Verified answer

Problem:

An Antique ring is purchased for $1000 and is expected to grow in value by 5% per year round your answers to the nearest cent.

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Formula:

[tex]{\underline{\boxed{{\tt \small{F=P(1 + \frac{r}{100}) {}^{t} }}}}}[/tex]

Whereas:

• [tex]\tt F = Future \: amount/value[/tex]

• [tex]\tt P= present \: value[/tex]

• [tex]\tt r = rate \: of \: increasing[/tex]

• [tex]\tt t = time \: in \: year[/tex]

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Solving:

a. So, in the first year value at the end of year is:

• [tex]\tt F = 1000(1 + \frac{5}{100} ) {}^{1} [/tex]

• [tex]\tt F = 1000 \times (0.05 + 1) {}^{2} [/tex]

• [tex]\tt F = 1000 \times 1.05=1050[/tex]

So increase in value is:

• = $(1050 - 1000) = $50 (Answer)

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Solving:

b. in the first year value at the end of year is:

• [tex]\tt F = 1000(1 + \frac{5}{100} ) {}^{1} [/tex]

• [tex]\tt F = 1000 \times (0.05 + 1) {}^{2} [/tex]

• [tex]\tt F = 1000 \times 1.05={\underline{\boxed{{\tt \small\green{1050}}}}}[/tex]

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Solving:

c. at the end of the 2nd year value is:

• [tex]\tt F = 1000(1 + \frac{5}{100} ) {}^{2} [/tex]

• [tex]\tt F = 1102.50[/tex]

Increase in value is:

• = $(1102.50 - 1050) = $52.5 (Answer)

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Solving:

d. at the end of the 2nd year value is:

• [tex]\tt F = 1000(1 + \frac{5}{100} ) {}^{2} [/tex]

• [tex]\tt F = {\underline{\boxed{{\tt \small\green{1102.50}}}}}[/tex]

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Solving:

e. at the end of fifth year the value is:

• [tex]\tt F = 1000(1 + \frac{5}{100} ) {}^{5} [/tex]

• [tex]\tt F = 1000(1.05) {}^{5} [/tex]

• [tex]\tt F = {\underline{\boxed{{\tt \small\green{1276.28}}}}}[/tex]

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