,\frac{0.0525t}{0.0525}

0.0525

0.0525t

Divide by 0.0525

20.9259=

How much longer for Noah to triple:

24.2989-20.9259

24.2989−20.9259

3.373

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<h3><u>Option 1</u></h3>

Earn $50 every month.

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Let y = the amount in the account
Initial amount = $1,000

$\implies y = 50x + 1000$

This is a <u>linear function</u>.

<h3><u>Option 2</u></h3>

Earn 3% interest each month.

(Assuming the interest earned each month is <u>compounding interest</u>.)

Let x = number of months the money is left in the account
Let y = the amount in the account
Initial amount = $1,000

$\implies y = 1000(1.03)^x$

This is an <u>exponential function</u>.

<h3><u>Table of values</u></h3>

<u />

$\large \begin{array}{| c | l | l |}\cline{1-3} & \multicolumn{2}{|c|}{\sf Account\:Balance} \\ \cline{1-3} & \sf Option\:1 & \sf Option\:2 \\\sf Month & \sf \$50\:per\:mth & \sf 3\%\:per\:mth \\\cline{1-3} 0 & \$1000 & \$1000 \\\cline{1-3} 1 & \$1050 & \$1030 \\\cline{1-3} 2 & \$1100 & \$1060.90 \\\cline{1-3} 3 & \$1150 & \$1092.73 \\\cline{1-3} 4 & \$1200 & \$1125.51 \\\cline{1-3} 5 & \$1250 & \$1159.27 \\\cline{1-3} 6 & \$1300 & \$1194.05 \\\cline{1-3} 7 & \$1350 & \$1229.87 \\\cline{1-3}\end{array}$

From the table of values, it appears that <u>Account Option 1</u> is the best choice, as the accumulative growth of this account is higher than the other account option.

However, there will be a point in time when Account Option 2 starts accruing more than Account Option 2 each month. To find this, graph the two functions and find the <u>point of intersection</u>.

From the attached graph, Account Option 1 accrues more until month 32. From month 33, Account Option 2 accrues more in the account.

<h3><u>Conclusion</u></h3>

If the money is going to be invested for less than 33 months then Account Option 1 is the better choice. However, if the money is going to be invested for 33 months or more, then Account Option 2 is the better choice.