Math, asked by amitdwivedibhaiyan, 7 days ago

At what rate per annum will 15,625 amount to 17,576 in 9 months, when compounded quarterly ?​

Answers

Answered by Itzintellectual
4

Step-by-step explanation:

 \tt \red{amount} =   17576 \:  \:  \:  \:  \:  \\  \\  \tt \red{principle} = 15625 \\  \\  \tt \red{time} = 9 \: month \:  \:  \:  \:  \:  \:  \:  \:  \\  \\   \tt \pink{\frac{a}{p} } =  \tt \orange{(1 +  \frac{r}{100} ) {}^{n} } \\  \\ \tt \pink{\frac{17576}{15625} } =  \tt \orange{(1 +  \frac{r}{100} ) {}^{ \frac{3}{4} } } \\  \\ \tt \pink{\frac{26}{25} } =  \tt \orange{(1 +  \frac{r}{100} ) {}^{3} } \\  \\

Answered by jaiswalkshitij456
1

Step-by-step explanation:

\begin{gathered} \tt \red{amount} = 17576 \: \: \: \: \: \\ \\ \tt \red{principle} = 15625 \\ \\ \tt \red{time} = 9 \: month \: \: \: \: \: \: \: \: \\ \\ \tt \pink{\frac{a}{p} } = \tt \orange{(1 + \frac{r}{100} ) {}^{n} } \\ \\ \tt \pink{\frac{17576}{15625} } = \tt \orange{(1 + \frac{r}{100} ) {}^{ \frac{3}{4} } } \\ \\ \tt \pink{\frac{26}{25} } = \tt \orange{(1 + \frac{r}{100} ) {}^{3} } \\ \\ \end{gathered}</p><p></p><p>

Answered by jaiswalkshitij456
1

Step-by-step explanation:

\begin{gathered} \tt \red{amount} = 17576 \: \: \: \: \: \\ \\ \tt \red{principle} = 15625 \\ \\ \tt \red{time} = 9 \: month \: \: \: \: \: \: \: \: \\ \\ \tt \pink{\frac{a}{p} } = \tt \orange{(1 + \frac{r}{100} ) {}^{n} } \\ \\ \tt \pink{\frac{17576}{15625} } = \tt \orange{(1 + \frac{r}{100} ) {}^{ \frac{3}{4} } } \\ \\ \tt \pink{\frac{26}{25} } = \tt \orange{(1 + \frac{r}{100} ) {}^{3} } \\ \\ \end{gathered}</p><p></p><p>

Answered by jaiswalkshitij456
1

Step-by-step explanation:

\begin{gathered} \tt \red{amount} = 17576 \: \: \: \: \: \\ \\ \tt \red{principle} = 15625 \\ \\ \tt \red{time} = 9 \: month \: \: \: \: \: \: \: \: \\ \\ \tt \pink{\frac{a}{p} } = \tt \orange{(1 + \frac{r}{100} ) {}^{n} } \\ \\ \tt \pink{\frac{17576}{15625} } = \tt \orange{(1 + \frac{r}{100} ) {}^{ \frac{3}{4} } } \\ \\ \tt \pink{\frac{26}{25} } = \tt \orange{(1 + \frac{r}{100} ) {}^{3} } \\ \\ \end{gathered}</p><p></p><p>

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