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Ken’s sister borrowed $530 from the bank at 8.4% per year.
A) how much interest did she pay in one year
b) how much interest did she pay in three years

Respuesta :

[tex]~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$530\\ r=rate\to 8.4\%\to \frac{8.4}{100}\dotfill &0.084\\ t=years\dotfill &1 \end{cases} \\\\\\ I=(530)(0.084)(1)\implies I=44.52 \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]

[tex]~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$530\\ r=rate\to 8.4\%\to \frac{8.4}{100}\dotfill &0.084\\ t=years\dotfill &3 \end{cases} \\\\\\ I=(530)(0.084)(3)\implies I=133.56[/tex]

AadhPrit

Answer:

A) She pays $44.52 in one year.

B) She pays $133.56 in three years.

Step-by-step explanation:

Ken’s sister borrowed $530 from the bank at 8.4% per year.

A) how much interest did she pay in one year

Here's the required formula to find the interest :

[tex]\longrightarrow{\tt{I = \dfrac{PRT}{100}}}[/tex]

  • [tex]\purple\star[/tex] P = Principle
  • [tex]\purple\star[/tex] R = Rate
  • [tex]\purple\star[/tex] T = Time

Substituting all the given values in the formula to find the interest for one year :

[tex]\begin{gathered} \qquad\longrightarrow{\sf{I = \dfrac{PRT}{100}}} \\ \\ \qquad\longrightarrow{\sf{I = \dfrac{P \times R \times T}{100}}} \\ \\ \qquad\longrightarrow{\sf{I = \dfrac{530 \times 8.4 \times 1}{100}}} \\ \\ \qquad\longrightarrow{\sf{I = \dfrac{530 \times 8.4}{100}}} \\ \\ \qquad\longrightarrow{\sf{I = \dfrac{4452}{100}}} \\ \\ \qquad\longrightarrow{\sf{\underline{\underline{\red{I = \$44.52}}}}} \end{gathered}[/tex]

Hence, the interest is $44.52.

[tex]\begin{gathered}\end{gathered}[/tex]

b) how much interest did she pay in three years

Here's the required formula to find the interest :

[tex]\longrightarrow{\tt{I = \dfrac{PRT}{100}}}[/tex]

  • [tex]\pink\star[/tex] P = Principle
  • [tex]\pink\star[/tex] R = Rate
  • [tex]\pink\star[/tex] T = Time

Substituting all the given values in the formula to find the interest for three years :

[tex]\begin{gathered} \qquad\longrightarrow{\sf{I = \dfrac{PRT}{100}}} \\ \\ \qquad\longrightarrow{\sf{I = \dfrac{P \times R \times T}{100}}} \\ \\ \qquad\longrightarrow{\sf{I = \dfrac{530 \times 8.4 \times 3}{100}}} \\ \\ \qquad\longrightarrow{\sf{I = \dfrac{530 \times 25.2}{100}}} \\ \\ \qquad\longrightarrow{\sf{I = \dfrac{13356}{100}}} \\ \\ \qquad\longrightarrow{\sf{\underline{\underline{\red{I = \$133.56}}}}} \end{gathered}[/tex]

Hence, the interest is $133.56.

[tex]\rule{300}{2.5}[/tex]

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