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aalyn [17]
4 years ago
5

If Alexis invests $2,000 into a fund that earns 5.5% interest compounded annually, how long will it take for her investment to g

row into $100,000?
Group of answer choices

About 62 years

About 58 years

About 73 years

About 108 years
Mathematics
1 answer:
scZoUnD [109]4 years ago
6 0

Answer: About 73 years

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = 2000

r = 5.5% = 5.5/100 = 0.055

n = 1 because it was compounded once in a year.

A = 100000

Therefore,.

100000 = 2000(1 + 0.055/1)^1 × t

100000/2000 = (1.055)^t

50 = 1.055^t

Taking log of both sides, it becomes

Log 50 = log 1.055^t

1.699 = t × log 1.055

1.699 = 0.0234t

t = 1.699/0.0234

t = 72.6

Approximately 73 years

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<u>Solution: </u>

Given that  Jane’s buying clothes.

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She can get 2 shirts & 3 sweaters for \$128

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And assume cost of 1 sweater = \$y

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\begin{array}{l}{\text { cost of 5 shirts }=5 \times \text { cost of 1 shirt }=5 \times x=5 x} \\\\ {\text { cost of 4 sweater }=4 \times \text { cost of 1 sweater }=4 \times y=4 y}\end{array}

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Second condition is Cost of 5 shirt + cost of 4 sweater = \$128

\begin{array}{c}{\text { cost of } 2 \text { shirts }=2 \times \text { cost of } 1 \text { shirt }=2 \times x=2 x} \\\\ {\text { cost of } 3 \text { sweater }=3 \times \text { cost of } 1 \text { sweater }=3 \times y=3 y}\end{array}

So using second condition equation which we get is as follows

2 \mathrm{x}+3 \mathrm{y}=128\quad\rightarrow (2)

Now we have two equations to be solved

\begin{array}{l}{5 x+4 y=229 \rightarrow (1)} \\\\ {2 x+3 y=128 \rightarrow (2)}\end{array}

On multiplying equation (1) by 2 and equation (2) by 5 to make coefficients of x equal in both equations, we get

\begin{array}{l}{2 x(5 x+4 y)=2\times 229} \\\\ {10 x+8 y=458 \rightarrow (3)} \\\\ {5 x(2 x+3 y)=5 \times 128} \\\\ {10 x+15 y=640 \rightarrow (4)}\end{array}

On subtracting (3) from (4), we get

\begin{array}{l}{(10 \mathrm{x}-10 \mathrm{x})+(15 \mathrm{y}-8 \mathrm{y})=640-458} \\\\ {\Rightarrow 7 \mathrm{y}=182} \\\\ {\Rightarrow \mathrm{y}=\frac{182}{7}=26}\end{array}

On substituting y = 26 in equation (1) we get

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